solucionario beer, johnton - octava edicion parcial ultimo

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

Chapter 14, Solution 1.

The masses are 1350 kg and 5400 kg.A B C

m m m= = =

Let , , and be the sought after final velocities, positive to the left.A B Cv v v

Initial velocities: ( ) ( ) ( )0 0 0

0, 8 km/h 2.2222 m/sA B Cv v v= = = =

. Truck strikes car . Plastic impact: 0First collision C B e =

( )0

Let be the common velocity of and after impact.BCv B C

:Conservation of momentum for B and C

( ) ( ) ( )0 0B C BC B B C C

m m v m v m v+ = +

( )( )6750 0 5400 2.2222BCv = + 1.77778 m/s

BCv =

Car-truck strikes carSecond collision. BC A.

Elastic impact. 1e =

( ) ( )0 0

1.77778 m/sA BC A BCv v e v v − = − − = − (1)

Conservation of momentum for A, B, and C.

( )( ) ( ) ( )( )0 0A A B C BC A A B C BC

m v m m v m v m m v+ + = + +

( )( )1350 6750 0 6750 1.77778A BCv v+ = + (2)

Solving (1) and (2) simultaneously for and ,A BCv v

2.9630 m/s, 1.18519 m/sA BC B Cv v v v= = = =

10.67 km/hA

=v

4.27 km/hB

=v

4.27 km/hC

=v






Conservation of linear momentum for block, cart, and bullet together.

components : ( )0B A B C fm v m m m v= + +

( )( )

00.028 550

1.7058 m/s5 0.028 4

Bf

A B C

m v

v

m m m

= = =+ + + +

(a) 1.706 m/sfv =

Consider block and bullet alone.

Principle of impulse and momentum.

components ( ) ( ): N t mg t N mg∆ − ∆ =

components : ( ) ( )0B k A Bm v N t m m v'µ− ∆ = +

Just after impact, t∆ is negligible. The velocity then is

( )( )0

0

0.028 5503.0628 m/s

5 0.028

B

A B

'm v

v

m m

= = =+ +

Also, just after impact, the velocity of the cart is zero.

Accelerations after impact.





Block and bullet: :F ma=∑

( ) ( )k A B A B ABm m g m m aµ + = +

( )( )0.50 9.81AB k

a gµ= =

2

4.905 m/s=

Cart:

:CF ma=∑

( ) :k A B C Cm m g m aµ + =

( ) ( )( )( )0.50 5.028 9.81

4

k A B

C

C

m m ga

m

µ += =

2

6.1656 m/s=

Acceleration of block relative to cart.

( ) 2

/4.905 6.1656 11.0706 m/s

AB Ca = − − =

Motion of the block relative to the cart.

( ) ( ) ( )( )

2 2

/

/ /2

2 2

AB C

AB C AB C

v v'a s− =

In the final position, /

0AB Cv =

( ) ( )

( )( )2 2

/

/C

3.06280.424 m

2 2 11.0706AB C

AB

v's

a= − = − = −

The block moves 0.424 to the left relative to the cart.

(b) This places the block 1.000 0.424 0.576 m− = from the left end of the cart.

0.576 m from left end of cart






( )( )22 20004000 3700The masses are 124.2 slugs, 114.9 slugs, and 1366.5 slugs

32.2 32.2 32.2A B F

m m m= = = = = =

Let , , and be the sought after velocities in ft/s, positive to the right.A B Fv v v

Initial values: ( ) ( ) ( )0 0 0

0.A B Fv v v= = =

Initial momentum of system: ( ) ( ) ( )0 0 0

0.A A B B F F

m v m v m v+ + =

There are no horizontal external forces acting during the time period under consideration. Momentum is

conserved.

0A A B B F F

m v m v m v= + +

124.2 114.9 1366.5 0A B Fv v v+ + = (1)

The relative velocities are given as

/

/

7 ft/s

3.5 ft/s

A F A F

B F B F

v v v

v v v

= − = −

= − = −

(2)

(3)

Solving (1), (2), and (3) simultaneously,

6.208 ft/s, 2.708 ft/s, 0.7919 ft/s= − = − =A B Fv v v

0.792 ft/sF

=v






The masses are , , and .A B F

A B F

W W Wm m m

g g g= = =

Let the final velocities be , , and 0.34 ft/s, positive to the right.=A B Fv v v


0A B Fv v v= = =


0A A B B F F

m v m v m v+ + =


conserved.

0A B F

A A B B F F A B F

W W Wm v m v m v v v v

g g g= + + = + +

Solving for ,F

W A A B B

F

F

W v W vW

v

+= − (1)

From the given relative velocities,

/

/

1.02 7.65 6.63 ft/s

1.02 7.5 6.48 ft/s

A F A F

B F B F

v v v

v v v

= + = − = −

= + = − = −

Substituting these values in (1),

( )( ) ( )( )4000 6.63 3700 6.48

49506 lb1.02

FW

− + −= − =

24.8 tonsF

W =






( ) ( )( )

3 3

3

The masses are the engine 80 10 kg , the load 30 10 kg , and the flat car

20 10 kg .

A B

C

m m

m

= × = ×

= ×

Initial velocities: ( ) ( ) ( )0 0 0

6.5 km/h 1.80556 m/s, 0.A B Cv v v= = = =

No horizontal external forces act on the system during the impact and while the load is sliding relative to the flat

car. Momentum is conserved.

Initial momentum: ( ) ( ) ( ) ( )0 0

0 0A A B C A A

m v m m m v+ + = (1)

(a) Let v′ be the common velocity of the engine and flat car immediately after impact. Assume that the

impact takes place before the load has time to acquire velocity.

Momentum immediately after impact:

( ) ( )0A B C A C

m v m m v m m v′ ′ ′+ + = + (2)

Equating (1) and (2) and solving for ,v′

( ) ( )( )

( )3

0

3

80 10 1.80556

1.44444 m/s

100 10

A A

A C

m v

v

m m

×′ = = =

+ ×

5.20 km/h′ =v

(b) Let fv be the common velocity of all three masses after the load has slid to a stop relative to the car.

Corresponding momentum:

( )A f B f C f A B C fm v m v m v m m m v+ + = + + (3)

Equating (1) and (3) and solving for ,fv

( ) ( )( )( )

3

0

3

80 10 1.80556

1.11111m/s

130 10

A Af

A B C

m v

v

m m m

×= = =

+ + ×

4.00 km/hf =v






The masses are m for the bullet and A

m and B

m for the blocks.

(a) The bullet passes through block A and embeds in block B. Momentum is conserved.

( ) ( )0 0Initial momentum: 0 0

A Bmv m m mv+ + =

Final momentum:B A A B B

mv m v m v+ +

0Equating,

B A A B Bmv mv m v m v= + +

( )( ) ( )( ) 3

0

3 3 2.5 543.434 10 kg

500 5

A A B B

B

m v m v

m

v v

−++= = = ×− −

43.4 gm =

(b) The bullet passes through block A. Momentum is conserved.

( )0 0Initial momentum: 0

Amv m mv+ =

1Final momentum:

A Amv m v+

0 1Equating,

A Amv mv m v= +

( )( ) ( )( )3

0

1 3

43.434 10 500 3 3

292.79 m/s43.434 10

A Amv m v

v

m

−

−

× −−= = =×

1293m/s=v






(a) Woman dives first.

Conservation of momentum:

( )1 1

120 300 18016 0v v

g g

+− − =

( )( )

1

120 163.20 ft/s

600v = =

Man dives next. Conservation of momentum:

( )1 2 2

300 180 300 18016v v v

g g g

+− = − + −

( )( )1

2

480 180 169.20 ft/s

480

v

v

+= =

29.20 ft/s=v

(b) Man dives first.


( )1 1

180 300 12016 v v

g g

+′ ′− −

( )( )

1

180 164.80 ft/s

600v′ = =

Woman dives next. Conservation of momentum:

( )1 2 2

300 120 300 12016v v v

g g g

+ ′ ′ ′− = − + −

( )( )1

2

420 120 169.37 ft/s

420

v

v

′ +′ = =

29.37 ft/s′ =v






(a) Woman dives first.


( )1 1

120 300 18016 0v v

g g

+− − + =

( )( )

1

120 163.20 ft/s

600v = =

Man dives next. Conservation of momentum:

( )1 2 2

300 180 300 18016v v v

g g g

+ = − + −

( )( )1

2

480 180 162.80 ft/s

480

− += =

v

v 2

2.80 ft/s=v

(b) Man dives first.


( )1 1

180 300 12016 0v v

g g

+′ ′− − =

( )( )

1

180 164.80 ft/s

600v′ = =

Woman dives next. Conservation of momentum:

( )1 2 2

300 120 300 12016v v v

g g g

+ ′ ′ ′− = + −

( )( )1

2

420 120 160.229 ft/s

420

v

v

′− +′ = = −

2

0.229 ft/s′ =v






The masses are 9 kg.A B C

m m m= = =

Position vectors (m): 0.9 , 0.6 0.6 0.9 , 0.3 1.2A B C

= = + + = +r k r i j k r i j

2In units of kg m /s,⋅ ( ) ( ) ( )O A A A B B B C C Cm m m= × + × + ×H r v r v r v

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

0 0 0.9 0.6 0.6 0.9 0.3 1.2 0

0 9 0 9 0 0 0 0 9

8.1 8.1 5.4 10.8 2.7

8.1 10.8 8.1 2.7 5.4

A B C

A B B C C

A C B C B

v v v

v v v v v

v v v v v

= + +

= − + − + −

= − + + − + −

i j k i j k i j k

i j k i j

i j k

But, O

H is given as ( )21.8 kg m /s− ⋅ k

Equating the two expressions for O

H and resolving into components,

: 8.1 10.8 0

: 8.1 2.7 0

: 5.4 1.8

A C

B C

B

v v

v v

v

− + =− =

− = −

i

j

k

(1)

(2)

(3)

( ) Solving for , , and ,A B C

a v v v 1.333 m/sAv = ( )1.333 m/s

A=v j

0.333m/sBv = ( )0.333m/s

B=v i

1.000 m/sCv = ( )1.000 m/s

C=v k

Coordinates of mass center G in m.

( )( ) ( )( ) ( )( )9 0.9 9 0.6 0.6 0.9 9 0.3 1.2

27

0.3 0.6 0.6

A A B B C C

A B C

m m m

m m m

+ +=+ +

+ + + + +=

= + +

r r rr

k i j k i j

i j k

Position vectors relative to the mass center in m.

( ) ( )( ) ( )( ) ( )

0.9 0.3 0.6 0.6 0.3 0.6 0.3

0.6 0.6 0.9 0.3 0.6 0.6 0.3 0.3

0.3 1.2 0.3 0.6 0.6 0.6 0.6

A A

B B

C C

′ = − = − + + = − − +

′ = − = + + − + + = +

′ = − = + − + + = −

r r r k i j k i j k

r r r i j k i j k i k

r r r i j i j k j k

( )( ) ( )( )( ) ( )( )( ) ( )

9 1.333 12 kg m/s

9 0.333 3 kg m/s

9 1 9 kg m/s

A A

B B

C C

m

m

m

= = ⋅

= = ⋅

= = ⋅

v j j

v i i

v k k





( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )

( )

0.3 0.6 0.3 12 0.3 0.3 3 0.6 0.6 9

3.6 3.6 0.9 5.4 1.8 0.9 3.6

G A A A B B B C C Cb m m m′ ′ ′= × + × + ×

= − − + × + + × + − ×

= − − + + = + −

H r v r v r v

i j k j i k i j k k

i k j i i j k

( ) ( ) ( )2 2 21.800 kg m /s 0.900 kg m /s 3.60 kg m /sG

= ⋅ + ⋅ − ⋅H i j k






The masses are 9 kg.A B C

m m m= = =

Position vectors (m): 0.9 , 0.6 0.6 0.9 , 0.3 1.2A B C

= = + + = +r k r i j k r i j

Coordinates of mass center G expressed in m.

( )( ) ( )( ) ( )( )9 0.9 9 0.6 0.6 0.9 9 0.3 1.2

27

0.3 0.6 0.6

A A B B C C

A B C

m m m

m m m

+ +=+ +

+ + + + +=

= + +

r r rr

k i j k i j

i j k

Position vectors relative to the mass center expressed in m.

( ) ( )( ) ( )( ) ( )

0.9 0.3 0.6 0.6 0.3 0.6 0.3

0.6 0.6 0.9 0.3 0.6 0.6 0.3 0.3

0.3 1.2 0.3 0.6 0.6 0.6 0.6

A A

B B

C C

′ = − = − + + = − − +

′ = − = + + − + + = +

′ = − = + − + + = −

r r r k i j k i j k

r r r i j k i j k i k

r r r i j i j k j k

Angular momenta.

( ) ( ) ( )( ) ( ) ( )

O A A A B B B C C C

G A A A B B B C C C

m m m

m m m

= × + × + ×

′ ′ ′= × + × + ×

H r v r v r v

H r v r v r v

Subtracting,

( ) ( ) ( ) ( )O G A A A A B B B B C C C Cm m m′ ′ ′− = − × + − × + − ×H H r r v r r v r r v

( ) ( ) ( )( )

0A A B B C C

A A B B C C

m m m

m m m

= × + × + ×

= × + + = ×

r v r v r v

r v v v r L

is parallel to .L r 2λ λ= ⋅ = ⋅L r L L r r

( )( )

2

2 2

2

4550 , 50 N s/m

0.9λ λ⋅= = = = ± ⋅

⋅L L

r r

( )( ) ( )( ) ( )( ) ( )9 9 9 50 0.3 0.6 0.6

A A B B C C

A B C

m m m

v v v

v v v r

j i k i j k

λ+ + =

+ + = ± + +





( ) Resolve into components and solve for , , and .A B C

a v v v

3.333 m/sAv = ( )3.33 m/s

A=v j

1.6667 m/sBv = ( )1.667 m/s

B=v i

3.333 m/sCv = ( )3.33 m/s

C=v k

(b) Angular momentum about O expressed in 2kg m /s.⋅

( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

0 0 0.9 0.6 0.6 0.9 0.3 1.2 0

0 9 0 9 0 0 0 0 9

8.1 8.1 5.4 10.8 2.7

8.1 10.8 8.1 2.7 5.4

9 4.5 9

O A A A B B B C C C

A B C

A B B C C

A C B C B

m m m

v v v

v v v v v

v v v v v

= × + × + ×

= + +

= − + − + −

= − + + − + −

= + −

H r v r v r v

i j k i j k i j k

i j k i j

i j k

i j k

( ) ( ) ( )2 2 29.00 kg m /s 4.50 kg m /s 9.00 kg m /sO

= ⋅ + ⋅ − ⋅H i j k






Position vectors expressed in ft.

3 6 , 6 3 , 3 3A B C

= + = + = +r i j r j k r i k

Momentum of each particle expressed in lb s.⋅

( ) ( )4 142 63 168 252

A AW

g g g= + = +v

i j i j

( ) ( )4 142 63 168 252

B BW

g g g= − + = − +v

i j i j

( ) ( )28 19 6 252 168

C CW

g g g= − = − −v

j k j k−−−−

Angular momentum of the system about O expressed in ft lb s.⋅ ⋅

( ) ( ) ( )

( )

13 6 0 0 6 3 3 0 3

168 252 0 168 252 0 0 252 168

1252 756 504 1008 756 504 756

10 0 0

A A B B C C

O A B C

W W W

g g g

g

g

g

= × + × + ×

= + + − − −

= − + − − + + + −

= + +

v v vH r r r

i j k i j k i j k

k i j k i j k

i j k

zeroO

=H






(a) 4 4 28 36 lbA B C

W W W W= + + = + + =

( )( ) ( )( ) ( )( ) 14 3 6 4 6 3 28 3 3

36

A A B B C C A A B B C Cm m m W W m

m W

+ + + += =

= + + + + +

r r r r r rr

i j j k i k

2.667 1.333 2.667= + +i j k

( ) ( ) ( )2.67 ft 1.333 ft 2.67 ft= + +r i j k (b) Linear momentum

( )1

A A B B C C A A B B C Cm m m m W W W

g= + + = + +v v v v v v v

( )( ) ( )( ) ( )( )14 42 63 4 42 63 28 9 6

g = + + − + + − − i j i j j k

( )1252 168

32.2= −j k

( ) ( )7.83 lb s 5.22 lb sm = ⋅ − ⋅v j k (c) Position vectors relative to the mass center G (ft).

( ) ( )

( ) ( )

( ) ( )

3 6 2.667 1.333 2.667

0.333 4.667 2.667

6 3 2.667 1.333 2.667

2.667 4.667 0.333

3 3 2.667 1.333 2.667

0.333 1.333 0.333

A A

B B

C C

′ = − = + − + +

= + −′ = − = + − + +

= − + +′ = − = + − + +

= − +

r r r i j i j k

i j k

r r r j k i j k

i j k

r r r i k i j k

i j k

Angular momentum about the mass center.

( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( )

10.333 4.667 2.667 2.667 4.667 0.333 0.333 1.333 0.333

4 42 4 63 0 4 42 4 63 0 0 28 9 28 6

A A B B C C

G A B C

W W W

g g g

g

′ ′ ′= × + × + ×

= − + − + − − − −

v v vH r r r

i j k i j k i j k

( ) ( ) ( )

( )

1672 448 700 84 56 112 308 56 84

1896 448 672 27.827 13.913 20.870

32.2

g= − − + − − + + + −

= − − = − −

i j k i j k i j k

i j k i j k

( ) ( ) ( )27.8 ft lb s 13.91 ft lb s 20.9 ft lb sG

= ⋅ ⋅ − ⋅ ⋅ − ⋅ ⋅H i j k





From Problem 14.28, O Gm= × +H r v H

2.667 1.333 2.667

0 7.83 5.22

27.826 13.913 20.870 27.827 13.913 20.870

0 0 0

O=

−

= − + + + − −= + +

i j k

H

i j k i j k

i j k

From Prob 14.11,

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( )

1

13 6 0 0 6 3 3 0 3

168 252 0 168 252 0 0 252 168

252 756 504 1008 756 504 756

10 0 0

O A A A B B B C C C

A A A B B B C C C

m m m

W W Wg

g

g

g

= × + × + ×

= × + × + ×

= + + − − −

1= − + − − + + + −

= + +

H r v r v r v

r v r v r v

i j k i j k i j k

k i j k i j k

i j k






Linear momentum of each particle expressed in kg m/s.⋅

3 2 4

8 6

6 15 9

A A

B B

C C

m

m

m

= − += += + −

v i j k

v i j

v i j k

Position vectors, (meters): 3 , 3 2.5 , 4 2A B C

= + = + = + +r j k r i k r i j k

( )2Angular momentum about , kg m /s .O ⋅

( ) ( ) ( )

( ) ( ) ( )

0 3 1 3 0 2.5 4 2 1

3 2 4 8 6 0 6 15 9

14 3 9 15 20 18 33 42 48

34 65 57

O A A A B B B C C Cm m m= × + × + ×

= + +− −

= + − + − + + + − + +

= − + +

H r v r v r v

i j k i j k i j k

i j k i j k i j k

i j k

( ) ( ) ( )2 2 234 kg m /s 65 kg m /s 57 kg m /sO

= − ⋅ + ⋅ + ⋅H i j k






Position vectors, (meters): 3 , 3 2.5 , 4 2A B C

= + = + = + +r j k r i k r i j k

( ) Mass center:a ( )A B C A A B B C Cm m m m m m+ + = + +r r r r

( )( ) ( )( ) ( )( )6 1 3 2 3 2.5 3 4 2

3 1.5 1.5

= + + + + + +

= + +

r j k i k i j k

r i j k

( ) ( ) ( )3.00 m 1.500 m 1.500 m= + +r i j k

Linear momentum of each particle, ( )kg m/s .⋅

3 2 4

8 6

6 15 9

A A

B B

C C

m

m

m

= − += += + −

v i j k

v i j

v i j k

(b) Linear momentum of the system, ( )kg m/s.⋅

17 19 5A A B B C C

m m m m= + + = + −v v v v i j k

( ) ( ) ( )17.00 kg m/s 19.00 kg m/s 5.00 kg m/sm = ⋅ + ⋅ − ⋅v i j k

Position vectors relative to the mass center, (meters).

3 1.5 0.5

1.5

0.5 0.5

A A

B B

C C

′ = − = − + −′ = − = − +′ = − = + −

r r r i j k

r r r j k

r r r i j k

(c) Angular momentum about G, ( )2kg m /s .⋅

( ) ( ) ( )

3 1.5 0.5 0 1.5 1 1 0.5 0.5

3 2 4 8 6 0 6 15 9

5 10.5 1.5 6 8 12 3 6 12

2 24.5 25.5

G A A A B B B C C Cm m m′ ′ ′= × + × + ×

= − − + − + −− −

= + + + − + + + + +

= + +

H r v r v r v

i j k i j k i j k

i j k i j k i j k

i j k

( ) ( ) ( )2 2 22.00 kg m /s 24.5 kg m /s 25.5 kg m /sG

= ⋅ + ⋅ + ⋅H i j k

( ) ( ) ( )2 2 2

3 1.5 1.5

17 19 5

36 kg m /s 40.5 kg m /s 31.5 kg m /s

m× =−

= − ⋅ + ⋅ + ⋅

i j k

r v

i j k

( ) ( ) ( )2 2 234 kg m /s + 65 kg m /s 57 kg m /sG

m+ × = − ⋅ ⋅ + ⋅H r v i j k





Angular momentum about O.

( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( )2 2 2

0 3 1 3 0 2.5 4 2 1

3 2 4 8 6 0 6 15 9

14 3 9 15 20 18 33 42 48

34 kg m /s + 65 kg m /s 57 kg m /s

O A A A B B B C C Cm m m= × + × + ×

= + +− −

= + − + − + + + − + +

= − ⋅ ⋅ + ⋅

H r v r v r v

i j k i j k i j k

i j k i j k i j k

i j k

Note that

O Gm= + ×H H r v






The mass center moves as if the projectile had not exploded.

( ) ( )( ) ( )( )

( ) ( )

22

0

1 160 2 9.81 2

2 2

120 m 19.62 m

t gt = − = −

= −

r v j i j

i j

( )A B A A B Bm m m m+ = +r r r

( )

( )( ) ( )

1

120 120 19.62 8 120 10 20

12

120 26.033 13.333

B A B A A

B

m m m

m

= + −

= − − − −

= − +

r r r

i j i j k

i j k

( ) ( ) ( )120.0 m 26.0 m 13.33 mB

= − +r i j k






There are no external forces. The mass center moves as if the explosion had not occurred.

( )( ) ( )0

450 4 1800 mt= = =r v i i

( )A B C A A B B C Cm m m m m m+ + = + +r r r r

( )

( )( ) ( )( )

( )( )

1

1500 1800 300 1200 350 600

50

150 2500 450 900

3300 750 900

C A B C A A B B

C

m m m m m

m

= + + − −

= − − −

− + +

= + +

r r r r

i i j k

i j k

i j k

( ) ( ) ( )3300 m 750 m 900 mC

= + +r i j k






Mass center at time of first collision.

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( )( ) ( )( ) ( )( )( ) ( )

1 1 1 1

1 1 1 1

1

1

9600 2800 27.8 3600 38.4 3200 120

40 ft 22.508 ft

A B C A A B B C C

A B C A A B B C C

m m m m m m

W W W W W W

+ + = + +

+ + = + +

= − + − +

= −

r r r r

r r r r

r j j i

r i j

Mass center at time of photo.

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( )( ) ( )( )( )( )( ) ( )

2 2 2 2

2 2 2 2

2

2

9600 2800 30.3 50.7 3600 30.3 61.2

3200 59.4 45.6

40 ft 22.5375 ft

A B C A A B B C C

A B C A A B B C C

m m m m m m

W W W W W W

+ + = + +

+ + = + +

= − + + − +

+ − −

= − +

r r r r

r r r r

r i j i j

i j

r i j

Since no external horizontal forces act, momentum is conserved, and the mass center moves at constant

velocity.

( ) ( ) ( ) ( )1 1 1A B C A A B B C C

m m m m m m+ + = + +v v v v (1)

2 1t− =r r v (2)

Combining (1) and (2), ( )( ) ( ) ( ) ( )2 1 1 1 1A B C A A B B C Cm m m m m m t + + − = + + r r v v v

( )( ) ( ) ( ) ( )2 1 1 1 1A B C A A B B C CW W W W W W t + + − = + + r r v v v

( )( ) ( )( ) ( )( )1

9600 80 45.0455 0 3600 3200 66Bv t − + = + + − i j j i

Components. : 768000 211200t− = −i 3.64 st =

: 432440 3600Bv t=j

( )

( )( )432440

30.0343600 3.6363

Bv = = 30.0 ft/s

Bv =






Mass center at time of first collision.

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( )( ) ( )( ) ( )( )( ) ( )

1 1 1 1

1 1 1 1

1

1

9600 2800 27.8 3600 38.4 3200 120

40 ft 22.508 ft

A B C A A B B C C

A B C A A B B C C

m m m m m m

W W W W W W

+ + = + +

+ + = + +

= − + − +

= −

r r r r

r r r r

r j j i

r i j

Mass center at time of photo.

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( )( ) ( )( )( )( )( ) ( )

2 2 2 2

2 2 2 2

2

2

9600 2800 30.3 50.7 3600 30.3 61.2

3200 59.4 45.6

40 ft 22.5375 ft

A B C A A B B C C

A B C A A B B C C

m m m m m m

W W W W W W

+ + = + +

+ + = + +

= − + + − +

+ − −

= − +

r r r r

r r r r

r i j i j

i j

r i j

Since no external horizontal forces act, momentum is conserved, and the mass center moves at constant

velocity.

( ) ( ) ( ) ( )1 1 1A B C A A B B C C

m m m m m m+ + = + +v v v v (1)

2 1t− =r r v (2)

Combining (1) and (2), ( )( ) ( ) ( ) ( )2 1 1 1 1A B C A A B B C Cm m m m m m t + + − = + + r r v v v

( )( ) ( ) ( ) ( )2 1 1 1 1A B C A A B B C CW W W W W W t + + − = + + r r v v v

( )( ) ( )( ) ( )( ) ( )1 1

9600 80 45.0455 0 3600 3200 3.4B Cv v − + = + + i j j i

Components.

( ) ( )1 1

: 432440 12240 , 35.33 ft/s,B Bv v= =j

24.1mi/hBv =

( ) ( )1 1

: 768000 10880 , 70.588 ft/s,C Cv v− = − =i

48.1mi/hCv =






Projectile motion 2

0, 9.81 m/s , 0x y z

a a g a= = − = − =

( ) ( ) ( )00 0

165 m/s, 0, 0x y zv v v= = =

After the chain breaks the mass center continues the original projectile motion.

At 1.5 s,t =

( ) ( )( )0 00 165 1.5 247.5 m

xx x v t= + = + =

( ) ( )( )22

00

1 115 0 9.81 1.5 3.9638 m

2 2y

y y v t gt= + − = + − =

( )0 00

zz z v t= + =

Position of first cannon ball at this time is

1 1 1240 m, 0, 7 mx y z= = =

Definition of mass center: ( )1 2 1 1 2 2m m m m+ = +r r r

( )1 2

2 1

2 2

m m m

m m

1+= −r r r

( ) ( )

( ) ( )

30 15247.5 3.9638 240 7

15 15

255 m . m 7

= + − +

= + 7 9276 −

i j i k

i j k

Position of second cannon ball: 2 2 2

255 m, 7.93 m, 7 mx y z= = = −






Place the vertical y axis along the initial vertical path of the rocket. Let the x axis be directed to the right (east).

Motion of the mass center: 0, 0, 0x x

a v x= = =

2

9.81 m/sy

a g= − = −

028 9.81

y yv v a t t= + = −

2 2

0 0

160 28 4.905

2

At 5.85 s, 0, 55.939 m

yy y v t a t t t

t x y

= + + = + −

= = =

:A A B B

Definition of mass center m m m= +r r r

component: 3 1 2

0 74.4 2 37.2 m

A B

B B

x x x x

x x

= += − + =

( )( ) component: 3 1 2

3 55.939 0 2 83.9 m

A B

B B

y y y y

y y

= +

= + =

.Position of part B 37.2 m(east), 83.9 m(up)






Velocities of pieces C and D after impact and fracture.

( ) ( )

( ) ( )

2.13 m/s, 3tan30 m/s

0.7

2.12.333 m/s, 2.3333tan m/s

0.9

CC Cx y

C

DD Dx y

D

xv v

t

xv v

t

θ

′ ′= = = = °

′ ′= = = = −

Assume that during the impact the impulse between spheres A and B is directed along the x axis. Then, the y

component of momentum of sphere A is conserved.

( )0 A ym v′=

Conservation of momentum of system:

( ) ( ) ( )0: 0

A B A A C C D D xxm v m m v m v m v′ ′ ′+ = + +

( ) ( ) ( )4.8 0 3 2.33332 2

A

m m

m mv′+ = + +

( )a 2.13 m/sA′ =v

( ) ( ) ( ) ( ) ( ): 0 0A B A A C C D Dy yym m m v m v m v′ ′ ′+ = + +

( ) ( )0 0 0 3tan30 2.3333tan2 2

m m θ+ = + ° −

( )b 3

tan tan30 0.74232.3333

θ = ° = 36.6θ = °

( ) ( ) ( ) ( )22 2 23 3tan30C C Cx y

v v v= + = + o

3.46 m/sCv =

( ) ( ) ( ) ( )22 2 22.3333 2.3333tan36.6D D Dx y

v v v= + = + o

2.91m/sDv =






( )( )( )( )

0 0Velocity vectors: cos30 sin 30

sin 7.4 cos7.4

sin 49.3 cos 49.3

cos 45 sin 45

A A

B B

C C

v

v

v

v

= ° + °

= ° + °

= ° − °

= ° + °

v i j

v i j

v i j

v i j


0A A A B B C Cm m m m= + +v v v v

Divide by and substitute data.A B C

m m m= =

( ) ( ) ( )( )

4 cos30 sin30 sin 7.4 cos7.4 sin 49.3 cos49.3

2.1 cos45 sin 45

A Bv v° + ° = ° + ° + ° − °

+ ° + °

i j i j i j

i j

Resolve into components and rearrange.

( ) ( )( ) ( )

: sin 7.4 sin 49.3 4cos30 2.1cos 45

: cos7.4 cos 49.3 4sin 30 2.1sin 45

A B

A B

v v

v v

° + ° = − °

° − ° = − °

i

j

o

o

Solving simultaneously,

(a) 2.01m/sAv =

(b) 2.27 m/sBv =






( ) ( )

0190 mi/h east 278.67 ft/s

Place orgin at point of impact.

A= =v i

0 0 00, 0, 0x y z= = =

After impact the motion is projectile motion.

2

0 0

1

2t gt= + −r r v j

0

0

1

2gt

t

−= +r rv j

where ( ) ( ) ( )1600 ft 2400 ft 400 ft= − +r i j k

0

0=r

( )( )

( ) ( ) ( )

0

1600 2400 400 132.2 12

12 12 12 2

133.333 ft/s ft/s 33.333 ft/s

= − + +

= − 6.80 +

v i j k j

i j k

Impact: Conservation of momentum.

( ) ( ) ( ) 00 0A A B B A Bm m m m+ = +v v v

( ) ( )00 0

A B A

B A

B B

m m m

m m

+= −v v v

( ) ( )

( ) ( ) ( )

23000 10000133.333 6.80 33.333 278.67

13000 13000

21.537 ft/s 12.031 ft/s 58.975 ft/s

= − + −

= − +

i j k i

i j k

Components: ( )21.537 ft/s 14.69 mi/h east =i

( )58.974 ft/s 40.2 mi/h south =k

( )12.031 ft/s 8.20 mi/h down − =j






Weight of arrow: 2 oz 0.125 lb.

Weight of bird: 6 lb.

A

B

W

W

= =

=

Conservation of momentum: Let v be velocity immediately after impact.

A B A B

A B

W W W W

g g g

++ =v v v

( )( ) ( )( )0.125 180 240 6 30

6.125

29.388 3.6735 4.8980

A A B B

A B A B

W W

W W W W

+ += + =

+ += + +

j k iv vv

i j k

( ) 2

00

1Vertical motion:

2y

y y v t gt= + −

( ) 2 210 45 3.6735 32.2 or 0.22817 2.7950 0

2t t t t= + − − − =

Solving for , 1.7898 st t =

Horizontal motion: ,x z

x v t z v t= =

( )( )( )( )29.388 1.7898 52.6 ft

4.8980 1.7898 8.77 ft

x

z

= =

= =

( ) ( )52.6 ft 8.77 ftP

= +r i k






( )( )( )

1 2 1 2

1 2 1 2

1 2 1 2

Position vectors (mm): 80 80 40 120

33 70 10 78.032

48 15 50.289

A A A A

B B B B

C C C C

= + + =

= − + − =

= − =

i j k

i j k

j k

1 2

1 2

1 2

Unit vectors: Along , 0.66667 0.66667 0.33333

Along , 0.42290 0.89707 0.12815

Along , 0.95448 0.29828

A

B

C

A A

B B

C C

= + += − + −= −

i j k

i j k

j k

λλλλλλλλλλλλ

Velocity vectors after the collisions:

A A A B B B C C Cv v v= = =v v vλλλλ λλλλ λλλλ


0 0 04 4 4 4

A B Cm m m m m m+ + = + +u v v v v v

Divide by m and substitute data.

( )600 750 800 2400 2400 4 4A A B B C Cv v v− + − + + = + +i j k j j λλλλ λλλλ λλλλ

Resolving into components,

: 600 0.66667 1.69160

: 5550 0.66667 3.58828 3.81792

: 800 0.33333 0.51260 1.19312

A B

A B C

A B C

v v

v v v

v v v

− = −= + +

− = − −

i

j

k

Solving the three equations simultaneously,

919.26 m/s, 716.98 m/s, 619.30 m/sA B Cv v v= = =

919 m/sAv =

717 m/sBv =

619 m/sCv =






(ft):Position vectors 18D

=r k

/ /

/ /

/ /

7.5 7.5 18 19.5

18 9 18 9 18 27

13.5 13.5 18 22.5

A A D A D

B B D B D

C C D C D

r

r

r

= − = − − =

= + = + − =

= − = − − =

r i r i k

r i j r i j k

r j r j k

( )

( )

( )

/

/

/

1 : Along , 7.5 18

19.5

1Along , 18 9 18

27

1Along , 13.5 18

22.5

A D A

B D B

C D C

Unit vectors = − −

= + −

= − −

r i k

r i j k

r j k

λλλλ

λλλλ

λλλλ

Assume that elevation changes due to gravity may be neglected. Then, the velocity vectors after the

explosition have the directions of the unit vectors.


0Conservation of momentum:

A A B B C Cm m m m= + +v v v v

( ) ( ) ( ) ( )18 8 6 460 45 1800 7.5 18 18 9 18 13.5 18

19.5 27 22.5

A B Cv v v

g g g g

− − = − − + + − + − −

i j k i k i j k j k

Multiply by g and resolve into components.

1080 60 10819.5 27

810 54 5227 22.5

32400 144 108 7219.5 27 22.5

A B

B C

A B C

v v

v v

v v v

= − +

− = −

− = − − −

Solving, 119.94419.5

Av = 2340 ft/s

Av =

76.63527

Bv = 2070 ft/s

Bv =

95.16022.5

Cv = 2140 ft/s

Cv =






(ft):Position vectors 18D

=r k

/ /

/ /

/ /

7.5 7.5 18 19.5

18 9 18 9 18 27

13.5 13.5 18 22.5

A A D A D

B B D B D

C C D C D

r

r

r

= − = − − =

= + = + − =

= − = − − =

r i r i k

r i j r i j k

r j r j k

( )

( )

( )

/

/

/

1 : Along , 7.5 18

19.5

1Along , 18 9 18

27

1Along , 13.5 18

22.5

A D A

B D B

C D C

Unit vectors = − −

= + −

= − −

r i k

r i j k

r j k

λλλλ

λλλλ

λλλλ

Assume that elevation changes due to gravity may be neglected. Then, the velocity vectors after the explosition

have the directions of the unit vectors.


/19.5

where 1950 ft/s0.010

A D

A

A

rv

t

= = =

/

271500 ft/s

0.018

B D

B

B

rv

t

= = =

C/

22.51875 ft/s

0.012

D

C

C

rv

t

= = =

( ) ( )so that 750 ft/s 1800 ft/sA

= − −v i k

( ) ( ) ( )1000 ft/s + 500 ft/s 1000 ft/sB

= −v i j k

( ) ( )1125 ft 1500 ft/sC

= − −v j k

Conservation of momentum: 0 A A B B C C

m m m m= + +v v v v

( ) ( ) ( )0750 1800 1000 500 1000 1125 1500

A B CW W W W

vg g g g

− = − − + + − + − −

k i k i j k j k





Divide by g and resolve into components.

0

: 0 750 1000

: 0 500 1125

: 1800 1000 1500

A B

B C

A B C

W W

W W

Wv W W W

= − += −

− = − − −

i

j

k

(1)

(2)

(3)

Since mass is conserved, 6 lbA B C

W W W W= + + = (4)

Solving equations (1), (2), and (4) simultaneously,

(a) 2.88 lb, 2.16 lb, 0.96 lbA B C

W W W= = =

substituting into (3),

( )( ) ( )( ) ( )( )06 1800 2.88 1000 2.16 1500 0.96v− = − − −

(b) 0

1464 ft/sv =






( )

( )

( ) ( )

1

1

1 1

From Eq. (14.7),n

O i i i

i

n

i i i

i

n n

i i i i i

i i

G

m

m

m v r m

m

=

=

= =

= ×

′ = + ×

′= × + ×

= × +

∑

∑

∑ ∑

H r v

r r v

r v

r v H






( )

( )

( )

( )

( )( )

1

1

1 1

1

1

if, and only if, 0

i A i

n

A i i i

i

n

i i A i

i

n n

i i A i i i

i i

n

i i A A

i

n

i i A A A

i

A A A

A A A A

m

m

m m

m

m

m

m

=

=

= =

=

=

′= +

′= ×

′ ′= × +

′ ′= × + ×

′ ′= × +

′= − × +

′= − × +

′= − × =

∑

∑

∑ ∑

∑

∑

v v v

H r v

r v v

r v r v

r v H

r r v H

r r v H

H H r r v

This condition is satisfied if,

( ) 0 Point has zero velocity.

or ( ) Point coincides with the mass center.

or ( ) is parallel to . Velocity is directed along line .

A

A

A A A

a A

b A

c AG

==

−

v

r r

v r r v






From equation (1), ( )1

n

A i i i

i

m

=′ ′ ′= ×∑H r v

( ) ( )1

n

A i A i i A

i

m

=′ = − × − ∑H r r v v

Differentiate with respect to time.

( ) ( ) ( ) ( )1 1

n n

A i A i i A i A i i A

i i

m m

= =′ = − × − + − × − ∑ ∑H r r v v r r v v& & & & &

But , , , andi i i i A A A A

= = = =r v v a r v v a& && &

( ) ( )

( ) ( )

( ) ( )

( )

1

1

1 1

Hence, 0n

A i A i i A

i

n

i A i i A

i

n n

i A i i i A A

i i

A A A

m

m

m

m r

=

=

= =

′ = + − × −

= − × −

= − × − − ×

= − − ×

∑

∑

∑ ∑

H r r a a

r r F a

r r F r r a

M r a

&

( )if, and only if, 0A A A A

M m′ = − × =H r r a&

This condition is satisfied if

( ) 0 The frame is newtonian.

or ( ) Point coincides with the mass center.

or ( ) is parallel to . Acceleration is directed along line .

A

A

A A A

a

b A

c AG

==

−

a

r r

a r r a






The masses are m for the bullet and A

m and B

m for the blocks.

The bullet passes through block A and embeds in block B. Momentum is conserved.

( ) ( )0 0Initial momentum: 0 0

A Bmv m m mv+ + =

Final momentum:B A A B B

mv m v m v+ +

0Equating,

B A A B Bmv mv m v m v= + +

( )( ) ( )( ) 3

0

3 3 2.5 543.434 10 kg

500 5

A A B B

B

m v m v

m

v v

−++= = = ×− −

The bullet passes through block A. Momentum is conserved.

( )0 0Initial momentum: 0

Amv m mv+ =

1Final momentum:

A Amv m v+

0 1Equating,

A Amv mv m v= +

( )( ) ( )( )3

0

1 3

43.434 10 500 3 3

292.79 m/s43.434 10

A Amv m v

v

m

−

−

× −−= = =×

(a) Bullet passes through block A. Kinetic energies.

( )( )22 3

0 0

1 1Before: 43.434 10 500 5429 J

2 2T mv

−= = × =

( )( ) ( )( )2 22 2 3

1 1

1 1 1 1After: 43.434 10 292.79 3 3 1875 J

2 2 2 2A A

T mv m v−= + = × + =

0 1Lost: 5429 1875 3554 JT T− = − = energy lost 3550 J=

(b) Bullet becomes embedded in block B. Kinetic energies.

( )( )22 3

2 1

1 1Before: 43.434 10 292.79 1861.7 J

2 2T mv

−= = × =

( ) ( )( )22

3

1 1After: 2.54343 5 31.8 J

2 2B B

T m m v= + = =

2 3Lost: 1862 31.8 1830 JT T− = − = energy lost 1830 J=






Data and results from Prob. 14.1.

Masses: 1350 kg,A B

m m= = 5400 kgc

m =

Initial velocities: 0 0

( ) ( ) 0,A Bv v= =

0( ) 8 km/h = 2.2222 m/s

cv =

Velocities after first collision:

1

( ) 0,Av =

1 1( ) ( ) 1.77778 m/s

B cv v= =

Velocities after second collision

2.9630 m/s,Av = 1.18519 m/s

B cv v= =

Initial kinetic energy: 2 2 2

0 0 0 0

1 1 1( ) ( ) ( )

2 2 2A A B B c B

T m v m v m v= + +

2 3

0

10 0 (5400)(2.2222) 13.3333 10 J

2T = + + = ×

Kinetic energy after the first collision:

( )22 2

1 1 11

1 1 1( ) ( )

2 2 2A A B B c c

T m v m v m v= + +

2 2 31 10 (1350) (1.77778) (5400)(1.77778) 10.6667 10 J

2 2= + + = ×

Kinetic energy after the second collision:

2 2 2

2

1 1 1

2 2 2A A B B c c

T m v m v m v= + +

2 2 2 31 1 1(1350)(2.9630) (1350)(1.18519) (5400)(1.18519) 10.6668 10 J

2 2 2= + + = ×

Kinetic energy lost in first collision: 3

0 12.6667 10 JT T− = ×

2.67 kJ

Kinetic energies before and after second collision:

2 110.67 kJT T= =






( )( )22 20004000 3700The masses are 124.2 slugs, 114.9 slugs, and 1366.5 slugs

32.2 32.2 32.2A B F

m m m= = = = = =

Let , , and be the sought after velocities in ft/s, positive to the right.A B Fv v v


0.A B Fv v v= = =


0.A A B B F F

m v m v m v+ + =


conserved.

0A A B B F F

m v m v m v= + +

124.2 114.9 1366.5 0A B Fv v v+ + = (1)

The relative velocities are given as

/

/

7 ft/s

3.5 ft/s

A F A F

B F B F

v v v

v v v

= − = −

= − = −

(2)

(3)


6.208 ft/s, 2.708 ft/s, 0.7919 ft/sA B Fv v v= − = − =

( ) ( ) ( )22 2

1 0 0 0

1 1 1Initial kinetic energy: 0

2 2 2A A B B C

T m v m v v= + + =

2 2 2

2

1 1 1Final kinetic energy:

2 2 2A A B B C C

T m v m v m v= + +

( )( ) ( )( ) ( )( )2 2 2

2

1 1 1124.2 6.208 114.9 2.708 1366.5 0.7919

2 2 2

3243 ft lb

T = + +

= ⋅

Work done by engines: 1 1T U+

2 2T=

1U

2 2 13243 ft lbT T= − = ⋅ 1

U2

3240 ft lb= ⋅






From the solution to Prob. 14.27,

Initial velocity of 6-lb shell: 0

1464 ft/sv =

Weights of fragments: 2.88 lb,A

W = 2.16 lb,B

W = 0.96 lbc

W =

Speeds of fragments: 1950 ft/s,Av = 1500 ft/s,

Bv = 1875 ft/s

cv =

Total kinetic energy before the explosion.

( )22 3

0 0

1 1 61464 199.69 10 ft lb

2 2 32.2

WT v

g

= = = × ⋅

Total kinetic energy after the explosion.

2 2 2

1

1 1 1

2 2 2

A B c

A B c

W W WT v v v

g g g= + +

( ) ( ) ( )2 2 21 2.88 1 2.16 1 0.961950 1500 1875

2 32.2 2 32.2 2 32.2

= + +

3297.92 10 ft lb= × ⋅

Increase in kinetic energy. 3

1 098.2 10 ft lbT T− = × ⋅






Velocity of mass center: ( )A B A A B Bm m m m+ = +v v v

A A B B

A B

m m

m m

+=+

v v

v

Velocities relative to the mass center:

( )

( )

B A BA A B B

A A A

A B A B

A A BA A B B

B B B

A B A B

mm m

m m m m

mm m

m m m m

−+′ = − = − =+ +

−+′ = − = − =+ +

v vv v

v v v v

v vv v

v v v v

Energies:

( ) ( )( )

( ) ( )( )

2

2

2

2

1

2 2

1

2 2

A B A B A B

A A A A

A B

A B A B A B

B B B B

A B

m mE m

m m

m mE m

m m

− ⋅ −′ ′= ⋅ =

+

− ⋅ −′ ′= ⋅ =

+

v v v v

v v

v v v v

v v

( ) :a Ratio / /A B B A

E E m m=

( ) 135 km/h 37.5 m/sA

b = =v , 90 km/h 25 m/sB

= =v

62.5 m/sA B

− =v v

( )( ) ( )

( )( )

2 2

3

2

2400 1350 62.5607.5 10 J

2 3750A

E = = × 608 kJA

E =

( ) ( )( )

( )( )

2 2

6

2

2400 1350 62.51.08 10 J

2 3750B

E = = × 1080 kJB

E =






( ) :A B A A B B

Velocity of mass center m m m m+ = +v v v

A A B B

A B

m m

m m

+=+

v v

v

Velocities relative to the mass center:

( )

( )

B A BA A B B

A A A

A B A B

A A BA A B B

B B B

A B A B

mm m

m m m m

mm m

m m m m

−+′ = − = − =+ +

−+′ = − = − =+ +

v vv v

v v v v

v vv v

v v v v

Energies:

( ) ( )( )

( ) ( )( )

2

2

2

2

1

2 2

1

2 2

A B A B A B

A A A A

A B

A B A B A B

B B B B

A B

m mE m

m m

m mE m

m m

− ⋅ −′ ′= ⋅ =

+

− ⋅ −′ ′= ⋅ =

+

v v v v

v v

v v v v

v v

( ) ( )2 2

0 00 0

1 1Energies from tests: ,

2 2A A B B

E m v E m v= =

( )( ) ( )

( )

( )( ) ( )

( )

2

2 2

0 0

2

2 2

0 0

Serverities:B A B A BA

A

A A B

A A B A BB

B

B A B

mES

E m m v

mES

E m m v

− ⋅ −= =

+

− ⋅ −= =

+

v v v v

v v v v

:Ratio

2

2

A B

B A

S m

S m=






(a) A strikes B and C simultaneously.

During the impact, the contact impulses make 30° angles with the velocity 0

v

( )( )

Thus, cos30 sin30

cos30 sin30

B B

C C

v

v

= ° + °

= ° − °

v i j

v i j

By symmetry,A A

v=v i

0Conservation of momentum:

A B Cm m m m= + +v v v v

0

component: 0 0 sin 30 sin 30

component: cos30 cos30

B C C B

A B C

y mv mv v v

x mv mv mv mv

= + ° − ° == + ° + °

( )0 0

0

2,

cos30 3 3

A A

B C A B C

v v v v

v v v v v v

− −+ = = − = =o

Conservation of energy: 2 2 2 2

0

1 1 1 1

2 2 2 2A B C

mv mv mv mv= + +

( )

( )( ) ( )

( )

22 2

0 0

22 2

0 0 0 0

0 0 0 0

0 0

2

3

2

3

2 1 5 1

3 3 3 5

6 2 3

55 3

A A

A A A A

A A A A

B C

v v v v

v v v v v v v v

v v v v v v v v

v v v v

= + −

− = − + = −

+ = − = − = −

= = =

00.200

Av=v

00.693

Bv=v 30°





(b) A strikes B before it strikes C.

First impact; A strikes B.

During the impact, the contact impulse makes a 30° angle with the velocity 0.v

( )Thus, cos30 sin30B B

v= ° + °v i j

0Conservation of momentum.

A Bm m m= +v v v

( ) ( )( ) ( )0 0

component: 0 sin30 sin30

component: cos30 cos30

A B A By y

A B A Bx x

y m v mv v v

x v m v mv v v v

′ ′= + ° = − °

′ ′= + ° = − °

Conservation of energy:

( ) ( )

( ) ( )

( )

2 22 2

0

2 2 2

0

2 2 2 2 2 2

0 0

1 1 1 1

2 2 2 2

1 1 1cos30 sin30

2 2 2

12 cos30 cos 30 sin 30

2

A A Bx y

B B B

B B B B

mv m v m v mv

m v v v v

m v v v v v v

′ ′= + +

= − ° + ° +

= − ° + ° + ° +

( )

( )

2

0 0 0 0

0 0

3 1cos30 , sin 30 ,

2 4

3cos30 sin 30

4

B A x

A y

v v v v v v

v v v

′= ° = = ° =

′ = − ° ° = −

Second impact: A strikes C.

During the impact, the contact impulse makes a 30o angle with the velocity 0.v

( )Thus, cos30 sin30C C

v= ° − °v i j

Conservation of momentum:A A C

m m m′ = +v v v





( ) ( )

( ) ( )

( ) ( )

( ) ( )

0

0

component: cos30 ,

1cos30 cos30

4

component: sin30

3sin30 sin30

4

A A Cx x

A A C Cx x

A A Cy y

A A C Cy y

x m v m v mv

v v v v v

y m v m v mv

v v v v v

′ = + °

′= − ° = − °

′ = − °

′= + ° = − + °


( ) ( ) ( ) ( )2 2 2 2 2

22

2 2 2

0 0 0 0

2 2 2

0 0

2 2 2 2

0 0

1 1 1 1 1

2 2 2 2 2

1 1 3 1 1 3cos30 sin30

2 16 16 2 4 4

1 1 1cos30 cos 30

2 16 2

3 3sin30 sin 30

16 2

A A A A Cx y x y

C C C

C C

C C C

m v m v m v m v mv

m v v m v v v v v

m v v v v

v v v v v

′ ′+ = + +

+ = − ° + − + ° +

= − ° + °

+ − ° + ° +

2

0

1 30 cos30 sin30 2

2 2C C

v v v

= − ° + ° +

( )

( )

0 0

0 0 0

0 0 0

1 3 3cos30 sin30

4 4 4

1 3 1cos30

4 4 8

3 3 3sin30

4 4 8

C

A x

A y

v v v

v v v v

v v v v

= ° + ° =

= − ° = −

= − + ° = −

00.250

Av=v 60°

00.866

Bv=v 30°

00.433

Cv=v 30°






(a) Velocity of B at maximum elevation. At maximum elevation ball B is at rest relative to cart A. B A

=v v

Use impulse-momentum principle.

components:x ( )00

A A A B B A B Bm v m v m v m m v+ = + = +

0A

B

A B

m v

v

m m

=+

(b) Conservation of energy:

( ) ( )

2

1 0 1

2 2

2 2 2 0

2

2

1, 0

2

1 1 1

2 2 2 2

A

A

A A B B A B B

A B

B

T m v V

m vT m v m v m m v

m m

V m gh

= =

= + = + =+

=

( )

2 2 1 1

2 2

20

0

1

2 2

A

B A

A B

T V T V

m vm gh m v

m m

+ = +

+ =+

2 2

2 0

0

1

2

A

A

B A B

m vh m v

m g m m

= − +

2

0

2

A

A B

m vh

m m g=

+






Velocity vectors: ( )0 0cos30 sin30v= ° − °v i j

015 ft/sv =

A A

v= −v j

( )( )sin30 cos30

cos30 sin30

B B

C C

v

v

= ° − °

= ° + °

v i j

v i j


0 A B C

m m m m= + +v v v v

Divide by m and resolve into components.

i: 0cos30 sin 30 cos30

B Cv v v° = ° + °

j: 0sin30 cos30 sin30

A B Cv v v v− ° = − − ° + °

Solving for and ,B Cv v

( ) ( )0 0

3 1

2 2B A C Av v v v v v= − = +


0

1 1 1 1

2 2 2 2A B C

mv mv mv mv= + +

Divide by 1

2m and substitute for and .

B Cv v

( ) ( )2 22 2

0 0 0

2 2

0 0

3 1

4 4

2

A A A

A A

v v v v v v

v v v v

= + − + +

= + −

0

17.5 ft/s

2Av v= = 7.50 ft/s

Av =

( )315 7.5 6.4952 ft/s

2Bv = − = 6.50 ft/s

Bv =

( )115 7.5 11.25 ft/s

2Cv = + = 11.25 ft/s

Cv =






Velocity vectors: ( )0 0cos45 sin 45v= ° + °v i j

015 ft/sv =

A A

v=v j

( )( )sin 60 cos60

cos60 sin 60

B B

C C

v

v

= ° − °

= ° + °

v i j

v i j


0 A B C

m m m m= + +v v v v


i: 0cos45 sin 60 cos60

B Cv v v° = ° + °

j: 0sin 45 cos60 sin 60

A B Cv v v v° = − ° + °

Solving for and ,B Cv v

0 00.25882 0.5 0.96593 0.86603

B A C Av v v v v v= − = −


0

1 1 1 1

2 2 2 2A B C

mv mv mv mv= + +

Divide by m and substitute for and .B Cv v

( ) ( )2 22 2

0 0 0

2 2

0 0

0.25882 0.5 0.96593 0.86603

1.4142 2

A A A

A A

v v v v v v

v v v v

= + + + −

= + +

0

0.70711 10.607 ft/sAv v= = 10.61 ft/s

Av =

0

0.61237 9.1856 ft/sBv v= = 9.19 ft/s

Bv =

0

0.35356 5.303 ft/sCv v= = 5.30 ft/s

Cv =






1sin ,

3θ =

8cos ,

3θ = 19.471θ = °

Velocity vectors 0 0

v= −v j

( )cos sinA A

v θ θ= −v i j

( )/sin cos

B A Bu θ θ= − −v i j

/B A B A

= +v v v

C C

v=v j

Conservation of momentum: 0 /

2A B C A B A C

m m m m m m m= + + = + +v v v v v v v


i: 0 2 cos sinA Bv uθ θ= −

−j : 0

2 sin cosA B C

v v u vθ θ= + + −

Solving for and ,A Bv u ( ) ( )0 0

1 0.94281

6A C B Cv v v u v v= + = +


0

1 1 1 1

2 2 2 2A B C

mv mv mv mv= + +

( )2 2 2 21 1 1

2 2 2A A B C

mv m v u mv= + + +

Divide by 1

2m and substitute for and .

A Bv u

( ) ( ) ( )2

2 222 2

0 0 0

12 0.94281

6C C C

v v v v v v

= + + + +

( )22 2

0 0 00.94445 0.02857

C C Cv v v v v v− = + =

00.0286

Cv=v

[0 00.17143 0.17143

A Av v v= =v ]19.471° ,

00.1714

Av=v 19.5°

[0 / 00.96975 0.96975

B B Au v v= v ]19.471°

/B A B A

= +v v v 0.985B

=v 80.1°






1 8sin , cos , 19.471

3 3θ θ θ= = = °

C strikes B.


0 0

or B C B C

m m m v v v′ ′= + = −v v v


( )22 2

0

1 1 1

2 2 2B C

mv m v mv′= +

( )22 2

0 0 C Cv v v v= − +

0Cv =

0B

v v′ =

Cord becomes taut.

Velocity vectors:

A A

v=v θ

/B A B

u=v θ

Conservation of momentum: /B A A B A

m m m m′ = + +v v v v


+ :θ sin 2B Av vθ′ =

0

1 1sin

2 6A Bv v vθ′= =





(a) 0

6A

v=v 19.5°

+ :θ cosB Bv uθ′ = 0

cos 83

B B

v

u v θ′= =

0

1 19.471

6B

v = °

v [ 00.94281v+ ]19.471°

[ 00.95743

Bv=v ]80.8°

00.957

Bv=v 80.5°

0C

=v

Initial kinetic energy: 2

1 0

1

2T mv=

Final kinetic energy: 2 2 2

2

1 1 1

2 2 2A B C

T mv mv mv= + +

( ) ( )2

22 2

0 0

1 1 1.95743 0 0.94444

2 6 2mv mv

= + + =

(b) Fraction lost: 1 2

1

1 0.944440.05555

1

T T

T

− −= =

Fraction of energy lost = 0.0556






(a) Use part (a) of sample Problem 14.4 with A B

m m m.= =

0 0

1

2

m

v v v

m m

= =+

0

1

2A Bv v v= =

(b) Consider initial position and position when 0θ = again.

Conservation of momentum

0 A A B B

mv m v m v= +

0B Av v v+ = (1)

Conservation of energy

2 2 2

0

1 1 1

2 2 2A B

mv mv mv= + (2)

Substituting (1) into (2),

2 2 21 1 1( )

2 2 2A B A B

m v v mv mv+ = +

0A B

mv v =

Either 0Av = with

0Bv v=

or 0Bv = with

0Av v=

(c) Consider positions when max

θ θ= and min

.θ θ=

Since /

0B Av = at these states

B Av v=

Conservation of momentum

0 A B

mv mv mv= +

0

1

2B Av v v= =





Conservation of energy would show that the elevation of B is the same for max

θ θ= and min

.θ θ=

Both A and B keep moving to the right with A and B stopping intermittently.






Relative velocity and acceleration.

/B A B A= + =v v v [vA ] + [vB/A 30° ]

/B A B A= + =a a a [aA ] + [aB/A 30° ]

Draw free body diagrams and apply Newton’s second law.

Block:

:F ma∑ = 1

cos30 sin30B B A

N m g m a− °= − ° (1)

:F ma∑ = 1 /sin30 cos30

B B A B AN m a m a°= °− (2)

Wedge:

:F ma∑ = 1sin30

A AN m a°= (3)

Rearranging (1), (2), and (3) and applying numerical data,

1

(6sin30 ) (6)(9.81)cos30A

N a+ ° = ° (1)

1 /

(sin30 ) 6 (6cos30 ) 0A B A

N a a° + − ° = (2)

1

(sin30 ) 10 0A

N a° − = (3)


1

44.325N,N = 22.2163 m/s ,

Aa = 2

/6.8243 m/s

B Aa =





Sliding motion of block relative to wedge.

2

/

/ /

( )

2

B A

B A B A

v

a s=

/ / /

2 (2)(6.8243)(1.0) 3.6944 m/sB A B A B Av a s= = =

v 3.6944

0.54136 s6.8243

B/A

B/A

t

a

= = =

Motion of wedge.

(2.2163)(0.54136) 1.1998 m/sA Av a t= = =

(a) Velocity of B relative to A. /

3.69 m/sB A

=v 30°

(b) Velocity of A . 1.200 m/sA

=v






There are no external forces. Momentum is conserved.

(a) Moments about D : ( )00.9 1.8 0.9

A C C A B Bm v m v m m v= + +

( ) ( )( ) ( )0

0.90.90.5 12 2.5 3.50

1.8 1.8

A BA

C B

C C

m mm

v v v

m m

+= − = − = 3.50 m/s

Cv =

Moments about C : ( )00.9 0.9 1.8

A A B B D Dm v m m v m v= + +

( ) ( )( ) ( )( )0

0.90.90.25 12 0.5 2.5 1.750m/s

1.8 1.8

A BA

D B

D D

m mm v

v v

m m

+= − = − = 1.750 m/s

Dv =

(b) Initial kinetic energy:

( )22

1 0

1 17.5 12 540 N m

2 2A

T m v= = = ⋅

Final kinetic energy:

( ) ( ) ( )

2 2 2

2

2 2 2

1 1 1( )

2 2 2

1 1 115 2.5 7.5 3.5 15 1.750 115.78 N m

2 2 2

A B B C C D DT m m v m v m v= + + +

= + + = ⋅

Energy lost: 540 115.78 424.22 N m− = ⋅

Fraction of energy lost 424.22

0.786540

= =

( )1 2

1

0.786T T

T

−=






There are no external forces. Momentum is conserved.

(a) Moments about D : 00.9 1.8 0.9

A C C B Bm v m v m v= +

( )( ) ( )( )0

0.9 0.90.5 12 0.5 3.5 4.25

1.8 1.8

A B

C B

C C

m m

v v v

m m

= − = − = 4.25 m/sCv =

Moments about C : 0

0.9 1.8 0.9A D D B B

m v m v m v= +

( )( ) ( )( )0

0.9 0.90.25 12 0.25 3.5 2.125 m/s

1.8 1.8

A B

D B

D D

m m

v v v

m m

= − = − = 2.13 m/sDv =

(b) Initial kinetic energy:

( )22

1 0

1 17.5 12 540 N m

2 2A

T m v= = = ⋅

Final kinetic energy:

( ) ( ) ( )

2 2 2

2

2 2 2

1 1 1

2 2 2

1 1 17.5 3.5 7.5 4.25 15 2.125 147.54 N m

2 2 2

B B C C D DT m v m v m v= + +

= + + = ⋅

Energy lost: 540 147.54 392.46 N m− = ⋅

Fraction of energy lost 392.46

0.727540

= =

( )1 2

1

0.727T T

T

−=






(a) Linear and angular momentum.

00

A A B Bm m mv= + = +L v v i

0mv=L i

0

2( ) ( ) 0

3 3G

l lmv= × + − ×H j i j i

0

2

3G

lmv= −H k

Motion of mass center G. Since there is no external force,

0constant

A A B Bm m mv= + = =L v v i

03m mv=v i

0

1

3v=v i

Motion about mass center.

( ) ( )G G i i i A A A B B B

m m m′ ′ ′ ′ ′ ′ ′= = Σ × = × + ×H H r v r v r v

where 2 1

3 3A B

l , l′ ′ ′ ′= = −r j r j

2 1,

3 3A B

l lθ θ′ ′ ′ ′= = −v i v i& &

Thus, 2 2 1 1 1

23 3 3 3 3

Gl ml l m lθ θ ′ ′ ′ ′= × + − × ⋅

H j i j i& &

22

3ml θ= − k&

But HG

is constant.

2 0

0

2 2

3 3

vml lmv

lθ θ− = − =k k &

0

2 2

3 3Av l vθ′ = =&

0

1 1

3 3Bv l vθ′ = =&





(b) After 180º rotation.

0 0

1 2

3 3A A

v v′= + = −v v v i i

0

1

3A

v= −v i

0 0

1 1

3 3B B

v v′= + = +v v v i i

0

2

3B

v=v i






Masses: 21253.882 lb s /ft, 2 , 3 .

32.2A B A C A

m m m m m= = ⋅ = =

Conservation of angular momentum about O.

240 240 2160 ( ) ( ) ( )A A A x A y A zv v v= + + = + +r i j k v i j k

600 1320 3240 500 1100 2200B B

= + + = + +r i j k v i j k

480 960 1920 400 ( ) ( )C C C y C zv v= − − + = − + +r i j k v i j k

Since the three parts pass through O, the angular momentum about O is zero. 0

0=H

00

A A A B B B C C Cm m m= × + × + × =H r v r v r v

[ 2 3 ] 0A A A B B C C

m × + × + × =r v r v r v

Dividing by mA and using determinant form,

240 240 2160 1200 2640 6480 1440 2880 5760

( ) ( ) ( ) 500 1100 2200 400 ( ) ( )A x A y A z C y C zv v v v v

+ + − −−

i j k i j k i j k

[240( ) 2160( ) ] [2160( ) 240( ) ]A z A y A x A zv v v v= − + −i j

6 6[240( ) 240( ) ] 1.320 10 0.6 10A y A xv v+ − − × + ×k i j

0 [ 2880( ) 5760( ) ]C z C yv v+ + − −k i

6 6[1440( ) 2.304 10 ] [ 1440( ) 1.152 10 ] 0C z C yv v+ − × + − − × =j k

Dividing by 240 and equating to zero the coefficients of i, j, and k,

: ( ) 9( ) 5500 12( ) 24( ) 0A z A y C z C yv v v vi − − − − = (1)

: 9( ) ( ) 7100 6( ) 0A x A z C zv v vj − − + = (2)

: ( ) ( ) 6( ) 4800 0A y A x C yv v vk − − − = (3)

Conservation of linear momentum.

0( )

A A B C C C A B Cm m m m m m+ + = + +v v v v

0( 2 3 ) 6 ( )

A A C C Am m+ + =v v v v





Dividing by mA and substituting given data,

( ) + ( ) ( ) (2)(500 1100 2200 ) (3)[ 400 + ( ) ( ) ] (6)(1500)A x A y A z C y C zv v v + v + v+ + + + − =i j k i j k i j k k

Separating into components,

: ( ) 1000 1200 0A xvi + − = (4)

: ( ) 2200 3( ) 0A y C yv vj + + = (5)

: ( ) 4400 3( ) 9000A z C zv vk + + = (6)

From (4), ( ) 200 ft/sA xv =

Solving (3) and (5) simultaneously,

( ) 200 ft/s ( ) 800 ft/sA y C yv v= = −


( ) 1300 ft/s ( ) 1100 ft/sA z C zv v= =

(200 ft/s) (200 ft/s) (1300 ft/s)A

= + +v i j k






Let the system consist of spheres A and B.

State 1. Instant cord DC breaks.

( ) 01

3 1

2 2A

m mv

= − −

v i j

( ) 01

3 1

2 2B

m mv

= −

v i j

( ) ( )1 01 1A Bm m mv= + = −L v v j

1

0

1

2 2v

m

= = −Lv j

Mass center lies at point G as shown.

( ) ( ) ( )1 11

0

3 3

2 2

3

2

G A Bl m l m

lmv

= × + − ×

=

H j v j v

k

2 2 2

1 0 0 0

1 1

2 2T mv mv mv= + =

State 2. The cord is taut. Conservation of linear momentum:

(a) 0

1

2D

v= = −v v j

00.500

Dv v=

Let ( )2

and A A B B

= + = +v v u v v u

2 12

A Bm m m= + + =L v u u L

B A B Au u= − =u u

( )2

2G A B A

lmu lmu lmu= + =H k k k





(b) Conservation of angular momentum:

( ) ( )2 1G G

=H H

0

32

2A

lmu lmv=k k

0

3

4A B

u u v= =

00.750u v=

( ) 2 2 2

2

2 2

0 0

1 1 12

2 2 2

1 1 9 9 13

2 2 16 16 16

A BT m v mu mu

mv mv

= + +

= + + =

(c) Fraction of energy lost: 13

1 2 16

1

1 3

1 16

T T

T

−− = =

1 2

1

0.1875T T

T

− =






The system is spheres A and B and the ring D.

Initial velocities: ( )0cos30 sin30

Av= − ° − °v i j

( )0cos30 sin30

0

B

D

v= − ° − °

=

v i j

v

Locate the mass center.

( ) ( )0

0

4

sin30 cos30 sin30 cos30

1

4

A Bm m m

ml ml

l

= +

= − ° + ° + − ° − °

= −

r r r

i j i j

r i

Velocity of mass center.

( ) ( )0 0

0

4

cos30 sin30 cos30 sin30

1

4

A Bm m m

mv mv

v

= +

= − ° − ° + ° − °

= −

v v v

i j i j

v j

(a) Motion of mass center 0

t= +r r v

0

1 1

4 4l v t= − −r i j

(b) / /G AG A B G B

m m= × + ×H r v r v

( )

( )

0

0

1cos30 cos30 sin 30

4

1 cos30 cos30 sin 30

4

l l mv

l l mv

= − + ° × − ° − °

+ − − ° × ° − °

i j i j

i j i j

0

7

4G

lmv=H k

(c) 2 2 2

0

1 1

2 2A B

T mv mv mv= + =

2

0T mv=






Let m be the mass of one ball.

Conservation of linear momentum: 0

( ) ( )m mΣ = Σv v

0 0 0( ) ( ) ( )

A B C A B Cm m m m m m+ + = + +v v v v v v

Dividing by m and applying numerical data,

(0.5 ft/s) [(3.75 ft/s) ( ) ] [( ) ( ) ] (6.5 ft/s) 0 0B y C x C yv v v+ + + + = + +i i j i j i

Components:

: 0.5 3.75 ( ) 6.5C x

x v+ = + ( ) 2.25 ft/sC xv =

: ( ) ( ) 0B y C yy v v+ = (1)

Conservation of angular momentum about O:

0[ ( )] [ ( )]m mΣ × = Σ ×r v r v

where rA = 0, rB = 0, (1.5 ft)(cos30 sin30 )C

= ° + °r i j

( )( )1.5 cos30 + sin 30 [ ( ) ( ) ] 0C x C ym v m vi j i j° ° × + =

Since their cross product is zero, the two vectors are parallel.

( ) ( ) tan30 2.25 tan30 1.2990 ft/sC y C xv v= ° = ° =

From (1), ( ) 1.2990 ft/sB yv = −

( ) 1.299 ft/sB yv = −

(2.25 ft/s) (1.299 ft/s)C

+=v i j






Let m be the mass of one ball.


( ) ( )m mΣ = Σv v

0 0 0( ) ( ) ( )

A B C A B Cm m m m m m+ + = + +v v v v v v

Dividing by m and applying numerical data,

0 [(6 ft/s) ( ) ] [( ) ( ) ] (8 ft/s) 0 0B y C x C yv v v+ + + + = + +i j i j i

Components:

: 6 ( ) 8C x

x v+ = ( ) 2 ft/sC xv =

: ( ) ( ) 0B y C yy v v+ = (1)


0[ ( )] [ ( )]m mΣ × = Σ ×r v r v

where rA = 0, rB = 0, (1.5 ft)(cos45 sin 45 )C

= ° + °r i j

(1.5)(cos45 sin 45 ) [ ( ) ( ) ] 0C x C ym v m v° + ° × + =i j i j

Since their cross product is zero, the two vectors are parallel.

( ) ( ) tan 45 2 tan 45 2 ft/sC y C xv v= ° = ° =

From (1), ( ) 2 ft/sB yv = −

( ) 2.00 ft/sB yv = −

(2.00 ft/s) (2.00 ft/s)C

+=v i j







A B A B

A B

W W W W

g g g g

+ = +

v v v

After multiplying by g, ( ) ( ) ( )7.2 5.76 1.44 4.8 2.4 2.4A B Bx yv v v+ = + +i j j i j

i: ( )41.472 2.4B xv= ( ) 17.28 ft/s

B xv =

j: ( )10.348 4.8 2.4A B yv v= − ( ) 4.32 2B Ay

v v= −

Speeds relative to the mass center: ( ) ( )( )1 13 8 8 ft/s

3 3A

u lω= = =

( ) ( )( )2 23 8 16 ft/s

3 3B

u lω= = =

Initial kinetic energy: ( ) ( )2 2 2 2

1 0 0

1 1 1

2 2 2

A B A BA Bx y

W W W WT v v u u

g g g g

= + + + +

( ) ( ) ( )2 22 2

1

1 7.2 1 4.8 1 2.45.76 1.44 8 16 18.2517 ft lb

2 32.2 2 32.2 2 32.2T

= + + + = ⋅

Final kinetic energy: ( ) ( )2 22

2

1 1 1

2 2 2

A B BA B Bx y

W W WT v v v

g g g= + +

( ) ( )2 22

2

1 4.8 1 2.4 1 2.417.28 4.32 2

2 32.2 2 32.2 2 32.2A A

T v v = + + −

20.2236 0.6440 11.8234

A Av v= − +

Conservation of energy: 1 2T T=

(a) 20.2236 0.6440 6.4283 0, 6.9919 ft/s

A A Av v v− − = = 6.99 ft/s

A=v

( ) ( )( ) ( ) ( )4.32 2 6.9919 9.6638 ft/s 17.28 ft/s 9.6638 ft/sB Byv = − = − = −v i j

19.80 ft/sB

=v 29.2°






( ) ( )( ) ( )

( ) ( ) ( )( ) ( )( )

0 0 0 01

2

3 3

7.2 4.8 2.47.44 5.76 1.0 8 2.0 16 6.0047 ft lb s

32.2 32.2 32.2

A B B B

O A B G A Bx x

W W W l W lH y m m v H y v u u

g g g g

= − + + = − + + +

= − + + = − ⋅ ⋅

( ) ( )2

( )A BO A A B B A B yy

W WH m v a m v b v a v b

g g= + = +

4.8 2.4(6.9919) ( 9.6638)(24) 1.0423 17.2868

32.2 32.2a a

= + − = −

( ) ( )2 1

1.0423 17.2868 6.0047O O

H H a= − = −

(b) 10.82 fta = 10.82 fta =







A B A B

A B

W W W W

g g g g

+ = +

v v v

( ) ( )0

9 6 37.68 10.8 6.72

32.2 32.2 32.2

= + −

v j i j

(a) ( ) ( )03.6 ft/s 2.88 ft/s= +v i j

04.61 ft/s=v 38.7°

Let Al be the distance from G to A and

Bl be the distance from G to B.

or 2A B A

A B B A A

B

W W Wl l l l l

g g W= = =

Let ω be the spin rate.

Initial kinetic energy: ( ) ( )2 22

1 0

1 1 1

2 2 2

A B A B

A B

W W W WT v l l

g g g gω ω

= + + +

( ) ( ) ( )

( )

2 22 2

1

2

1 9 1 6 1 33.6 2.88 2

2 32.2 2 32.2 2 32.2

2.9703 0.27950

A A

A

T l l

l

ω ω

ω

= + + +

= +

Final kinetic energy: 2 2

2

1 1

2 2

A B

A B

W WT v v

g g= +

( ) ( )2 2 2

2

1 6 1 37.68 10.8 6.72 13.0324

2 32.2 2 32.2T

= + + =

Conservation of energy: 2 1

T T= .

( ) ( )213.0324 2.9703 0.27950 6.000 ft/s

A Al lω ω= + =


( ) ( ) ( ) ( ) ( )

( )( ) ( ) ( )( )

( ) ( )( )

0 0 0 01

2

2

9 6 30 7.5 3.6 2 2

32.2 32.2 32.2

7.5466 0.55901 7.5466 3.3540

A B A BO A A B By x

A A A

A A

W W W WH x v y v l l l l

g g g g

l l l

l l

ω ω

ω ω

ω

= + − + +

= − + +

= − + = − +

( ) ( ) ( )( ) ( )( )

( ) ( )

2

2 1

6 37.68 5.58 6.72 21.6

32.2 32.2

5.5382 ft lb s

: 5.5382 7.5466 3.3540 0.600 ft

A BO A B y

O O A A

W WH v a v b

g g

H H l l

= + = + −

= − ⋅ ⋅

= − = − + =

(b) 2 1.200 ft B A A Bl l l l l= = = + 1.800 ftl =

(c) 6.00

10.000.600

A

A

l

l

ωω = = = 10.00 rad/sω =






Velocities in m/s. Lengths in meters. Assume masses are 1.0 for each ball.

Before impacts: ( ) ( ) ( )00 0 04 , 0

A B Cv= = = =v i i v v

After impacts: ( ) ( )1.92 , , A B B B C Cx yv v v= − = + =v j v i j v i

Conservation of linear momentum: 0 A B C

= + +v v v v

i: ( ) ( )4 0 4B C B Cx xv v v v= + + = −

j: ( ) ( )0 1.92 0 1.92B By yv v= − + + =


0

1 1 1 1

2 2 2 2A B C

v v v v= + +

( ) ( ) ( ) ( )22 2 2 21 1 1 1 14 1.92 1.92 4

2 2 2 2 2C Cv v= + + − +

24 3.6864 0

C Cv v− + =

( ) ( )( )24 4 4 3.6864

2 0.56 2.56 or 1.442

Cv

± −= = ± =

Conservation of angular momentum about :B′

( )( )( ) ( )( )

00.75 1.8

0.75 4 1.8 1.65 1.92 2.712

2.712

A C

C

C

v a v cv

cv

c

v

= − +

= − − =

=

If 1.44,Cv = 1.8833 off the table. Rejectc =

If 2.56,Cv = 1.059c =

Then, ( ) 4 2.56 1.44, 1.44 1.92B Bxv = − = = +v i j

Summary.

(a) 2.40 m/sB

=v 53.1°

2.56 m/sC

=v

(b) 1.059 mc =






Velocities in m/s. Lengths in meters. Assume masses are 1.0 for each ball.

Before impacts: ( ) ( ) ( )00 0 05 , 0

A B Cv= = = =v i i v v

After impacts: ( ) ( ), , 3.2A A B B B Cx yv v v= − = + =v j v i j v i

Conservation of linear momentum: 0 A B C

= + +v v v v

i: ( ) ( )5 0 3.2 1.8B Bx xv v= + + =

j: ( ) ( )0 0 A B B Ay yv v v v= − + + =


0

1 1 1 1

2 2 2 2A B C

v v v v= + +

( ) ( ) ( ) ( ) ( )2 2 2 2 21 1 1 1 15 1.8 3.2

2 2 2 2 2A Av v= + + +

(a) 25.76 2.4

A Av v= = 2.40 m/s

A=v

( ) 2.4B yv = 1.8 2.4

B= +v i j 3.00 m/s

B=v 53.1°

Conservation of angular momentum about :B′

( )00.75 1.8

A Cv a v cv= − +

0

1.8 0.75A A C

av v cv v= + −

( )( ) ( )( ) ( )( )1.8 2.4 1.22 3.2 0.75 5 4.474= + − =

(b) 4.474 4.474

2.4A

a

v

= = 1.864 ma =






Use a frame of reference that is translating with the mass center G of the system. Let

0v be its velocity.

0 0v=v i

The initial velocities in this system are ( ) ( )0 0,

A B′ ′v v and ( )

0,

C′v each having a magnitude of .lω They are

directed 120° apart. Thus,

( ) ( ) ( )0 0 0

0A B C′ ′ ′+ + =v v v

(a) Conservation of linear momentum:

( ) ( ) ( ) ( ) ( ) ( )0 0 00 0 0A B C A B Cm m m m m m′ ′ ′+ + = − + − + −v v v v v v v v v

( ) ( ) ( )0 0 00

A B Cv v v v v v= − + − − + −j i j i i i

Resolve into components.

i: ( )0 0

1 13 0 4.5

3 3C Cv v v v− = = =

01.500 m/s=v

j: 0 2.6 m/sA B B Av v v v− = = =

Conservation of angular momentum about G:

( ) ( ) ( )2

0 0 03

G A A B B C Cml m rω= = × − + × − + × −H k r v v v v r v v

( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )

( )( ) ( )( )

2

0

2 2

3

10.260 2.6 0.150 4.5 0.45033 m /s

3

A B A C C A B C

A C A C

l v v v

a d v av dv

l

ω

ω

= − × + × − + +

= × + − × = +

= + =

k r r j r i r r r i

i v j j i k

Conservation of energy: ( )2 2 2 2

1

1 332 2

T ml mlω ω= =

0 0

0 0

0 0

2.6 1.5 3.00 m/s

2.6 1.5 3.00 m/s

4.5 1.5 3.00 m/s

A A

B B

C C

− = − − =

− = − − − =

− = − − =

v v j i v v

v v j i v v

v v i i v v

( ) ( ) ( )2 2 2

2 0 0 0

1 1 1

2 2 2A B C

T m m m= − + − + −v v v v v v

1 2T T=

( ) ( ) ( )2 2 22 23 1 1 13 3 3

2 2 2 2ml m m mω = + +

3 m/slω =





(b) 2 2

0.45033 m /s0.1501 m

3 m/s

ll

l

ωω

= = = 150.1 mml =

(c) 3 m/s

0.1501

l

l

ωω = = 19.99 rad/sω =






Use a frame of reference that is translating with the mass center G of the system. Let 0

v be its velocity.

0 0v=v i

The initial velocities in this system are ( ) ( )0 0, ,

A B′ ′v v and ( )

0,

C′v each having a magnitude of .lω They are

directed 120° apart. Thus,

( ) ( ) ( )0 0 0

0A B C′ ′ ′+ + =v v v

Conservation of linear momentum:

( ) ( ) ( ) ( ) ( ) ( )0 0 00 0 0A B C A B Cm m m m m m

′ ′ ′+ + = − + − + −v v v v v v v v v

( ) ( ) ( )0 0 00

A B Cv v v v v v= − + − − + −j i j i i i


i: 0

3 0Cv v− = ( )( )0

3 3 0.4 1.2 m/sCv v= = =

j: 0A Bv v− =

B Av v=

Initial kinetic energy: 2 2 2

1 0

1 13 3

2 2T mv ml ω = +

Final kinetic energy: 2 2 2 2 2

2

1 1 1 1

2 2 2 2A B C A C

T mv mv mv mv mv= + + = +

Conservation of energy: 2 1

T T= Solve for 2

Av .

( ) ( ) ( ) ( )2 2 2 22 2 2

0

3 3 1 3 3 10.4 0.75 1.2

2 2 2 2 2 2A Cv v l vω= + − = + −

(a) 2 20.36375 m /s ,= 0.6031 m/s

Av = 0.603 m/s

A=v

0.603 m/sB

=v

1.200 m/sC

=v





Use a frame of reference moving with velocity 0.v

Conservation of angular momentum about G.

( ) ( ) ( )2

0 0 03

G A A B B C Cml m r m mω= = × − + × − + × −H k r v v v v r v v

( ) ( ) ( ) ( )( )2

03

A B A C C A B Cl v v vω = − × + × − + +k r r j r i r r r i

( ) ( ) ( ) ( ) ( )3A C A C

l l a d v av dvω = × + − × = +k i v j j i k

( )( )( ) ( )3 0.075 0.75 0.130 1.200Av d= +

(b) ( )( )0.1406 0.1083 0.603 0.0753 md = − = 75.3 mmd =






Mass flow rate. As the fluid moves from section 1 to section 2 in time ∆t, the mass ∆m moved is

( )m A lρ∆ = ∆

Then, 1

( )dm m A lAv

dt t t

ρ ρ∆ ∆= = =∆ ∆

Data: 3 2 6 2

11000 kg/m , 500 mm 500 10 m , 25 m/sA vρ = = = × =

6(1000)(500 10 )(25) 12.5 kg/sdm

dt

−= × =


:

1( ) 0m v P t∆ − ∆ =

m dmP v v

t dt

∆= =∆

(12.5)(25)P =

312 NP =






Consider velocities measured with respect to the plate, which is moving

with velocity V. The velocity of the stream relative to the plate is

1u v V= − (1)

Mass flow rate. As the fluid moves from section 1 to section 2 in time ∆t,

the mass ∆m moved is

( )m A lρ∆ = ∆

Then ( )dm m A l

Audt t t

ρ ρ∆ ∆= = =∆ ∆

(2)


( ) ( ) 0m u P t∆ − ∆ =

2m dmP u u Au

t dtρ∆= = =

∆

Pu

Aρ=

From (1), 1 1

PV v u v

Aρ= − = −

Data: 2 6 2400 N, 600 mm 600 10 mP A

−= = = ×

3

130 m/s, 1000 kg/mv ρ= =

6

40030

(1000)(600 10 )V −= −

× 4.18 m/sV =






Let F be the force that the wedge exerts on the stream. Assume that the fluid speed is constant. 60 ft/s.v =

Volumetric flow rate: 3475 gal/min ft /sQ = = 1.0583

Mass flow rate: ( )3 362.4slug/ft 1.0584 ft /s 2.051 slug/s

32.2

dmQ

dtρ = = =

Velocity vectors: ( )1, cos30 sin 30v v= = ° + °v i v i j

( )2cos45 sin 45v= ° − °v i j

Impulse – momentum principle:

( ) ( ) 1 22 2

m mm t

∆ ∆∆ + ∆ = +v F v v

( ) ( )

( )( )( )( ) ( )

1 2

1 1

2 2

1 1cos30 sin 30 cos 45 sin 45

2 2

2.051 60 ft/s 0.21343 0.10355

26.26 lb 12.74 lb

m

t

dmv

dt

∆ = + − ∆

= ° + ° + ° − ° −

= − −

= − −

F v v v

i j i j i

i j

i j

Force that the stream exerts on the wedge:

( ) ( )26.26 lb 12.74 lb− = +F i j drag 26.3 lb=

lift 12.74 lb=






For a fixed observer, the upstream velocity is ( )48 ft/s .=v i

Volumetric flow rate: 3500 gal/min ft /sQ = = 1.1141


32.2

dmQ

dtρ = = =

Use a frame of reference that is moving with the wedge to the left at 12 ft/s. In this frame of reference the

upstream velocity vector is

( ) ( )48 12 60 ft/s .= − − =u i i i

For the moving frame of reference the mass flow rate is ( )602.1590 2.6987 slug/s.

48

dm u dm

dt v dt

′ = = =

Velocity vectors: ( )1, cos30 sin 30u u= = ° + °u i u i j

( )2cos45 sin 45u= ° − °u i j

Let F be the force that the wedge exerts on the stream.

Impulse-momentum principle:

( ) ( ) 1 22 2

m mm t

∆ ∆∆ + ∆ = +u F u u

( ) ( )

( )( )( )( ) ( )

1 2

1 1

2 2

1 1cos30 sin30 cos45 sin 45

2 2

2.6987 60 0.21343 0.10355

34.6 lb 16.76 lb

m

t

dmu

dt

∆ = + − ∆

′ = ° + ° + ° − ° −

= − −

= − −

F u u u

i j i j i

i j

i j

Force that the stream exerts on the wedge

( ) ( )34.6 lb 16.76 lb− = +F i j drag 34.6 lb=

lift 16.76 lb=






Let F be the force exerted on the chips. Apply the impulse-momentum principle to the chips. Assume that

the feed velocity is negligible.

( ) ( ) Ct m∆ = ∆F v

( )cos25 sin 25C

m dmv

t dt

∆ = = ° + ° ∆ F v i j

( )( )1060 cos25 sin 25

32.2

= ° + °

i j

( ) ( )16.89 lb 7.87 lb= +i j

0: 0x x x

F D FΣ = − =

16.89 lbx

D =

Force on truck hitch at D:

16.89 lbx

D− = − 16.89 lbx

D− =






Initial momentum: ( ) 0.A

m∆ =v

Impulse – momentum principle.

( )( ) ( )x yF F t m+ ∆ = ∆i j v

( )o o

cos35 sin35x y

m dmF F v

t dt

∆ + = = + ∆ i j v i j

x component:

oEngine thrust cos35x

dmF v

dt= =

Data: 3 388 m /min m /s

60Q = = 31000 kg/mρ =

( ) 81000 133.333 kg/s

60

dmQ

dtρ = = =

( )( ) o

133.333 50 cos35 5461 Nx

F = =

5.46 kNx

F =






Weight: (600)(9.81) 5886 NW mg= = =


Moments about F:

1 1 1 2

( ) (3 ) (2 ) ( ) ( ) ( ) 3( )m v a mv a m v a W t c R t L m v h∆ + ∆ + ∆ + ∆ − ∆ = ∆

1 2

16 3

m mR cW av hv

L t t

∆ ∆ = + + ∆ ∆

Data: 6 m, 4 m, 1.5 m, 0.8 m, 40 kg/sm dm

L c a ht dt

∆= = = = =∆

[ ]1(4)(5886) (6) (1.5) (3) (40) (3) (0.8) (4) (40) 4040 N

6R = + − =

4040 N=R






Assume A B

u u u= =


Moments about O :

( ) ( ) ( )A c B

R m u x t R m u∆ + ∆ = ∆F kk r

( ) ( ) ( ) 0c B Ax t R m u u∆ = ∆ − =r F k

The line of action of F passes through point O.

Components : ( ) ( ) ( )sin cosA B

m u F t m uα θ∆ + ∆ = ∆

sin (1 cos )m

F ut

α θ∆= −∆

(1)

Components : ( ) ( )0 cos sinF t m uα θ+ ∆ = ∆

cos sinm

F ut

α θ∆=∆

(2)

Dividing (2) by (1),

22sin

1 cos 2tan tan

sin 22sin cos

2 2

θθ θα θ θθ

−= = =

.2

θα =

Thus point C lies at the midpoint of arc AB.






( ) ( )( )80013.333 L/s 1.000 kg/L 13.333 L/s 13.333 kg/s

60

dmQ Q

dtρ= = = = =

( ) ( )( )30 m/s 30 m/s sin 40 cos 40B C

= = ° + °v j v i j

Apply the impulse – momentum principle.

x components: ( ) ( )( )0 30sin 40x

A t m+ ∆ = ∆ °

( ) ( )( )30sin 40 13.333 30sin 40x

mA

t

∆= ° = °∆

257Nx

A =

y components: ( )( ) ( ) ( )( )30 30cos40y

m A t m∆ + ∆ = ∆ °

( ) ( )30cos40 30 13.333 30cos40 30y

mA

t

∆= ° − = ° −∆

93.6 N= − 93.6 Ny

A =

moments about :A ( )( )( ) ( )0.060 30

Am M t∆ + ∆

( )( )( ) ( )( )( )0.180 30cos40 0.300 30sin 40m m= ∆ ° − ∆ °

( ) ( ) ( )1.8 1.6484A

m M t m∆ = ∆ − ∆

( ) ( )( )3.4484 13.333 3.4484 46.0 N mA

mM

t

∆= − = − = − ⋅∆

46.0 N mA

= ⋅M

274 N=A 20.0°






Mass flow rate: 3 3

2

62.4 lb/ft 40 ft /min1.29193 lb s/ft

60 s/min32.2 ft/s

dmQ

dt g

γ= = = ⋅

75 ft/sA Bv v= =

Use impulse - momentum principle.

moments about :D ( )( ) ( )( ) ( )15 23 15

sin 60 cos6012 12 12

A Am v m v C t

− ∆ ° + ∆ ° − ∆

( ) 3

12B

m v

= ∆

( )( )( )15 15 23 3sin 60 cos60 1.29193 75 0.37420

12 12 12 12A

mC v

t

∆ = − ° + ° − = − ∆

29.006 lbC = − 0, 29.0 lbx y

C C= = −

x component: ( ) ( ) ( )cos60A x B

m v D t m v∆ ° + ∆ = ∆

( ) ( )( )cos60 1.29193 75 75cos60x B A

mD v v

t

∆ = − ° = − ° ∆ 48.4 lb

xD =

y component: ( ) ( ) ( )sin 60 0A ym v C t D t∆ ° + ∆ + ∆ =

( )( )sin 60 29.006 1.29193 75 sin 60y A

mD C v

t

∆= − − ° = + − °∆

54.9 lby

D = −






Volumetric flow rate: 3300 gal/min 0 ft /sQ = = .6684


32.2

dmQ

dtρ = = =

90 ft/sA Bv v= =

Use the impulse – momentum principle.

Moments about C: ( ) ( ) ( ) cosA P B

m v a W t l m v bθ∆ − ∆ = ∆

(a) ( ) ( )( ) ( )( )( )( )

90 4 /12 90 1/12cos 1.2954 0.7287

40 1

A B

p

m v a v b

t W lθ

−∆ −= = =∆

43.23θ = ° 43.2θ = °

x components: ( ) ( ) ( ) cosA x B

m v C t m v θ∆ + ∆ = ∆

( ) ( )( )cos 1.2954 90cos 90 31.63 lbx B A

mC v v

tθ θ∆= − = − = −

∆

y components: ( ) ( ) ( )0 siny p BC t W t m v θ+ ∆ − ∆ = − ∆

( ) ( )( )sin 40 1.2954 90 sin 39.84 lby p B

mC W v

tθ θ∆= − = − = −

∆

(b) [31.63 lb=C ] [39.84 lb+ ] 50.9 lb=C 51.6°






Volumetric flow rate: 3300 gal/min 0 ft /sQ = = .6684


32.2

dmQ

dtρ = = =

, 45A Av v v θ= = = °

Use the impulse-momentum principle.

moments about C: ( ) ( ) ( ) cosA p Bm v a W t l m v bθ∆ − ∆ = ∆

(a) ( )( ) ( )( )( )

( ) cos 40 1 cos 45

87.338 ft/s4 1/

1.295412 12

pA B

W lv v

a b m t

θ °= = = =

− ∆ ∆ −

87.3 ft/sv =

x components: ( ) ( ) ( ) cosA x B

m v C t m v θ∆ + ∆ = ∆

( ) ( )[ ]cos 1.2954 87.338cos45 87.338 33.137 lbx B A

mC v v

tθ∆= − = ° − = −

∆

y components: ( ) ( ) ( )0 siny p BC t W t m v θ+ ∆ − ∆ = − ∆

( )( )sin 40 1.2954 87.338 sin 45 40.0 lby p B

mC W v

tθ∆= − = − ° = −

∆

(b) [33.137 lb=C ] [40 lb+ ] 51.9 lb=C 50.4°






Symbols: mass flow ratedm

dt=

exhaust relative to the airplaneu =

speed of airplanev =

drag forceD =


( ) ( ) ( )m v D t m u∆ + ∆ = ∆

m dm D

t dt u v

∆ = =∆ −

Data: 900 km/h = 250 m/sv =

600 m/su =

35 kN 35000ND = =

35000

600 250

dm

dt=

− 100 kg/sdm

dt=






Let F be the force exerted on the slipstream of one engine. Then, the force exerted on the airplane is −2F as

shown.

Statics.

0B

MΣ =

( ) ( )( )0.9 4.8 2 0W F− =

( )( )

( )( )0.9 6000

2 4.8F =

562.5 lb=

Calculation of .

dm

dt

mass density volume density area length= × = × ×

( ) ( ) ( )B B

B B B

A v tm A l A v t

g

γρ ρ ∆∆ = ∆ = ∆ =

B B

m dm A v

t dt g

γ∆ = =∆

Force exerted on the slipstream: ( )B A

dmF v v

dt= −

Assume that Av , the speed far upstream, is negligible.

( ) 2 20

4

B B

B B

A vF v D v

g g

γ γ π = − =

( )( )( )

( ) ( )2 2 2

2 2

4 562.5 32.247058.9 ft /s

6.6 0.075B

Fgv

Dπ γ π= = =

84.0 ft/s B

=v






Let F be the force exerted on the slipstream of one engine.

( )B A

dmF v v

dt= −

Calculation of .

dm

dt mass density volume density area length= × = × ×

( ) ( ) ( )B B

B B B

A v tm A l A v t

g

γρ ρ ∆∆ = ∆ = ∆ =

2 or

4

B B

B

m A v dmD v

t g dt g

γ γ∆ π = = ∆

Assume that ,Av the velocity far upstream, is negligible.

( ) ( ) ( )2 22 0.0750 6.6 60 286.87 lb

4 32.2 4B B

F D v vg

γ π π = − = =

The force exerted by two slipstreams on the airplane is 2 .F− 2 573.74 lbF− =

Statics.

0:B

MΣ =

( )0.9 4.8 2 9.3 0W F A− − − =

( )( ) ( )( )10.9 6000 4.8 573.74

9.3A = −

284.5 lb= 285 lb=A

0: 2 0x x

F F B= − − =

2 573.74 lbx

B F= − =

( )0: 284.5 6000 0y y y

F A B W BΣ = + − = + − =

5715.5 lby

B =

[573.74 lb=B ] [5715.5 lb+ ] 5740 lb=B 84.3°






Use a frame of reference moving with the plane. Apply the impulse-momentum principle. Let F be the force that

the plane exerts on the air.

x components: ( ) ( ) ( )A Bm u F t m u∆ + ∆ = ∆

( ) ( )B A B A

m dmF u u u u

t dt

∆= − = −∆

(1)

moments about B: ( ) ( ) 0A B

e m u M t− ∆ + ∆ =

B A

dmM e u

dt= (2)

Let d be the distance that the line of action is below B.

B

Fd M= B A

B A

M eud

F u u= =

− (3)

Data: 90 kg/s,dm

dt= 600 m/s,

Bu = 4 me =

(a) 480 km/h 133.333 m/sA

u = =

From (1), ( )( ) 390 600 133.333 42 10 NF = − = × 42.0 kNF =

From (2), ( )( )

( )4 133.333

600 133.333d =

− 1.143 md =

(b) 960 km/h 266.67 m/sA

u = =

From (1), ( )( ) 390 600 266.67 30 10 NF = − = × 30.0 kNF =

From (2), ( )( )

( )4 266.67

600 266.67d =

− 3.20 md =






The thrust on the fluid is ( )B A

dmF v v

dt= −

Calculation of dm

dt. mass density volume density area length= × = × ×

( ) ( )B B Bm A l A v tρ ρ∆ = ∆ = ∆

B B

m dmA v

t dtρ∆ = =

∆

whereB

A is the area of the slipstream well below the helicopter and Bv is the corresponding velocity in the

slipstream. Well above the blade, 0.Av ≈

Hence, 2

BF Avρ=

( ) ( ) ( )2 23

3 2

1.21 kg/m 9 m 24 m/s4

44.338 10 kg m/s

π =

= × ⋅

44.3 kNF =

The force on the helicopter is 44.3 kN .

Weight of helicopter: 15kNH

=W

Weight of payload: P P

W=W

Statics: 0y H PF F W WΣ = − − =

44.3 15 29.3 kN.P H

W F W= − = − = 29.3 kNW =






Let dm

dt= mass flow rate, u = discharge velocity relative to the airliner, speed of airliner,v = and

thrustF = of the engines.

0F D− = ( ) 0dm

u v Ddt

− − =

Configuration before control surface malfunction:

1

720= 22.36 slug/s, 1860 ft/s, 560 mi/h 821.33 ft/s

32.2

dmu v

dt= = = =

( )( ) 1 122.36 1860 821.33 0 23225 lbD D− − = =

Drag force factor: ( )

2 2 21

1 1 1 1 2 2

1

23225 0.03443 lb s /ft

821.33

DD k v k

v= = = = ⋅

After control surface malfunction: 2 2

2 11.2 0.04131 lb s /ftk k= = ⋅

When the new cruising speed is attained,

( ) 2

2 2 20

dmu v k v

dt− − =

( )( ) 2

2 222.36 1860 0.04131 0v v− − =

Solving for v2, 2768.6 ft/sv =

2524 mi/hv =






Apply the impulse - momentum principle to the moving air. Use a frame of reference that is moving with the

airplane. Let F be the force on the air.

270 km/h 75 m/s

600 m/s

v

u

= ==

( ) ( ) ( )2 sin 20

2

mm v F t u

∆− ∆ + ∆ = °

( ) ( )

( )( ) 3

sin 20 sin 20

120 75 600sin 20 33.6 10 N

m dmF v u v u

t dt

F

∆= + ° = + °∆

= + ° = ×

Force on airplane is .−F 33.6 kN=F






Symbols: n = number engines operating

dm

dt= mass flow rate for one engine

u = discharge velocity relative to jetliner = 800 m/s

v = speed of jetliner

F = thrust force ( )dm

n u vdt

= −

D = drag force 2kv=

Force balance.

0F D− = 2( )dm

n u v kvdt

− =

2 1

( )

v dm

n u v k dt=

−

All 3 engines operating: 3, 900km/h = 250 m/sn v= =

2(250) 1

37.879 m/s3(800 250)

dm

k dt= =

−

(a) One engine lost: n = 2

2

37.8792(800 )

v

v

=−

275.758 60606 0v v+ − =

211.2m/sv = 760 km/hv =

(b) 2 engines lost: n = 1

2

37.879800

v

v

=−

237.879 30303 0v v+ − =

156.16 m/sv =

562 km/hv =






Let u be the velocity of the stream relative to the velocity of the blade. ( )u v V= −

Mass flow rate: A

mAv

tρ∆ =

∆


( ) ( ) ( ) cost

m u F t m u θ∆ − ∆ = ∆

( )1 cos ( ) (1 cos )t A A

mF u Av v V

tθ ρ θ∆= − = − −

∆

where Ft is the tangential force on the fluid.

The force Ft on the fluid is directed to the left as shown. By Newton’s law of action and reaction, the

tangential force on the blade is Ft to the right.

Output power: out( ) (1 cos )

t A AP FV Av v V Vρ θ= = − −

(a) V for maximum power output.

out ( 2 ) (1 cos ) 0

A

dPA v V

dVρ θ= − − =

1

2Av V=

(b) Maximum power.

out max

1 1( ) (1 cos )

2 2A A A A

P Av v v vρ θ = − −

3

out max

1( ) (1 cos )

4A

P Avρ θ= −

Input power = rate of supply of kinetic energy of the stream

2 2 3

in

1 1 1 1( )

2 2 2A A A

mP m v v Av

t tρ∆ = ∆ = = ∆ ∆





(c) Efficiency. out

in

p

pη =

3

( ) (1 cos )

1

2

A A

A

Av v V V

Av

ρ θηρ

− −=

2 1 (1 cos )A A

V V

v vη θ

= − −






Data: 240

slugs/s,32.2

dm

dt= = 7.4534 2200 ft/s,u = 570 mi/h 836 ft/sv = =

( ) ( )( )7.4534 2200 836 10166 lbdm

F u vdt

= − = − =

(a) Power used to propel airplane:

( )( ) 6

110166 836 8.499 10 ft lb/sP Fv= = = × ⋅

propulsion power 15450 hp=

Power of kinetic energy of exhaust:

( ) ( )( )22

1

2P t m u v∆ = ∆ −

( ) ( )( )2 2 6

2

1 17.4534 2200 836 6.934 10 ft lb/s

2 2

dmP u v

dt= − = − = × ⋅

(b) Total power: 6

1 215.433 10 ft lb/sP P P= + = × ⋅

total power 28060 hp=

(c) Mechanical efficiency: 6

1

6

8.499 100.551

15.433 10

P

P

×= =×

mechanical efficiency 0.551=






Kinetic energy of fluid in slipstream passing in time .t∆

2 21 1 mass speed density volume speed

2 2T∆ = × = × ×

21

density area length speed2

= × × ×

( ) ( )2 21 1

2 2A l v Av t vρ ρ= ∆ = ∆

31

2

TAv

tρ∆ =

∆

Input power 31

2

dTAv

dtρ= =

Data: 31.2 kg/mρ =

( )22 26.5 33.183 m

4 4A d

π π= = =

30 km/h 8.333 m/sv = =

(a) ( )( )( )3 311.2 33.183 8.333 11.521 10 N m/s

2

dT

dt= = × ⋅

11.52 kJ/s

Input power 11.521 kWdT

dt= =

(b) Output power ( )( )0.4 11.521 4.61 kW= =

output power 4.61 kW=






Kinetic energy of fluid in slipstream passing in time .t∆

2 21 1 mass speed density volume speed

2 2T∆ = × = × ×

21

density area length speed2

= × × ×

( ) ( )2 21 1

2 2A l v Av t vρ ρ= ∆ = ∆

31

2

TAv

tρ∆ =

∆ (1)

Data: output power 3.5 kW 3500 W= =

3500input power 10000 W

0.35= =

input power 10,000 W, 36 km/h 10 m/sdT

vdt

= = = =

Using (1), ( )( )( )( )

2

3 3

2 10000216.667 m

1.2 10

dTA

dtvρ= = =

(a) ( )( )24 16.6674

4

Ad A d

ππ π

= = =

4.61 md =

(b) From above, 10.00 kJ/sdT

dt=






Mass flow rate:

mass density volume

density area length

= ×= × ×

( ) ( )m bd l bdv tρ ρ∆ = ∆ = ∆

dm m

bdvdt t

ρ∆= =∆

1 dm

Q bdvdtρ

= =

Continuity of flow: 1 2

Q Q Q= =

1 2

1 2

,

Q Qv v

bd bd=

Resultant pressure forces:

1 1 2 2 p d p dγ γ= =

2

1 1 1 1

1 1

2 2F p bd bdγ= =

2

2 2 2 2

1 1

2 2F p bd bdγ= =

Apply impulse-momentum principle to water between

sections 1 and 2.





( ) ( ) ( ) ( )1 1 2 2m v F t F t m v∆ + ∆ − ∆ = ∆

( )1 2 2 1

mv v F F

t

∆ − = −∆ ( )2 2

2 1

1 2

1

2

Q QQ b d d

bd bdρ γ

⋅ − = −

( ) ( )( )

2

2 1

1 2 2 1

1 2

1

2

Q d db d d d d

bd d

ργ

−= + −

Noting that ,gγ ρ=

( )1 2 1 2

1

2Q b gd d d d= +






Mass flow rate:

mass density volume

density area length

= ×= × ×

( ) ( )m bd l bdv tρ ρ∆ = ∆ = ∆

dm m

bdvdt t

ρ∆= =∆

1 dm

Q bdvdtρ

= =

Continuity of flow: 1 2

Q Q Q= =

1 2

1 2

,

Q Qv v

bd bd=

Resultant pressure forces:

1 1 2 2 p d p dγ γ= =

2

1 1 1 1

1 1

2 2F p bd bdγ= =

2

2 2 2 2

1 1

2 2F p bd bdγ= =

Apply impulse-momentum principle to water between

sections 1 and 2.





( ) ( ) ( ) ( )1 1 2 2m v F t F t m v∆ + ∆ − ∆ = ∆

( )1 2 2 1

mv v F F

t

∆ − = −∆ ( )2 2

2 1

1 2

1

2

Q QQ b d d

bd bdρ γ

⋅ − = −

( ) ( )( )

2

2 1

1 2 2 1

1 2

1

2

Q d db d d d d

bd d

ργ

−= + −

Noting that ,gγ ρ=

( )1 2 1 2

1

2Q b gd d d d= +

Data: 1 2

9.81 m/s, 3 m, 1.25 m, 1.5 mg b d d= = = =

( )( )( )( ) 313 9.81 1.25 1.5 1.25 1.5 15.09 m /s

2Q = + =

315.09 m /sQ =






From hydrostatics, the pressure at section 1 is 1

.p rh ghρ= =

The pressure at section 2 is 2

0.p =

Calculate the mass flow rate using section 2.


( ) ( )2 2A Am l v tρ ρ∆ = ∆ = ∆

2A

dm mv

dt tρ∆= =

∆

Apply the impulse-momentum principle to fluid between sections 1 and 2.

( ) ( ) ( )1 1 1m v p A t m v∆ + ∆ = ∆

1 1 1

dm dmv p A v

dt dt+ =

( ) ( )1 1 1 2 1

dmp A v v A v v v

dtρ= − = −

But 1v is negligible,

1,p ghρ= and 2v gh=

( )1 22ghA A ghρ ρ= or

1 22A A=

2 22

4 4D d

π π =

2

Dd =






The flow through each arm is 1.25 gal/min.

3

3 3gal 1 ft 1 min5 11.1408 10 ft /s

min 7.48 gal 60 sQ − = = ×

( )3

3 3

2

3

62.4 lb/ftQ 11.1408 10 ft /s

32.2 ft/s

21.590 10 lb s/ft

dmQ

dt g

γρ

−

= = = ×

= × ⋅

Consider the moment about O exerted on the fluid stream of one arm.

Apply the impulse-momentum principle. Compute moments about O.

First, consider the geometry of triangle OAB. Using first the law of

cosines,

( ) ( ) ( ) ( )2 2 2 o6 4 2 6 4 cos120OA = + −

76 in. 0.72648 ftOA = =

Law of sines. o

sin sin120

4 76

β =

o

23.413 ,β =

o o

60 36.587α β= − =

Moments about O:

( )( )( ) ( ) ( )( ) ( )( )( )0 sinO O s

m v M t OA m v OA m OAα ω∆ + ∆ = ∆ − ∆

( ) ( )

( ) ( )( ) ( )( )

2

23

sin

21.590 10 0.72648 60 sin 36.587 0.72648

0.56093 0.011395 lb ft

O s

mM OA v OA

tα ω

ω

ω

−

∆ = − ∆ = × −

= − ⋅

Moment that the stream exerts on the arm is .

OM−

Friction couple for one arm:

( )10.275 0.06875 lb ft

4F

M = = ⋅

Balance of moments on one arm:

0 F O F O

M M M M− = =

0.06875 0.56093 0.011395ω= −

43.19 rad/s 412 rpmω = =

412 rpmω =






Consider the conservation of the horizontal component of momentum of the railroad car of mass 0

m and the

sand mass .qt

( ) 0 0

0 0 0

0

m vm v m qt v v

m qt= + =

+ (1)

0 0

0

dx m vv

dt m qt= =

+

Integrating, using 0

0x = and x L= when ,L

t t=

( )0 0 0 0

0 00 0

0

0 0 0

0

ln ln

ln

L Lt t

L

L

m v m vL vdt dt m qt m

m qt q

m v m qt

q m

= = = + − ++=

∫ ∫

0

0 0 0

lnL

m qt qL

m m v

+ =

0 0/0

0

qL m vLm qte

m

+ =

(a) Final mass of railroad car and sand 0 0/

0 0

qL m vLm qt m e+ =

(b) Using (1), 0 0/0 0 0 0

0 0

qL m vL

L

m v m vv e

m qt m

−= =+

0 0/

0

qL m vLv v e

−=






Apply the impulse-momentum principle.

Moments about C : ( )( ) ( ) ( ) ( )( )0.9 3 1.8 1.65A B

m v D t W t m v− ∆ + ∆ − ∆ = − ∆

( )

( )( ) ( ) ( )( ) ( )( )

1.8 10.9 1.65

3 3

1.8 4000 1100 0.9 4.5 1.65 4.5 2287.5 N

3 3

A B

mD W v v

t

∆= + −∆

= + − =

2.29 kN=D

x components: ( ) ( ) ( )A x Bm v C t m v∆ + ∆ = ∆

( ) ( )( )100 4.5 4.5 0x B A

mC v v

t

∆= − = − = ∆

y components: ( ) ( ) ( )0 0y

C t D t W t+ ∆ + ∆ − ∆ =

4000 2287.5 1712.5 Ny

C W D= − = − =

1.712 kN=C






Let ρ be the mass per unit length of chain. Apply the impulse - momentum to the entire chain. Assume that the

reaction from the floor it equal to the weight of chain still in contact with the floor.

Calculate the floor reaction.

( )R g l yρ= − 1y

R mgl

= −

Apply the impulse-momentum principle.

( ) ( ) ( ) ( )yv P t R t gl t y y vρ ρ ρ+ ∆ + ∆ − ∆ = + ∆

( ) ( ) ( )P t y v gl t R tρ ρ∆ = ∆ + ∆ − ∆

(a) ( )yP v gl l y g

tρ ρ ρ∆= + − −

∆ Let

y dyv

t dt

∆ = =∆

2P v gyρ ρ= + ( )2mP v gy

l= +

(b) From above, 1y

mgl

= −

R






(a) Let ρ be the mass per unit length of chain. The force P supports the weight of chain still off the floor.

P gyρ= mgy

Pl

=

(b) Apply the impulse-momentum principle to the entire chain.

( ) ( ) ( ) ( )yv P t R t gl t g y y vρ ρ ρ− + ∆ + ∆ − ∆ = − + ∆

( ) ( ) ( ) ( )R t gl t P t g y vρ ρ∆ = ∆ − ∆ − ∆

y

R gl gy vt

ρ ρ ρ ∆= − −∆

Let 0.t∆ → Then, y dy

vt dt

∆ = = −∆

( ) 2R g l y vρ ρ= − + ( ) 2mg l y v

l = − + R






Let ρ be the mass per unit length of chain. Consider the impulse-momentum applied to the link being brought to

rest at point C.

Calculation of .m∆

( ) ( )m l v tρ ρ∆ = ∆ = ∆


( ) 0m v C t− ∆ + ∆ =

( ) 0v t v C tρ− ∆ + ∆ =

2C vρ=

Impulse – momentum applied to the moving portion of the chain. Consider only the changes in momentum and

forces contributing to moments about O in the diagram.

Moments about O:

( ) [ ] ( )m v gh C t m vρ∆ + − ∆ = ∆

C ghρ=

Equating the two expressions for C,

2v ghρ ρ=

2vh

g=






Let ρ be the mass per unit length of chain. Assume that the weight of any chain above the hole is supported by

the floor. It and the corresponding upward reaction of the floor are not shown in the diagrams.

Case 1. Apply the impulse-momentum principle to the entire chain.

( )( )( ) ( ) ( )( )

( )( )

yv gy t y y v v

yv y v y v y v

y vy vgy v y

t t t

ρ ρ ρ

ρ ρ ρ ρ

ρ ρ ρ ρ

+ ∆ = + ∆ + ∆

= + ∆ + ∆ + ∆ ∆

∆ ∆∆ ∆= + +∆ ∆ ∆

Let 0.t∆ → ( )dy dv dgy v y yv

dt dt dtρ ρ ρ ρ= + =

Multiply both sides by .yv ( )2 dgy v yv yv

dtρ ρ=

Let dy

vdt

= on left hand side. ( )2 dy dgy yv yv

dt dtρ ρ=

Integrate with respect to time. ( ) ( )2g y dy yv d yvρ ρ=∫ ∫

( )231 1

3 2gy yvρ ρ= or 2 2

3v gy= (1)

Differentiate with respect to time. 2 2

23 3

dv dyv g gvdt dt

= =





(a)

1

3

dva g

dt= = 0.333g=a

(b) Set in (1)y l= 2 2

3v gl= 0.817 gl=v

Case 2. Apply conservation of energy using the floor as the level for from which the potential energy is

measured.

1 10, 0T V= =

2

2 2

1,

2 2

yT mv V gyρ= = −

1 1 2 2T V T V+ = +

2 21 10

2 2mv gyρ= −

2 2

2 gy gyv

m l

ρ= = (1)

Differentiating with respect to y, 2

2dv gy

vdy l

=

(a) Acceleration. dv gy

a vdy l

= = gy

l=a

(b) Setting in (1),y l= 2v gl= gl=v

Note: The impulse-momentum principle may be used to obtain the force that the edge of the hole exerts

on the chain.






750 lb/s,dW

dt=

1 75023.292 lb s/ft

32.2

dm dW

dt g dt= = = ⋅

Thrust of one engine: ( )( ) 3 12500 23.292 291.15 10 lb

dmP u

dt= = = ×

For 3 engines, 33 873.4 10 lbP = ×

Thrust 873 kips=






Thrust of one engine: 3

3 1200 10400 10 lb.

3P

×= = ×

But, dm

P udt

=

3400 10

32 lb s/ft12500

dm P

dt u

×= = = ⋅

( )( )32.2 32 1030 lb/sdW dm

gdt dt

= = = 1030 lb/sdW

dt=






Thrust, .

dmP u

dt=

F P mg maΣ = − =

P u dm

a g gm m dt

= − = −

Data: 15 kg/sdm

dt=

As rocket is fired: 1500 kgm =

0

159.81 0.01 9.81

1500

ua u= − = − (1)

As all the fuel is consumed: 1

1500 1200 300 kg.m = − =

1

159.81 0.05 9.81

300

u

a u= − = − (2)

From the given data, 2

1 0220 m/sa a− = (3)

Using (1) and (2) for 1a and

0a and substituting into (3),

0.04 220u = 5500 m/su =






Thrust: dm

P u uqdt

= =

Since u and dm

dt are constant, P is also constant.

:F maΣ =

P mg ma− =

( )P m a g= +

( )min maxm a g= +

( )( )1500 25 9.81= +

352.215 10 N= ×

(a) Fuel consumption rate.

3

52.215 10

450

Pq

u

×= =

116.0 kg/sq =

(b) Mass of fuel consumed.

( )( )fuel116.0 15.6m qt= =

fuel

1810 kgm =






Apply the principle of impulse and momentum to the satellite plus the fuel expelled in time ∆t.

: ( )( ) ( )( )mv m m v v m v v v= − ∆ + ∆ + ∆ + ∆ −

( ) ( ) ( )( ) ( ) ( )( ) ( )mv m v m v m v m v m v m v= + ∆ − ∆ − ∆ ∆ + ∆ + ∆ ∆ − ∆

( ) ( ) 0m v u m∆ − ∆ =

( )dmm t

dt∆ = − ∆

v dv u dm

t dt m dt

∆ = = −∆

1 1 1

0 00

v t m

v m

u dm dmdv dt u

m dt m= − = −∫ ∫ ∫

1 0

1 0

0 1

ln lnm m

v v u u

m m

− = − =

0 1 0

1

expm v v

m u

− =

Data: 1 0

2430 m/s,v v− = 0

4200 m/s 5000 kgu m= =

1

5000 2430exp 1.7835

4200m

= =

1

2800 kgm =

fuel 0 1

5000 2800m m m= − = − fuel

2200 kgm =






Data from Problem 14.97: 0

5000 kg, 4200 m/sm u= =

1 0 fuel

5000 1500 3500 kgm m m= − = − =

Apply the principle of impulse and momentum to the satellite plus the fuel expelled in time ∆t.

: ( )( ) ( )( )mv m m v v m v v v= − ∆ + ∆ + ∆ + ∆ −

( ) ( ) ( )( ) ( ) ( )( ) ( )mv m v m v m v m v m v m v= + ∆ − ∆ − ∆ ∆ + ∆ + ∆ ∆ − ∆

( ) ( ) 0m v u m∆ − ∆ =

( )dmm t

dt∆ = − ∆

v dv u dm

t dt m dt

∆ = = −∆

1 1 1

0 00

v t m

v m

u dm dmdv dt u

m dt m= − = −∫ ∫ ∫

1 0

1 0

0 1

ln lnm m

v v u u

m m

− = − =

1 0

50004200 ln

3500v v v∆ = − = 1498 m/sv∆ =






Apply conservation of momentum to the rocket plus the fuel.

( )( ) ( )( )( ) ( ) ( )( ) ( ) ( )( ) ( )

mv m m v v m v v v

mv m v m v m v m v m v m v

= − ∆ + ∆ + ∆ + ∆ −

= + ∆ − ∆ − ∆ ∆ + ∆ + ∆ ∆ − ∆

( ) ( ) 0m v u m∆ − ∆ =

( )dmm t

dt∆ = − ∆

v dv u dm

t dt m dt

∆ = = −∆

1 1 1

0 00

v t m

v m

u dm dmdv dt u

m dt m= − = −∫ ∫ ∫

1 0 0

1 0

0 1 1

ln ln lnm m W

v v u u um m W

− = − = =

Data: 1 0

360 ft/sv v− =

( )0 111,600 lb, 11,600 1000 10,600 lbW W= = − =

11,600

360 ln10,600

u= 3990 ft/su =






Apply conservation of momentum to the rocket plus the fuel.

( )( ) ( )( )( ) ( ) ( )( ) ( ) ( )( ) ( )

mv m m v v m v v v

mv m v m v m v m v m v m v

= − ∆ + ∆ + ∆ + ∆ −

= + ∆ − ∆ − ∆ ∆ + ∆ + ∆ ∆ − ∆

( ) ( ) 0m v u m∆ − ∆ =

( )dmm t

dt∆ = − ∆

v dv u dm

t dt m dt

∆ = = −∆

1 1 1

0 00

v t m

v m

u dm dmdv dt u

m dt m= − = −∫ ∫ ∫

1 0

1 0

0 1

ln lnm m

v v u u

m m

− = − =

0 1 0 0

1 1

expm v v W

m u W

− = =

Data: 1 0

450 ft/s,v v− = 5400 ft/su =

0 1 fuel 1

1200 lbW W W W= + = +

1

1

1200 450exp 1.08690

5400

W

W

+ = =

1

12000.08690

W=

113810 lbW =






See sample Problem 14.8 for derivation of

0 0

0 0

ln lnm m qt

v u gt u gtm qt m

−= − = − −−

(1)

Note that g is assumed to be constant.

Set dy

vdt

= in (1) and integrate with respect to time.

0

0 0 0

0

20

0

0

ln

1ln

2

h t t

t

mh dy vdt u gt dt

m qt

m qtu dt gt

m

= = = − −

−= − −

∫ ∫ ∫

∫

Let 0

0

m qtz

m

−= 0

qdz dt

m= − or 0

mdt dz

q= −

( )0 0

2 20 0

20 0 0 0 0

0 0 0 0

20 0

0 0

0 0 0

0 0

1 1ln ln

2 2

1ln 1 ln 1

2

11 ln 1 1

2

ln 1 1 ln 1

zz

z z

m u m uh z dz gt z z z gt

q q

m u m qt m qt m mgt

q m m m m

m u qt m qtgt

q m m

m u m qt m qtut

q m m

= − = + −

− −= − − − −

−= − − + −

− −= − + − −

∫

2

20 0

0

1

2

1ln

2

gt

m u m qtut ut gt

q m

−

−= + − −

20 0

0

1ln

2

m mh u t t gt

q m qt

= − − − −

(2)





Data: 20

0

7300226.7 lb s /ft

32.2

Wm

g= = = ⋅

260

8.0745 lb s/ft32.2

wq

g= = = ⋅

&

2fuel

fuel

4000124.22 lb s /ft

32.2

Wm

g= = = ⋅

fuel124.22

15.385 s8.0745

mt

q= = =

2

0226.7 124.22 102.48 lb s /ftm qt− = − = ⋅

1500 ft/su =

( )( )2226.7 226.7 11500 15.385 15.385 ln 32.2 15.385

8.0745 102.48 2h

= − − −

4150 fth =






Thrust force dm

P u uqdt

= =

Mass of rocket plus unspent fuel 0

m m qt= −

Acceleration: 0

P uqa

m m qt= =

−

Integrating with respect to time to obtain the velocity,

0 00 0

0

t t qv v adt v u dt

m qt= + = +

−∫ ∫

( ) 0

0 0 0 0

0

ln ln lnm qt

v u m qt m v um

− = − − − = − (1)

Integrating again to obtain the displacement,

0

0 0 0

0

lnt m qt

s s v t u dtm

−= + − ∫

Let 0 0

0 0

or

m qt q mz dz dt dt dz

m m q

−= = − = −

( ) ]0 0

0 0

0 0

00 0 0 0

0 0

0 0 0 0

0 0

0 0

0

0 0 0

0 0

0 0

ln ln

ln 1 ln 1

1 ln 1 1

ln 1 1 ln 1

zz

z z

o

m u m us s v t zdz z z z

q q

mm u m qt m qt ms v t

q m m m m

m u qt m qts v t

q m m

m u m qt m qts v t ut

q m m

= + + = −

− −= + + − − −

−= + + − − +

− −= + + − + − −

∫

0 0

0 0

0

lnm u m qt

s v t ut utq m

−= + + + −

0 0

0 0

0

lnm m

s s v t u t tq m qt

= + + − − −

(2)





Data: 0

7500 ft/s 7500 360 7860 ft/sv v= = + =

( )( )fuel fuel

100060 s 0.5176 slug/s

32.2 60

m Wt q

t gt= = = = =

0

0

11600360.25 slugs

32.2

Wm

g= = =

0

11600 1000329.19 slugs

32.2 32.2m qt− = − =

0

0s =

From (1), 329.19

7860 7500 ln360.25

u= −

360.25

360 ln329.19

u= 3993 ft/su =

From (2), ( )( ) 360.25 360.250 7500 60 3993 60 60 ln

0.5176 329.19s

= + + − −

6460.6 10 ft= × 87.2 mis =






See sample Problem 14.8 for derivation of

0 0

0 0

ln lnm m qt

v u gt u gtm qt m

−= − = − −−

(1)

Note that g is assumed to be constant.

Set dy

vdt

= in (1) and integrate with respect to time.

0

0 0 0

0

20

0

0

ln

1ln

2

h t t

t

mh dy vdt u gt dt

m qt

m qtu dt gt

m

= = = − −

−= − −

∫ ∫ ∫

∫

Let 0 0

0 0

or


m m q

−= = − = −

( ) ]0 0

2 20 0

20 0 0 0 0

0 0 0 0

20 0

0 0

0 0 0

0 0

1 1ln ln

2 2

1ln 1 ln 1

2

11 ln 1 1

2

1ln 1 1 ln 1

2

zz

z z

m u m uh zdz gt z z z gt

q q

m u m qt m qt m mgt

q m m m m

m u qt m qtgt

q m m

m u m qt m qtut

q m m

= − = − −

− −= − − − −

−= − − + −

− −= − + − − −

∫

2

20 0

0

1ln

2

gt

m u m qtut ut gt

q m

−= + − −

20 0

0

1ln

2

m mh u t t gt

q m qt

= − − − −

(2)

Data: 2

0960 kg, 10 kg/s, 3600 m/s, 9.81 m/sm q u g= = = =

fuel

fuel

800800 kg, 80 s

10

mm t

q= = = =





(a) ( )( )2 3960 960 13600 80 80 ln 9.81 80 153.4 10 m

10 960 800 2h

= − − − = × −

153.4 kmh =

(b) From equation (1), ( )( )9603600ln 9.81 80

960 800v = −

−

5670 m/s=v






Data from Problem 14.97: 0

5000 kg, 4200 m/sm u= =

0 0

0, 0v s= =

Thrust force dm

P u uqdt

= =

Mass of satellite plus unspent fuel 0

m m qt= −

Acceleration: 0

P uqa

m m qt= =

−

Integrating with respect to time to obtain the velocity,

0 00 0

0

t t qv v adt v u dt

m qt= + = +

−∫ ∫ (1)

( ) 0

0 0 0 0

0

ln ln lnm qt

v u m qt m v um

− = − − − = −

Integrating again to obtain the displacement,

0

0 0 0

0

lnt m qt

s s v t u dtm

−= + − ∫

Let 0 0

0 0

or


m m q

−= = − = −

( ) ]0 0

0 0

0 0

00 0 0 0

0 0

0 0 0 0

0 0

0 0

0

0 0 0

0 0

0 0

ln ln

ln 1 ln 1

1 ln 1 1

ln 1 1 ln 1

zz

z z

o

m u m us s v t zdz z z z

q q

mm u m qt m qt ms v t

q m m m m

m u qt m qts v t

q m m

m u m qt m qts v t ut

q m m

= + + = −

− −= + + − − −

−= + + − − +

− −= + + − + − −

∫

0 0

0 0

0

lnm u m qt

s v t ut utq m

−= + + + −

0 0

0 0

0

lnm m

s s v t u t tq m qt

= + + − − −

(2)

Using the data,

( )( )5000 5000

0 0 4200 80 80 ln18.75 5000 18.75 80

s

= + + − − −

356.367 10 m= × 56.4 kms =






Let F be the thrust force, and dm

dt be the mass flow rate.

Absolute velocity of exhaust: ev u v= −

Thrust force: ( )dmF u v

dt= −

Power of thrust force: ( )1

dmP Fv u v v

dt= = −

Power associated with exhaust: ( ) ( ) ( )( )22

2

1 1

2 2e

P t m v m u v∆ = ∆ = ∆ −

( )22

1

2

dmP u v

dt= −

Total power supplied by engine: 1 2

P P P= +

( ) ( ) ( )2 2 21 1

2 2

dm dmP u v v u v u v

dt dt

= − + − = −

Mechanical efficiency: 1useful power

total power

P

Pη = =

( )2 2

2 u v v

u v

η−

=−

( )2v

u v

η =+

1η = when .u v= The exhaust, having zero velocity, carries no power away.






Let F be the thrust force and dm

dt be the mass flow rate.

Absolute velocity of exhaust: ev u v= −

Thrust force: dm

F udt

=

Power of thrust force: 1

dmP Fv uv

dt= =

Power associated with exhaust: ( ) ( ) ( )( )22

2

1 1

2 2e

P t m v m u v∆ = ∆ = ∆ −

( )22

1

2

dmP u v

dt= −

Total power supplied by engine: 1 2

P P P= +

( ) ( )2 2 21 1

2 2

dm dmP uv u v u v

dt dt

= + − = +

Mechanical efficiency: 1useful power

total power

P

Pη = =

( )2 2

2uv

u v

η =+

1η = when .u v= The exhaust, having zero velocity, carries no power away.






The weights are 30 lb, 40 lb, and 50 lb.A B C

W W W= = =

Initial velocities: ( )0

9 ft/sAv = , ( )

06 ft/s

Bv = , ( )

0and 0.

Cv =

There are no horizontal external forces acting during the impacts, and the baggage carrier is free to coast

between the impacts.

(a) Suitcase A is thrown first.

Let 1v be the common velocity of suitcase A and the carrier after the first impact and

2v be the common

velocity of the two suitcases and the carrier after the second impact.

Initial momenta: ( ) ( )0 0, , and 0.A B

A B

W Wv v

g g

Suitcase A impacts carrier. Conservation of momentum:

( ) ( ) ( )( )0

1 10

30 90 3.375 ft/s

80

A AA A C

A

A C

W vW W Wv v v

g g W W

++ = = = =+

Suitcase B impacts on suitcase A and carrier. Conservation of momentum:

( ) 1 20

B A C A B C

B

W W W W W Wv v v

g g g

+ + ++ =

( ) ( ) ( )( ) ( )( )10

2

40 6 80 3.3754.25 ft/s

120

B B A C

A B C

W v W W vv

W W W

+ + += = =

+ + 24.25 ft/s=v

(b) Suitcase B is thrown first.

Let 3v be the common velocity of suitcase B and the carrier after the first impact and

4v be the common

velocity of all after the second impact.

Suitcase B impacts the carrier. Conservation of momentum:

( ) ( ) ( )( )0

3 30

40 60 2.6667 ft/s

90

B BB B C

B

B C

W vW W Wv v v

g g W W

++ = = = =+

Suitcase A impacts on suitcase B and carrier. Conservation of momentum:

( ) 3 40

A B C A B C

A

W W W W W Wv v v

g g g

+ + ++ =

( ) ( ) ( )( ) ( )( )30

4

30 9 90 2.66674.25 ft/s

120

A A B C

A B C

W v W W vv

W W W

+ + += = =

+ + 44.25 ft/s=v






The weights are 30 lb, ?, and 50 lb.A B C

W W W= = =

Initial velocities: ( ) ( )0 0

7.2 ft/sA Bv v= = ( )

0, 0

Cv =

Final velocity: 3.6 ft/sfv =

(a) Conservation of momentum:

( ) ( )0 0

0A B A B C

A B f

W W W W Wv v v

g g g

+ ++ + = (1)

( )( ) ( ) ( )( )30 7.2 7.2 30 50 3.6B B

W W+ = + +

20.0 lbB

W =

(b) Equation (1) shows that the final velocity is independent of the order in which the suitcases are thrown.

3.60 ft/sf =v






Linear momentum of each particle expressed in kg m/s.⋅

12 6 6

8 6

8 16 8

A A

B B

C C

m

m

m

= + += += − + +

v i j k

v i j

v i j k

Position vectors, (meters): 3 , 1.2 2.4 3 , 3.6A B C

= = + + =r j r i j k r i

( )2Angular momentum about , kg m /s .O ⋅

( ) ( ) ( )

( ) ( ) ( )

0 3 0 1.2 2.4 3 3.6 0 0

12 6 6 8 6 0 8 16 8

18 36 18 24 12 28.8 57.6

0 4.8 9.6

O A A A B B B C C Cm m m= × + × + ×

= + +−

= − + − + − + − +

= − +

H r v r v r v

i j k i j k i j k

i k i j k j k

i j k

( ) ( )2 24.80 kg m /s 9.60 kg m /sO

= − ⋅ + ⋅H j k






Position vectors, (meters): 3 , 1.2 2.4 3 , 3.6A B C

= = + + =r j r i j k r i

( ) Mass center:a ( )A B C A A B B C Cm m m m m m+ + = + +r r r r

( )( ) ( )( ) ( )( )9 3 3 2 1.2 2.4 3 4 3.6

1.86667 1.53333 0.66667

= + + + +

= + +

r j i j k i

r i j k

( ) ( ) ( )1.867 m 1.533m 0.667 m= + +r i j k

Linear momentum of each particle, ( )2kg m /s .⋅

12 6 6

8 6

8 16 8

A A

B B

C C

m

m

m

= + += += − + +

v i j k

v i j

v i j k

(b) Linear momentum of the system, ( )kg m/s.⋅

12 28 14A A B B C C

m m m m= + + = + +v v v v i j k

( ) ( ) ( )12.00 kg m/s 28.0 kg m/s 14.00 kg m/sm = ⋅ + ⋅ + ⋅v i j k

Position vectors relative to the mass center, (meters).

1.86667 1.46667 0.66667

0.66667 0.86667 2.33333

1.73333 1.53333 0.66667

A A

B B

C C

′ = − = − + −′ = − = − + +′ = − = − −

r r r i j k

r r r i j k

r r r i j k





(c) Angular momentum about G, ( )2kg m /s .⋅

( ) ( )( )

1.86667 1.46667 0.66667 0.66667 0.86667 2.33333

12 6 6 8 6 0

1.73333 1.53333 0.66667

8 16 8

12.8 3.2 28.8 14 18.6667 10.9333

1.6 8.5333 15.4667

2.8 13.3333

G A A A B B B C C Cm m m′ ′ ′= × + × + ×

= − − + −

+ − −−

= + − + − + −

+ − − +

= − + −

H r v r v r v

i j k i j k

i j k

i j k i j k

i j k

i j 24.2667k

( ) ( ) ( )2 2 22.80 kg m /s 13.33 kg m /s 24.3 kg m /sG

= − ⋅ + ⋅ − ⋅H i j k

( ) ( ) ( )2 2 2

1.86667 1.53333 0.66667

12 28 14

2.8 kg m /s 18.1333 kg m /s 33.8667 kg m /s

m× =

= ⋅ − ⋅ + ⋅

i j k

r v

i j k

( ) ( )2 24.8 kg m /s 9.6 kg m /sG

m+ × = − ⋅ + ⋅H r v j k


( ) ( ) ( )

( ) ( ) ( )( ) ( )2 2

0 3 0 1.2 2.4 3 3.6 0 0

12 6 6 8 6 0 8 16 8

18 36 18 24 12 28.8 57.6

4.8 kg m /s 9.6 kg m /s

O A A A B B B C C Cm m m= × + × + ×

= + +−

= − + − + − + − +

= − ⋅ + ⋅

H r v r v r v

i j k i j k i j k

i k i j k j k

j k

Note that

O Gm= + ×H H r v






Choose x axis pointing east, y axis north, and z axis vertical.

Velocities before collision:

( )0

5.5 mi 5280 ft/miHelicoptor: 121 ft/s

4 min 60 s/minHv = ⋅ =

( )0

10 mi 5280 ft/miAirplane: 220 ft/s

4 min 60 s/minA xv = ⋅ =

( )0

7.5 mi 5280 ft/mi165 ft/s

4 min 60 s/minA yv

= − ⋅ = −

( ) ( ) ( ) ( ) ( )

( ) ( )

0 0 0 0

Mass center:

6000 3000121 220 165

9000 9000

154 ft/s 55 ft/s

H AH A Ax y

A H A H

m m

v v v

m m m m

= + + + +

= + −

= −

v i i j

i i j

i j

No external forces act during impact. Assume that only gravity acts after the impact. Motion of mass center

after impact:

( )2 2

0 0

1154 55 3600 16.1

2t z gt t t t

= + − = − + −

r v k i j k

Time of fall. 2 3600

14.953 s16.1

t t= =

( )( ) ( )( ) ( ) ( )154 14.953 55 14.953 2302.8 ft 822.42 ft= − = −r i j i j

( ) ( ) ( ) ( ) ( )1 1 2 2H A H H H H A A

m m m m m+ = + +r r r r

( ) ( ) ( ) ( ) ( )

( )( )

( )( ) ( )( )

1 1 2 2

1

19000 2302.8 822.42

3000

2000 1500 300 4000 1800 1500

A H A H H H H

A

m m m m

m

= + − −

= −

− − − −

r r r r

i j

i j i j

( ) ( )3510 ft 267 ft= −i j

( )Coordinates of point : 3510 ft, 267 ftA −






Choose x axis pointing east, y axis north, and z axis vertical.

Velocities before collision:

( )0

5.5 mi 5280 ft/miHelicoptor: 121 ft/s

4 min 60 s/minHv = ⋅ =

( )0

10 mi 5280 ft/miAirplane: 220 ft/s

4 min 60 s/minA xv = ⋅ =

( )0

7.5 mi 5280 ft/mi165 ft/s

4 min 60 s/minA yv

= − ⋅ = −

( ) ( ) ( ) ( ) ( )

( ) ( )

0 0 0 0

Mass center:

6000 3000121 220 165

9000 9000

154 ft/s 55 ft/s

H AH A Ax y

A H A H

m m

v v v

m m m m

= + + + +

= + −

= −

v i i j

i i j

i j

No external forces act during impact. Assume that only gravity acts after the impact. Motion of mass center

after impact:

( )2 2

0 0

1154 55 3600 16.1

2t z gt t t t

= + − = − + −

r v k i j k

Time of fall. 2 3600

14.953 s16.1

t t= =

( )( ) ( )( ) ( ) ( )154 14.953 55 14.953 2302.8 ft 822.42 ft= − = −r i j i j

( ) ( ) ( ) ( ) ( )1 1 2 2H A H H H H A A

m m m m m+ = + +r r r r

( ) ( ) ( ) ( ) ( )

( )( )

( )( ) ( )( )

2 1 1

2

1

19000 2302.8 822.42

4000

3000 3600 240 2000 1200 600

H H A A A H H

H

r m m m m

m

= + − −

= −

− + − −

r r r

i j

i j i j

( ) ( )1881 ft 1730 ft= −i j ( )2

Coordinates of point : 1881 ft, 1730 ftH −






Use a frame of reference moving with the mass center.


0A A B B

B

A B

A

m v m v

m

v v

m

′ ′= − +

′ ′=


( ) ( ) ( )

( ) ( )

( )

2

2 2 2

2

1 1 1 1

2 2 2 2

2

2

B

A A B B A B B B

A

B A B

B

A

A

B

B A B

mV m v m v m v m v

m

m m mv

m

m Vv

m m m

′ ′ ′ ′= + = +

+′=

′ =+

Data: 2 25 30.15528 lb s /ft, 0.09317 lb s /ft

32.2 32.2A B

m m= = ⋅ = = ⋅

90 ft lbV = ⋅

( )( )( )( )( )

2 0.15528 9034.75 34.75 ft/s

0.09317 0.24845B Bv v′ ′= = = 30°

( )0.0931734.75 20.85 20.85 ft/s

0.15528A Av v′ ′= = = 30°

Velocities of A and B:

[24 ft/sA

=v ] [20.85 ft/s+ ]30° 12.00 ft/sA

=v 60.3°

[24 ft/sB

=v ] [34.75 ft/s+ ]30° 56.8 ft/sB

=v 17.8°






Locate the mass center.

Let l be the distance between A and B.

( )B A B Am l m m l= +

2

7

5

7

B

A

A B

B

ml l l

m m

l l

= =+

=

(a) Linear momentum.

( )( )02.5 3.5 8.75

AL m v= = = 8.75 kg m/s= ⋅L

Angular momentum about G:

( )( )( )0 0

2 20.210 2.5 3.5 0.525

7 7G A A A

H l m v lm v= = = =

20.525 kg m /s

G= ⋅H

(b) There are no resultant external forces acting on the system;

therefore, L and G

H are conserved.

:L A A B B

m v m v L+ = 2.5 1.0 8.75A Bv v+ = (1)

:G

H B B B A A A Gl m v l m v H− =

( )( ) ( )( )5 20.210 1.0 0.210 2.5 0.525

7 7B Av v− =

0.15 0.15 0.525B Av v− = (2)

Solving (1) and (2) simultaneously, 1.5 m/s, 5 m/sA Bv v= =

1.500 m/sA

=v

5.00 m/sB

=v






For steady flow 1 2

Q Q Q+ = (1)

Assume that the fluid speed is constant.

Velocity vectors: ( ) 1 2sin cos , , v v vθ θ= − = − =v i j v i v i

Let Pj be the force that plate C exerts on the fluid.


( ) ( ) ( ) ( )1 21 2m P t m m∆ + ∆ = ∆ + ∆v j v v

( ) ( )

1 2

1 2

m m mP

t t t

∆ ∆ ∆= + −∆ ∆ ∆

j v v v

( ) ( ) ( )1 2

sin cosdm dm dm

v v v vdt dt dt

θ θ = − + − −

i i i j

( )1 2sin cosQ v Q v Qvρ ρ ρ θ θ= − + − −i i i j


i: 1 2 2 1

0 sin sinQ v Q v Qv Q Q Qρ ρ ρ θ θ= − + − = − (2)

j: cosP Qvρ θ= (3)

Data: 2 462.41.93789 lb s /ft , 90 ft/s

32.2v

g

γρ = = = ⋅ =

3 3 3 3

1 226 gal/min 57.932 10 ft /s, 130 gal/min 289.66 10 ft /sQ Q− −= = × = = ×





From (1), 3 3347.59 10 ft /sQ −= ×

(a) From (2), 2 1sin 0.66667 41.810

Q Q

Qθ θ−= = = °

41.8θ = °

From (3), ( )( )( )31.93789 347.59 10 90 cos 41.810 45.2 lb.P

−= × ° =

(b) Force that stream exerts on plate C:

45.2 lbP− =j






Calculation of or .

m dm

t dt

∆∆


( ) ( ) mm A l Av t Av

tρ ρ ρ∆∆ = ∆ = ∆ =

∆

3 2

2 2 2

62.4 lb/ft 1.5 in60 ft/s 1.21118 lb s/ft

32.2 ft/s 144 in /ft

dmAv

dtρ= = ⋅ ⋅ = ⋅

Apply the principle of impulse-momentum.

moments about D: ( ) ( ) ( ) ( )12 16 10 4

12 12 12 12A B

m v C t W t m v − ∆ − ∆ + ∆ = − ∆

( )16 10 4 12B A

mC W v v

t

∆= + −∆

( )( ) ( ) ( )( ) ( )( )10 10 1.21118 4 60 12 60 481.37 = + − = −

30.085 lbC = − 30.1 lb=C

x components: ( ) ( ) ( )A x Bm v D t m v∆ + ∆ = ∆

( ) ( ) ( )( )1.21118 60 60 0x B A B A

m dmD v v v v

t dt

∆= − = − = − =∆

0x

D =

y components: ( ) ( ) ( )0 0y

C t D t W t+ ∆ + ∆ − ∆ =

( )10 30.085 40.085 lby

D W C= − = − − = 40.1 lb=D






Calculation of dm

dt at a section in the airstream:


( )m A l Av tρ ρ∆ = ∆ = ∆

m dm

Avt dt

ρ∆ = =∆

(a) ( )ThrustB A

dm

dt= −v v where

Bv is the velocity just downstream

of propeller and A

v is the velocity far upstream. Assume A

v is

negligible.

( ) 2 2Thrust

4Av v D v

πρ ρ = =

( ) ( )2 2 23600 1.21 2 3.801

4v v

π = =

30.774m/sv = 30.8 m/sv =

(b) ( ) ( )2212 30.774

4 4

dmQ Av D v

dt

π πρ

= = = =

3

96.7 m /sQ =

(c) Kinetic energy of mass :m∆

( ) ( ) ( )2 2 21 1 1

2 2 2T m v A l v Av t vρ ρ∆ = ∆ = ∆ = ∆

3 2 31 1

2 2 4

T dTAv D v

t dt

πρ ρ∆ = = = ∆

( ) ( ) ( )2 3 311.21 2 30.774 55.4 10 N m/s

2 4

π = = × ⋅

55.4 kWdT

dt=






From Eq. (14.44) of the text book, the thrust is

( )( ) 3 2 310 kg/s 3600 m/s 36 10 kg m/s 36 10 Ndm

P udt

= = = × ⋅ = ×

F maΣ =

P mg ma− = P

a gm

= − (1)

(a) At the start of firing, 2

0960 kg, 9.81 m/s m m g= = =

From (1), 3

236 109.81 27.69 m/s

960a

×= − = 227.7 m/s=a

(b) As the last particle of fuel is consumed,

960 800 160 kg,m = − = ( )29.81 m/s assumedg =

From (1), 3

236 109.81 215.19 m/s

160a

×= − = 2215 m/s=a


Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.


5.4 kg, 600 mm 0.6m, 300 mm 0.3 mm a b= = = = =

Total length of rod. 2 2 1.8 ml a b= + =

Mass per unit length. 5.4

3 kg/m1.8

m

lρ = = =

Use principal axes ,x z′ ′ as shown.

tan 0.5 26.565b

aβ β= = = °

10 cos rad/s 0x yω β ω′ = =

10 sin rad/szω β′ = −



Moments of Inertia:

Part , kgm 2, kg mxI ′ ⋅ 2, kg mzI ′ ⋅

1.8aρ = 2

0.04052

bm =

210.054

12ma =

0.9bρ = 210.00675

12mb =

2

0.0812

am =

1.8aρ = 2

0.04052

bm =

210.054

12m a =

0.9bρ = 210.00675

12mb =

2

0.0812

am =

Σ 5.4lρ = 0.0945xI ′ = 0.270zI ′ =

Angular momentum .GH G x x y y z zI I Iω ω ω′ ′ ′ ′ ′ ′= + +′ ′ ′H i j k

( )( ) ( )( )0.0945 10 cos 0 0.270 10 sinG β β′ ′= + + −H i k

( ) ( ) ( ) ( )0.945cos cos sin 2.70sin sin cosβ β β β β β= + + − − +i k i k

( ) ( )2 20.945 cos 2.70sin 0.945 2.70 sin cosβ β β β= + + −i k

1.296 0.702= −i k

( ) ( )2 21.296 kg m /s 0.702 kg m /sG = ⋅ − ⋅H i k !




Use principal axes , y z′ ′ as shown.

cos 45 , sin 45y zω ω ω ω′ ′= ° = °

0xω ′ =

2 21 1,

3 12x yI ma I ma′ ′= =

25

12z x yI I I ma′ ′ ′= + =

A x x y y z zI I Iω ω ω′ ′ ′ ′ ′ ′′ ′ ′= + +H i j k

( ) ( )2 21 50 cos 45 sin 45

12 12ma maω ω ′ ′= + ° + °

j k

( )( )21cos 45 cos 45 sin 45

12A ma ω = ° ° − °

H j k

( )( )25sin 45 sin 45 cos 45

12ma ω + ° ° + °

j k

( )2

3 212

ω= +H j kAma

!




23.60.1118 lb s /ft 24 in. 2 ft

32.2m l= = ⋅ = =

( )( )1200 2125.664 rad/s

60

πω = =

Use principal axes , x y′ ′ as shown.

sin 20 42.980 rad/sxω ω′ = − ° = −

cos 20 118.085 rad/syω ω′ = ° =

For the rod, 0′ ≈xI

( )( )22 21 10.1118 2 0.037267 lb s ft

12 12yI ml′ = = = ⋅ ⋅

G x x y y z zI I Iω ω ω′ ′ ′ ′ ′ ′′ ′ ′= + +H i j k

( )( )0 0.037267 118.085 0′= + +j

4.40068 ′= j

( )4.40068 sin20 cos 20= ° + °i j

1.5051 4.1353= +i j

( ) ( )1.505 lb s ft 4.14 lb s ftG = ⋅ ⋅ + ⋅ ⋅H i j!




Use principal axes as shownx , y ,z .′ ′ ′

cos sinβ β′ = +i i j

sin cosβ β= − +′j i j

cos sinβ β′ ′= −i i j

sin cosβ β′ ′= +j i j

cos sinω ω β ω β= = −′ ′i i jωωωω

Pr incipal moments of inertia:

2 21 1,

2 4x y zI mr I I mr′ ′ ′= = =

Angular momentum:

G x x y y z zI I Iω ω ω′ ′ ′ ′ ′ ′′ ′ ′= + +H i j k

2 21 1cos sin

2 4mr mrω β ω β′ ′= −i j

( ) ( )2 21 1cos cos sin sin sin cos

2 4G mr mrω β β β ω β β β = + − − + H i j i j

2 2 2 21 1 1cos sin sin cos

2 4 4mr mrω β β ω β β = + +

i j

For 20 ,β = ° ( ) ( )2 20.47076 0.080348G mr mrω ω= +H i j

( )( )

0.080348tan

0.47076

G y

G x

H

Hθ = =

9.7θ = ° !




Body diagonal: ( )22 22 6= + + =d a a a a

( ) 226 6 6

a a adω ω ω ω= − + − = − + −ω i j k i j k

( )2 2 21 5212 12xI m a a ma = + =

2 2 21 112 6yI m a a ma = + =

( )22 21 5212 12zI m a a ma = + =

(a) G x x y y z zI I Iω ω ω= + +H i j k

2 2 25 1 2 512 6 126 6 6

ma ma maω ω ω = − + + −

i j k

( )2

5 4 512 6ma ω= − + −i j k

2 2

2 2 2 115 4 51212 6

ω ω= + + =Gma maH 20.276 ω=GH ma !

(b) ( )( ) ( ) ( )2 2

5 4 5 212 6

ω⋅ = − + − ⋅ − + −H ω i j k i j kGma

( )( )2 2

218 112 6 4

ω ω= =ma ma

2 21112GH maω ω=

12cos 0.904534 11

θω⋅= = =H ωG

GH 25.2θ = °!




Body diagonal ( )22 22 6= + + =d a a a a

( ) 226 6 6

ω ω ω ω= − + − = − + −ω i j k i j ka a ad

Total area ( )2 2 2 22 2 2 10= + + =a a a a

For each square plate: 110

′ =m m

2 2 2 21 13 1312 12 120xI m a m a m a ma′ ′ ′= + = =

2 21 16 60yI m a ma′= =

213120z xI I ma= =

For each plate parallel to the yz plane: 15

′ =m m

( )22 2 21 5 1212 12 12xI m a a m a ma ′ ′= + = =

2

2 2 21 1 112 2 3 15y

aI m a m m a ma ′ ′ ′= + = =

( )2

2 2 21 7 7212 2 12 60z

aI m a m m a ma ′ ′= + = =

For each plate parallel to the xy plane: 15

′ =m m

( )2

2 2 21 7 7212 2 12 60x

aI m a m m a ma ′ ′ ′= + = =



2

2 2 21 1 112 2 3 15y

aI m a m m a ma ′ ′ ′= + = =

( )22 2 21 5 1212 12 12zI m a a m a ma ′ ′= + = =

Total moments of inertia:

2 213 1 7 372120 12 60 60xI ma ma = + + =

2 21 1 1 3260 15 15 10yI ma ma = + + =

2 213 7 1 372120 60 12 60zI ma ma = + + =

(a) ( )2

37 36 3760 6G x x y y z zmaI I I ωω ω ω= + + = − + −H i j k i j k

( ) ( ) ( )2

2 2 2 237 36 37 0.4321660 6

ω ω= + + =GmaH ma

20.432 ω=GH ma !

(b) ( )( ) ( ) ( )2

2 237 36 37 2 0.4055660 6Gma maω ω⋅ = − + − ⋅ − + − =H ω i j k i j k

0.40556cos 0.938440.43216

θω⋅= = =H ωG

GH

20.2θ = °!




( ) ( )2 1 4 rad/s 12 rad/sω ω= + = +ω j k j k

For axes , , x y z′ ′ ′ parallel to x, y, z with origin at A,

( )( )22 21 18 0.100 0.02 kg m

4 4xI mr′ = = = ⋅

2 20.02 kg m , 0.04 kg my x z x yI I I I I′ ′ ′ ′ ′= = ⋅ = + = ⋅

A x x y y z zI I Iω ω ω′ ′ ′ ′ ′ ′= + +H i j k

( )( ) ( )( )0 0.02 4 0.04 12 0.08 0.48= + + = +j k j k

( ) ( )2 20.0800 kg m /s 0.480 kg m /sA = ⋅ + ⋅H j k!




( ) ( )1 2 8 rad/s 16 rad/sω ω= + = +ω j i i j


( )( )22 21 16 0.160 0.0384 kg m

4 4xI mr′ = = = ⋅

2 20.0384 kg m , 0.0768 kg mz x y x zI I I I I′ ′ ′ ′ ′= = ⋅ = + = ⋅


( )( ) ( )( )0.0384 8 0.0768 16 0= + +i j

0.3072 1.2288= +i j

( ) ( )2 20.307 kg m /s 1.229 kg m /sA = ⋅ + ⋅H i j!




260 1.86335 lb s /ft, 2.4 in. 0.2 ft, 10 in. 0.83333 ft.32.2

= = ⋅ = = = =x ym k k

( )( )22 21.86335 0.2 0.074534 lb s ftx xI mk= = = ⋅ ⋅

( )( )22 21.86335 0.83333 1.29400 lb s fty z yI I mk= = = = ⋅ ⋅

( ) ( ) ( )G G G G x x y y z zx y zH H H I I Iω ω ω= + + = + +H i j k i j k

( ) 0.640 8.5867 rad/s

0.074534G x

xx

HI

ω = = =

( ) 0.018 0.013910 rad/s1.29400

G yy

y

H

Iω −= = = −

0zω =

sinω θ ω= −p y

0.013910 0.1596 rad/ssin sin 5

ωω

θ−

= = =°

yp

(a) Rate of spin. cosx s pω ω ω θ= +

coss x pω ω ω θ= −

8.5867 0.1596cos5= − °

8.43 rad/ssω = !

(b) Rate of precession. 0.1596 rad/spω = !




2/

60 111.86335 lb s /ft, ft

32.2 12G Am = = ⋅ =

r i

cos sinv vθ θ= −v i j

/= + ×H H r vA G G A m

( )110.640 0.018 1950cos5 1950sin 5

12 = − + × ° − °

i j i i j

( )110.640 0.018 1950sin 5

12 = − − °

i j k

( ) ( ) ( )0.640 lb s ft 0.018 lb s ft 155.8 lb s ftA = ⋅ ⋅ − ⋅ ⋅ − ⋅ ⋅H i j k !




Using Equation (18.11)

= × +H r v HO Gm

( ) ( ) 21 1

1

2 2ω ω = × + −

i k i j

rL mr mr

L

3

21 1 1

1 1

2 4

rmrL mr m

Lω ω ω= − + −j i j

( )2 2 21 1

1 1/

2 4ω ω = − +

H i jO mr m L r r L

which is the answer obtained in part b of Sample Problem 18.2.




(a) Angular momenta HA and HB are related to HG and mv by

/ / and = × + = × +H r v H H r v HA G A G B G B Gm m

Subtracting, / /− = × − ×H H r v r vB A G B G Am m

( )/ /= + − ×H H r r vB A G B G A m

( )/ /= + + ×H r r vA G B A G m

/= + ×H H r vB A A B m

(b) It follows that =H HA B if, and only if

/ 0× =r vA B m

With points A and B located on the fixed axis, ω= λω

Where λ is a unit vector along the fixed axis, and

/ /ω= × = ×v r rG A G Aω λ

Then, ( )/ / 0ω× × =r rA B G Am λ

but, /rA B is parallel to λ, hence,

( )/ 0× × =rG Aλ λ

Let /G A= ×u rλ , so that 0.× =uλ

Note that u must be either perpendicular to λ or equal to zero. But, if u is perpendicular to λ, × uλ cannot be equal to zero.

Hence, / 0G A= × =u rλ

/rG A is parallel to λ and point G lies on the fixed axis.






( )( )22 21 1 8 0.100 0.02 kg m4 4xI mr′ = = = ⋅



( )( ) ( )( ) ( ) ( )2 20 0.02 4 0.04 12 0.08 kg m /s 0.48 kg m /s= + + = ⋅ + ⋅j k j k

Point A is the mass center of the disk.

( ) ( )/ 0.32 m 0.2 mA O = −r i j

( ) ( )2 / 4 0.32 0.2 1.28 m/sA A Oω= = × = × − = −v v j r j i j k

( )( ) ( )8 1.28 10.24 kg m/sm = − = − ⋅v k k

( ) ( )/ 0.32 0.2 10.24A O m× = − × −r v i j k

( ) ( )2 22.048 kg m /s 3.2768 kg m /s= ⋅ + ⋅i j

/O A A O m= + ×H H r v

( ) ( ) ( )2 2 22.05 kg m /s 3.36 kg m /s 0.480 kg m /sO = ⋅ + ⋅ + ⋅H i j k !




( ) ( )2 1 8 rad/s 16 rad/sω ω= + = +ω i j i j


( )( )22 21 16 0.160 0.0384 kg m

4 4xI mr′ = = = ⋅



( )( ) ( )( )0.0384 8 0.0768 16 0= + +i j

( ) ( )2 20.3072 kg m /s 1.2288 kg m /s= ⋅ + ⋅i j


( ) ( ) ( )/ 0.24 m 0.18 m 0.18 m= + −r i j kA D

( )2 / 8 0.24 0.18 0.18A A Dω= = × = × + −v v i r i i j k

( ) ( )1.44 m/s 1.44 m/s= +j k

( ) ( )8.64 kg m/s 8.64 kg m/sm = ⋅ + ⋅v j k

/ 0.24 0.18 0.18

0 8.64 8.64A D m× = −

i j k

r v

( ) ( ) ( )2 2 23.1104 kg m /s 2.0736 kg m /s 2.0736 kg m /s= ⋅ − ⋅ + ⋅i j k

/D A A D m= + ×H H r v

( ) ( ) ( )2 2 23.42 kg m /s 0.845 kg m /s 2.07 kg m /sD = ⋅ − ⋅ + ⋅H i j k!




5 kg, 200 mm 0.2 m= = =m a

( )( )2 36012 rad/s. 0, 0, 12 rad/s

60π

ω π ω ω ω π= = = = =x y z

Use parallel axes x′, y′, z′ with origin at the mass center as shown.

( )G xzxH I ω= −

( )G yzyH I ω= −

( )G zzH I ω=

Segments 1, 2, 3, and 4, each of mass 2.5 kg,′ =m contribute to , and .xz yz zI I I

Part xzI yzI zI

21

2′− m a 21

4′m a 21 1 1

12 4 ′+ +

m a

21

4′− m a 0 21

3′m a

21

4′− m a 0 21

3′m a

21

2′− m a 21

4′m a 21 1 1

12 4 ′+ +

m a

Σ 232

′− m a 212

′m a 2103

′m a

continued



( ) 232G xH m a ω ′= − −

( )( ) ( )23 2.5 0.2 122

π =

25.6549 kg m /s= ⋅

( ) 212G yH m a ω ′= −

( )( ) ( )21 2.5 0.2 122

π = −

21.8850 kg m /s= − ⋅

( ) 2103G zH m a ω ′=

( )( ) ( )210 2.5 0.2 123

π =

212.5664 kg m /s= ⋅

Velocity of the mass center: 0=v

(a) /= + × =H H r v HA G G A Gm

( ) ( ) ( )2 2 25.65 kg m /s 1.885 kg m /s 12.57 kg m /s= ⋅ − ⋅ + ⋅H i j kA !

(b) ( ), = ⋅ = ⋅ =k H k HAB AB A A A zHλ λ

( ) ( ) ( )2 2 2 25.6549 1.8850 12.5664 13.9085 kg m /s= + + = ⋅AH

( ) 12.5664cos

13.9085θ = =A z

A

HH

25.4θ = °!




5 kg, 200 mm 0.2 m= = =m a

( )( )2 36012 rad/s. 0, 0, 12 rad/s

60π

ω π ω ω ω π= = = = =x y z

Use parallel axes x′, y′, z′ with origin at the mass center as shown.

( )G xzxH I ω= −

( )G yzyH I ω= −

( )G zzH I ω=

Segments 1, 2, 3, and 4, each of mass 2.5 kg,′ =m contribute to , and .xz yz zI I I

Part xzI yzI zI

21

2′− m a 21

4′m a 21 1 1

12 4 ′+ +

m a

21

4′− m a 0 21

3′m a

21

4′− m a 0 21

3′m a

21

2′− m a 21

4′m a 21 1 1

12 4 ′+ +

m a

Σ 232

′− m a 212

′m a 2103

′m a



( ) 232G xH m a ω ′= − −

( )( ) ( )23 2.5 0.2 122

π =

25.6549 kg m /s= ⋅

( ) 212G yH m a ω ′= −

( )( ) ( )21 2.5 0.2 122

π = −

21.8850 kg m /s= − ⋅

( ) 2103G zH m a ω ′=

( )( ) ( )210 2.5 0.2 123

π =

212.5664 kg m /s= ⋅

Velocity of the mass center: 0=v

(a) /= + × =H H r v HB G G B Gm

( ) ( ) ( )2 2 25.65 kg m /s 1.885 kg m /s 12.57 kg m /s= ⋅ − ⋅ + ⋅H i j kB !

(b) ( ), = − ⋅ = − ⋅ = −k H k HBA BA A A A zHλ λ

( ) ( ) ( )2 2 2 25.6549 1.8850 12.5664 13.9085 kg m /s= + + = ⋅AH

( ) 12.5664cos

13.9085θ = − = −A z

A

HH

154.6θ = °!




2100.31056 lb s /ft

32.2

Wm

g= = = ⋅

( )12 rad/s , 0ω ω ω= = =i y z

( )G xxH I ω=

( )G xyyH I ω= −

( )G xzzH I ω= −

The shaft is comprised of 8 sections, each of length

10

ft 12

a = and of mass 20.03882 lb s /ft.8

mm′ = = ⋅

( ) ( )( ) ( )2

2 2 2 21 10 10 104 2 0.03882 0.089861 lb s ft

3 3 3 12xI m a m a m a ′ ′ ′= + = = = ⋅ ⋅

0xyI =

( ) ( )( )2

2 2104 2 2 0.03882 0.053916 lb s ft

2 12xza

I m a m a ′ ′= = = = ⋅ ⋅

( ) ( )( )0.089861 12 1.0783 lb s ftG xH = = ⋅ ⋅

( ) 0G yH =

( ) ( )( )0.053916 12 0.6470 lb s ftG zH = − = − ⋅ ⋅

(a) ( ) ( )1.078 lb s ft 0.647 lb s ftG = ⋅ ⋅ − ⋅ ⋅H i k!

( ) ( )2 21.0783 0.6470 1.2575 lb s ftGH = + = ⋅ ⋅

(b) ( )1.0783 0.6470 12 12.940 lb ftG ⋅ = − ⋅ = ⋅H i k iωωωω

( )( )12.940

cos 0.857521.2575 12

G

GHθ

ω⋅= = =H ωωωω

31.0θ = °!




2100.31056 lb s /ft

32.2

Wm

g= = = ⋅

( )12 rad/s , 0ω ω ω= = =i y z

( )G xxH I ω=

( )G xyyH I ω= −

( )G xzzH I ω= −

The shaft is comprised of 8 sections, each of length

10

ft 12

a = and of mass 20.03882 lb s /ft.8

mm′ = = ⋅

( ) ( )( ) ( )2

2 2 2 21 10 10 104 2 0.03882 0.089861 lb s ft

3 3 3 12xI m a m a m a ′ ′ ′= + = = = ⋅ ⋅

0xyI =

( ) ( )( )2

2 2104 2 2 0.03882 0.053916 lb s ft

2 12xza

I m a m a ′ ′= = = = ⋅ ⋅

( ) ( )( )0.089861 12 1.0783 lb s ftG xH = = ⋅ ⋅

( ) 0G yH =

( ) ( )( )0.053916 12 0.6470 lb s ftG zH = − = − ⋅ ⋅

( ) ( )1.078 lb s ft 0.647 lb s ftG = ⋅ ⋅ − ⋅ ⋅H i k

Since point G lies on the axis of rotation, its velocity is zero.

0G= =v v

(a) /= + × =H H r v HA G G A Gm

( ) ( )1.078 lb s ft 0.647 lb s ftA = ⋅ ⋅ − ⋅ ⋅H i k !

(b) /= + × =H H r v HB G G B Gm

( ) ( )1.078 lb s ft 0.647 lb s ftB = ⋅ ⋅ − ⋅ ⋅H i k !




Angular velocity: ω ω= j

Angular momentum about the mass center: G xy y yzI I Iω ω ω= − + −H i j k (1)

Since the body lies in the xy plane, 0.yzI =

Calculation of and : Letxy yI I 2

mass per unit area.m m

A abρ = = =

( ) ( )mass area.I Iρ=

For the pairs of area elements shown,

2el el2 2

2xyx

dI x y dA yxdy x ydy= = =

2

20 0

a axy

byI x ydy b ydy

a = = −

∫ ∫

2 2

2 2 320 0 0

2a a ab bb ydy y dy y dy

a a= − +∫ ∫ ∫

2 2 2 3 2 2 2 21 2 1 1

2 3 4 12b a b a b a b a= − + =

3 31 12

12 6yI ab ab = =

For mass, 21 1,

6 3xy yI mab I mb= =

Data: 8 kg, 750 mm 0.75 m, 6 rad/sm a b ω= = = = =

( )( )( ) 218 0.75 0.75 0.75 kg m

6xyI = = ⋅

( )( )2 218 0.75 1.50 kg m

3yI = = ⋅

From (1), ( ) ( )2 24.50 kg m /s 9.00 kg m /sG = − ⋅ + ⋅H i j!




Mass of each sheet 1.6

0.8 kg2

m = =

Required moment and products of inertia.

( ) ( )( )22 3 21 12 2 0.8 0.120 7.68 10 kg m

3 3xI mb − = = = × ⋅

( )( )( ) 3 20.8 0.080 0.060 3.84 10 kg m2 2xya b

I m − = − = − = − × ⋅

( )( )( ) 3 20.8 0.080 0.060 3.84 10 kg m2 2xza b

I m − = = = × ⋅

Angular velocity. 20 rad/s, 0x y zω ω ω= = =

Angular momentum about G.

( )G x x xy y xz zxH I I Iω ω ω= − −

( )( )3 27.68 10 20 0 0 0.1536 kg m /s−= × − − = ⋅

( )G xy x y y yz zyH I I Iω ω ω= − + −

( )( )3 23.84 10 20 0 0 0.0768 kg m /s−= − − × + − = ⋅

( )G xz x yz y z zzH I I Iω ω ω= − − +

( )( )3 23.84 10 20 0 0 0.0768 kg m /s−= − × − + = − ⋅

( ) ( ) ( )2 2 20.1536 kg m /s 0.0768 kg m /s 0.0768 kg m /sG = ⋅ + ⋅ − ⋅H i j k !




mass per unit length 1.2 kg/mρ = =

225 mm 0.225 m, 300 mm 0.3 m.r l= = = =

Moments and products of inertia: Formulas

m xI yI zI yzI

Upper ring 2 rπ ρ 2 2 21 12 4

m r r l + +

( )2 2m r r+ 2 21 12 4

m r l +

12

m rl −

Lower ring 2 rπ ρ 2 2 21 12 4

m r r l + +

( )2 2m r r+ 2 21 12 4

m r l +

12

m rl −

Rod AB lρ 2112

ml 0 2112

ml 0

Moments and products of inertia: Values

( )kgm ( )2kg mxI ⋅ ( )2kg myI ⋅ ( )2kg mzI ⋅ ( )2kg myzI ⋅

Upper ring 1.69646 0.166995 0.171767 0.081112 −0.057256

Lower ring 1.69646 0.166995 0.171767 0.081112 −0.057256

Rod AB 0.36 0.0027 0 0.0027 0

Totals 3.75292 0.336691 0.343533 0.164924 0.114511



Impulse-momentum principle: Before impact, 0, 0G= =v H

(a) ( ) 5.4 3.75292

F tt m

m∆

∆ = = =F v v j

( )1.439 m/s=v j!

(b) Moments about G: /E G Gt× ∆ =r F H

( ) ( ) ( ) ( ) ( )0.225 0.15 0.225 5.4 G G Gx y zH H H− + × = + +i j k j i j k

( ) ( )1.215 1.215 x x y y yz z yz y z zI I I I Iω ω ω ω ω− + = + − + − +i k i j k

Substitute numerical values for moments and products of inertia and resolve into components.

: 1.215 0.336691 xω− =i

: 0 0.343533 0.114511y zω ω= +j

: 1.215 0.114511 0.164924y zω ω= +k

Solving, 3.6087 rad/s, 3.1952 rad/s, 9.5855 rad/sx y zω ω ω= − = − =

( ) ( ) ( )3.61 rad/s 3.20 rad/s 9.59 rad/s= − − +i j kω !




mass per unit length 1.2 kg/mρ = =

225 mm 0.225 m, 300 mm 0.3 m.r l= = = =

Moments and products of inertia: Formulas

m xI yI zI yzI

Upper ring 2 rπ ρ 2 2 21 12 4

m r r l + +

( )2 2m r r+ 2 21 12 4

m r l +

12

m rl −

Lower ring 2 rπ ρ 2 2 21 12 4

m r r l + +

( )2 2m r r+ 2 21 12 4

m r l +

12

m rl −

Rod AB lρ 2112

ml 0 2112

ml 0

Moments and products of inertia: Values

( )kgm ( )2kg mxI ⋅ ( )2kg myI ⋅ ( )2kg mzI ⋅ ( )2kg myzI ⋅

Upper ring 1.69646 0.166995 0.171767 0.081112 −0.057256

Lower ring 1.69646 0.166995 0.171767 0.081112 −0.057256

Rod AB 0.36 0.0027 0 0.0027 0

Totals 3.75292 0.336691 0.343533 0.164924 0.114511




(a) 5.4 3.75292

tt mm∆ −∆ = = =F iF v v

( )1.439 m/s= −v i!

(b) Moments about G: /E G Gt× ∆ =r F H

( ) ( ) ( ) ( ) ( )0.225 0.15 0.225 5.4 G G Gx y zH H H− + × − = + +i j k i i j k

( ) ( )1.215 0.81 x x y y yz z yz y z zI I I I Iω ω ω ω ω− − = + − + − +j k i j k

Substitute numerical values for moments and products of inertia and resolve into components.

: 0 0.336691 xω=i

: 1.215 0.343533 0.114511y zω ω− = +j

: 0.81 0.114511 0.164924y zω ω− = +k

Solving, 0, 2.4717 rad/s, 3.1952 rad/sx y zω ω ω= = − = −

( ) ( )2.47 rad/s 3.20 rad/s= − −j kω !




Total mass. 3m m m m+ + =

Moments and products of inertia.

2 2 21 1 1012 12 6xI md md md= + + =

2 2

2 2 21 1 22 12 12 2 3yd dI m md md m md = + + + =

2 2

2 2 21 1 212 2 12 2 3z

d dI md m md m md = + + + =

0yz zx xyI I I= = =

Angular momentum about mass center G.

G x x y y z zI I Iω ω ω= + +H i j k

Constraint of cable at G. 0yv =




Linear momentum components.

: 0 3 xx mv= 0xv =

: 3 0yy F t T t mv− ∆ + ∆ = = T t F t∆ = ∆

2: 0 3z mv= 2 0v =

Angular momentum components.

( ) ( )2 2A Gd dF t F t × − ∆ = + × − ∆ =

r j i k j H

( ) ( )1 12 2 x x y y z zdF t dF t I I Iω ω ω∆ = − ∆ = + +i k i j k

( ) 21 1:2 6 xdF t md ω∆ =i ( )3

xF tmd

ω∆

=

22: 03 ymd ω=j 0yω =

( ) 21 2:2 3 zdF t md ω− ∆ =k ( )3

4zF t

mdω

∆= −

(a) Velocity of the mass center. x y zv v v= + +v i j k

0=v !

(b) Angular velocity. x y zω ω ω= + +ω i j k

3 34

F t F tmd md

∆ ∆ = −

ω i k !




Total mass. 3m m m m+ + =

Moments and products of inertia.

2 2 21 1 10

12 12 6xI md md md= + + =

2 2

2 2 21 1 2

2 12 12 2 3yd d

I m md md m md = + + + =

2 2

2 2 21 1 2

12 2 12 2 3zd d

I md m md m md = + + + =

0yz zx xyI I I= = =

Angular momentum about mass center G.


Constraint of cable at G. 0yv =




Linear momentum components.

: 0 3 xx mv= 0xv =

: 3 yy T t mv∆ = 0T t∆ =

: 3 zz F t mv− ∆ = 3zF t

vm

∆= −

Angular momentum components.

( ) ( )2 2B Gd d

F t F t × − ∆ = − + × − ∆ =

r k i j k H

( ) ( )1 1

2 2 x x y y z zdF t dF t I I Iω ω ω− ∆ − ∆ = + +i j i j k

( ) 21 1:

2 6 xdF t md ω− ∆ =i ( )3

x

F t

mdω

∆= −

( ) 21 2:

2 3 ydF t md ω− ∆ =j ( )3

y

F t

mdω

∆= −

22: 0

3 zmd ω=k 0zω =

(a) Velocity of the mass center. x y zv v v= + +v i j k

3

F t

m

∆ = −

v k !

(b) Angular velocity. x y zω ω ω= + +ω i j k

3 3

4

F t F t

md md

∆ ∆ = − −

ω i j!




Moments and products of inertia:

Part m xI yI yI xyI

OA 15 m

( )( )15

2 2 21 1 112 4 4

m

a a a× + +

( )( )15

2 2 21 112 4

m

a a a× + + ( )( )2 21 1

5 4m a a+ ( )( )21 15 2m a−

AB 15 m ( )( )21 1

5 4m a ( )( )21 15 3m a

( )( )15

2 2 21 1 112 4 4

m

a a a× + + ( )( )21 1

5 4m a−

BC 15 m ( )( )21 1

5 12m a 0 ( )( )21 15 12m a 0

CD 15 m ( )( )21 1

5 4m a ( )( )21 15 3m a

( )( )15

2 2 21 1 112 4 4

m

a a a× + + ( )( )21 1

5 4m a−

DE 15 m

( )( )15

2 2 21 1 112 4 4

m

a a a× + +

( )( )15

2 2 21 112 4

m

a a a× + + ( )( )2 21 1

5 4m a a+ ( )( )21 15 2m a−

Σ m 20.35ma 20.66667ma 20.75ma 20.3ma−

2 2 21 1 1 1 0.25 2 5 2xzI m a m a ma = + =

2 2 21 1 1 1 0.15 4 5 4yzI m a m a ma = − + − = −


( ) 2 2 20.35 0.3 0.2G x x xy y xz z x y zxH I I I ma ma maω ω ω ω ω ω= − − = + −

( ) 2 2 20.3 0.66667 0.1G xy x y y yz z x y zyH I I I ma ma maω ω ω ω ω ω= − + − = + +

( ) 2 2 20.2 0.1 0.75G xz x yz z z z x y zzH I I I ma ma maω ω ω ω ω ω= − − + = − + +



Constraint of the supporting cable: 0yv =

Impulse-momentum principle: Before impact, 0, 0.G= =v H

(a) Linear momentum: ( )t T t mv∆ + ∆ =F j Resolve into components.

0 , 0, 0x zmv F t T t mv= − ∆ + ∆ = =j

0, , 0x zv T t F t v= ∆ = ∆ = 0=v !

(b) Angular momentum, moments about G: /A G Gt× ∆ =r F H

( ) ( ) ( ) ( ) ( )2 G G Gx y za a F t aF t H H H − × − ∆ = ∆ = + +

j i j k i j k

Using expressions for ( ) ( ), ,G Gx yH H and ( )G zH and resolving into components,

2 2 2: 0 0.35 0.3 0.2x y zma ma maω ω ω= + −i

2 2 2: 0 0.3 0.66667 0.1x y zma ma maω ω ω= + +j

2 2 2: 0.2 0.1 0.75x y zaF t ma ma maω ω ω∆ = − + +k

Solving, ( ) ( ) ( )z2.5 , 1.454 , 2.19x y

F t F t F tma ma ma

ω ω ω∆ ∆ ∆

= = − =

( )2.50 1.454 2.19F tma

∆ = − +

i j kω !





Part m xI yI yI xyI

OA 15 m

( )( )15

2 2 21 1 112 4 4

m

a a a× + +

( )( )15

2 2 21 112 4

m

a a a× + + ( )( )2 21 1

5 4m a a+ ( )( )21 15 2m a−

AB 15 m ( )( )21 1

5 4m a ( )( )21 15 3m a

( )( )15

2 2 21 1 112 4 4

m

a a a× + + ( )( )21 1

5 4m a−

BC 15 m ( )( )21 1

5 12m a 0 ( )( )21 15 12m a 0

CD 15 m ( )( )21 1

5 4m a ( )( )21 15 3m a

( )( )15

2 2 21 1 112 4 4

m

a a a× + + ( )( )21 1

5 4m a−

DE 15 m

( )( )15

2 2 21 1 112 4 4

m

a a a× + +

( )( )15

2 2 21 112 4

m

a a a× + + ( )( )2 21 1

5 4m a a+ ( )( )21 15 2m a−

Σ m 20.35ma 20.66667ma 20.75ma 20.3ma−

2 2 21 1 1 1 0.25 2 5 2xzI m a m a ma = + =

2 2 21 1 1 1 0.15 4 5 4yzI m a m a ma = − + − = −


( ) 2 2 20.35 0.3 0.2G x x xy y xz z x y zxH I I I ma ma maω ω ω ω ω ω= − − = + −

( ) 2 2 20.3 0.66667 0.2G xy x y y yz z x y zyH I I I ma ma maω ω ω ω ω ω= − + − = + +

( ) 2 2 20.2 0.1 0.75G xz x yz z z z x y zzH I I I ma ma maω ω ω ω ω ω= − − + = − + +



Constraint of the supporting cable: 0yv =

Impulse-momentum principle: Before impact, 0, 0.G= =v H

(a) Linear momentum: ( )t T t m∆ + ∆ =F j v Resolve in components.

0 , 0 0, x zm T t F t m= + ∆ = − ∆ =v v

0, 0, x zF tT Tm∆= ∆ = = −v v ( )/F t m= − ∆v k !

(b) Angular momentum, moments about G: /A G Gt× ∆ =r F H

( ) ( ) ( ) ( )12 2 G G Gx y za F t aF t H H H − × − ∆ = ∆ = + +

j k i i j k

Using expressions for ( ) ( ), ,G Gx yH H and ( )G zH and resolving into components,

2 2 21: 0.35 0.3 0.22 x y zaF t ma ma maω ω ω∆ = + −i

2 2 2: 0 0.3 0.66667 0.1x y zma ma maω ω ω= + +j

2 2 2: 0 0.2 0.1 0.75x y zma ma maω ω ω= − + +k

Solving, ( ) ( ) ( )3.75 , 1.875 , 1.25x y zF t F t F tm m ma

ω ω ω∆ ∆ ∆

= = − =

( )3.75 1.875 1.250F tma

∆ = − +

i j kω !





m xI yI zI yzI

AB 14

m

2 2 21 1 14 12 4

m a a a + +

214

ma

21 14 3

m a

21 14 2

m a −

BC 12

m

( )21 1 22 12

m a

( )21 1 22 12

m a

0 0 0xyI =

CD 14

m

2 2 21 1 14 12 4

m a a a + +

214

ma

21 14 3

m a

21 14 2

m a −

0xzI =

Σ m 256

ma

223

ma

216

ma

214

ma−


(a) ( )t m∆ =F v

tm∆= Fv F t

m∆ =

v i !



(b) Moments about G: /A G Gt× ∆ =r F H

( ) ( ) ( ) ( ) ( )G G Gx y za a F t H H H− + × ∆ = + +j k i i j k

( ) ( ) ( ) ( )x x y y yz z yz y z zaF t aF t I I I I Iω ω ω ω ω∆ + ∆ = + − + − +j k i j k

Substitute expressions for moments and products of inertia and resolve into components.

25: 06 xma ω=i

2 22 1: 3 4y zaF t ma maω ω∆ = +j

2 21 1: 4 6y zaF t ma maω ω∆ = +k

Solving, ( ) ( )z0, 1.714 , 8.57x y

F t F tma ma

ω ω ω∆ ∆

= = − =

( )1.714 8.57F tma

∆ = − +

j kω !





m xI yI zI yzI

AB 14

m

2 2 21 1 14 12 4

m a a a + +

214

ma

21 14 3

m a

21 14 2

m a −

BC 12

m

( )21 1 22 12

m a

( )21 1 22 12

m a

0 0 0xyI =

CD 14

m

2 2 21 1 14 12 4

m a a a + +

214

ma

21 14 3

m a

21 14 2

m a −

0xzI =

Σ m 256

ma

223

ma

216

ma

214

ma−

Impulse-Momentum Principle: Before impact, 0, 0G= =v H

(a) ( )t m∆ =F v

tm∆= Fv F t

m∆ =

v i !



(b) Moments about G: /B G Gt× ∆ =r F H

( ) ( ) ( ) ( )G G Gx y za F t H H H× ∆ = + +k i i j k

( ) ( ) ( )x x y y yz z yz y z zaF t I I I I Iω ω ω ω ω∆ = + − + − +j i j k

Substitute expressions for moments and products of inertia and resolve into components.

25: 06 xma ω=i

2 22 1: 3 4y yaF t ma maω ω∆ = +j

2 21 1: 04 6y zma maω ω= +k

Solving, ( ) ( )z0, 3.43 , 5.14x y

F t F tma ma

ω ω ω∆ ∆

= = = −

( )3.43 5.14F tma

∆ = −

j kω !




Point of impact: ( ) ( )2.7 m 0.225 mA = +r i k Initial linear momentum of the meteorite, ( )kg m/s :⋅

( )( )0 0.14 720 900 960 100.8 126 134.4m = − + = − +v i j k i j k

Its moment about the origin, ( )2kg m /s :⋅

0 2.7 0 0.225 28.35 340.2 340.2100.8 126 134.4

A m× = = − −−

i j kr v i j k

Final linear momentum of meteorite and its moment about the origin, ( ) ( )2kg m/s and kg m /s :⋅ ⋅

00.8 80.64 100.8 107.5m = − +v i j k

( )00.8 22.68 272.16 272.16A m× = − −r v i j k

Let AH be the angular momentum of the probe and m′ be its mass. Conservation of angular momentum about the origin for a system of particles consisting of the probe plus the meteorite:

( )0 00.8A A Am m× = + ×r v H r v

( ) ( ) ( )2 2 25.67 kg m /s 68.04 kg m /s 68.04 kg m /sA = ⋅ − ⋅ − ⋅H i j k

( ) ( )( )( )2 2

5.67, 0.012496 rad/s1500 0.55

A xx x A xx

x

HI H

m kω ω= = = =

′

( )( )

( )( )2 268.04 0.139612 rad/s

1500 0.57

A yy y A yy

y

HI H

m kω ω −= = = = −

′

( ) ( )( )( )2 2

68.04 0.18144 rad/s1500 0.50

A zz z A zz

z

HI H

m kω ω −= = = = −

′

( ) ( ) ( )0.0125 rad/s 0.1396 rad/s 0.1814 rad/s= − −i j kω !




Point of impact: ( ) ( )2.7 m 0.225 mA = +r i k Initial linear momentum of the meteorite, ( )kg m/s :⋅

( )( )0 0.14 0.14 0.14 0.14x y z x y zm v v v v v v= + + = + +v i j k i j k


( ) 00 2.7 0 0.2250.14 0.14 0.14

A A

x y z

mv v v

= × =i j k

H r v

( )0.0315 0.378 0.0315 0.378y z x yv v v v= − − − +i j k

Final linear momentum of the meteorite, ( )kg m/s :⋅

00.75 0.105 0.105 0.105x y zm v v v= + +v i j k


( ) ( )00.75 0.023625 0.2835 0.023625 0.2835A y z x ym v v v v× = − − − +r v i j k

Initial linear momentum of the space probe, ( )kg m/s :⋅ 0 0m′ ′ =v

Final linear momentum of the space probe, ( )kg m/s :⋅

( ) ( )1500 0.017x y z y zm v v v v v′ ′ ′ ′ ′ ′+ + = − + +i j k i j k

Final angular momentum of space probe, ( )2kg m /s :⋅

( )2 2 2A x x y y z zm k k kω ω ω′= + +H i j k

( ) ( ) ( ) ( ) ( )2 2 21500 0.55 0.05 0.57 0.12 0.50 zω = + − +

i j k

22.6875 58.482 375 zω= − +i j k



Conservation of linear momentum of the probe plus the meteorite, ( )kg m/s :⋅

0.14 0.14 0.14 0.105 0.105 0.105x y z x y zv v v v v v+ + = + +i j k i j k

( )( )1500 0.017 1500 1500y zv v′ ′+ − + +i j k

: 0.035 25.5xv = −i 728.57 m/sxv = −

: 0.035 1500y yv v′=j

: 0.035 1500z zv v′=k

Conservation of angular momentum about the origin, ( )2kg m /s :⋅

( )0.0315 0.378 0.0315 0.378y z x yv v v v− − − +i j k

( )0.023625 0.2835 0.023625 0.2835y z x yv v v v= − − − +i j k

22.6875 58.482 375 zω+ − +i j k

: 0.007875 22.6875yv− =i 2880.95 m/syv = −

: 0.0945 0.007875 58.482 0.083333 618.86 558.15 m/sz x z xv v v v− + = − = + =j

: 0.0945 375y zv ω− =k

(a) 6252 10z yvω −= − × 0.726 rad/szω = − !

(b) ( ) ( ) ( )0 729 m/s 2880 m/s 558 m/s= − − +v i j k !




The disk is constrainted to rotate about the fixed point A. Point B of the disk does not rebound after the impact.

( ) 0B zv =

Kinematics. /B B A= ×v rω

( ) ( )0

B B x y zx zv va a

ω ω ω+ =−

i j ki k

( ) ( ), , 0 0B z B z x y x yx yv a v a a aω ω ω ω ω ω= − = = − − + = (1)

Moments of inertia:

2 2 2 2 2 2 21 5 1 1 3, , 4 4 4 2 2x y zI ma ma ma I ma I ma ma ma= + = = = + =


Moments about :A ( ) ( ) ( ) ( )0 1A Aa a B t+ − × ∆ =H i j k H

( ) ( )0y x x y y z zI aB t a B t I I Iω ω ω ω+ − ∆ − ∆ = + +j i j i j k



Substitute values for , , and x y zI I I and resolve into components.

( ) 25: 04 xaB t ma ω− ∆ =i (2)

( )2 20

1 1: 4 4 yma aB t maω ω− ∆ =j (3)

23: 02 zma ω=k 0zω =

Solving (1), (2) and (3) simultaneously for , and ,x y B tω ω ∆

0 0 01 1 5, , 6 6 24x y B t maω ω ω ω ω= − = ∆ =

(a) Angular velocity: 0 01 16 6

ω ω= − +i jω !

(b) Velocity of mass center: /G G A= ×v rω

( )0 00

16 6 6G a aω ω ω = − + × − =

v i j j k

016G aω=v k !




The disk is constrainted to rotate about the fixed point A. Point B of the disk does not rebound after the impact.

( ) 0B zv =


( ) ( )0

B B x y zx zv va a

ω ω ω+ =−

i j ki k

( ) ( ), , 0 0B z B z x y x yx yv a v a a aω ω ω ω ω ω= − = = − − + = (1)

Moments of inertia:

2 2 2 2 2 2 21 5 1 1 3, , 4 4 4 2 2x y zI ma ma ma I ma I ma ma ma= + = = = + =


Moments about A: ( ) ( ) ( ) ( )0 1A Aa a B t+ − × ∆ =H i j k H




Substitute values for , , and x y zI I I and resolve into components.

( ) 25: 04 xaB t ma ω− ∆ =i (2)

( )2 20

1 1: 4 4 yma aB t maω ω− ∆ =j (3)

23: 02 zma ω=k 0zω =

Solving (1), (2) and (3) simultaneously for , and ,x y B tω ω ∆

0 0 01 1 5, , 6 6 24x y B t maω ω ω ω ω= − = ∆ =

(a) Impulse at B: 00.208t maω∆ =Β k !

Velocity of the mass center: /G G A= ×v rω

( )0 0 01 1 16 6 6G a aω ω ω = = − + × − =

v v i j j k

(b) Linear momentum: ( )0m t B t m+ ∆ + ∆ =v A k v

( )0t m m B t∆ = − − ∆A v v k

0 0 01 506 24

ma maω ω= − −k k

0124

t maω∆ = −A k !




Principal moments of inertia.

( )( )22 21200 1.120 1505.28 kg mx xI mk= = = ⋅

( )( )22 21200 1.200 1728 kg my yI mk= = = ⋅

( )( )22 21200 0.900 972 kg mz zI mk= = = ⋅

Initial angular velocity.

( ) ( ) ( )11 10.050 rad/s, 0, 0.075 rad/sω ω ω= = =x y z

Initial angular momentum.

( )1G x x y y z zI I Iω ω ω= + +H i j k

( ) ( )2 275.264 kg m / s 72.9 kg m /s= ⋅ + ⋅i k

Final angular momentum if ( )20.=ωωωω ( ) 0G z

=ΗΗΗΗ

Positions of jets A and B relative to mass center.

( ) ( )/ 1.2 m 1.8 m= +A Gr i k ( ) ( )/ 1.2 m 1.6 m= −B Gr i k

Impulses from jets A and B. ( ) ( )andA B

F t F t∆ ∆j j

Principle of impulse and momentum. Moments about G:

( ) ( ) ( ) ( )/ / G1G A G B GA B zF t F t+ × ∆ + × ∆ =H r j r j Η

( ) ( ) ( ) ( )75.264 72.9 1.2 1.8 1.2 1.6A B

F t F t+ + + × ∆ + − × ∆ = 0i k i k j i k j

( ) ( ) ( ) ( )75.264 +1.2 1.8 1.2 1.6 0+ 72.9 ∆ − ∆ + ∆ + ∆ =A A B B

F t F t F t F ti k k i k i

Components: i: ( ) ( )75.264 1.8 1.6 0A B

F t F t− ∆ + ∆ = (1)

k: ( ) ( )72.9 1.2 1.2 0A B

F t F t+ ∆ + ∆ = (2)




( ) 6.4518 N s,∆ = − ⋅A

F t ( ) 54.298 N s∆ = − ⋅B

F t

Since both impulse components are negative, the 50-N thrusts of jets A and B act in the negative y direction.

50 N= −F

(a) Operating times of jets.

( ) 6.4518

50

∆ −∆ = =−

AA

F tt

F 0.1290 sAt∆ = "

( ) 54.298

50

∆ −∆ = =−

BB

F tt

F 1.086 sBt∆ = "

Principle of impulse and momentum. Linear momentum:

( ) ( )A BF t F t m∆ + ∆ = ∆j j v

6.4518 54.298 1200− − = ∆j j v

( )310 m/s−∆ = − 50.6 ×v j

(b) Change in velocity of mass center. ( )50.6 mm/s∆ = −v j "




Data from Problem 18.33.

Mass of satellite: 1200 kg.m =

Initial angular velocity: ( ) ( )0 0.050 rad/s 0.075 rad/s= +i jωωωω

Principal radii of gyration: 1.120 m, 1.200 m, 0.900 mx y zk k k= = =

Jet thrust 50 N= parallel to the y axis.

Principal moments of inertia.

( )( )22 21200 1.120 1505.28 kg mx xI mk= = = ⋅

( )( )22 21200 1.200 1728 kg my yI mk= = = ⋅

( )( )22 21200 0.900 972 kg mz zI mk= = = ⋅

Initial angular velocity.

( ) ( ) ( )11 10.050 rad/s, 0, 0.075 rad/sω ω ω= = =x y z

Initial angular momentum.

( )1G x x y y z zI I Iω ω ω= + +H i j k

( ) ( )2 275.264 kg m / s 72.9 kg m /s= ⋅ + ⋅i k

Final angular momentum if ( )20.x =ωωωω

( ) ( ) ( ) ( )22 2ω ω ω= + +G x x y y z zI I IH i j k

( ) ( )220 1728 972y zω ω= + +j k

Positions of jet B relative to mass center. ( ) ( )/ 1.2 m 1.6 m= −B Gr i j

Impulse from jet B. ( )BF t∆ j



Principle of impulse and momentum. Moments about G:

( ) ( ) ( )/ G1 2G B G BF t+ × ∆ =H r j Η

( ) ( ) ( ) ( )2275.264 72.9 1.2 1.6 1728 972y zB

F t ω ω+ + − × ∆ = +i k i k j j k

( ) ( ) ( ) ( )2275.264 +1.2 1.6 1728 972ω ω+ 72.9 ∆ + ∆ = +y zB B

F t F ti k k i j k

Components: i: ( )75.264 1.6 0B

F t+ ∆ =

( ) 47.04 N s∆ = − ⋅B

F t

j: ( ) ( )2 2

0 1728 0y yω ω= =

k: ( ) ( )272.9 1.2 972 zB

F t ω+ ∆ =

( ) ( )( )2

72.9 1.2 47.040.01693 rad/s

972ω

+ −= =z

Since ( )BF t∆ is negative, the 50-N thrust of jet B acts in the negative y direction.

(a) Operating time of jet B.

( ) 47.04

50

∆ −∆ = =−

BB

F tt

F 0.941 sBt∆ = "

(b) Final angular velocity. ( ) ( ) ( )22 2x y zω ω ω= + +i j kωωωω

( )0.01693 rad/s= kωωωω "

Principle of impulse and momentum. Linear components:

( ) ( )BF t m∆ = ∆j v

47.04 1200− = ∆j v ( )0.0392 m/s∆ = −v j

(c) Change in velocity of mass center. ( )39.2 mm/s∆ = −v j "




Principle of impulse-momentum:

Linear momentum: ( )0 x y zm C t m m m− + ∆ = + +v j j v i v j v k : 0 xmv=i 0xv =

0: ym C t mv− + ∆ =j v ( )0 yC t m v v∆ = + (1)

: 0 zmv=k 0zv =

Since the impact is perfectly plastic, ( ) 0.C yv =

( ) ( )/C G c G y x y zr v a cω ω ω= + × = + + + × −v v j i j k i kω

( ) ( ) ( )C C y y x z yx zv v c v c a aω ω ω ω+ = − + + + −i k i j k

: 0C x zv c aω ω+ + =j y x zv c aω ω= − − (2)

Moments about C: ( ) ( ) ( ) ( )0 0 G G G yx y zcmv amv H H ai c mv+ = + + + − + ×i k H i j k k j

( )2 2 2 22

2 2 23 3 3x y y ymc m c a ma mcv mavω ω ω= + + + − −i j k i k

20

1: 3 x ycmv mc mcvω= −i (3)

( )2 21: 03 ym c a ω= +j 0yω =

20

1: 3 z yamv ma mavω= −k (4)



After using (2) to eliminate ,yv (3) and (4) become 043 x zc a vω ω+ =

043x zc a vω ω+ =

from which 0 03 3, 7 7x zc v a vω ω= =

03 1 17

vc a = +

i kω !





Linear momentum: ( )0 x y zm C t m m m− + ∆ = + +v j j v i v j v k

: 0 xmv=i 0xv =

0: ymv C t mv− + ∆ =j ( )0 yC t m v v∆ = +

: 0 zmv=k 0zv =

Since the impact is perfectly plastic, ( ) 0.C yv =

( ) ( )/C G c G y x y zr v a cω ω ω= + × = + + + × −v v j i j k i kωωωω


: 0C x zv c aω ω+ + =j y x zv c aω ω= − − (2)

Moments about C:

( ) ( ) ( ) ( )0 0 G G G yx y z

cmv amv H H ai c mv+ = + + + − + ×i k H i j k k j

( )2 2 2 22

2 2 2

3 3 3x y y ymc m c a ma mcv mavω ω ω= + + + − −i j k i k

20

1:

3 x ycmv mc mcvω= −i (3)

( )2 21: 0

3 ym c a ω= +j 0yω =

20

1:

3 z yamv ma mavω= −k (4)



After using (2) to eliminate ,yv (3) and (4) become 04

3 x zc a vω ω+ =

04

3x zc a vω ω+ =

from which 0 03 3

, 7 7x zc v a vω ω= =

(a) Velocity of mass center:

From (2), 0 0 03 3 6

7 7 7yv v v v= − − = −

06

7v= −v j"

(b) Impulse at C:

From (1), ( )0 0 06

7yC t m v v m v v ∆ = + = −

01

7t mv∆ =C j"




(a) ( ) ( )0 x x xy y xz z xy x y y yz zI I I I I Iω ω ω ω ω ω= − − + − + −H i j

( )xz x yz y z zI I Iω ω ω+ − − + k

x y zω ω ω= + +i j kω

2 20 x x xy y x xz z x xy x y y y yz z yI I I I I Iω ω ω ω ω ω ω ω ω ω⋅ = − − − + −H ω

2xz x z yz y z z zI I Iω ω ω ω ω− − +

( ) ( )2 2 212 2 2 22 x x y y z z xy x y yz y z xz x zI I I I I Iω ω ω ω ω ω ω ω ω = + + − − −

2T=

(b) 0 0 cosH ω θ⋅ =H ω

02 cosT H ω θ=

0

2cos TH

θω

=

But 00, 0, 0T H ω> > >

cos 0 90θ θ> < °




(a) ( )2 2 21 2 2 22 x x y y z z xy x y yz y z xz x zT I I I I I Iω ω ω ω ω ω ω ω ω= + + − − −

Let cosx x xω ω θ ωλ= =

cosy y yω ω θ ωλ= =

cosz z zω ω θ ωλ= =

( )2 2 2 21 2 2 22 x x y y z z xy x y yz y z xz x zT I I I I I Iλ λ λ λ λ λ λ λ λ ω= + + − − −

212 OLI ω= 21

2 OLT I ω= !

(b) Each particle of mass ( )im∆ describes a circle of radius .iρ

The speed of the particle is i iv ρ ω=

Its kinetic energy is ( ) ( ) ( )2 2 21 12 2i i i iiT m v m ρ ω∆ = ∆ = ∆

The kinetic energy of the entire body is

( ) ( ) 2 212 i iiT T m ρ ω= Σ ∆ = Σ ∆

but ( ) 2OL i iI m ρ= Σ ∆

Hence, 212 OLT I ω= !




From the solution of Prob. 18.1,

26.565β = °

20.0945 kg m / sxI ′ = ⋅ 20.270 kg m / szI ′ = ⋅

10cos 8.9443 rad/sxω β′ = =

0yω ′ =

10sin 4.4721 rad/szω β′ = − = −

Kinetic energy. 2 2 21 1 1

2 2 2x x y y z zT I I Iω ω ω′ ′ ′ ′ ′ ′= + +

( )( ) ( )( )2 21 10.0945 8.9443 0 0.270 4.4721

2 2= + + −T

6.48 J=T "





cos 45 , sin 45y zω ω ω ω′ ′= ° = °

0xω ′ =

2 21 1,

3 12x yI ma I ma′ ′= =

25

12z x yI I I ma′ ′ ′= + =

2 2 21 1 1

2 2 2x y zx y zT I I Iω ω ω′ ′ ′′ ′ ′= + +

( ) ( ) ( )2 22 2 21 1 1 1 1 50 cos 45 sin 45

2 3 2 12 2 12ma ma maω ω = + +

o o

2 2 2 21 50

48 48ma maω ω= + +

2 21

8T ma ω= "




23.6 = 0.1118 lb s /ft 24 in. 2 ft

32.2m l= ⋅ = =

( )( )1200 2125.664 rad/s

60

πω = =


sin 20 42.980 rad/sxω ω′ = − ° = −

cos 20 118.085 rad/syω ω′ = ° =

For the rod, 0xI ′ ≈

( )( )22 21 10.1118 2 0.037267 lb s ft

12 12yI ml′ = = = ⋅ ⋅

2 2 2 21 1 1 1

2 2 2 2x x y y z zT mv I I Iω ω ω′ ′ ′ ′ ′ ′= + + +

( )( )210 0 0.037267 118.085 0

2= + + +

260 ft lb= ⋅

260 ft lbT = ⋅ "




Use principal axes , , as shown′ ′ ′x y z .


sin cosβ β′ = − +j i j



cos sinω ω β ω β′ ′= = −i i jωωωω

cos sinω ω β ω ω β′ ′= = −x y

Pr incipal moments of inertia:

2 21 1,

2 4x y zI mr I I mr′ ′ ′= = =

Kinetic energy: 2 2 2 21 1 1 1

2 2 2 2x x y y z zT mv I I Iω ω ω′ ′ ′ ′ ′ ′= + + +

( ) ( )2 22 21 1 1 10 cos sin 0

2 2 2 4ω β ω β = + + − +

mr mr

( )2 2 2 212cos sin

8mr ω β β= +

20 ,For β = ° ( )2 2 212cos 20 sin 20

8ω ° + °T = mr

2 20.235T mr ω= "




Body diagonal: ( )22 22 6d a a a a= + + =

( ) 22

6a a a

d

ω ω ω ω= − + − = − + −6 6

ω i j k i j k

( )2 2 21 52

12 12xI m a a ma = + =

2 2 21 1

12 6yI m a a ma = + =

( )22 21 52

12 12zI m a a ma = + =

Axis of rotation passes through the mass center, hence 0.=v


2 2 2 2x y zx y zT mv I I Iω ω ω= + + +

2 2 22 2 2 2 21 5 1 1 2 1 5 1

02 12 2 6 2 12 86

T ma ma ma maω ω ω ω = + + + = 6 6

2 20.1250 T ma ω= "




Body diagonal ( )22 22 6d a a a a= + + =

( ) 22

6a a a

d

ω ω ω ω= − + − = − + −6 6

ω i j k i j k

( )2 2 2 2Total area 2 2 2 10a a a a= + + =

For each square plate: 1

10m m′ =

2 2 2 21 13 13

12 12 120xI m a m a m a ma′ ′ ′= + = =

2 21 1

6 60yI m a ma′= =

213

120z xI I ma= =

For each plate parallel to the yz-plane: 1

5m m′ =

( )22 2 21 5 12

12 12 12xI m a a m a ma ′ ′= + = =

2

2 2 21 1 1

12 2 3 15ya

I m a m m a ma ′ ′ ′= + = =

( )2

2 2 21 7 72

12 2 12 60za

I m a m m a ma ′ ′= + = =

For each plate parallel to the xy plane: 1

5m m′ =

( )2

2 2 21 7 72

12 2 12 60xa

I m a m m a ma ′ ′ ′= + = =



2

2 2 21 1 1

12 2 3 15ya

I m a m m a ma ′ ′ ′= + = =

( )22 2 21 5 12

12 12 12zI m a a m a ma ′ ′= + = =

Total moments of inertia:

2 213 1 7 372

120 12 60 60xI ma ma = + + =

2 21 1 1 32

60 15 15 10yI ma ma = + + =

2 213 7 1 372

120 60 12 60zI ma ma = + + =

Axis of rotation passes through the mass center, hence 0.=v


2 2 2 2x y zx y zT mv I I Iω ω ω= + + +

2 2 22 2 21 37 1 3 2 1 37

02 60 2 10 2 60 6

T ma ma maω ω ω = + + + 6 6

2 273

360ma ω=

2 20.203 T ma ω= "





For axes , , x y z′ ′ ′ parallel to , , x y z with origin at A,

( )( )22 21 18 0.100 0.02 kg m

4 4xI mr′ = = = ⋅



( ) ( )/ 0.32 m 0.2 mA O = −r i j

( ) ( )2 / 4 0.32 0.2 1.28 m/sA A Oω= = × = × − = −v v j r j i j k

1.28 m/sv =


2 2 2 2x y zx y zT mv I I Iω ω ω′ ′ ′= + + +

( )( ) ( )( ) ( )( )2 2 21 1 18 1.28 0 0.02 4 0.04 12

2 2 2T = + + +

6.5536 0.16 2.88= + +

2 29.5936 kg m /s 9.5935 N m 9.5935 J= ⋅ = ⋅ =

9.59 JT = "




( ) ( )2 1 8 rad/s 15 rad/sω ω= + = +ω i j i j

For axes , , x y z′ ′ ′ parallel to , , x y z with origin at A,

( )( )22 21 16 0.160 0.0384 kg m

4 4xI mr′ = = = ⋅



( ) ( )/ 0.18 m 0.18 mA C = −r j k

( )2 / 8 0.18 0.18A A Cω= = × = × −v v i r i j k

( ) ( )1.44 m/s 1.44 m/s= +j k

( ) ( )2 21.44 1.44 2.0365 m/sv = + =


2 2 2 2x y zx y zT mv I I Iω ω ω′ ′ ′= + + +

( )( ) ( )( ) ( )( )2 2 21 1 16 2.0365 0.0384 8 0.0768 16 0

2 2 2T = + + +

12.4416 1.2288 9.8304= + +

2 223.5 kg m /s 23.5 N m 23.5 J= ⋅ = ⋅ =

23.5 JT = !




5 kg, 200 mm 0.2 mm a= = =

( )( )2 36012 rad/s.

60π

ω π= =

0, 0, 12 rad/sx y zω ω ω π= = =

Use parallel axes , ,x y z′ ′ ′ with origin at the mass center as shown.

2 2 21 1 12 2 2x zx y y z xy x y yz y z xz x zT I I I I I Iω ω ω ω ω ω ω ω ω= + + − − −

212 xI ω=

Part zI

21 1 1

12 4m a ′+ +

21

3m a′

21

3m a′

21 1 1

12 4m a ′+ +

Σ 2103

m a′

Segments 1, 2, 3, and 4, each of mass 2.5 kg,m′ = contribute to zI .

( )( )210 2.5 0.23zI =

20.33333 kg m= ⋅

( )( )21 0.33333 122

T π=

2 2237 kg m /s 237 N m= ⋅ = ⋅

237 JT = !




210 0.31056 lb s /ft32.2

Wmg

= = = ⋅

( )12 rad/s , 0y zω ω ω= = =i

2 2 21 1 12 2 2x y zx y z xy x y yz y z xz x zT I I I I I Iω ω ω ω ω ω ω ω ω= + + − − −

12 xI ω 2=

The shaft is comprised of eight sections, each of length

10 ft12

a = and of mass 20.03882 lb s /ft.8mm′ = = ⋅

( ) ( )( ) ( )2

2 2 2 21 10 10 104 2 0.3882 0.089861 lb s ft3 3 3 12xI m a m a m a ′ ′ ′= + = = = ⋅ ⋅

( )( )21 0.089861 12 6.47 ft lb2

T = = ⋅

6.47 ft lbT = ⋅ !




Angular velocity: ( )= 6 rad/sω= j jωωωω

0, 6 rad/s, 0x y zω ω ω= = =

Velocity of mass center G. = 0.v

From the solution to Prob. 18.19, 21.50 kg myI = ⋅

Kinetic energy. 2 2 2 21 1 1 1

2 2 2 2x x y y z zT mv I I Iω ω ω= + + +

xy x y yz y z xz x zI I Iω ω ω ω ω ω− − −

( )( )210 1.50 6 0 0 0 0 0

2T = + + + + + +

27.0 JT = !




From the solution given in the textbook for Sample Prob. 18.1,

2 21 1, ,

12 12x yI mb I ma= =

Initial state: = 0, 0 0Τ1= , =v ωωωω

Final state: =F t

m

∆ −

v k

( )6F ta b

mab

∆ = −

i jωωωω

or 0, 0,x y zF t

v v vm

∆= = = −

6

, 6 , 0x y zF t

F tmb

ω ω ω∆= = − ∆ =

Additional moments and products of inertia.

( )2 21, 0

12z xy yz zxI m a b I I I= + = = =

Final kinetic energy.

2 2 2 22

1 1 1 1


2 2 2

2 21 1 1 6 1 1 60

2 2 12 2 12

F t F t F tm mb ma

m mb ma

∆ ∆ ∆ = + + +

( ) ( ) ( ) ( )2 2 2 2

1 3 3 7

2 2 2 2

F t F t F t F t

m m m m

∆ ∆ ∆ ∆= + + =

Kinetic energy imparted. 2 1T T T∆ = −

( )27

2

F tT

m

∆∆ = !




Point of impact: ( ) ( )2.7 m 0.225 mA = +r i k

Initial linear momentum of the meteorite, ( )kg m/s :⋅

( )( )0 0.14 720 900 960 100.8 126 134.4m = − + = − +v i j k i j k


0 2.7 0 0.225 28.35 340.2 340.2

100.8 126 134.4A m× = = − −

−

i j k

r v i j k

Final linear momentum of meteorite and its moment about the origin, ( ) ( )2kg m/s and kg m /s .⋅ ⋅

00.8 80.64 100.8 107.5m = − +v i j k

( )00.8 22.68 272.16 272.16A m× = − −r v i j k

Let AH be the angular momentum of the probe and m′ be its mass. Conservation of angular momentum about the origin for a system of particles consisting of the probe plus the meteorite:

( )0 00.8A A Am m× = + ×r v H r v

( ) ( ) ( )2 2 25.67 kg m /s 68.04 kg m /s 68.04 kg m /s ,A = ⋅ − ⋅ − ⋅H i j k

( ) ( )( )( )2 2

5.67, 0.012496 rad/s

1500 0.55

A xx x A xx

x

HI H

m kω ω= = = =

′

( )( )

( )( )2 2

68.04, 0.139612 rad/s

1500 0.57

A yy y A yy

y

HI H

m kω ω −= = = = −

′

( ) ( )( )( )2 2

68.04, 0.18144 rad/s

1500 0.50

A zz z A zz

z

HI H

m kω ω −= = = = −

′

Kinetic energy of motion of the probe about its mass center:

( ) ( )2 2 2 2 2 2 2 2 21

2 2x x y y z z x x y y z zm

T I I I k k kω ω ω ω ω ω= + + = + +

( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2 215000.55 0.012496 0.57 0.139612 0.50 0.18144

2 = + +

2 210.96 kg m /s 10.96 N m= ⋅ = ⋅ 10.96 JT = !




Point of impact: ( ) ( )2.7 m 0.225 mA = +r i k

Initial linear momentum of the meteorite, ( )kg m/s :⋅

( )( )0 0.14 0.14 0.14 0.14x y z x y zm v v v v v v= + + = + +v i j k i j k


( ) 002.7 0 0.225

0.14 0.14 0.14A A

x y z

m

v v v

= = =i j k

H r v

( )0.0315 0.378 0.0315 0.378y z x yv v v v= − − − +i j k

Final linear momentum of the meteorite, ( )kg m/s :⋅

00.75 0.105 0.105 0.105x y zm v v v= + +v i j k


( ) ( )00.75 0.023625 0.2835 0.023625 0.2835A y z x ym v v v v× = − − − +r v i j k

Initial linear momentum of the space probe, ( )kg m/s :⋅ 0 0m′ ′ =v

Final linear momentum of the space probe, ( )kg m/s :⋅

( ) ( )1500 0.017x y z y zm v v v v v′ ′ ′ ′ ′ ′+ + = − + +i j k i j k

Final angular momentum of space probe, ( )2kg m /s :⋅

( )2 2 2A x x y y z zH m k k kω ω ω′= + +i j k

( ) ( ) ( ) ( ) ( )2 2 21500 0.55 0.05 0.57 0.12 0.50 zω = + − +

i j k

22.6875 58.482 375 zω= − +i j k

Conservation of linear momentum of the probe plus the meteorite, ( )2kg m /s :⋅

0.14 0.14 0.14 0.105 0.105 0.105x y z x y zv v v v v v+ + = + +i j k i j k

( ) ( )1500 0.017 1500 1500y zv v′ ′+ − + +i j k



: 0.035 25.5xv = −i 728.57 m/sxv = −

: 0.035 1500y yv v′=j

: 0.035 1500z zv v′=k

Conservation of angular momentum about the origin.

( ) ( )0.0315 0.378 0.0315 0.378 0.023625 0.2835 0.023625y z x y y z xv v v v v v v− − − + = − − −i j k i j

0.2835 22.6875 58.482 375y zv ω+ + − +k i j k

: 0.007875 22.6875yv− =i 2880.95 m/syv = −

: 0.0945 375y zv ω− =k

6252 10z yvω −= − × 0.726 rad/szω = −

Kinetic energy of motion of probe relative to its mass center:

( ) ( )2 2 2 2 2 2 2 2 21 1

2 2x x y y z z x x y y z zT I I I m k k kω ω ω ω ω ω′ = + + = + +

( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2 211500 0.55 0.05 0.57 0.12 0.50 0.726

2 = + − + −

102.9 JT ′ = !




The disk is constrainted to rotate about the fixed point A. Point B of the disk does not rebound after the impact. ( ) 0B zv =


( ) ( )0

B B x y zx zv va a

ω ω ω+ =−

i j ki k

( ) ,B zxv aω= − ( ) ,B zyv aω= 0 x ya aω ω= − − 0x yω ω+ = (1)

Moments of inertia: 2 2 21 5 ,4 4xI ma ma ma= + = 21 ,

4yI ma= 2 2 21 32 2zI ma ma ma= + =

Impulse-Momentum Principle:

Moments about A: ( ) ( ) ( ) ( )0 1A Aa a B t+ − × ∆ =H i j k H


Substitute values for , ,x yI I and zI and resolve into components.

( ) 25: 04 xaB t ma ω− ∆ =i (2)

( )2 20

1 1: 4 4 yma aB t maω ω− ∆ =j (3)

21: 02 zma ω=k 0zω =



Solving (1), (2) and (3) simultaneously for , x yω ω and B t∆

01 ,6xω ω= − 0

1 ,6yω ω= 0

524

B t maω∆ =

Angular velocity: 0 01 16 6

ω ω ω= − +i j

Kinetic energy before impact: 2 2 20 0 0

1 12 8yT I maω ω= =

Kinetic energy after impact: 2 2 21

1 1 12 2 2x x y z zT I I Iω ω ω= + +

2 22 2 2 2

1 0 0 01 5 1 1 1 1 102 4 6 2 4 6 48

T ma ma maω ω ω = − + + =

Energy lost: 2 20 1 0

548

T T ma ω− = 2 20

548

ma ω "





Linear momentum: ( )0 x y zmv C t mv mv mv− + ∆ = + +j j i j k

: 0 xmv=i 0xv =

0: ymv C t mv− + ∆ =j ( )0 yC t m v v∆ = + (1)

: 0 zmv=k 0zv =

Since impact is perfectly plastic, ( ) 0.C yv =

( ) ( )/C G C G y x y zr v a cω ω ω= + × = + + + × −v v j i j k i kωωωω


: 0C x zv c aω ω+ + =j y x zv c aω ω= − −

Moments about C:

( ) ( ) ( ) ( )0 0 G G G yx y zcmv amv H H a c mv+ = + + + − + ×i k H i j k i k j

( )2 2 2 22 2 2

3 3 3x y z y ymc m c a ma mcv mavω ω ω= + + + − −i j k i k

20

1:

3 x ycmv mc mcvω= −i (3)

( )2 21: 0

3 ym c a ω= +j 0yω =

20

1:

3 z yamv ma mavω= −k (4)



After using (2) to eliminate ,yv (3) and (4) become

04

3 x zc a vω ω+ =

04

3x zc a vω ω+ =

from which 03

,7xc vω = 0

3

7za vω =

From (2), 0 0 03 3 6

7 7 7yv v v v= − − = −

Kinetic energy before impact: 20 0

1

2T mv=

Kinetic energy after impact: 2 2 2 21

1 1 1 1


22 2 2 2

1 01 6 1 1 1 1

02 7 2 3 2 3x zT m v mc maω ω = + + +

2 2 2

20 0 0 0

1 6 1 3 1 3 3

2 7 3 7 3 7 7m v v v mv = + + =

Energy lost: 20 1 0

1

14T T mv− = 2

0 1 01

14T T mv− = !




From the solution to Prob. 18.1

( ) ( )2 21.296 kg m /s 0.702 kg m /sG = ⋅ − ⋅H i k

( )10 rad/s= iωωωω

Let the frame of reference , ,x y z′ ′ ′ be rotating with angular velocity.

( )10 rad/s= iΩΩΩΩ

Rate of change ofG GH H& .

( )G G GGx y z′ ′ ′= + ×H H H& & ΩΩΩΩ

( )0 10 1.296 0.702 7.02= + × − =i i k j

( )7.02 N mG = ⋅H j& !





cos 45 , sin 45y zω ω ω ω′ ′= ° = °

0xω ′ =

2 21 1,

3 12x yI ma I ma′ ′= =

25

12z x yI I I ma′ ′ ′= + =


( ) ( )2 21 50 cos 45 sin 45

12 12ma maω ω ′ ′= + ° + °

j k

( ) ( )21cos 45 cos 45 sin 45

12A ma ω = ° ° − °

H j k

( ) ( )25sin 45 sin 45 cos 45

12ma ω + ° ° + °

j k

( )2

3 212

ω= +H j kAma

Let the frame of reference Axyz be rotating with angular velocity

ω= jΩΩΩΩ

Then, ( )A AAxyz= + ×H H Η& & ΩΩΩΩ

( )2 2 2

0 3 212 6

ma maω ωω= + × + =j j k i

2 216A ma ω=H i& !




23.60.1118 lb s /ft 24 in. 2 ft

32.2m l= = ⋅ = =

( ) ( )1200 2125.664 rad/s

60

πω = =


sin 20 42.980 rad/sxω ω′ = − ° = −

cos 20 118.085 rad/syω ω′ = ° =

For the rod, 0xI ′ ≈

( ) ( )22 21 10.1118 2 0.037267 lb s ft

12 12yI ml′ = = = ⋅ ⋅

G x x y y z zI I Iω ω ω′ ′ ′ ′ ′ ′= + +′ ′ ′H i j k

( ) ( )0 0.037267 118.085 0= + +′j

4.40068= ′j

( )4.40068 sin20 cos 20= ° + °i j

( ) ( )1.5051 lb s ft 4.1353 lb s ft= ⋅ ⋅ + ⋅ ⋅i j

Let the frame of reference Gxyz be rotating with angular velocity

( )125.664 rad/sω= j jΩ =Ω =Ω =Ω =

Then, ( )G G GGxyz= + ×H H Η& & ΩΩΩΩ

( ) ( ) ( )0 125.664 1.5051 4.1353 189.14 lb ftG = + × + = − ⋅H j i j k&

( )189.1 lb ftG = − ⋅H k& !




Use principal axes , , x y z′ ′ ′ as shown.

cos sinβ β= +′i i j

sin cosβ β= − +′j i j

cos sinβ β= −′ ′i i j

sin cosβ β= +′ ′j i j

( ) ( )cos sinω ω β ω β= = −′ ′i i jωωωω

( ) ( )cos sin 0ω ω β ω β= = − =′ ′i i j& & &αααα

Principal moments of inertia:

2 21 1,

2 4x y zI mr I I mr′ ′ ′= = =

Angular momentum:

G x x y y z zI I Iω ω ω′ ′ ′ ′ ′ ′= + +′ ′ ′H i j k

2 21 1cos sin

2 4mr mrω β ω β= −′ ′i j

Let the frame of reference Gx y z′ ′ ′ be rotating with angular velocity

ω=ΩΩΩΩ

( ) 0G G G GGx y z′ ′ ′= + × = + ×H H Η H& & Ω ΩΩ ΩΩ ΩΩ Ω

( ) 2 21 1cos sin cos sin

2 4mr mrω β ω β ω β β = − × −′ ′ ′ ′

i j i j

2 21sin cos

4mr ω β β= k

2 21sin cos

4G mr ω β β=H k& !




( ) ( )2 1 4 rad/s 12 rad/sω ω= + = +j k j kωωωω


( ) ( )22 21 18 0.100 0.02 kg m

4 4xI mr′ = = = ⋅



( )( ) ( )( ) ( ) ( )2 20 0.02 4 0.04 12 0.08 kg m /s 0.48 kg m /s= + + = ⋅ + ⋅j k j k


( )2 4 rad/sω= =j jΩΩΩΩ

Then, ( ) 20A A A AAxyzω= + × = + ×H H Η j Η& & ΩΩΩΩ

( ) ( )2 24 0.08 0.48 1.92 kg m /sA = × + = ⋅H j j k i&

( )1.920 N mA = ⋅H i& !




( ) ( )2 1 8 rad/s 16 rad/sω ω= + = +i j i jωωωω


( ) ( )22 21 16 0.160 0.0384 kg m

4 4xI mr′ = = = ⋅


A x x y y z zI I Iω ω ω′ ′ ′ ′ ′ ′′= + +H i j k

( ) ( ) ( ) ( )0.0384 8 0.0768 16 0= + +i j

( ) ( )2 20.3072 kg m /s 1.2288 kg m /s= ⋅ + ⋅i j


( )2 8 rad/sω= =i iΩΩΩΩ

Then, ( ) 0A A A AAxyzω= + × = + ×H H Η i Η& &

2222ΩΩΩΩ

( ) ( )2 28 0.3072 1.2288 9.8304 kg m /sA = × + = ⋅H i i j k&

( )9.83 N mA = ⋅H k& !





cos 45 , sin 45y zω ω ω ω′ ′= ° = °

0xω ′ =

2 21 1,

3 12x yI ma I ma′ ′= =

25

12z x yI I I ma′ ′ ′= + =


( ) ( )2 21 50 cos 45 sin 45

12 12ma maω ω ′ ′= + ° + °

j k

( )( )21cos 45 cos 45 sin 45

12A ma ω = ° ° − °

H j k

( )( )25sin 45 sin 45 cos 45

12ma ω + ° ° + °

j k

( )2

3 212

ω= +H j kAma

Let the frame Axyz be rotating with angular velocity .ω=Ω j

Then ( )A A AAxyzH= + ×H Ω H& &

where ( ) ( ) ( )2 2

3 2 3 212 12A Axyz

ma maH

ω α= + = +j k j k&&

and ( )2 2 2

3 212 6A

ma maω ωω

× = × + =

Ω H j j k i

So ( )2

22 3 212A

ma ω α α= + +H i j k& "




Use principal axes , ,x y z′ ′ ′ as shown.


sin cosβ β′ = − +j i j



( ) ( )cos sinω ω β ω β′ ′= = −ω i i j

( ) ( )cos sinα α β α β′ ′= = −α i i j

Principal moments of inertia:

2 21 1,2 4x y zI mr I I mr′ ′ ′= = =

Angular momentum:

G x x y y z zI I Iω ω ω′ ′ ′ ′ ′ ′′= + +H i j k

2 21 1cos sin2 4

mr mrω β ω β= −i j

Let the frame of reference Gx y z′ ′ ′ be rotating with angular velocity

ω=Ω

( )G G GGx y z′ ′ ′= + ×H H H! ! Ω

x x y y z zI I Iα α α′ ′ ′ ′ ′ ′′ ′= + +i j k

( ) 2 21 1cos sin cos sin2 4

mr mrω β ω β ω β β ′ ′ ′ ′+ − × −

i j i j



( )( )21 cos cos sin2

mr α β β β = +

i j

( )( )21 sin sin cos 04

α β β β + − − + +

mr i j

2 21 sin cos4

mr ω β β+ k

( )2 2 21 2cos sin4G mr α β β= +H i! 2 2 21 1sin cos sin cos

4 4mr mrα β β ω β β+ +j k !




Mass of sheet metal: 1.5 kg.=m

Sheet metal dimension: 0.120 m=b

Area of sheet metal: 2 2 2 2 2 21 1 3 0.0432 m2 2

= + + + = =A b b b b b

Let

21.5 34.722 kg/m mass per unit area.0.0432

ρ = = = =mA

Moments and products of inertia: mass areaρ=I I

xy plane (rectangles)

4 4 41 1 23 3 3

= + =xI b b b

( )( )442 2 34.722 0.1203 3

ρ= =xI b

3 24.8 10 kg m−= × ⋅

( ) ( )2 2 43 1 5 1 12 2 2 2 2

= + − = −

xyI b b b b b b b

( )( )441 1 34.722 0.1202 2xyI bρ= − = −

3 23.6 10 kg m−= − × ⋅

xz plane (triangles)

4 4 41 1 112 12 6

= + =xI b b b

( )( )441 1 34.722 0.1206 6

ρ= =xI b

3 21.2 10 kg m−= × ⋅



For calculation of ,xzI use pairs of elements 1dA and 2 :dA 2 1.=dA dA

( ) ( ) ( ) 21 2 1 04 2 2

2 2 = + − − = − − = − −

∫ ∫ ∫ ∫b

xzz zI x dA b x dA b x zdA b x z dx

but .z x=

Hence, ( )2 3 4 4 40

2 1 523 4 12

= − − = − − = −

∫a

xzI bx x dx b b b

( )( )44 3 25 5 34.722 0.120 3.0 10 kg m12 12

ρ −= − = − = − × ⋅xzI b

Total for :xI 3 3 3 24.8 10 1.2 10 6.0 10 kg m− − −= × + × = × ⋅xI

The mass center lies on the rotation axis, therefore 0=a

0 mΣ = + = = = −F A B a A B

, A x xy xzI I Iω ω ω ω α= − − = =H i j k i iω α


ω= = iΩ ω

( )A A A AAxyzΣ = = + ×M H H H! ! Ω

( ) ( )0 4 α α α ω ω ω ω+ × + = − − + × − −y z x xy xz x xy xzM b B B I I I I I Ii i j k i j k i i j k

( ) ( )2 20 4 4 α α ω α ω− + = − − − +z y x xy xz xz xyM bB bB I I I I Ii j k i j k

Resolve into components and solve for yB and .zB

0: xM I α=i

( )2

: 4

xy xzz

I IB

b

α ω−=j

( )2

: 4

xz xyy

I IB

b

α ω+= −k

Data: ( )0

2 2400, 25.133 rad/s, 0.120 m 0

60π

α ω= = = = =b M

( )( )( )( )

230 3.0 10 25.1333.95 N

4 0.120

−− − ×= =zB



( )( )

( )( )

230 3.6 10 25.1334.74 N

4 0.120

−+ − ×= − =yB

4.74 N= − = −y yA B

3.95 N= − = −z zA B

( ) ( )4.74 N 3.95 N= − −A j k !

( ) ( )4.74 N 3.95 N= +B j k !




Mass of sheet metal: 1.5 kg.=m

Sheet metal dimension: 0.120 m=b

Area of sheet metal: 2 2 2 2 2 21 13 0.0432 m

2 2= + + + = =A b b b b b

Let

21.534.722 kg/m mass per unit area.

0.0432ρ = = = =m

A

Moments and products of inertia: mass areaρ=I I


4 4 41 1 2

3 3 3= + =xI b b b

( )( )442 234.722 0.120

3 3ρ= =xI b

3 24.8 10 kg m−= × ⋅

( ) ( )2 2 43 1 5 1 1

2 2 2 2 2 = + − = −

xyI b b b b b b b

( )( )441 134.722 0.120

2 2xyI bρ= − = −

3 23.6 10 kg m−= − × ⋅


4 4 41 1 1

12 12 6= + =xI b b b

( )( )441 134.722 0.120

6 6ρ= =xI b

3 21.2 10 kg m−= × ⋅



For calculation of ,xzI use pairs of elements 1dA and 2 :dA 2 1 =dA dA

( ) ( ) ( ) 21 2 1 0

4 2 22 2

= + − − = − − = − −

∫ ∫ ∫ ∫b

xzz z

I x dA b x dA b x zdA b x z dx

But ,=z x

Hence, ( )2 3 4 4 40

2 1 52

3 4 12 = − − = − − = −

∫a

xzI bx x dx b b b

( )( )44 3 25 534.722 0.120 3.0 10 kg m

12 12ρ −= − = − = − × ⋅xzI b

Total for :xI 3 3 3 24.8 10 1.2 10 6.0 10 kg m− − −= × + × = × ⋅xI

The mass center lies on the rotation axis, therefore 0=a

0 Σ = + = = = −F mA B a A B

, ω ω ω ω α= − − = =A x xy xzI I IH i j k i iω αω αω αω α


ω= = iΩ ωΩ ωΩ ωΩ ω

( )A A A AAxyzΣ = = + ×M H H H& & ΩΩΩΩ

( ) ( )0 4 α α α ω ω ω ω+ × + = − − + × − −y z x xy xz x xy xzM b B B I I I I I Ii i j k i j k i i j k

( ) ( )2 20 4 4 α α ω α ω− + = − − − +z y x xy xz xz xyM bB bB I I I I Ii j k j k

Resolve into components and solve for yB and ,zB

0: xM I α=i

( )2

: 4

xy xzz

I IB

b

α ω−=j

( )2

: 4

xz xyy

I IB

b

α ω+= −k

Data: ( )2 360

0, 37.699 rad/s60

πα ω= = =

0 0=M



( ) ( )

( )( )

230 3.0 10 37.6998.88 N

4 0.120

− − − × = =zB

( ) ( )( )( )

230 3.6 10 37.69910.66 N

4 120

− + − × = − =yB

10.66 N= − = −y yA B

8.88 N= − = −z zA B

( ) ( )10.66 N 8.88 N= − −A j k"

( ) ( )10.66 N 8.88 N= +B j k"




Use principal axes x′, y′, z′ as shown.

( ) ( )cos sinω θ ω θ′ ′= −i jωωωω

0=αααα Moments of inertia:

2 21 1,

2 4′ ′ ′= = =x y zI mr I I mr

Angular momentum:

ω ω ω′ ′ ′ ′ ′ ′′= + +&G x x y y z zI I IH i j k

2 21 1cos sin

2 4ω θ ω θ= −mr mri j

Let the frame of reference Gx′y′z′ be rotating with angular velocity =Ω ωΩ ωΩ ωΩ ω

( ) ′ ′ ′= + ×& &

G G GGx y zH H HΩΩΩΩ

( ) 2 21 1cos sin cos sin

2 4x x y y z zI I I mr mrα α α ω θ ω θ ω θ θ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′= + + + − × −

i j k i j i j

2 21sin cos

4ω θ θ= mr k

Since the mass center G lies on the rotation axis, 0, 0= =v a

: 0 m mΣ = + = = = −F a A B a B A

( ): 2G G y zM a a a A AΣ = − × + × = − × +H i A i B i j k&

2 212 2 sin cos

4ω θ θ− =z yaA aA mrj k k

( )

( )

22

2 2

8 1010 sin 20 cos 20

32.2 121 sin cos0.4159 lb

8 208

12

ymr

Aa

ω θ θ

° ° = − = − = −

0=zA ( )0.416 lb= −A j""""

0.4159 lb, 0yy zB A B= − = = ( )0.416 lb=B j""""




Length of rod: 8 8

2.76106 ft12 12

L π= + =

Angular velocity: ( )20 rad/sω= =i iωωωω

Angular acceleration: α= iαααα

Angular momentum: ω ω= −G x xzI IH i k

Let the reference frame Gxyz be rotating with angular velocity


Rate of change of angular momentum: ( )= + ×& &G G GGxyz

H H HΩΩΩΩ

( )G x x xz x x xzI I I Iα α ω ω ω= − + × −H i k i i k&

2x xz xzI I Iα ω α= + −i j k

Since the mass center G lies on the rotation axis, 0, 0 8 in. + 4 in. = 1 ftb= = =v a

: m mΣ = + = = −F a A B a B A

( ) ( )0 0 2Σ = + − × + × = − ×G M b b M bM i i A i B i i A

( ) ( )0 02 2 2= − × + = + −y z z yM b A A M bA bAi i j k i j k

Set Σ = &G GM H and resolve into components.

0: xM I α=i

22: 2

2xz

z xz z zI

bA I A Bb

ωω= = = −j

: 2 2xz

y xz y yI

bA I A Bb

αα− = − = = −k

Calculation of xI and xzI :

Let mass per unit length .ρ = = m

L

For portions AC and DB,

0, 0= =x xzI I

For portion CG use polar coordinate .θ



( )1 cos , sin θ θ ρ θ= − − = − =x r z r dm rd

2 2 2 30

sin2

π πθρ θ ρ= = =∫ ∫xI z dm r rd r

( )2 30

1 cos sin 2π θ θρ θ ρ= = − =∫ ∫xzI xzdm r rd r

Likewise, for portion GD, 3 3, 22

π ρ ρ= =x xzI r I r

Total: 3 3

3 3 4, 4

ππρ ρ= = = =x xzmr mr

I r I rL L

Data: 21.6 40.049689 lb s /ft 4 in. ft

32.2 12

Wm r =

g= = = ⋅ =

( )

3

2

40.049689

12 0.0020940 lb s ft2.76106xI

π = = ⋅ ⋅

( )( )

3

2

44 0.049689

12 0.0026661 lb s ft2.76106xzI

= = ⋅ ⋅

Reactions: Since 0M is zero, 0α = .

( )( )

( )( )2

0.0026661 200.53322

2 1z zA B= = = −

0y yA B= = − ( )0.533 lb=A k "

( )0.533 lb= −B k "




The mass center G lies on fixed axis of rotation so that 0.=a

Resultant force on assembly: = 0mΣ =F a

0+ =A B = −A B

Angular velocity: ( )20 rad/s=ω k

20 rad/s, 0z x yω ω ω= = =

Angular momentum about .G

G xz yz zI I Iω ω ω= − − +H i j k

Rate of change of .GH Let the reference frame Gxyz be rotating with angular velocity ( )20 rad/s= =Ω ω k

( ) 0G G G GGxyz= + × = + ×H H Ω H ω H& &

( )xz yz zI I Iω ω ω ω= × − − −k i j k

2 2yz xzI Iω ω= −i j



Calculation of products of inertia yzI and xzI

( )( )( ) ( )( )( ) 21 1 1 2 2 2 2.4 0.100 0.100 2.4 0.100 0.100 0.048 kg myzI m y z m y z= + = + − = ⋅

( )( )( ) ( )( )( ) 21 1 1 2 2 2 2.4 0.100 0.100 2.4 0.100 0.100 0.048 kg mxzI m x z m x z= + = − + − = − ⋅

Couple formed by reactions A and B.

( )/ 0.600 0.6 0.6B A x y y xB B B B= × = − × + = −M r B k i j i j

G=M H& Resolve into components.

( )( )22: 0.6 0.048 20 19.2y yzB I ω= = =i

( )( )22: 0.6 0.048 20 19.2x xzB I ω− = − = − − =j

32 N, 32 N, 32 N, 32 Nx y x yB B A A= − = = = −

Dynamic reactions. ( ) ( )32.0 N 32.0 N= −A i j "

( ) ( )32.0 N 32.0 N= − +B i j"




By symmetry, the mass center is at G as shown. Angular velocity and angular acceleration: , ω α ω= = = !j j jω α

Angular momentum about the mass center: ω ω ω= − + −G xy y yzI I IH i j k

Since the body lies in the xy plane, 0.=yzI

Let the reference frame Gxyz be rotating with angular velocity ω= jΩ

( ) ( )ω ω ω ω ω= + × = − + + × − +! ! ! !G G G xy y xy yGxyzI I I IH H H i j j i jΩ

2α α ω= − + +xy y xyI I Ii j k

Calculation of xyI and :yI Let 2 mass per unit area.ρ = = =m mA ab

( ) ( )mass area.ρ=I I


2el el2 2

2= = =xy

xdI x y dA yxdy x ydy

2

20 0a a

xybyI x ydy b ydya

= = −

∫ ∫

2 2

2 2 320 0 02a a ab bb ydy y dy y dy

a a= − +∫ ∫ ∫

2 2 2 3 2 2 2 21 2 1 12 3 4 12

= − + =b a b a b a b a

3 31 1212 6 = =

yI ab ab

For mass, 21 1, 6 3

= =xy yI mab I mb

Since the mass center lies on the rotation axis, 0.=a



effΣ = ΣF F

0+ = =mA B a

= −B A

Reactions A and B form a couple.

Σ = !G GM H

( )0 + × + = !x z GM c A Aj j i k H

0+ − = !z x GcA M cAi j k H


: xyz xy z

IcA I A

cα

α= − = −i (1)

0: yM I α=j (2)

2

2: xyx xy x

IcA I A

cω

ω− = = −k (3)

Data: 5 kg, 900 mm 0.9 m, 600 mm 0.6 m, 2 1.8 m= = = = = = =m a b c a

8 rad/s, 0ω α= =

( )( ) ( )( )( )2 2 21 15 0.6 0.60 kg m , 5 0.9 0.6 0.45 kg m3 6

= = ⋅ = = ⋅y xyI I

From (2), 0 0=M

From (3), ( )( )20.45 816 N

1.8xA = − = −

From (1), 0=zA ( )16.00 N= −A i !




By symmetry, the mass center is at G as shown.

Angular velocity and angular acceleration: , ω α ω= = = &j j jω αω αω αω α

Angular momentum about the mass center: ω ω ω= − + −G xy y yzI I IH i j k

Since the body lies in the xy plane, 0.=yzI

Let the reference frame Gxyz be rotating with angular velocity ω= jΩΩΩΩ

( ) ( )ω ω ω ω ω= + × = − + + × − +& & & &G G G xy y xy yGxyzI I I IH H H i j j i jΩΩΩΩ

2α α ω= − + +xy y xyI I Ii j k v

Calculation of xyI and :yI Let 2

mass per unit area.ρ = = =m m

A a

( ) ( )mass area.ρ=I I


2el el2 2

2= = =xy

xdI x y dA yxdy x ydy

2

20 0

a axy

byI x ydy b ydy

a = = −

∫ ∫

2 2

2 2 320 0 0

2a a ab bb ydy y dy y dy

a a= − +∫ ∫ ∫

2 2 2 3 2 2 2 21 2 1 1

2 3 4 12= − + =b a b a b a b a

3 31 12

12 6 = =

yI ab ab

For mass, 21 1,

6 3= =xy yI mab I mb

Since the mass center lies on the rotation axis, 0.=a



effΣ = ΣF F

0+ = =mA B a

= −B A Reactions A and B form a couple.

Σ = &G GM H

( )0 + × + = &x z GM c A Aj j i k H

0+ − = &z x GcA M cAi j k H


: xyz xy z

IcA I A

c

αα= − = −i (1)

00: y

y

MM I

Iα α= =j (2)

2

2: xyx xy x

IcA I A

c

ωω− = = −k (3)

Data: 5 kg, 900 mm 0.9 m, 600 mm 0.6 m, 2 1.8 m= = = = = = =m a b c a

00, 36 N mω = = ⋅M

( )( ) ( )( )( )2 2 21 15 0.6 0.60 kg m , 5 0.9 0.6 0.45 kg m

3 6= = ⋅ = = ⋅y xyI I

(a) From (2), 23660 rad/s

0.60α = = 260.0 rad/sα = "

From (3), 0=xA

(b) From (1), ( )( )0.45 60

15.00 N1.8

= − = −zA ( )15.00 N= −A k"

( )15.00 N=B k"




The sheet metal component rotates about the fixed z axis, so that equations (18.29) of the textbook apply. These are relisted below as equations (1), (2), and (3).

2x xz yzM I Iα ωΣ = − + (1)

2y yz xzM I Iα ωΣ = − − (2)

z zM I αΣ = (3)

Calculation of the required moment and products of inertia. Total mass: 600 g 0.6 kgm = =

Total area: ( ) ( )( ) ( )2 2 3 21 10.075 0.150 0.075 0.075 16.875 10 m2 2

A −= + + = ×

Let ρ be the mass per unit area. 235.556 kg/mmA

ρ = =

The component is comprised of 3 parts: triangle 1, triangle 2, and rectangle 3 as shown. Let the lengths of 75 mm be labeled b. Triangle 1. The equation of the upper edge is

2by z= −

Use elements of width dz and height y.

( ) 2el

112zdm ydz dI y dmρ= =

( ) ( )el el0 0xz yzdI dI= =

Coordinates of the element mass center:

el el el1, ,2

x b y y z z= − = =

( ) el elelxz xzdI dI x z dm= +

( ) ( )02zbb z yd bz z dzρ ρ = + − = − −



42

2

12 12

b

xz bbI b z z dz bρ ρ

−

= − − =

∫

( ) el elelyz yzdI dI y z dm= +

( )21 10

2 2 2by z ydz z zdzρ ρ = + = −

2

42

2

1 12 2 24

b

yz bbI z zdz bρ ρ

−

= − = −

∫

( ) ( )2 2el elelz zdI dI x y dm= + +

2 2 2 2 31 1 112 4 3

y b y ydz b y y dzρ ρ = + + = +

3 2 2 313 5 1 124 4 2 3

b b z bz z dzρ = − + −

3 2 2 3 42

2

13 3 1 1 724 4 2 3 12

b

z bI b b z bz z dz bρ ρ−

= − + − =

∫

Applying the data,

( )( )4 6 21 35.556 0.075 93.75 10 kg m12xzI −= = × ⋅

( )( )4 6 21 35.556 0.075 46.875 10 kg m24yzI −= − = − × ⋅

( )( )4 6 27 35.556 0.075 656.25 10 kg m12zI −= = × ⋅

Triangle 2. The equation of the lower edge is .2by z = − −


( ) ( ) ( )2el el el

1, , 0, 012z zx yzdm ydz dI y dm dI dIρ= − = = =


el el el1, ,2

x b y y z z= = =

The integrals for , ,xz yzI I and zI turn out to be the same as those of triangle 1.

6 2 6 2 6 293.75 10 kg m , 46.875 10 kg m , 656.25 10 kg mxz yz zI I I− − −= × ⋅ = − × ⋅ = × ⋅

continued



Rectangle 3. Area: ( )( ) 3 20.150 0.075 11.25 10 mA −= = ×

Mass: 3400 10 kgm Aρ −= = ×

0, 0xz yzI I= =

( ) ( )( )2 23 6 21 12 400 10 0.150 750 10 kg m12 12zI m b − −= = × = × ⋅

Totals. 6 2187.5 10 kg mxz xzI I −= Σ = × ⋅

6 293.75 10 kg myz yzI I −= Σ = − × ⋅

6 22062.5 10 kg mz zI I −= Σ = × ⋅

Since the mass center lies on the fixed axis, the acceleration a of the mass center is zero.

0mΣ = =F a

The reactions at A and B form a couple.

= −B A

Let x yA A= +A i j

Resultant couple acting on the body:

( ) 0x yc A A M= × + +M k i j k

0y xcA cA M= − + +i j k

(a) Moment 0.M Using equation (3),

( )( )60 2062.5 10 12z zM M I α −Σ = = = ×

30 24.8 10 N mM −= × ⋅ !

(b) Reactions at A and B for the case 0.ω =

2x y xz yz xzM cA I I Iα ω αΣ = − = − − = −

( )( )6

3187.5 10 12

15 10 N0.150

xzy

IAcα

−−

×= = = ×

2y x yz xz yzM cA I I Iα ω αΣ = = − − = −

( )( )6

393.75 10 12

7.5 10 N0.150

yzx

IA

cα −

−− ×

= − = − = ×

( ) ( )3 37.50 10 N 15.00 10 N− −= × + ×A i j !

( ) ( )3 37.50 10 N 15.00 10 N− −= − × − ×B i j !




Use principal axes x′, y′, z′ as shown.

0=ωωωω

( ) ( )cos sinα θ α θ′ ′= −i jαααα

Moments of inertia:

2 21 1,

2 4′ ′ ′= = =x y zI mr I I mr

Angular momentum:

ω ω ω′ ′ ′ ′ ′ ′′= + +&G x x y y z zI I IH i j k

2 21 1cos sin

2 4ω θ ω θ= −mr mri j

Let the frame of reference Gx′y′z′ be rotating with angular velocity

= =Ω ω 0Ω ω 0Ω ω 0Ω ω 0

( ) ′ ′ ′= + ×& &

G G GGx y zH H HΩΩΩΩ

0α α α′ ′ ′ ′ ′ ′′ ′= + + +x x y y z zI I Ii j k

2 21 1cos sin

2 4α θ α θ′ ′= −mr mri j

( ) ( )2 21 1cos cos sin sin sin cos

2 4α θ θ θ α θ θ θ= + − − +mr mri j i j

2 2 2 21 1 1cos sin sin cos

2 4 4α θ θ α θ θ = + +

mr mri j

Since the mass center G lies on the rotation axis, 0, 0= =v a

: 0 m mΣ = + = = = −F a A B a B A



( ) ( ) ( ) ( )0Σ = + − × + + × +G y z y zM a A A a B BM i i j k i j k

( )0 02 2 2y z z yM a A A M aA aA= − × + = + −i i j k i j k

Resolve into components.Σ = &G GM H

(a) 2 2 20

1 1: cos sin

2 4M mr α θ θ = +

i

2

2 28 10 1 118 cos 20 sin 20

32.2 12 2 4α = ° + °

221.618 rad/sα =

( )2222 rad/s= iαααα "

(b) 21: 2 sin cos

4zaA mr α θ θ=j

( )

( )

2

28 10

221.618 sin 20 cos20sin cos 32.2 12

0.92167 lb208 812

zmr

Aa

α θ θ ° ° = = =

: 2 0 0y yaA A− = =k ( )0.922 lb=A k"

( )0.922 lb= −B k "




Length of rod: 8 8

2.76106 ft12 12

L π= + =

Angular velocity: 0ω= =iωωωω

Angular acceleration: α= iαααα

Angular momentum: ω ω= −G x xzI IH i k



Rate of change of angular momentum: ( )= + ×& &G G GGxyz

H H HΩΩΩΩ

( )G x x xz x x xzI I I Iα α ω ω ω= − + × −H i k i i k&

2x xz xzI I Iα ω α= + −i j k

Since the mass center G lies on the rotation axis, 0, 0 8 in. 4 in. 1 ftb= = = + =v a

Σ = + = = −m mF a A B a B A

( ) ( )0 0 2Σ = + − × + × = − ×G M b b M bM i i A i B i i A

( ) ( )0 02 2 2= − × + = + −y z z yM b A A M bA bAi i j k i j

Set Σ = &G GM H and resolve into components.

0: α= xM Ii 2

2: 2 2

ωω= = = −xzz xz z z

IbA I A B

bj

: 2 2

αα− = − = = −xzy xz y y

IbA I A B

bk

Calculation of xI and :xzI

Let mass per unit lengthρ = = m

L

For portions AC and DB,

0, 0= =x xzI I

For portion CG use polar coordinate .θ

continued



( )1 cos , sin θ θ ρ θ= − − = − =x r z r dm rd

2 2 2 30

sin2xI z dm r rd r

π πθρ θ ρ= = =∫ ∫

( )2 30

1 cos sin 2xzI xzdm r rd rπ θ θρ θ ρ= = − =∫ ∫

Likewise, for portion GD 3 3, 22

π ρ ρ= =x xzI r I r

Total: 3 3

3 4,

ππρ= = =x xzmr mr

I r IL L

Data: 21.6 40.049689 lb s /ft, 4 in. ft

32.2 12

Wm r =

g= = = ⋅ =

( )

3

2

40.049689

12 0.0020940 lb s ft2.76106xI

π = = ⋅ ⋅

( ) ( )

3

2

44 0.049689

120.0026661 lb s ft

2.76106xzI

= = ⋅ ⋅

(a) ( )( )0 0.0020940 150xM I α= = 0 0.314 lb ftM = ⋅ "

(b)2

02xz

z zI

A Bb

ω= = = −

( ) ( )

( ) ( )0.0026661 150

0.2002 1y yA B= = = −

( )0.200 lb=A j"

( )0.200 lb= −B j"




Fixed axis rotation with constant angular acceleration. , α α ω ω= =i i

2 20 0

1 12 2

t t tθ θ ω α α= + + = ( ) 22 2

2 22 3.1416 rad/s2t

πθα = = =

0ω =

Use centroidal axes x, y, z with origin at C. C x xy xzI I Iω ω ω= − −H i j k

Let the reference frame Cxyz rotate with angular velocity ω= = iΩ ω

( ) 2 2C C C x xy xz xz xyCxyz

I I I I Iα α α ω ω= + × = − − + −H H Ω H i j k j k! !

Required moments and products of inertia. Let mass per unit areamA

ρ = =

mass area 200mm 0.2mI I rρ= = =

Part A xI xyI xzI

21

2rπ 41

8rπ 42

3r− 0

21

2rπ 41

8rπ 0 42

3r−

21

2rπ 41

8rπ 0 42

3r−

21

2rπ 41

8rπ 42

3r− 0

Σ 22 rπ 412

rπ 443

r− 443

r−



( )2

2 26 23.873 kg/m

2 2 0.2mr

ρπ π

= = =

( ) ( )4 2123.873 0.2 0.06 kg m2xI π = = ⋅

( ) ( )4 2423.873 0.2 0.05093 kg m3xyI = − = − ⋅

( ) ( )4 2423.873 0.2 0.05093 kg m3xzI = − = − ⋅

Since the mass center lies on the rotation axis, 0=a

eff 0mΣ = + = Σ = =F A B F a = −A B

( ) ( )0 0 2C y zM b b M b B BΣ = + − × + × = + × +M i i A i B i i j k

0 2 2z yM bB bB= − +i j k where 2 1.0 mb =

C CMΣ = H! Resolve into components.

( )( )0: 0.06 3.1416xM I α= =i (a) ( )0 0.1885 N m= ⋅M i!

2: 2 z xy xzb B I Iα ω− = − +j

( )( )0.05093 3.1416 00.1600 N

1.0zB− − +

= − = − 0.1600 NzA =

2: 2 y xz xybB I Iα ω= − −k

( )( )0.05093 3.1416 00.1600 N

1.0yB− +

= − = 0.1600 NyA = −

(b) ( ) ( )0.1600 N 0.1600 N= − +A j k !

( ) ( )0.1600 N 0.1600 N= −B j k!




Fixed axis rotation with constant angular acceleration. , α α ω ω= =i i

2 20 0

1 1

2 2θ θ ω α α= + + =t t t

( ) 22 2

2 223.1416 rad/s

2t

πθα = = =

( )( )0 3.1416 2 6.2832 rad/st tω ω α α= + = = =

Use centroidal axes x, y, z with origin at C. C x xy xzI I Iω ω ω= − −H i j k

Let the reference frame Cxyz rotate with angular velocity ω= = iΩ ωΩ ωΩ ωΩ ω

( ) 2 2C C C x xy xz xz xyCxyz

I I I I Iα α α ω ω= + × = − − + −H H H i j k j k& & ΩΩΩΩ

Required moments and products of inertia. Let mass per unit aream

Aρ = =

mass area 200 mm 0.2 mI I rρ= = =

Part A xI xyI xzI

21

2rπ 41

8rπ 42

3r− 0

21

2rπ 41

8rπ 0 42

3r−

21

2rπ 41

8rπ 0 42

3r−

21

2rπ 41

8rπ 42

3r− 0

∑ 22 rπ 41

2rπ 44

3r− 44

3r−



( )2

2 2

623.873 kg/m

2 2 0.2

m

rρ

π π= = =

( ) ( )4 2123.873 0.2 0.06kg m

2xI π = = ⋅

( ) ( )4 2423.873 0.2 0.05093kg m

3xyI = − = − ⋅

( ) ( )4 2423.873 0.2 0.05093 kg m

3xzI = − = − ⋅

Since the mass center lies on the rotation axis, 0=a

eff 0mΣ = + = Σ = =F A B F a = −A B

( ) ( )0 0 2C y zM b b M b B BΣ = + − × + × = + × +M i i A i B i i j k

0 2 2z yM bB bB= − +i j k where 2 1.0 mb =

C CMΣ = H& Resolve into components.

( )( )0: 0.06 3.1416xM I α= =i 0 0.1885 N mM = ⋅

2: 2 z xy xzb B I Iα ω− = − +j

( )( ) ( )( )20.05093 3.1416 0.05093 6.2832

1.851 N,1.0zB

− − + −= − = 1.851 NzA = −

2: 2 y xz xybB I Iα ω= − −k

( )( ) ( )( )20.05093 3.1416 0.05093 6.2832

2.17 N,1.0yB

− + −= − = 2.17 NyA = −

( ) ( )2.17 N 1.851 N= − −A j k"

( ) ( )2.17 N 1.851 N= +B j k"




Mass of sheet metal: 1.5kg.m =

Sheet metal dimension: 0.120 mb =

Area of sheet metal: 2 2 2 2 2 21 13 0.432m

2 2A b b b b b= + + + = =

Let

21.534.722 kg/m mass per unit area

0.0432

m

Aρ = = = =

Moments and products of inertia: ( ) ( )areamassI Iρ=


4 4 41 1 2

3 3 3xI b b b= + =

( )( )442 234.722 0.120

3 3xI bρ= =

3 24.8 10 kg m−= × ⋅

( ) ( )2 2 43 1 5 1 1

2 2 2 2 2xyI b b b b b b b = + − = −

( )( )441 134.722 0.120

2 2xyI bρ= − = −

3 23.6 10 kg m−= − × ⋅




4 4 41 1 1

12 12 6xI b b b= + =

( )( )441 134.722 0.120

6 6xI bρ= =

3 21.2 10 kg m−= × ⋅

For calculation of ,xzI use pairs of elements 1dA and 2:dA 2 1dA dA=

( ) ( ) ( ) 21 2 1 0

4 2 22 2

bxz

z zI x dA b x dA b x zdA b x z dx

= ∫ + ∫ − − = − ∫ − = − −

∫

But .z x=

Hence, ( )2 3 4 4 40

2 1 52

3 4 12a

xzI bx x dx b b b = − − = − − = −

∫

( )( )44 3 25 534.722 0.120 3.0 10 kg m

12 12xzI bρ −= − = − = − × ⋅

Total for xI 3 3 3 24.8 10 1.2 10 6.0 10 kg mxI − − −= × + × = × ⋅

The mass center lies on the rotation axis, therefore, 0=a

0F mΣ = + = =A B a = −A B

A x xy xzI I Iω ω ω= − −H i j k ,ω α= =i α iωωωω

Let the frame of reference Axyz be rotating with angular velocity .ω= = iΩ ωΩ ωΩ ωΩ ω

( )A A A AAxyzΣ = = + ×M H H H& & ΩΩΩΩ

( ) ( )0 4 y z x xy xz x xy xzM b B B I I I I I Iα α α ω ω ω ω+ × + = − − + × − −i i j k i j k i i j k

( ) ( )2 20 4 4z y x xy xz xz xyM bB bB I I I I Iα α ω α ω− + = − − − +i j k i j k

Resolve into components and solve for and .y zB B



0: xM I α=i

( )2

: 4

xy xzz

I IB

b

α ω−=j

( )2

: 4

xz xyy

I IB

b

α ω+= −k

Data: 2150 rad/s , 0, 0.120 mbα ω= = =

(a) ( )( )30 6.0 10 150 0.9 N mM −= × = ⋅ ( )0 0.900 N m= ⋅M i"

(b) ( )( )

( )( )

33.6 10 150 01.125 N

4 0.120zB−− × −

= = −

( )( )( )( )

33.0 10 150 00.9375 N

4 0.120yB−− × +

= − =

0.9375 Ny yA B= − = −

1.125 Nz zA B= − =

( ) ( )0.938 N 1.125 N= − +A j k "

( ) ( )0.938 N 1.125 N= −B j k"




The sheet metal component rotates about the fixed z axis with angular acceleration ( )212 rad/s .kα =

(a) Angular velocity at t = 0.6 s.

( )( )0 0 12 0.6 7.2 rad/stω ω α= + = + =

( )7.20 rad/s kω = !

(b) Dynamic reactions. Equations (18.29) of the textbook apply. These are relisted below as equations (1), (2), and (3).

2x xz yzM I Iα ωΣ = − + (1)

2y yz xzM I Iα ωΣ = − − (2)

z zM I αΣ = (3)

Calculation of the required moment and products of inertia. Total mass: 600g 0.6kgm = =

Total area: ( ) ( )( ) ( )2 2 3 21 10.075 0.150 0.075 0.075 16.875 10 m2 2

A −= + + = ×

Let ρ be the mass per unit area. 235.556 kg/mmA

ρ = =

The component is comprised of 3 parts: triangle 1, triangle 2, and rectangle 3 as shown. Let the lengths of 75 mm be labeled b. Triangle 1. The equation of the upper edge is

2by z= −

Use elements of width dz and height y

dm y dzρ= ( ) 2el

112zdI y dm=

( )el0xzdI = ( )el

0yzdI =




el ,x b= − el1 ,2

y y= elz z=

( ) el elelxz xzdI dI x z dm= +

( ) ( )02bb z ydz b z z dzρ ρ = + − = − −

42

2

12 12

b

xz bbI b z z dz bρ ρ

−

= − − =

∫

( ) el elelyz yzdI dI y z dm= +

( )21 10

2 2 2by z ydz z z dzρ ρ = + = −

2

42

2

1 12 2 24

b

yz bbI z zdz bρ ρ

−

= − = −

∫

( ) ( )2 2el elelz zdI dI x y dm= + +

2 2 2 2 31 1 112 4 3iy b y y dz b y y dzρ ρ = + + = +

3 2 2 313 5 1 124 4 2 3

b b z bz z dzρ = − + −

3 2 2 3 42

2

13 3 1 1 724 4 2 3 12

b

z bI b b z bz z dz bρ ρ−

= − + − =

∫

Applying the data,

( )( )4 6 21 35.556 0.075 93.75 10 kg m12xzI −= = × ⋅

( )( )4 6 21 35.556 0.075 46.875 10 kg m24yzI −= − = − × ⋅

( )( )4 6 27 35.556 0.075 656.25 10 kg m12zI −= = × ⋅

Triangle 2. The equation of the lower edge is .2by z = − −


( ) ( ) ( )2el el el

1, , 0, 012z zx yzdm ydz dI y dm dI dIρ= − = = =

continued




el ,x b= el1 ,2

y y= elz z=

The integrals for Ixz, Iyz, and Iz turn out to be the same as those of triangle 1.

6 2 6 2 6 293.75 10 kg m , 46.875 10 kg , 656.25 10 kg mmxz yz zI I I− − −= × ⋅ = − × = × ⋅⋅

Rectangle 3. Area: ( )( ) 3 20.150 0.075 11.25 10 mA −= = ×

Mass: 3400 10 kgm Aρ −= = ×

0,xzI = 0yzI =

( ) ( )( )2 23 6 21 12 400 10 0.150 750 10 kg m12 12zI m b − −= = × = × ⋅

Totals. 6 2187.5 10 kg mxz xzI I −= Σ = × ⋅

6 293.75 10 kg myz yzI I −= Σ = − × ⋅

6 22062.5 10 kg mz zI I −= Σ = × ⋅

Since the mass center lies on the fixed axis, the acceleration a of the mass center is zero. 0mΣ = =F a

The reactions at A and B form a couple. = −B A

Let x yA A= +A i j

Resultant couple acting on the body:

( ) 0+x yc A A M= × +M k i j k

0+ +y xcA cA M= − i j k

From (1), 2

y xz yzcA I Iα ω− = − +

( )( ) ( )( )26 623

187.5 10 12 93.75 10 7.247.4 10 N

0.150 0.150yzxz

yIIA

c cωα

− −−

× − ×= − = − = ×

From (2), 2x yz xzcA I Iα ω= − −

( )( ) ( )( )26 623

93.75 10 12 187.5 10 7.257.3 10 N

0.150 0.150yz xz

xI IA

c cα ω

− −−

− × ×= − − = − − = − ×

( ) ( )3 357.3 10 N + 47.3 10 N− −= − × ×A i j !

( ) ( )3 357.3 10 N 47.3 10 N− −= × − ×B i j !




For each wheel,

65 mph 95.333 ft/s, 0xv ω= = =

95.333

0.21185 rad/s450y

vωρ

= = =

23 23

in. 0.95833 ft2 2 24

dr = = = =

( )( )2

2 2 248 0.750.83851 lb s ft

32.2= = = = ⋅ ⋅z

WI mk k

g

95.333

99.478 rad/s0.95833

= − = − = −zv

rω

G x x y y z z y y z zI I I I Iω ω ω ω ω= + + = +H i j k j k

Let reference frame Gxyz be rotating with angular velocity .yω= jΩΩΩΩ

( ) ( )0G G G y y y z z z z yGxyzI I Iω ω ω ω ω= + × = + × + =H H H j j k i& & ΩΩΩΩ

( )( )( ) ( )0.83851 99.478 0.21185 17.671 lb ft= − = − ⋅&GH i i

Let O be the point at the center of the axle. For the two wheels plus the axle,

( )35.342 lb ftO G G= + = − ⋅H H H i& & &

The distance between the wheels is 4.5 ft.

4.5 4.5O y yF F= × = −M k j i

Set O O=M H& and solve for .yF

35.342

7.85 lb4.5yF

−= =−

7.85 lbyF = !




Let the x axis lie along the axle AB and the y axis be vertical.

( )2 100001047.2 rad/s, 500 mph 733.33 ft/s

60= = = =x v

πω

733.332250 ft, 0.32593 rad/s 02250

= = − = − = − =y zvρ ω ωρ

Angular momentum: G x x y y z z x x y yI I I I Iω ω ω ω ω= + + = +H i j k i j

Let the reference frame Gxyz be turning about the y axis with angular velocity .yω= jΩ

( ) ( )G G G y x x y yGxyzH I Iω ω ω= + × = × +H j i j! ! Ω Η

x x yI ω ω= − k

Data for the disk: 7

3 216 13.587 10 lb s /ft32.2

m −= = × ⋅

( )2

2 3 6 21 1 213.587 10 188.708 10 lb s ft2 2 12xI mr − − = = × = × ⋅ ⋅

( )( )( ) ( )6188.708 10 1047.2 0.32593 0.06441 lb ftG G−= = − × − = ⋅M H k k!

The spring forces AF and BF exerted on the yoke provide the couple .GM The force exerted by spring B is upward.

Let . Then .B AF F= = −F j F j

/5 0.41667

12G B A F F F= × = × =M r i j i j k

From , 0.41667 0.06441 0.15458 lb.G G F F= = =M H!



Compression of spring B:

0.15458 0.0038645 ft 0.04637 in.40B

Fk

δ = = = =

Point B moves 0.0464 in. down. Point A moves 0.0464 in. up

Turning angle for yoke:

0.4637 0.4637 0.01855 rad5

θ += =

1.063θ = ° !

Point moves .A up !




1 1ω= iωωωω ( )

12 1800

188.5 rad/s60

π= =ωωωω

Angular velocity: 1 2ω ω= +i jωωωω

Angular momentum: 1 2G x yI Iω ω= +H i j

Let the reference frame be rotating with angular velocity 2 .ω= jΩΩΩΩ

( ) ( )2 1 2 1 20G G G x y xGxyzI I Iω ω ω ω ω= + × = + × + = −H H H j i j k& & ΩΩΩΩ

Assume that the acceleration of the mass center is negligible. Then the dynamic reactions at A and B reduce to a couple.

= −A B

( )y z z yb b B B b B b B= × = × + = − +M i B i j k j k

G=M H& Resolve into components.

: 0zb B− =j 0,zB = 0zA =

1 2: y xb B I ω ω= −k 2

1 2 1 2x xy

I mkB

b b

ω ω ω ω= − = −

( )( ) ( )( )230.300 75 10 188.5 0.61.527 N,

0.125yB−×

= − = − 1.527 NyA =

( )1.527 N=A j!

( )1.527 N= −B j!




Choose axes x, y, z as shown.

Angular velocity of airplane: 2vω ρ=

where 540 km/h 150 m/s, 600 mv ρ= = =

2 2 2150

0.25 rad/s, 600

ω ω= = = kωωωω

Angular velocity of propeller relative to the airplane: 1 1ω= iωωωω

( )( )1

2 1600167.552 rad/s

60

πω = =

Angular momentum of propeller: 1 2G x zI Iω ω= +H i k

Let the reference frame Gxyz be rotating with angular velocity 2 .ω= kΩΩΩΩ

( ) ( )2 1 2 1 20G G G x z xGxyzI I Iω ω ω ω ω= + × = + × + = −H H H k i k j& & ΩΩΩΩ

Couple exerted on the propeller: 1 2G xI ω ω= = −M H j&

Couple exerted by the propeller: 1 2xI ω ω′ = − =M M j (1)

Data for the propeller: 160 kgm =

( ) ( )22 2160 0.8 102.4 kg mx xI mk= = = ⋅

Using Equation (1), ( ) ( ) ( ) ( )102.4 167.552 0.25 4289 N m= = ⋅′M k k

Magnitude of couple: 4290 N mM = ⋅ !




Let the x axis be a horizontal axis directed along the engine mounting, i.e., longitudinally for rear-wheel drive and transversely for front-wheel drive.

Let the y axis be vertical.

The angular velocity of the automobile, 2,ωωωω is equal to ( )/ ,v ρ j where 60 mph 88 ft/s and 600 ft.v ρ= = =

( )288

0.146667 rad/s600

= =j jωωωω

Angular velocity of the fly wheel relative to the automobile: 1 1ω= iωωωω

where ( )

12 2700

282.74 rad/s60

πω = =

Angular momentum of fly wheel: 1G x y yI Iω ω= +H i j

Let the reference frame Gxyz be rotating with angular velocity 2ω= jΩΩΩΩ

( ) ( )2 1 2 1 20G G G x y xGxyzI I Iω ω ω ω ω= + × = + × + = −H H H j i j k& & ΩΩΩΩ

Couple exerted by the shaft on the fly wheel: 1 2xI ω ω= −M k

Couple exerted by the fly wheel on the shaft: 1 2xI ω ω′ = − =M M k (1)

Data for fly wheel: ( ) ( ) ( )22 30.284 lb/in 20 in. 0.75 in. 66.916 lb4 4

W d tπ πγ = = =

266.9162.0781 lb s /ft

32.2

Wm

g= = = ⋅

For a circular plate, ( )2

2 21 1 102.0781 0.72157 lb s ft

2 2 12xI mr = = = ⋅ ⋅

Using Equation (1), ( )( )( ) ( )0.72157 282.74 0.146667 29.9 lb ft′ = = ⋅M k k

(a) Magnitude of couple for rear-wheel drive: 29.9 lb ftM ′ = ⋅ !

(b) For front-wheel drive: 29.9 lb ftM ′ = ⋅ !




Angular velocity: 1,xω ω= 0,yω = 2.zω ω=

2+ω ω1i kω =ω =ω =ω =

Angular momentum: + +O x x y y z zI I Iω ω ω=H i j k

1 2+x zI Iω ω= i k

Let frame Oxyz be rotating with angular velocity 1 .ω= iΩΩΩΩ

Rate of change of angular momentum.

( )O O OOxyz= + ×H H& & Ω ΗΩ ΗΩ ΗΩ Η

( )1 2 1 1 2

21 2 1 2

+ + +

10 0 0 =

2

x z x z

z

I I I I

I mr

ω ω ω ω ω

ω ω ω ω

= ×

= + + − −

i k i i k

j j

& &

Dynamic reaction couple: = OM H&

( )( )2

21 2

1 1 5 43 6

2 2 32.2 12 = −

mr =ω ω −M j j

( )= 0.1553 lb ft− ⋅M j!




Let point O be the midpoint of axle AB. Choose principal axes , ,x y z′ ′ ′ with origin at point O.

2 21 1, ,

12 3x zI ma I ma′ ′= =

25

12y x zI I I ma′ ′ ′= + =

sin cosω β ω β′ ′= − +i jωωωω

O x x y y z zI I Iω ω ω′ ′ ′ ′ ′′ ′= + +H i j k

sinxI ω β= − i

Let the reference frame Ox y z′ ′ ′ be rotating about the fixed point O with

angular velocity .Ω = ωΩ = ωΩ = ωΩ = ω

( ) 0O O O OOx y z′ ′ ′= + × = + ×H H Η H& & Ω ωΩ ωΩ ωΩ ω

( ) ( )sin cos sin cosx yI Iω β ω β ω β ω β′ ′′ ′ ′ ′= − + × − +i j i j

( ) 2 sin cosy xI I ω β β′ ′= − − k

21sin cos

3ma β β= − k

cos2O Oa

W β= − =M k H&

22sin

3W ma mgω β= =

22sin

3g aω β=



(a) ( )( )

( )( )( )2 2

3 9.813sin 0.45417

2 2 0.225 12

g

aβ

ω= = = 27.0β = °!

(b) Set 90 , sin 1β β= ° =

( )( )

( )( )2 3 9.813

65.42 2 0.225

g

aω = = = 8.09 rad/sω = !




Let point O be the midpoint of axle AB. Choose principal axes

, ,x y z′ ′ ′ with origin at point O.

2 21 1, ,12 3x zI ma I ma′ ′= =

2512y x zI I I ma′ ′ ′= + =

sin cosω β ω β′ ′= − +i jω

O x x y y z zI I Iω ω ω′ ′ ′ ′ ′ ′′ ′= + +H i j k

sinxI ω β= − i

Let the reference frame Ox y z′ ′ ′ be rotating about the fixed point O with angular velocity .Ω = ω

( ) 0O O O OOx y z′ ′ ′= + × = + ×H H Η H! ! Ω ω

( ) ( )sin cos sin cosx yI Iω β ω β ω β ω β′ ′′ ′ ′ ′= − + × − +i j i j

( ) 2 sin cosy xI I ω β β′ ′= − − k

21 sin cos3

ma β β= − k

cos2O OaW β= − =M k H!

22 sin3

W ma mgω β= =

22 sin3

g aω β=



(a) ( )( )( )( ) ( )22 3 9.813 56.638 rad/s

2 sin 2 0.3 sin 60g

aω

β= = =

°

7.53 rad/sω = !

(b) Set 90 , sin 1β β= ° =

( )( )( )( ) ( )22 3 9.813 49.05 rad/s

2 2 0.3ga

ω = = = 7.00 rad/sω = !




Choose principal axes x, y, z with origin at the fixed point A.

2 2 2 2 21 32 , ,

2 2x y zI mr mr mr I mr I mr= + = = =

sin cosω β ω β= +i jωωωω

A x x y y z zI I Iω ω ω= + +H i j k

sin cosx yI Iω β ω β= +i j

Let the reference frame Axyz be rotating with angular velocity .Ω = ωΩ = ωΩ = ωΩ = ω

( ) 0A A A AAxyz= + × = + ×H H Η H& & Ω ωΩ ωΩ ωΩ ω

( ) ( )sin cos sin sinx yI Iω β ω β ω β ω β= + × +i j i j

( ) 2 2 23sin cos sin cos

2y xI I mrω β β ω β β= − = −k k

( )sinA Amgr βΣ = − =M k H&

23cos

2g rω β=

(a) ( )

( ) ( )2 2

2 32.22cos 0.322

83 3 1012

g

rβ

ω= = =

71.2β = °!

(b) Set 0 , cos 1β β= ° =

( )( )( )

2 2 32.2232.2

83 312

g

rω = = =

5.67 rad/sω = !




Choose principal axes x, y, z with origin at the fixed point A.

2 2 2 2 21 32 , ,

2 2x y zI mr mr mr I mr I mr= + = = =

sin cosω β ω β= +i jωωωω


sin cosx yI Iω β ω β= +i j

Let the reference frame Axyz be rotating with angular velocity .Ω = ωΩ = ωΩ = ωΩ = ω

( ) 0A A A AAxyz= + × = + ×H H Η H& & Ω ωΩ ωΩ ωΩ ω

( ) ( )sin cos sin sinx yI Iω β ω β ω β ω β= + × +i j i j

( ) 2 2 23sin cos sin cos

2y xI I mrω β β ω β β= − = −k k

( )sinA Amgr βΣ = − =M k H&

23cos

2g rω β=

(a) ( )( )

( )( )22 2 32.22

30.358 rad/s123 cos 3 cos 4512

g

rω

β= = =

°

5.51 rad/sω = !

(b) Set 0, cos 1β β= =

( )( )( )

( )22 2 32.2221.467 rad/s

123 312

g

rω = = =

4.63 rad/sω = !




20 430 , 20 in. ft, 4 in. ft

12 12L rβ = ° = = = =

Choose principal axes x, y, z as shown.

Kinematics: 1 1 1 2 2sin cosω β ω β ω= + =i j jω ωω ωω ωω ω

( )1 2 1 1 2sin cosω β ω β ω= + = + +i jω ω ωω ω ωω ω ωω ω ω

1 1 2sin cosx yω ω β ω ω β ω= = +

1 / 1 / 1 sinG G D G D Lω β= × = × = −v r j r kω ωω ωω ωω ω

21 1 sinG G Lω β= × =a vωωωω

/C G C G= + ×v v rωωωω

( ) ( )10 sin x yL rω β ω ω= − + + × −k i j i

1 sin yL rω β ω= − +k k

1 sinyL

rω ω β=

Angular momentum: G x x y yI Iω ω= +H i j



Let the reference frame Dxyz be rotating with angular velocity 1.=Ω ωΩ ωΩ ωΩ ω

( ) ( ) ( )

( )1 1

2 21 1 1

0 sin cos

1 1sin cos sin cos

2 4

G G G x x y yGxyz

y y x x y x

I I

I I mr mr

ω β ω β ω ω

ω ω β ω ω β ω β ω β ω

= + × = + + × +

= − = −

H H Η i j i j

k k

& & ΩΩΩΩ

Moments about D:

sin ,D Fd mgL β= −M k k

where 2 2d L r= +

( ) /effD G G D Gm= + ×M H r a&

2 21 sin cosG m Lω β β= +H k&

Equating ( )effD D=M M and taking the z component,

2 2 2 2 2 21 1

1 1sin sin sin cos sin cos

2 4CLF d mgL mr mr m Lr

β β β β ω ω β β − = − +

2 2 21

1 1sin sin cos cos

2 4m rL r Lω β β β β = − −

Set 210 and solve for :F ω=

( )21 2 2

2 2

2032.2

121 1 20 1 4 1 4 20cos cos sin cos30 cos30 sin 304 2 12 4 12 2 12 12

gL

L r rLω

β β β

= =

+ − ° + ° − °

23.427= 1 4.84 rad/sω = !




20 430 , 20 in. ft, 4 in. ft

12 12L rβ = ° = = = =

Choose principal axes x, y, z as shown.

Kinematics: 1 1 1 2 2sin cosω β ω β ω= + =i j jω ωω ωω ωω ω

( )1 2 1 1 2sin cosω β ω β ω= + = + +i jω ω ωω ω ωω ω ωω ω ω

1 1 2sin cosx yω ω β ω ω β ω= = +

1 / 1 / 1 sinG G D G D Lω β= × = × = −v r j r kω ωω ωω ωω ω

21 1 sinG G Lω β= × =a vωωωω

/C G C G= + ×v v rωωωω

( ) ( )10 sin x yL rω β ω ω= − + + × −k i j i

1 sin yL rω β ω= − +k k

1 sinyL

rω ω β=

Angular momentum: G x x y yI Iω ω= +H i j



Let the reference frame Dxyz be rotating with angular velocity 1.=Ω ωΩ ωΩ ωΩ ω

( ) ( ) ( )1 10 sin cosG G G x x y yGxyzI Iω β ω β ω ω= + × = + + × +H H Η i j i j& & ΩΩΩΩ

( ) 2 21 1 1

1 1sin cos sin cos

2 4y y x x y xI I mr mrω ω β ω ω β ω β ω β ω = − = −

k k

Moments about D:

sin ,D Fd mgL β= −M k k

where 2 2d L r= +

( ) /effD G G D Gm= + ×M H r a&

2 21 sin cosG m Lω β β= +H k&

Equating ( )effD D=M M and taking the z component,

2 2 2 2 2 21 1

1 1sin sin sin cos sin cos

2 4CLF d mgL mr mr m Lr

β β β β ω ω β β − = − +

2 2 21

1 1sin sin cos cos

2 4m rL r Lω β β β β = − −

Solving for ,CF

2 2 21

sin 1 1cos cos sin

4 2Cm

F gL L r rLd

β ω β β β = − + −

Additional data: 21

20.062112 lb s /ft, 4 rad/s

32.2m ω= = ⋅ =

2 220 4

1.69967 ft12 12

d = + =

continued



( ) ( ) ( )2 2

20.061112 sin 30 20 20 1 4 1 20 432.2 4 cos30 cos30 sin 30

1.69967 12 12 4 20 2 12 12CF ° = − ° + ° − °

0.311 lb=

1 1 4tan 30 tan 18.7

20

r

Lα β α− − = − = ° − = °

0.311 lb 18.7°!




Let 0.3 m, and 0.1 m.AB L BC b= = = =

Choose , ,x y z axes as shown. 210,12x y zI I I mL≈ = =

Angular velocity: sin 30 cos30ω ω= ° + °i jω

Angular momentum of rod AB about its mass center G:

cos30G x x y y z z yI I I Iω ω ω ω= + + = °H i j k j

Let the reference frame Gxyz be rotating with angular velocity .=Ω ω

( )G G GGxyz= + ×H H Η! ! Ω

( )0 sin 30 cos30 cos30yIω ω ω= + ° + ° × °i j j

2 2 23sin 30 cos3048yI mLω ω= ° ° =k k

Radius of circular path of point G: cos30 0.22990 m2Lr b= ° + =

Acceleration of the mass center: 2rω=a Equations of motion:



2cos30 sin 30

2 2B GL Lmg mrω Σ = ° = ° +

M k k H!

2 23 1 34 4 48

mgL mLr mL ω

= +

k k

23 1 34 4 48

g r L ω

= +

( ) ( ) ( ) 23 1 39.81 0.22990 0.34 4 48

ω

= +

2 62.194ω = 7.89 rad/sω = !




Let 0.3 m, and 0.1 m.AB L BC b= = = =

Choose , ,x y z axes as shown. 210,

12x y zI I I mL≈ = =

Angular velocity: sin 30 cos30ω ω= − ° + °i jωωωω

Angular momentum of rod AB about its mass center G:

cos30G x x y y z z yI I I Iω ω ω ω= + + = °H i j k j

Let the reference frame Gxyz be rotating with angular velocity .Ω = ωΩ = ωΩ = ωΩ = ω

( )G G GGxyz= + ×H H Η& & ΩΩΩΩ

( )0 sin 30 cos30 cos30yIω ω ω= + − ° + ° × °i j j

2 2 23sin 30 cos30

48yI mLω ω= − ° ° = −k k

Radius of circular path of point G: cos30 0.029904 m2

Lr b= ° − =

Acceleration of the mass center: 2rω=a

Equations of motion:



2cos30 sin 30

2 2B GL L

mg mrω Σ = − ° = − ° +

M k k H&

2 23 1 3

4 4 48mgL mLr mL ω

− = − +

k k

23 1 3

4 4 48g r L ω

= +

( ) ( ) ( ) 23 1 39.81 0.029904 0.3

4 4 48ω

= +

2 232.11ω = 15.24 rad/sω = !




Angular velocity: 1 2ω ω= +ω i k

Angular momentum: 1 2C x x y y z z x zI I I I Iω ω ω ω ω= + + = +H i j k i

Let the reference frame Cxyz be rotating with angular velocity 1ω=Ω i

( ) ( )1 1 2 1 20C C C x z zCxyzI I Iω ω ω ω ω= + × = + × + = −H H Ω H i i k j& &

Acceleration of mass center: 0=a

Equivalence of external and effective forces.



0mΣ = + = = −F a A B B A

The resultant of A and B is a couple. Its moment is

( ) ( )/ 2 2 2A B y z z ya A A aA aA× = − × + = −r A i j k j k

/A B C× =r A H&

components: 1 21 2: 2

2z

z z zI

aA I Aa

ω ωω ω= − = −j

: 2 0yaA− =k 0yA =

Data : 21 2

1 1

2 4z zI ma A maω ω= = −

1 21

4maω ω= −A k !

1 21

4maω ω=B k !




Angular velocity of the disk. ( ) ( )1 2 5 rad/s 15 rad/sω ω= + = +ω j k j k

Mass of disk. 210 0.31056 lb s /ft32.2

= = = ⋅Wmg

Moments of inertia about principal axes passing through the mass center.

( )2

2 21 1 60.31056 0.019410 lb s ft4 4 12x yI I mr = = = = ⋅ ⋅

( )2

2 21 1 60.31056 0.038820 lb s ft2 2 12zI mr = = = ⋅ ⋅

Angular momentum about mass center C.

( )( ) ( )( )0 0.019410 5 0.038820 15C x x y y z zI I Iω ω ω= + + = + +H i j k j k

( ) ( )0.09705 lb s ft 0.5823 lb s ft= ⋅ ⋅ + ⋅ ⋅j k

Rate of change of HC . Let the frame Axyz be turning with angular velocity 1 .ω=Ω j

( ) 0C C C CAxyz= + × = + ×H H Ω H Ω H! !

( ) ( )5 0.09705 0.5823 2.9115 lb ft= × + = ⋅j j k i

Position vector of point C. ( ) ( )/18 9 1.5 ft 0.75 ft12 12C A = + = +r i j i j

Velocity of point C, the mass center of the disk.

( ) ( )1 / 5 1.5 0.75 7.5 ft/sC C Aω= × = × + = −v j r j i j k

Acceleration of point C.

( ) ( )21 / 1 0 5 7.5 37.5 ft/sC C A Cα ω= × + × = + × − = −a j r j v j k i



Effective force at C. ( )( ) ( )0.31056 37.5 11.646 lbCm = − = −a i i


Linear components: Cm=A a ( )11.65 lb= −A i !

Moments about A. /A C A C Cm= × +M r a H!

( ) ( )1.5 0.75 11.646 2.9115A = + × − +M i j i i

( ) ( )2.91 lb ft 8.73 lb ftA = ⋅ + ⋅M i k !




Angular velocity: 2 1ω ω= −i kωωωω

Angular momentum: 2 1C x x y y z z x zI I I I Iω ω ω ω ω= + + = −H i j k i k

Let the reference frame Cxyz be rotating with angular velocity 2 .ω= iΩΩΩΩ

( ) ( )2 2 1 2 10C C C x z zCxyzI I Iω ω ω ω ω= + × = + × − =H H H i i k j& & ΩΩΩΩ

Acceleration of mass center:

0=a

mΣ =F a

0A B− =k k

A B=

C C=M H&

2 12 1 z

zI

LB I BL

ω ωω ω= =j j

Data: ( )( )22 6 21 10.3 kg, 0.3 0.050 375 10 kg m

2 2zm I mr −= = = = × ⋅

( )1 2

2 75025 rad/s, 6 rad/s, 0.200 m

60L

πω π ω= = = =

( )( )( )6375 10 6 250.88357 N

0.200A B

π−×= = =

( )0.884 N=A k!

( )0.884 N= −B k!







( ) ( )2 2 1 2 10C C C x z zCxyzI I Iω ω ω ω ω= + × = + × − =H H H i i k j& & ΩΩΩΩ

Acceleration of mass center:

0=a

mΣ =F a

0A B− =k k

A B=

C C=M H&

2 12 1 z

zI

LB I BL

ω ωω ω= =j j

Data: ( )( )22 6 21 10.3 kg, 0.3 0.050 375 10 kg m

2 2zm I mr −= = = = × ⋅

( )1

2 75025 rad/s, 0.200 m

60L

πω π= = =

1 NA B= =

( )( )

( )( )2 61

0.200 1.0

375 10 25z

LB

Iω

ω π−= =

× 2 6.79 rad/sω = !




For disk A: 1 2A ω ω= −j kω

1 20A x x y y z z y zI I I I Iω ω ω ω ω= + + = + −H i j k j j

2 21 2

1 12 4

mr mrω ω= −j k

Let the reference frame Oxyz be rotating with angular velocity 2 .ω= − kΩ

( ) ( ) 2 22 1 2

1 102 4A A AOxyz

mr mrω ω ω = + × = + − × −

H H H k j k! ! Ω

22 1

12

mr ω ω= i

Acceleration of its mass center: 22A bω= −a j

For disk B: 22, , , B A B A B A B bω= = = =H H H H a j! !ω ω

Let the system of particles be disks A and B together with axle AB and shaft CD. Neglect the mass of the axle and the shaft.

A Bm mΣ = +F a a 2 22 2C D b bω ω− + = − +j j j j

D C=



22 2 1

0 2 1 A BmrcC mr C

cω ωω ωΣ = = + = =M i H H i! !

(a) Data: 5 kg, 0.3 m, 0.6 m,m r c= = =

( ) ( )1 2

2 1200 2 6040 rad/s 2 rad/s

60 60π π

ω π ω π= = = =

( )( ) ( )( )25 0.3 40 2592 N

0.6C

π π= = ( )592 N= −C j!

( )592 N=D j!

(b) For disk A: 1 2 1 2, A A y zI Iω ω ω ω= − − = − −j k H j kω

22 1

1 , 02A A Bmr ω ω= − + =H i H H! ! ! 0= =C D !




For disk A: 1 2A ω ω= −j kωωωω

1 20A x x y y z z y zI I I I Iω ω ω ω ω= + + = + −H i j k j j

2 21 2

1 1

2 4mr mrω ω= −j k

Let the reference frame Oxyz be rotating with angular velocity 2 .ω= − kΩΩΩΩ

( ) ( ) 2 22 1 2

1 10

2 4A A AOxyzmr mrω ω ω = + × = + − × −

H H H k j k& & ΩΩΩΩ

22 1

1

2mr ω ω= i

Acceleration of its mass center: 22A bω= −a j

For disk B: 22, , , B A B A B A B bω= = = =H H H H a j& &ω ωω ωω ωω ω

Let the system of particles be disks A and B together with axle AB and shaft CD.

Neglect the mass of the axle and the shaft.

A Bm mΣ = +F a a

2 22 2C D b bω ω− + = − +j j j j

D C=



22 2 1

0 2 1 A Bmr

cC mr Cc

ω ωω ωΣ = = + = =M i H H i& &

Solving for 2,ω 2 21

cC

mrω

ω=

Data: 5 kg , 0.3 m, 0.6 mm r c= = =

( )1

2 120040 rad/s, 350 N

60C D

πω π= = = =

( )( )

( )( ) ( )2 2

0.6 3503.7136 rad/s

5 0.3 40ω

π= =

2 35.5 rpmω = !




Use principal axes , , x y z′ ′ as shown.

Moments of inertia: ( )2 213xI m a b′ = +

2 21 1, 3 3y zI ma I mb′ ′= =

Kinematics: 1θ ω=!

2 2 1sin , cos , x y zω ω θ ω ω θ ω ω′ ′ ′= = =

1 2 1 2cos , sin , 0x y zω ω ω θ ω ω ω θ ω′ ′ ′= = − =! ! !

Since the acceleration of the mass center is zero, the resultant force acting on the column CD is zero.

0=R !

Eulers equations of motion for the plate:

( ) ( )1 2 1 2cos cosx x x y z y z x y zM I I I I I Iω ω ω ω ω θ ω ω θ′ ′ ′ ′ ′ ′ ′ ′ ′ ′Σ = − − = − −!

( ) 21 2 1 2

2cos cos3x z yI I I mbω ω θ ω ω θ′ ′ ′= + − =

( ) ( )1 2 1 2sin siny y y z x x z y z xM I I I I I Iω ω ω ω ω θ ω ω θ′ ′ ′ ′ ′ ′ ′Σ = − − = − − −!

( ) 1 2 sin 0x y zI I I ω ω θ′ ′ ′= − − =

( ) ( ) 220 sin cosz z z x y x y x yM I I I I Iω ω ω ω θ θ′ ′ ′ ′ ′ ′ ′Σ = − − = − −!

2 22

1 sin cos3

mb ω θ θ= −



2 2 21 2 2

2 1cos sin cos3 3

mb mbω ω θ ω θ θ′Σ = −M i k

( )2 2 21 2 2

2 1cos cos sin sin cos3 3

mb mbω ω θ θ θ ω θ θ= + −i j k

Resolve into components: 2 21 2

2 cos3xM mb ω ω θΣ =

2 2 21 2 2

2 1sin cos , sin cos3 3y zM mb M mbω ω θ θ ω θ θΣ = Σ = −

Data: 2100 3.1056 lb s /ft, 2.4 ft32.2

m b= = ⋅ =

1 22 21.0472 rad/s, 0.5236 rad/s6 12π πω ω= = = =

( )( ) ( )( )2 2 22 3.1056 2.4 1.0472 0.5236 cos 6.54cos lb ft3xM θ θΣ = = ⋅

( )( ) ( )( )22 3.1056 2.4 1.0472 0.5236 sin cos 6.54sin cos lb ft3yM θ θ θ θΣ = = ⋅

( )( ) ( )2 21 3.1056 2.4 0.5236 sin cos 1.635sin cos lb ft3zM θ θ θ θΣ = − = − ⋅

( ) ( ) ( )26.54 lb ft cos 6.54 lb ft sin cos 1.635 lb ft sin cosD θ θ θ θ θ= ⋅ + ⋅ − ⋅M i j k !




Use principal axes , , x y z′ ′ as shown.

Moments of inertia: ( )2 21

3xI m a b′ = +

2 21 1,

3 3y zI ma I mb′ ′= =

Kinematics: 1θ ω=&

2 2 1sin , cos , x y zω ω θ ω ω θ ω ω′ ′ ′= = =

1 2 1 2cos , sin , 0x y zω ω ω θ ω ω ω θ ω′ ′ ′= = − =& & &

Euler’s equations of motion for the plate:

( ) ( )1 2 1 2cos cosx x x y z y z x y zM I I I I I Iω ω ω ω ω θ ω ω θ′ ′ ′ ′ ′ ′ ′ ′ ′ ′Σ = − − = − −&

( ) 21 2 1 2

2cos cos

3x z yI I I mbω ω θ ω ω θ′ ′ ′= + − =

( ) ( )1 2 1 2sin siny y y z x x z y z xM I I I I I Iω ω ω ω ω θ ω ω θ′ ′ ′ ′ ′ ′ ′Σ = − − = − − −&

( ) 1 2 sin 0x y zI I I ω ω θ′ ′ ′= − − =

( ) ( ) 220 sin cosz z z x y x y x yM I I I I Iω ω ω ω θ θ′ ′ ′ ′ ′ ′ ′Σ = − − = − −&

2 22

1sin cos

3mb ω θ θ= −



2 2 21 2 2

2 1cos sin cos

3 3mb mbω ω θ ω θ θ′Σ = −M i k

( )2 2 21 2 2

2 1cos cos sin sin cos

3 3mb mbω ω θ θ θ ω θ θ= + −i j k

Resolve into components: 2 21 2

2cos

3xM mb ω ω θΣ =

2 2 21 2 2

2 1sin cos , sin cos

3 3y zM mb M mbω ω θ θ ω θ θΣ = Σ = −

(a) ,D = ΣM M which is independent of the dimension 2a.

(b) 22 1 2

2sin cos

3zM M mb ω ω θ θ= Σ =

2 21 2

1sin cos

3yM M mb ω θ θ= Σ =

ratio 1 2

2 12

M

M

ωω

=




Angular velocity: 1 2ω ω= +ω i k

Angular momentum: 1 2C x x y y z z x zI I I I Iω ω ω ω ω= + + = +H i j k i

Let the reference frame Cxyz be rotating with angular velocity 1ω=Ω i

( ) 1 2 1C C C x z CCxyzI Iω ω ω= + × = + + ×H H Ω H i k i H! ! ! !

( )1 1 1 2 1 1 20x x z x zI I I I Iα ω ω ω α ω ω= + + × + = −i i i k i j





0mΣ = + = = −F a A B B A

The resultant of A and B is a couple. Its moment is

( ) ( )/ 2 2 2A B y z z ya A A aA aA× = − × + = −r A i j k j k

Resultant moment on rod AB and disk.

1 2 2z y CM aA aA= + − =M i j k H!

components: 1 1: xM I α=i

1 21 2: 2

2z

z z zIaA I A

aω ωω ω= − = −j

: 2 0yaA− =k 0yA =

Data : 2 21 14 2x zI ma I ma= =

(a) Moment 1 .M i

21 1

14

M ma α=i i !

(b) Reactions at A and B. 1 214

mω ω= −A k !

1 214

mω ω=B k !







( ) ( )1 2 2 1C C C z x zCxyzI I Iω ω ω ω= + × = − + × −H H H k i i k& & &ΩΩΩΩ

1 1 2z zI Iω ω ω= − +k j&


mΣ =F a 0+ =A B = −A B

Moments about C : C C=M H&

( ) 1 1 2y z z zL B B I Iω ω ω× + = − +i j k k k&

1 1 2y z z zLB LB I Iω ω ω− = − +k j k k&



Data: ( )( )22 6 21 10.3 kg, 0.3 0.050 375 10 kg m

2 2zm I mr −= = = = × ⋅

21 2 112 rad/s, 6 rad/s, 4 rad/sω ω ω= = = −&

0.200 mL =

( )( ) ( )( )( )6 60.200 0.200 375 10 4 375 10 12 6y zB B − −− = − × − + ×k j k j

( )( )6375 10 4

0.0075 N0.200yB

−− × −= = 0.0075 NyA = −

( )( )( )6375 10 12 6

0.1350 N0.200zB

−×= = 0.1350 NzA =

( ) ( )0.0075 N 0.1350 N= − +A j k !

( ) ( )0.0075 N 0.1350 N= −B j k !




Angular velocity of the shaft and arm ABC: 2ω= jΩΩΩΩ

Angular velocity of disk C: 2 1 1, 0ω ω ω= + =j k &ωωωω

Its angular momentum about C: 2 1C x x y y z z y zI I I I Iω ω ω ω ω= + + = +H i j k j k

Let the reference frame Axyz be turning with angular velocity .ΩΩΩΩ

( ) ( )2 2 2 1C C C y y zAxyzI I Iω ω ω ω= + × = + × +H H H j j j k& & &ΩΩΩΩ

2 22 1 2 2 1 2

1 1

2 4z yI I mr mrω ω ω ω ω ω= + = +i j i j& &

Velocity and acceleration of the mass center C of the disk:

( )/ 2 2 2C C A b c c bω ω ω= × = × + = −v r j i k i kΩΩΩΩ

( ) ( )2 22 / 2 2 2 2 2C C A C c b b cω ω ω ω ω ω= × + × = − + +a j r j v i k& & &

eff : Resolve into components.x z CA A mΣ = Σ + =F F i k a

( ) ( )2 22 2 2 2 x zA m c b A m b cω ω ω ω= − = − +& &

( )/A A C C A C C Cm b c mΣ = = + × = + + ×M H H r a H i k a& & &

2 2 2 22 1 2

1 1 Resolve into components.

2 4mr m r b cω ω ω = + + +

i j&



( ) ( )2 2 2 22 1 0 2

1 1

2 4A Ax yM mr M M m r b cω ω ω = = = + +

&

Solving for 2 :ω& 02 2 2 2

2 2 2

1.51 80 1 10 16 124 32.2 4 12 12 12

M

m r b cω = =

+ + + +

&

20.204565 rad/s=

At 5 s,t = ( )( )2 2 0.204565 5 1.022824 rad/stω ω= = =&

After 5 s, let 2 0ω =& so that 2ω is constant.

( )222

80 161.022824 3.4656 lb

32.2 12xA mbω = − = − = −

( )222

80 121.022824 2.5992 lb

32.2 12zA mcω = − = − = −

( ) ( )3.47 lb 2.60 lb= − −R i k !

( ) 0A yM =

( ) ( )( )2

22 1

1 1 80 101.022824 80 70.588 lb ft

2 2 32.2 12A xM mr ω ω = = = ⋅

( )70.6 lb ftA = ⋅M i!




Angular velocity of the shaft and arm ABC: 2ω= jΩ

Angular velocity of disk C: 2 1 1, 0ω ω ω= + =j k !ω

Its angular momentum about C: 2 1C x x y y z z y zI I I I Iω ω ω ω ω= + + = +H i j k j k

Let the reference frame Axyz be turning with angular velocity .Ω

( ) ( )2 2 2 1C C C y y zAxyzI I Iω ω ω ω= + × = + × +H H H j j j k! ! !Ω

2 22 1 2 2 1 2

1 12 4z yI I mr mrω ω ω ω ω ω= + = +i j i j! !

Velocity and acceleration of the mass center C of the disk:

( )/ 2 2 2C C A b c c bω ω ω= × = × + = −v r j i k i kΩ

( ) ( )2 22 / 2 2 2 2 2C C A C c b b cω ω ω ω ω ω= × + × = − + −a j r j v i k! ! !

eff : Resolve into components.x z CA A mΣ = Σ + =F F i k a

( ) ( )2 22 2 2 2 x zA m c b A m b cω ω ω ω= − = − +! !

( )/A A C C A C C Cm b c mΣ = = + × = + + ×M H H r a H i k a! ! !

2 2 2 22 1 2

1 1 Resolve into components.2 4

mr m r b cω ω ω = + + +

i j!



( ) ( )2 2 2 22 1 0 2

1 1 2 4A Ax yM mr M M m r b cω ω ω = = = + +

!

Since the couple 0M is given as constant, 2ω! is constant.

222

2.4 0.8 rad/s3t

ωω = = =!

(a) ( )2 2 2

080 1 10 16 12 0.8

32.2 4 12 12 12M

= + +

( )0 5.87 lb ft= ⋅M j!

(b) ( ) ( )280 12 160.8 2.4 17.09 lb32.2 12 12xA = − = −

( ) ( )280 16 120.8 2.4 16.96 lb32.2 12 12zA = − + = −

( ) ( )17.09 lb 16.96 lbA = − −R i k!

( ) ( )( )21 80 10 2.4 80 165.6 lb ft

2 32.2 12A xM = = ⋅

( )165.6 lb ftA = ⋅M i!




Angular velocity of shaft DCE and arm CBA: 2ω= kΩΩΩΩ

Angular velocity of disk A: 1 2ω ω= +j kωωωω

Its angular momentum about A: 1 2A x x y y z z y zI I I I Iω ω ω ω ω= + + = +H i j k j k

Let the reference frame Cxyz be rotating with angular velocity .ΩΩΩΩ

( ) ( )1 2 2 1 2A A A y z y zAxyzI I I Iω ω ω ω ω= + × = + + × +H H H j k k j k& & & &ΩΩΩΩ

2 2 22 1 1 2 2 1 1 2

1 1 1

2 2 4y y zI I I mr mr mrω ω ω ω ω ω ω ω= − + + = − + +i j k i j k& & & &

Velocity and acceleration of the mass center A of the disk:

( )2 / 2 2 2A A C b c c bω ω ω= × = × − = +v k r k i j i jωωωω

( ) ( )2 22 / 2 2 2 2 2A A C A c b b cω ω ω ω ω ω= × + × = − + +a k r k v i j& & &

effΣ = ΣF F

x y x y AD D E E m+ + + =i j i j a


( )22 2x xD E m c bω ω+ = −&

( )22 2y yD E m b cω ω+ = +&



( )/E E A A E A A Am b c l mΣ = = + × = + − + ×M H H r a H i j k a& & &

( ) ( ) ( ) ( )2 2 2 20 2 2 2 2 22 x y Al D D M m bl cl m cl bl m b cω ω ω ω ω× + + = + − + + + +k i j k H i j k& & & &

2 20 2 1 2 2

12 2

2y xlD lD M m r bl clω ω ω ω − + + = − + −

i j k i&

2 2 2 2 21 2 2 2

1 1

2 4m r cl bl m r b cω ω ω ω + + − + + +

j k& & &

2 2 20 2

1:

4M m r b c ω = + +

k &

2 2 2 21 2 2 1 2 2

1 1:

2 2 2 2x xm m

D r cl bl E r cl bll l

ω ω ω ω ω ω = + − = − + −

j & & & &

2 2 2 22 1 2 2 2 1 2 2

1 1:

2 2 2 2y ym m

D r bl cl E r bl cll l

ω ω ω ω ω ω ω ω = + + = − + +

i & &

Data: 2.5 kg, 0.1 m,m r= =

0.15 m, 0.075 m, 0.18 mb c l= = =

21 1 2 250 rad/s, 15 rad/s , 12 rad/s, 0ω ω ω ω= = − = =& &

( )( ) ( ) ( ) ( )( )( )2 22.5 10.1 15 0 0.15 0.18 12 27.52 N

2 0.18 2xD = − + − = −

( )( ) ( ) ( )( ) ( )( )( )2 22.5 10.1 12 50 0 0.075 0.18 12 34.33 N

2 0.18 2yD = + + =

( ) ( )27.5 N 34.3 N= − +D i j!

( )( ) ( ) ( ) ( )( )( )2 22.5 10.1 15 0 0.15 0.18 12 26.48 N

2 0.18 2xE = − − + − = −

( )( ) ( ) ( )( ) ( )( )( )2 22.5 10.1 12 50 0 0.075 0.18 12 7.333 N

2 0.18 2yE = − + + = −

( ) ( )26.5 N 7.33 N= − −E i j!




Angular velocity of shaft DCE and arm CBA: 2ω= kΩ

Angular velocity of disk A: 1 2ω ω= +j kω

Its angular momentum about A: 1 2A x x y y z z y zI I I I Iω ω ω ω ω= + + = +H i j k j k

Let the reference frame Cxyz be rotating with angular velocity .Ω

( ) ( )1 2 2 1 2A A A y z y zAxyzI I I Iω ω ω ω ω= + × = + + × +H H H j k k j k! ! ! !Ω

2 2 22 1 1 2 2 1 1 2

1 1 12 2 4y y zI I I mr mr mrω ω ω ω ω ω ω ω= − + + = − + +i j k i j k! ! ! !

Velocity and acceleration of the mass center A of the disk:

( )2 / 2 2 2A A C b c c bω ω ω= × = × − = +v k r k i j i jω

( ) ( )2 22 / 2 2 2 2 2A A C A c b b cω ω ω ω ω ω= × + × = − + +a k r k v i j! ! !

effΣ = ΣF F

x y x y AD D E E m+ + + =i j i j a


( )22 2x xD E m c bω ω+ = −!

( )22 2y yD E m b cω ω+ = +!



( )/E E A A E A A Am b c l mΣ = = + × = + − + ×M H H r a H i j k a! ! !

( ) ( ) ( ) ( )2 2 2 20 2 2 2 2 22 x y Al D D M m bl cl m cl bl m b cω ω ω ω ω× + + = + − + + + +k i j k H i j k! ! ! !

2 20 2 1 2 2

12 22y xlD lD M m r bl clω ω ω ω − + + = − + −

i j k i!

2 2 2 2 21 2 2 2

1 12 4

m r cl bl m r b cω ω ω ω + + − + + +

j k! ! !

2 2 20 2

1: 4

M m r b c ω = + +

k !

2 2 2 21 2 2 1 2 2

1 1: 2 2 2 2x xm mD r cl bl E r cl bll l

ω ω ω ω ω ω = + − = − + −

j ! ! ! !

2 2 2 22 1 2 2 2 1 2 2

1 1: 2 2 2 2y ym mD r bl cl E r bl cll l

ω ω ω ω ω ω ω ω = + + = − + +

i ! !

Data: 2.5 kg, 0.1 m,m r= =

0.15 m, 0.075 m, 0.18 mb c l= = =

21 1 2 2 250 rad/s, 0, 12 rad/s, 8 rad/sω ω ω ω α= = = = =! !

(a) ( ) ( ) ( ) ( ) ( )2 2 20

12.5 0.1 0.15 0.075 84

M = + + ( )0 0.613 N m= ⋅M k"

(b) ( )( ) ( )( )( ) ( )( )( )22.5 0 0.075 0.18 8 0.15 0.18 12 26.25 N2 0.18xD = + − = −

( )( ) ( ) ( )( ) ( )( )( ) ( )( )( )2 22.5 1 0.1 12 50 0.15 0.18 8 0.075 0.18 12 35.83 N2 0.18 2yD = + + =

( ) ( )26.3 N 35.8 N= − +D i j"

( )( ) ( )( )( ) ( )( )( )22.5 0 0.075 0.18 8 0.15 0.18 12 26.25 N2 0.18xE = − + − = −

( )( ) ( ) ( )( ) ( )( )( ) ( )( )( )2 22.5 1 0.1 12 50 0.15 0.18 8 0.075 0.18 12 5.833 N2 0.18 2yE = − + + = −

"

( ) ( )26.3 N 5.83 N= − −E i j"




Use principal axes x, y, z with origin at O.

Angular velocity: ( )sin cosφ θ ψ φ θ= + +i k& &&ωωωω

sin , 0, cosx y zω φ θ ω ω ψ φ θ= = = +& &&

Angular momentum about O:

sinO x x y y z z zI I I I Iω ω ω φ θ ω′= + + = +H i j k i k&

Let the reference frame Oxyz be rotating with angular velocity sin cos .φ θ φ θ= +i k& &ΩΩΩΩ

O Ο= ×H H& ΩΩΩΩ

( ) ( )sin cos sin zI Iφ θ φ θ φ θ ω′= + × +i k k& & &

( )sin cos sinzI Iφ θ θ ω θ φ′= − j& &

O OΣ =M H&

( )sin sin cos sinzWc I Iθ φ θ θ ω θ′− = −j j&

( )coszWc I Iω φ θ φ′= − & &

( ) cosWc I I Iψ φ θ φ ′= − − & && (1)

Data: 3

lb 0.1875 lb16

W = =

3 20.18755.823 10 lb s ft

32.2m −= = × ⋅ ⋅



( )2

2 3 6 21.055.823 10 44.582 10 lb s ft

12zI mk − − = = × = × ⋅ ⋅

( )2

2 3 6 22.255.823 10 204.715 10 lb s ft

12xI mk − − ′ = = × = × ⋅ ⋅

1.875 in. 0.15625 ft, 1800 rpm 188.496 rad/s,c ψ= = = =&

30 .θ = °

Substituting into (1),

( ) ( ) ( ) ( )60.1875 0.15625 44.582 10 188.496−= ×

( )6160.133 10 cos30φ φ− − × °& &

6 2 3 3138.679 10 8.4035 10 29.297 10 0φ φ− − −× − × + × =& &

30.298 26.584 3.714 rad/s, 56.882 rad/sφ φ= ± =& &

35.5 rpm, 543 rpmφ =& !




Use principal axes x, y, z with origin at O.

Angular velocity: ( )sin cosφ θ ψ φ θ= + +i k! !!ω

sin , 0, cosx y zω φ θ ω ω ψ φ θ= = = +! !!


sinO x x y y z z zI I I I Iω ω ω φ θ ω′= + + = +H i j k i k!

Let the reference frame Oxyz be rotating with angular velocity sin cos .φ θ φ θ= +i k! !Ω

O Ο= ×H H! Ω

( ) ( )sin cos sin zI Iφ θ φ θ φ θ ω′= + × +i k k! ! !

( )sin cos sinzI Iφ θ θ ω θ φ′= − j! !

O OΣ =M H!

( )sin sin cos sinzWc I Iθ φ θ θ ω θ′− = −j j!

(a) ( )coszWc I Iω φ θ φ′= − ! !!

( ) cosWc I I Iψ φ θ φ ′= − − ! !! (1)

(b) For Wc Iφ ψ ψφ≈! !! !" (2) !

(c) Data: 3 lb 0.1875 lb16

W = =

3 20.1875 5.823 10 lb s ft32.2

m −= = × ⋅ ⋅



( )2

2 3 6 21.055.823 10 44.582 10 lb s ft12zI mk − − = = × = × ⋅ ⋅

( )2

2 3 6 22.255.823 10 204.715 10 lb s ft12xI mk − − ′ = = × = × ⋅ ⋅

1.875 in. 0.15625 ft, 1800 rpm 188.496 rad/s,c ψ= = = =!

30 .θ = °


( ) ( ) ( ) ( )60.1875 0.15625 44.582 10 188.496−= ×

( )6160.133 10 cos30φ φ− − × °! !

6 2 3 3138.679 10 8.4035 10 29.297 10 0φ φ− − −× − × + × =! !

30.298 26.584 3.714 rad/s, 56.882 rad/sφ φ= ± =! !

From (2), ( )( )( )( )6

0.1875 0.156253.486 rad/s

44.582 10 188.496WcI

φψ −

= = =×

!!

3.486 3.714% error 100%3.714

−= × % error 6.14%= − !




Use principal axes x, y, z with origin at A.

Angular velocity: ( )sin cosφ β ψ φ β= + −i k! !!ω

sin , 0, cosx y zω φ β ω ω ψ φ β= = = −! !!

Angular momentum about A:


( )sin cosI Iφ β ψ φ β′= + −i k! !!

Let the reference frame Axyz be rotating with angular velocity sin cos .φ β φ β= −i k! !Ω

A Α= ×H H! Ω

( )sin sin cosI I Iφ ψ β φ β β ′= − + − j! !!

A AΣ =M H!

( )sin sin sin cosmgc I I Iβ φ ψ β φ β β ′− = − + − j j! !!

( ) 2 cosmgc I I Iφψ φ β′= + −! !!

( ) 2 cos1 0

I IImgc mgc

φ βφψ ′ −+ − =

!! ! (1)

Data: 3 in. 0.25 ft, 24 in. 2 ft,r c= = = =

232.2 ft/sg =

continued



( )22 21 0.25 0.03125 ft2 2

I rm

1= = =

( ) ( )2 22 2 21 1 0.25 2 4.015625 ft4 4

I r cm′

= + = + =

60 , 36 rpm 3.7699 rad/sβ φ= ° = =!


( )( )( )( )

( )( )( )( )

20.03125 3.7699 4.015625 0.03125 3.7699 cos601 0

32.2 2 32.2 2ψ − °

+ − =!

31.82934 10 0.43965 1 0ψ−× + − =!

306.3 rad/s,ψ =! 2930 rpmψ =! !




Use principal axes x, y, z with origin at A.

Angular velocity: ( )sin cosφ β ψ φ β= + −i k! !!ω

sin , 0, cosx y zω φ β ω ω ψ φ β= = = −! !!



( )sin cosI Iφ β ψ φ β′= + −i k! !!

Let the reference frame Axyz be rotating with angular velocity sin cos .φ β φ β= −i k! !Ω

A Α= ×H H! Ω

( )sin sin cosI I Iφ ψ β φ β β ′= − + − j! !!

A AΣ =M H!

( )sin sin sin cosmgc I I Iβ φ ψ β φ β β ′− = − + − j j! !!

( ) 2 cosmgc I I Iφψ φ β′= + −! !!

( ) 2 cos1 0

I IImgc mgc

φ βφψ ′ −+ − =

!! ! (1)

Data: 3 in. 0.25 ft, 24 in. 2 ft,= = = =r c

232.2 ft/sg =



( )22 21 0.25 0.03125 ft2 2

I rm

1= = =

( ) ( )2 22 2 21 1 0.25 2 4.015625 ft4 4

I r cm′

= + = + =

45 , 2100 rpm 219.91 rad/sβ ψ= ° = =!


( )( )( )( )

( )( )( )

20.03125 219.91 4.015625 0.03125 cos 451 0

32.2 2 32.2 2φ φ− °

+ − =! !

( ) 20.106711 0.043748 1 0φ φ+ − =! !

6.1537 rad/s or 3.7145 rad/sφ φ= − =! !

58.8 rpm, 35.5 rpmφ = −! !




Use principal axes xyz with origin at A as shown.

For the solid cone, 0.080 m,r =

0.240 m,h = 3 0.180 m4hc = =

23

10I mr=

2 23 15 4

I m h r ′ = +

2 23 15 4

I I m h r ′− = −

Angular velocity. spin: ψ! about negative z axis

precession: φ! about positive z axis

φ ψ= −K k! !ω

( )cos sinφ β β ψ= + −k j k! !

0, sin , cosx y zω ω φ β ω φ β ψ= = = −! ! !

Angular momentum about fixed point A.

A x y zI I Iω ω ω′ ′= + +H i j k

( )sin cosI Iφ β φ β ψ′= + −j k! ! !

Let frame Axyz be rotating with angular velocity Ω .

cos sinφ φ β φ β= = +K j k! ! !Ω Rate of change of .AH

( )

0 sin cos

0 sin cosA A

I I

φ β φ β

φ β φ β ψ

= × =

′ −

i j kH H ! !!

! ! !

Ω

( ) 2 cos sin sinI I Iφ β β φψ β ′= − − + i! ! !



Momentum about A. sin β= −A mgcM i

A A=M H! leads to

2 cosI I Igcm m

φ β φψ′ −= +! ! !

2 2 2 23 1 3cos5 4 10

h r rφ β φψ = − +

! ! !

( )2 2 2 220 12 3 cos 6gc h r rφ β φψ= − +! ! !

( )2

2 2 220 6cos12 3

gc rh r

φψβφ

−=−

! !!

Data: 1600 rpm = 167.552 rad/sψ =!

40 rpm = 4.1888 rad/sφ =!

( )( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )

2

2 2 2

20 9.81 0.180 6 0.080 167.552 4.1888cos

12 0.240 3 0.080 4.1888β

−=

−

0.70947= 44.8β = ° !




Use principal axes xyz with origin at A as shown.

For the solid cone, 0.080 m,r =

0.240 m,h = 3 0.180 m4hc = =

23

10I mr=

2 23 15 4

I m h r ′ = +

2 23 15 4

I I m h r ′− = −

Angular velocity. spin: ψ! about negative z axis

precession: φ! about positive z axis

φ ψ= −K k! !ω

( )cos sinφ β β ψ= + −k j k! !

0, sin , cosx y zω ω φ β ω φ β ψ= = = −! ! !

Angular momentum about fixed point A.

A x y zI I Iω ω ω′ ′= + +H i j k

( )sin cosI Iφ β φ β ψ′= + −j k! ! !

Let frame Axyz be rotating with angular velocity Ω .

cos sinφ φ β φ β= = +K j k! ! !Ω

Rate of change of .AH

( )

0 sin cos

0 sin cosA A

I I

φ β φ β

φ β φ β ψ

= × =

′ −

i j kH H ! !!

! ! !

Ω

( ) 2 cos sin sinI I Iφ β β φψ β ′= − − + i! ! !



Momentum about A. sin β= −A mgcM i

A A=M H! leads to

2 cosI I Igcm m

φ β φψ′ −= +! ! !

2 2 2 23 1 3cos5 4 10

h r rφ β φψ = − +

! ! !

( )2 2 2 220 12 3 cos 6gc h r rφ β φψ= − +! ! !

( )2 2 2

2

20 12 3 cos

6

gc h r

r

φ βψ

φ

− −=

!! !

Data: 30β = °

30 rpm = rad/sφ π=!

( )( )( ) ( )( ) ( )( )

( )( )

2 2 2

2

20 9.81 0.180 12 0.240 3 0.080 cos30

6 0.080

πψ

π

− − ° =!

245.13 rad/s= 2340 rpmψ =! !




Choose principal centroidal axes xyz as shown.

The symmetry (spin) axis is the z axis.

Reduce the drag force to a force-couple system at the mass center.

( ) ( )sin cos cos sinD W D Wβ β β β= − − −F i k

sinG Dc β=M j

For the occurrence of steady precession, the precession axis must be parallel to the drag force. Thus, 3 .θ β= = °

(a) Using Equation (18.44), G GΣ =M H!

( )sin cos sinzDc I Iβ ω φ θ φ θ′= − ! !

Using and coszβ θ ω ψ φ θ= = + !! gives

( ) 2 cosI I I Dcψφ φ θ′− − =! !! (1)

Neglecting the quadratic term in ,φ!

DcI DcI

ψφ φψ

≈ ≈! !!!



Data: 18 kgm =

( )( )22 218 0.06 0.0648 kg mzI mk= = = ⋅

6000 rpm 628.32 rad/s, 0.175 mcψ = = =!

( )( )( )( )

110 0.1750.4728 rad/s

0.0648 628.32φ ≈ =! 4.51 rpmφ ≈! !

(b) ( )( )22 218 0.25 1.125 kg mxI mk′ = = = ⋅

21.0602 kg mI I′ − = ⋅

Substituting into Equation (1),

( )( ) ( ) ( )( )20.0648 628.32 1.0602 cos3 110 0.175φ φ− ° =! !

21.0587 40.7151 19.25 0φ φ− + =! !

19.2288 18.7501 0.4787 rad/s, or 37.98 rad/sφ = ± =!

4.57 rpm, 363 rpmφ =! !




Use principal centroidal axes x, y, z as shown.

Angular velocity: ( )sin cosφ θ ψ φ θ= − + +i k! !!ω

Angular momentum about the mass center G:


( )sin cosI Iφ θ ψ φ θ′= − + +i k! !!


sin cosφ θ φ θ= − +i k! !Ω

G G= ×H H! Ω

( ) 2sin sin cosI I Iφψ θ φ θ θ ′= − − j! !!

Acceleration of the mass center:

( ) 2sin sinl cβ θ φ= +a !

eff :Σ = ΣF F

: cos 0, cosmgT mg Tβ

β− = = (1)

: sinT maβ =

( ) 2tan sin sing l cβ β θ φ= + ! (2)

continued



: ( )sinG GM Tc Hβ θΣ = − = !

( ) ( ) 2sinsin sin cos

cosmgc

I I Iβ θ

ψφ θ φ θ θβ

−′= − −! !!

For alignment of axis BC with cord AB, .θ β=


0 cosI I Im m

ψ φ β′ −= − !! ( )cos II I

ψβφ

=′−!!

For a cone, 34

c h= 2 23 310 40

I mr md= =

2 2 2 2 23 3 3 320 5 80 80

I mr mh mc md mh′ = + − = −

2 23 380 80

I I md mh′ − = +

2

2 22I I d

I d h′ − =

+

( )2

2 22cos d

d hψβ

φ=

+

!! ( )

21

2 22cos d

d hψβ

φ− = +

!! "











G G= ×H H! Ω




eff :Σ = ΣF F


β− = = (1)

: sinT maβ =




cosmgc

I I Iβ θ

ψφ θ φ θ θβ

−′= − −! !! (3)



Data: 3 in. 0.25 ft, 12 in. 1 ft,r h= = = =

23 0.75 ft, 32.2 ft/s4

c h g= = =

( )22 23 3 0.25 0.01875 ft10 10

I rm

= = =

( ) ( ) ( )2 2 22 2 23 3 3 30.25 1 0.7520 5 20 5

I r h cm′

= + − = + −

20.046875 ft=

20.028125 ftI Im

′ − =

45 , 30 15β θ β θ= ° = ° − = °

(a) Dividing Eq. (3) by m and substituting numerical data,

( )( ) ( )( ) ( )( )232.2 0.75 sin150.01875 8 sin 30 0.028125 8 sin 30 cos30

cos 45ψ

°= ° − ° °

°!

8.8395 0.075 0.77942ψ= −!

128.252 rad/sψ =! 128.3 rad/sψ =! !

(b) Substituting data into Eq. (2),

( )( )232.2 tan 45 sin 45 0.75sin 30 8l° = ° + °

32.2 45.255 24, 0.1812 ftl l= + =

2.17 in.l = !




( )( )( )( ) 925800 years 25800 yr 365.24 day/yr 24 h/day 3600 s/h 814.16 10 s= = ×

129

27.7173 10 rad/s

814.16 10

πφ −= = ××

&

( )( )62

72.935 10 rad/s23.93 h 3600 s/h

πψ −= = ×&

mass density of Earth: 2 434410.6832 lb s /ft

32.2g

γρ = = = ⋅

radius of Earth: ( )( ) 63960 mi 5280 ft/mi 20.9088 10 ftR = = ×

mass of Earth: ( ) ( )33 6 21 24 420.9088 10 10.6832 409.05 10 lb s /ft

3 3Rπ ρ π= × = × ⋅

( )( )22 21 6 36 22 2409.05 10 20.9088 10 71.531 10 lb s ft

5 5I mR= = × × = × ⋅ ⋅

Using Equation (18.44),

( )cos sinzM I Iω φ θ φ θ′= − & &

( ) cos sinI I Iψ φ θ φ θ ′= + − & &&

sinIψφ θ= &&

( )( )( )36 6 1271.531 10 72.935 10 7.7173 10 sin 23.45− −= × × × °

2116.02 10 lb ft= × ⋅

2116.02 10 lb ftM = × ⋅ !











G G= ×H H! Ω




eff :Σ = ΣF F


β− = = (1)

: sinT maβ =




cosmgc

I I Iβ θ

ψφ θ φ θ θβ

−′= − −! !! (3)



For a solid sphere, 22 ,5

I I mc′= = 0I I′− =

Substituting into Eg. (3),

( ) 2cos sin sin cos 2 sincos 5

mgcmc

θ β θ βψφ θ

β−

= !!

( ) 22tan tan tan5

mgc mcβ θ ψφ θ− = !!

5 tantan5 2

gg c

βθψφ

=+ !!

1 5 tantan5 2

gg c

βθψφ

− = + !!

!











G G= ×H H! Ω




eff :Σ = ΣF F


β− = = (1)

: sinT maβ =




cosmgc

I I Iβ θ

ψφ θ φ θ θβ

−′= − −! !! (3)



For a solid sphere, 22 ,5

I I mc′= = 0I I′− =

Substituting into Eg. (3),

( ) 2cos sin sin cos 2 sincos 5

mgcmc

θ β θ βψφ θ

β−

= !!

( ) 22tan tan tan5

mgc mcβ θ ψφ θ− = !!

5 tantan5 2

gg c

βθψφ

=+ !!

1 5 tantan5 2

gg c

βθψφ

− = + !!

Data: 40 mm = 0.040 m, = 9.81 m/sc g=

5 rad/s, = 30φ β= °!

(a) 0:ψ =! ( )( )( )( )

1 5 9.81 tan 30tan

5 9.81θ − °

=

30.0θ = ° !

(b) 30 rad/s:ψ =! ( )( )( )( ) ( )( )( )( )

1 5 9.81 tan 30tan

5 9.81 2 0.040 30 5θ − °

= +

24.9θ = ° !

(c) 30 rad/s:ψ = −! ( )( )( )( ) ( )( )( )( )

1 5 9.81 tan 30tan

5 9.81 2 0.04 30 5θ − °

= + −

37.4θ = ° !




By Equation (18.48), sin cos

, 0, G Gx y z

H H

I I

θ θω ω ω= − = =′

By Equation (18.35) with 0,θ =&

sin , 0, cosx y zω φ θ ω ω ψ φ θ= − = = +& &&

Eliminating , and x zω ω

GH

Iφ =

′&

coscos GH

I

θψ φ θ+ =&&

( )coscos cos coscos GG G G H I IH H H

I I I II

θθ θ θψ φ θ′ −

= − = − =′ ′

&&




(a) Angular velocity of the body: 2sinφ θ ω= − +i k!ω

Its angular momentum about G: 2sinG I Iφ θ ω′= − +H i k!

Let the reference frame Gxyz be rotating with angular velocity ,Ω where


( ) ( ) ( )20 sin cos sinG G GGxyzI Iφ θ φ θ φ θ ω′= + × = + − + × − +H H H i k i k! ! !! ! Ω

22 sin sin cosI Iφω θ φ θ θ′= −! !

For no force, 0G =H!

Hence, ( )2 2cos sin 0 or cos 0I I I Iω φ θ φ θ ω φ θ′ ′− = − =! ! ! (1)

Solving for ,φ! 2

cosI

Iωφ

θ=

′!

(b) Comparing Equation (1) with Equation (18.44) yields 0 0,Σ =M which is the condition for no force.




Angular velocity of the body: ( )sin cosφ θ φ θ ψ= − + +i k& & &ωωωω

Let the reference frame Gxyz be rotating with angular velocity ,ΩΩΩΩ where

sin cosφ θ φ θ= − +i j& &ΩΩΩΩ

Angular acceleration of the body: Gxyz= + ×&α ω Ω ωα ω Ω ωα ω Ω ωα ω Ω ω

0 sinφψ θ= + j& &αααα

The rate of change of angular velocity as observed from the body is .−αααα

Assume that −αααα may be represented as the angular velocity vector rotating with angular velocity .l m n+ +i j k

( ) ( ) ( )x z z x z xl m n m n l mω ω ω ω ω ω− = + + × + = + − −i j k i k i j kαααα

Matching components: : 0 0zm mω= =i

: sin x zn lφψ θ ω ω− = −j & & (1)

: 0 0xm mω= − =k

From Equation (1), ( )sin sin cosn lφψ θ φ θ φ θ ψ− = − − +& & && &

from which 0 and .l n ψ= = &

Using Equation (18.44) with 0 0Σ =M yields cos 0zI Iω φ θ′− =&

or cos zI

I

ωφ θ =′

&

But, cos zz

I

I

ωω ψ φ θ ψ= + = +′

&& &

Solving for ,ψ& zI I

Iψ ω′ −=&

Using 2 and z nω ω ψ= = & yields 2.I I

nI

ω′ −=′

!




For no force, 0G =M

(a) Using Equation (18.44), ( )0 cos sinzI Iω φ θ φ θ′= − ! !

cos 0zI Iω φ θ′− =!

Using coszω ψ φ θ= + !! yields ( ) cos 0I I Iψ φ θ′− − =!!

( )cosI

I Iψφ

θ=

′ −!!

( ) secII I

ψφ θ=′−!!

For a flat disk, 12

I I′ =

For any other shape, 12

I I′ >

Hence, 1 and 22

II I II I

′− < >′−

Also sec 1.θ > 2φ ψ>! !

(b) 1tan tan or tan tan tan2

I II I

γ θ θ γ γ′= = >

′

Angle of surface of space cone is .γ θ−

Angle of axis is .θ

( ) ( )( )

tan tan1 1tan tan2 2 1 tan tan

θ γ θθ θ γ θ

θ γ θ+ −

> + − = + −

( ) ( )1 11 tan tan tan tan tan2 2

θ γ θ θ θ γ θ + − > + −



( ) ( )2 1 1tan tan tan tan tan2 2

θ θ γ θ θ γ θ+ − > + −

( ) ( )21 1tan tan tan tan2 2

θ γ θ θ γ θ> − − −

( )1 1tan tan2 2

θ γ θ> −

( )tan tanθ γ θ> −

( )θ γ θ> −

The axis lies outside the space cone.




Angular velocity of the body: ( )sin cosφ θ φ θ ψ= − + +i k& & &ωωωω

Let the reference frame Gxyz be rotating with angular velocity ΩΩΩΩ, where

sin cosφ θ φ θ= − +i j& &ΩΩΩΩ

Angular acceleration of the body: ω= + ×&Gxyzα Ω ωα Ω ωα Ω ωα Ω ω

The rate of change of angular velocity as observed from the body is −αααα

Assume that α− may be represented as the angular velocity vector rotating with angular velocity

.+ +i j kl m n

( ) ( ) ( )ω ω ω ω ω ω− = + + × + = − + − −i j k i k i j kx z z x z xl m n m n l mαααα

Matching components: : 0 0zm mω= =i

: sin x zn lφψ θ ω ω− = −j & & (1)

: 0 0xm mω= − =k

From Equation (1), ( )sin sin cosn lφψ θ φ θ φ θ ψ− = − − +& & &&

from which 0 and .ψ= = &l n

Using Equation (18.44) with 0 0 yields cos 0zI Iω φ θ′Σ = − =M &

or cos zI

I

ωφ θ =′

&

But, cos zz

I

I

ωω ψ φ θ ψ= + = +′

&& &

Solving for ,ψ& ψ ω′ −=& zI I

I



Using 2 2 and yields ω ω ψ ω′ −= = =′

&zI I

n nI

Data for Earth: 0.9967 , 0.0033 ,′ ′= − = −I I I I I

2 20.0033

0.0033110.9967

ω ω= − = −n

( )( )2 2

2 1 2 2period 302.03 302.03 24 h 7248 h

0.003311

π π πω ω

= = = = =n

period 302 days= !




tan ωγω

= − x

z

15γ = °

For steady precession with no force,

tan tan 3tan15θ γ′= = °I

I

38.794θ = °

(a) 38.794 15β θ γ= − = − °

23.8β = °!

(b) sin sinxω φ θ ω γ= − = −!

( )( )

200 rpm sin15sinsin sin 38.794

ω γφθ

°= =

°!

82.621 rpm=

precession: 82.6 rpmφ =! !

cos coszω ψ φ θ ω γ= + =!!

cos cos 200cos15 82.621cos38.794ψ ω γ φ θ= − = ° − °!!

spin: 128.8 rpmψ =! !




Moments of inertia: 2 21 1, 2 4

I mr I mr′= =

Euler angle θ for steady precession: 15θ = °

For axisymmetric body under no force, Equation (18.49) gives for the body cone angle:

tan tan 2 tan15 28.187γ θ γ= = ° = °′

II

(a) Angular velocity ( )sin cosω γ γ= − +i kω

Its projection onto the vertical direction is

( ) ( )cos sin cos sin cosω β ω γ γ θ θ= ⋅ = − + ⋅ − +k i k i kω

( ) ( )sin sin cos cos cosω γ θ γ θ ω γ θ= + = −

( )cos cos 28.187 15 13.187β γ θ β γ θ= − = − = ° − ° = °

Angle between ω and vertical direction GD: 13.19β = ° !

(b) sintancos

x

z

ω φ θγω φ θ ψ

= − =+

!! !

from which ( )sin sin 28.187 2.0705

sin sin13.187ψ γ ψφ ψ

θ γ°= = = −

− − °! !! !

With ( )( )600 rpm, 2.0705 600ψ φ= = −!! 1242 rpmφ =! (retrograde)!




Mass of satellite: 2800 24.845 lb s /ft32.2

Wmg

= = = ⋅

Principal moments of inertia: ( )2

2 228.824.845 143.106 lb s ft12x xI mk = = = ⋅ ⋅

( )2

2 232.424.845 181.118 lb s ft12y yI mk = = = ⋅ ⋅

2143.106 lb s ftz xI I= = ⋅ ⋅

Mass of meteorite: ( ) ( )26 0.011649 lb s /ft

16 32.2m = = ⋅′

Initial momentum of meteorite: ( ) ( )0 0.011649 1600 1300 4000m = − + +′v i j k

( ) ( ) ( )18.633 lb s 15.140 lb s 46.584 lb s= − ⋅ + ⋅ + ⋅i j k

Assume that the position of the mass center of the satellite plus the meteorite is essentially that of the satellite alone. Position of point B relative to the mass center.

( ) ( )/42 20 3.5 ft 1.66667 ft12 12B G = − = −

r i j i j

Angular velocity of satellite before impact. ( )0 1.5 rad/s= jω ( ) ( ) ( )0 0 00, 1.5 rad/sx z yω ω ω= = =

Angular momentum of satellite-meteorite system before impact. ( ) ( )0 / 00G y B GI mω= + × ′H j r v

( ) ( )

( ) ( ) ( )

181.118 1.5 3.5 1.66667 018.633 15.140 46.584

77.64 lb s ft 108.637 lb s ft 21.935 lb s ft

= + −−

= − ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅

i j kj

i j k



Principle of impulse and momentum for satellite-meteorite system. Moments about G:

( ) ( )1 0G G G= =H H H

Angular velocity immediately after impact. x y zω ω ω= + +i j kω

Neglect the mass of the meteorite.


( ) 77.64 0.54253 rad/s

143.106G x

xx

HI

ω −= = = −

( ) 108.637 0.59981 rad/s

181.118G y

yy

H

Iω = = =

( ) 21.934 0.15327 rad/s143.106

G zz

z

HI

ω = = =

( ) ( ) ( )0.54253 rad/s 0.59981 rad/s 0.15327 rad/s= − + +i j kω

( ) ( ) ( )2 2 20.54253 0.59981 0.15327 0.82317 rad/sω = + + =

( ) ( ) ( )2 2 277.64 108.637 21.935 135.319 lb s ftGH = + + = ⋅ ⋅

Motion after impact. Since the moments of inertia xI and zI are equal, the body moves as an axisymmetrical body with the y axis as the symmetry axis.

Moment of inertia about the symmetry axis: 2181.118 lb s ftyI I= = ⋅ ⋅

Moment of inertia about a transverse axis through G:

2143.106 lb s ftx zI I I= = = ⋅ ⋅′

The motion is a steady precession φ! about the precession axis together with a steady spin ψ! about the spin or symmetry axis. Since ,I I> ′ the precession is retrograde.

Precession axis. The precession axis is directed along the angular momentum vector HG, which remains fixed. Immediately after impact its direction cosines relative to the body axes x, y, z are:

( ) 77.64cos 0.57376

135.319G x

xG

HH

θ −= = = − 125.0xθ = ° !

( ) 108.637cos 0.80282

135.319G y

yG

H

Hθ = = = 36.6yθ = ° !

( ) 21.935cos 0.16210

135.319G z

zG

HH

θ = = = 80.7zθ = ° !



The angle θ between the spin axis (y-axis) and the precession axis remains constant.

36.600yθ θ= = °

The angle γ between the angular velocity vector and the spin axis (y-axis) is

0.59981cos0.82317

yωγ

ω= = 43.226γ = °

The angle γ could also have been calculated from

181.118tan tan tan 36.600143.106

II

γ θ= = °′

The angle between the precession axis and the angular velocity vector is

6.626γ θ− = °

Rates of precession and spin.

Set up the triangle of vector addition for the components of angular velocity. Apply the law of sines.

( )sin sin sinω φ ψ

θ γ θ γ= =

−

! !

( )sin 0.82317 sin 6.626sin sin 36.600

ω γ θφ

θ− °= =

°!

( )0.1593 rad/s retrogradeφ =! !

sin 0.82317 sin 43.226sin sin 36.600

ω γψθ

°= =°

!

0.946 rad/sψ =! !




Mass of satellite: 2800 24.845 lb s /ft32.2

wmg

= = = ⋅

Principal moments of inertia: ( )2

2 228.824.845 143.106 lb s ft12x xI mk = = ⋅ ⋅

( )2

2 232.424.845 181.118 lb s ft12y yI mk = = = ⋅ ⋅

2143.106 lb s ftz xI I= = ⋅ ⋅

Mass of meteorite: ( ) ( )26 0.011649 lb s /ft

16 32.2m = = ⋅′

Initial momentum of meteorite: ( ) ( )0 0.011649 1600 1300 4000m = − + +′v i j k

( ) ( ) ( )18.633 lb s 15.140 lb s 46.584 lb s= − ⋅ + ⋅ + ⋅i j k

Assume that the position of the mass center of the satellite plus the meteorite is essentially that of the satellite alone.

Position of point A relative to the mass center.

( )/42 3.5 ft12A G = =r i i

Angular velocity of satellite before impact. ( )0 1.5 rad/s= jω ( ) ( ) ( )0 0 00, 1.5 rad/sx z yω ω ω= = =

Angular momentum of satellite-meteorite system before impact. ( ) ( )0 / 00G y A GI mω= + × ′H j r v

( ) ( )

( ) ( )

181.118 1.5 3.5 0 018.633 15.140 46.584

108.637 lb s ft 52.99 lb s ft

= +−

= ⋅ ⋅ + ⋅ ⋅

i j kj

j k



Principle of impulse and momentum for satellite-meteorite system. Moments about G:

( ) ( )1 0G G G= =H H H

Angular velocity immediately after impact. x y zω ω ω= + +ω i j k

Neglect the mass of the meteorite.


( )

0G xx

x

HI

ω = =

( ) 108.637 0.59981 rad/s

181.118G y

yy

H

Iω = = =

( ) 52.99 0.37028 rad/s

143.106G z

zz

HI

ω = = =

( ) ( )0.59981 rad/s 0.37028 rad/s= +j kω

( ) ( )2 20.59981 0.37028 0.70490 rad/sω = + =

( ) ( )2 2108.637 52.99 120.872 lb s ftGH = + = ⋅ ⋅

Motion after impact. Since the moments of inertia xI and zI are equal, the body moves as an axisymmetrical body with the y axis as the symmetry axis.

Moment of inertia about the symmetry axis: 2181.118 lb s ftyI I= = ⋅ ⋅

Moment of inertia about a transverse axis through G:

2143.106 lb s ftx zI I I= = = ⋅ ⋅′

The motion is a steady precession φ! about the precession axis together with a steady spin ψ! about the spin or symmetry axis. Since ,I I> ′ the precession is retrograde.

Precession axis. The precession axis is directed along the angular momentum vector HG, which remains fixed. Immediately after impact its direction cosines relative to the body axes x, y, z are:

( )

cos 0G xx

G

HH

θ = = 90.0xθ = ° !

( ) 108.637cos 0.89878

120.872G y

yG

H

Hθ = = = 26.0yθ = ° !

( ) 52.99cos 0.43840

120.872G z

zG

HH

θ = = = 64.0zθ = ° !

continued



The angle θ between the spin axis (y axis) and the precession axis remains constant.

26.002yθ θ= = °

The angle γ between the angular velocity vector and the spin axis (y axis) is

0.59981cos0.70490

yωγ

ω= = 31.689γ = °

The angle γ could also have been calculated from

181.118tan tan tan 26.002143.106

II

γ θ= = °′

The angle between the precession axis and the angular velocity vector is

5.687γ θ− = °

Rates of precession and spin.

Set up the triangle of vector addition for the components of angular velocity. Apply the law of sines.


θ γ θ γ= =

−

! !

( )sin 0.70490 sin 5.687sin sin 26.002

ω γ θφ

θ− °= =

°!

( )0.1593 rad/s retrogradeφ =! !

sin 0.70490 sin 31.689sin sin 26.002

ω γψθ

°= =°

!

0.844 rad/sψ =! !




For entire station: ( )2 2 21 13 ,

12 2I m a L I ma′ = + =

( )( )( )( ) ( )

22

2 2 2 2

6 360.0582524

3 3 3 30= = =

′ + +I a

I a L

From Equation (18.49), tan tan 0.0582524 tan 40 2.7984γ θ γ= = ° = °′

I

I

( )sin 2 rev/h sin 40 1.28557 rev/hxω φ θ= − = − ° = −&

1.28557

26.3005 rev/htan tan

ωωγ γ

= − = =xz

cos 26.3005 2cos 40zψ ω φ θ= − = − °&&

24.8 rev/hψ =& !




(a) Angular velocity after impact

From Sample Problem 18.6:

2 21 5,

2 4′= = = = =z x yI I ma I I I ma

Conservation of angular momentum: Angular momentum after impact:

( )0 0 0 0 0 0G C m I a a m v Iω ω= × + = − − × +H r v k j k i k

( )0 0 0 0 0am v I am vω= − + +j k

Data: 0800 mm 0.8 m, 60 rpm 2 rad/s,ω π= = = =a

0 02000 m/s, 1000

= = mv m

( )0ω = =G x

xx

H

I

( )0 0 0 0

2

45 54

ω = = − = −G y

yy

H am v m v

I m ama

4 1 2000

2 rad/s5 1000 0.8 = − = −

( ) 0 0 0 00 0

22

12

ω ω ω= = + = +G zz

z

H am v m v

I m ama

1 2000

2 2 11.2832 rad/s1000 0.8

π = + =



( ) ( )2.00 rad/s 11.28 rad/s= − +j kωωωω

( ) ( )2 22 11.2832 11.4591 rad/s 109.426 rpmω = + = =

109.4 rpω =

2tan 10.0515

11.2832

ωγ γ

ω= = = °y

z

90 , xγ γ= °

(b) Precession axis

( )5 2tan tan 2 23.900

4 11.2832θ γ θ′ = = = °

I

I

90 , xθ θ= °

(c) Rates of precession and spin

13.8484θ γ− = °

Law of sines

( )sin sin sin

φ ψ ωγ θ γ θ

= =−

& &

sin 109.4sin10.05

sin sin 23.9

ω γφθ

°= =°

&

precession: 47.1 rpmφ =& !

( )sin 109.4sin13.85

sin sin 23.9

ω θ γψ

θ− °= =

°&

spin: 64.6 rpmψ =& !




Immediately after the rod breaks, the angular velocity remains equal to that calculated in Sample Prob.18.6.

6.594 rad/sω = tan 0.3183γ = 17.7γ = °

For disk A alone, 21 ,2

I ma= 214

I ma′ =

From Eq. (18.49) of textbook,

1tan tan tan 0.15915 9.042

II

θ γ γ θ′= = = = °

The new precession axis makes angleθ with the z-axis. Its direction angles are:

90 99.04xθ θ= + ° = ° !

90yθ = ° !

9.04zθ θ= = ° !

Since ,γ θ> the precession is retrograde.

The angular velocity vector is the sum of an angular velocityφk! along the precession axis plusψk! along the spin axis (z-axis). This vector addition is performed using a triangle construction and applying the law of sines.


θ γ γ θ= =

−

! !

sin 6.594sin17.7 12.76 rad/ssin sin 9.04

ω γφθ

°= = =°

!

( )sin 6.594sin8.66 6.32 rad/ssin sin 9.04

ω γ θψ

θ− °= = =

°!

( )121.8 rpm retrogradeφ =! !

60.3 rpmψ =! !




Place the origin at the center of mass and let Oxyz be a principal axis frame of reference with the y axis directed along the moving axle AB. Let the Z axis lie along the fixed axle. Useful unit vectors are i, j and k along the x, y, z axes and K along the Z axis.

sin cosθ θ= − +K i k

Angular velocity: φ θ= +ω k j! !

sin cosω φ θ θ φ θ= − + +i j k! ! !

Moments of inertia: 1 2, x y z yI e I I e I= =


O x x y y z zI I Iω ω ω= + +H i j k

( )1 2sin cosyI e eφ θ θ φ θ= − + +i j k! ! !

The moment about the fixed Z axis is zero, hence, constantO ⋅ =H K

( )2 21 2 1sin cosO y yI e e I Cθ θ φ⋅ = + =H K !

( )2 211 1 0 2 0 02 2

1 2 sin cos

sin cosC C e e

e eφ θ θ φ

θ θ= = +

+! !

Twice the kinetic energy: 2 2 22 constantx x y y z zT I I Iω ω ω= + + =

( ) ( )2 2 2 2 21 2 1 22 sin cosy y yT I e e I C I Cθ θ φ θ φ θ = + + = + =

! ! ! !

2 22 1 2 0 1 0 C C C Cθ φ θ φ= − = +! ! ! !

Data: 2 2 21 2 1 21, 2, sin cos 1 cose e e eθ θ θ= = + = +

0 0 090 , 0, 16 rad/sθ θ φ= ° = =! !

( )( )21

161 cos 90 16 16 rad/s 1 cos

C φθ

= + ° = =+

!



(a) min16 8

1 cos0φ = =

+! min 8.00 rad/sφ =! !

( )( ) ( )22 0 16 16 256 rad/sC = + =

22 1C Cθ φ= −! !

(b) ( ) ( )( ) ( )22

2 1 minmax256 16 8 128 rad/sC Cθ φ= − = − =! ! max 11.31 rad/sθ =! !




Place the origin at the center of mass and let Oxyz be a principal axis frame of reference with the y axis directed along the moving axle AB. Let the Z axis lie along the fixed axle. Useful unit vectors are i, j, and k along the x, y, z axes and K along the Z axis.


Angular velocity: φ θ= +K j& &ωωωω

sin cosω φ θ θ φ θ= − + +i j k& & &



0 x x y y z zI I Iω ω ω= + +H i j k

( )1 2sin cosyI e eφ θ θ φ θ= − + +i j k& & &

The moment about the fixed Z axis is zero, hence constant.O ⋅ =H K

( )2 21 2 1sin cosO y yI e e I Cθ θ φ⋅ = + =H K &

( )2 211 1 0 2 0 02 2

1 2

sin cossin cos

CC e e

e eφ θ θ φ

θ θ= = +

+& &


( )2 2 2 21 22 sin cosyT I e eθ θ φ θ = + +

& &

( )2 21 2y yI C I Cφ θ= + =& &

continued



2 22 1 2 0 1 0 C C C Cθ φ θ φ= − = +& & & &

Data: 2 2 21 2 1 21, 2 sin cos 1 cosθ θ θ= = + = +e e e e

0 0 045 , 0, 8 rad/sθ θ φ= ° = =& &

( ) ( )21 0 2

1 121 cos 45 1 8 12 rad/s

2 1 cosC φ φ

θ = + ° = + = = +

& &

( )( ) ( )22 0 12 96 96 rad/s= + =C

(a) 22

14496 12 96 0

1 cosθ φ

θ= − = − ≥

+& &

2cos 0.5 cos 0.70711θ θ> >

45 45θ− ° < < °!

(b) min12

6 rad/s1 cos 0

φ = =+

&

min 6.00 rad/sφ =& !

(c) ( ) ( )( ) ( )2 2max 2 1 min 96 12 6 24 rad/sC Cθ φ= − = − =& &

max 4.90 rad/sθ =& !




Place the origin at the center of mass and let Oxyz be a principal axis frame of reference with the y axis directed along the moving axle AB. Let the Z axis lie along the fixed axle. Useful unit vectors are i, j, and k along the x, y, z axes and K along the Z axis.


Angular velocity: φ θ= +K j! !ω

sin cosω φ θ θ φ θ= − + +i j k! ! !



0 x x y y z zI I Iω ω ω= + +H i j k

( )1 2sin cosyI e eφ θ θ φ θ= − + +i j k! ! !

The moment about the fixed Z axis is zero, hence constant.O ⋅ =H K

( )2 21 2 1sin cosO y yI e e I Cθ θ φ⋅ = + =H K !

( )2 211 1 0 2 0 02 2

1 2 sin cos

sin cosC C e e

e eφ θ θ φ

θ θ= = +

+! !


( )2 2 2 21 22 sin cosyT I e eθ θ φ θ = + + +

! !

( )2 21 2y yI C I Cφ θ= + =! !



2 2

2 1 2 0 1 0 C C C Cθ φ θ φ= − = +! ! ! !

Data: 2 2 21 2 1 21, 2 sin cos 1 cosθ θ θ= = + = +e e e e

0 0 090 , 0, 8 rad/sθ θ φ= ° = =! !

( )( )21 2

81 cos 90 8 8 rad/s 1 cos

C φθ

= + ° = =+

!

(a) min 28 4

1 cosφ

θ= =

+

min 4.00 rad/sφ =! !

( )( ) 22 0 8 8 64 rad/s= + =C

22 1C Cθ φ= −! !

(b) ( ) ( )( ) ( )2 2max 2 1 min 64 8 4 32 rad/sC Cθ φ= − = − =! !

max 5.66 rad/sθ =! !




Let the Z axis be vertical. For principal axes xyz with origin at A, the principal moments of inertia are

( )22 21 1724 4x yI I I m a a ma ′ = = = + =

212zI I ma= =

Angular velocity components:

sin , , cosx y zω φ θ ω θ ω ψ φ θ= = − = +! ! !!



sin zI I Iφ θ θ ω′ ′= − +i j k! !

Kinetic energy: 2 2 21 1 12 2 2x x y y z zT I I Iω ω ω= + +

( )2 2 2 21 1sin2 2 zT I Iφ θ θ ω′= + +! !

Potential energy: 2 cosV mga θ= − Conservation of angular momentum about fixed Z axis:

( )sin cosA A θ θ⋅ = ⋅ +H K H i k

2sin coszI Iφ θ ω θ′= +!

2 2 217 1sin cos4 2 zma maφ θ ω θ α= + =! (1)

where α is a constant.

continued



Conservation of energy: ,T V E+ = where E is a constant.

( )2 2 2 2 2 217 1sin 2 cos8 4 zma ma mga Eφ θ θ ω θ+ + − =! ! (2)

Constraint of clevis: 0 coszψ ω φ θ= = !!

(a) From (1), 2 2 2 2 2 2 2 20 0

17 1 17 1sin cos sin 90 cos 904 2 4 2m m m mma ma ma maφ θ φ θ φ φ+ = ° + °! ! ! !

2 2 017 1 17 17 1 17sin cos4 2 4 4 2 8m m

m

φθ θφ

+ = = =

!!

( )2 217 1 17sin 1 sin4 2 8m mθ θ+ − =

2 13sin , sin 0.65828 41.16930m m mθ θ θ= = = °

41.2mθ = °!

(b) At the minimum value of ,θ 0θ =!

From (2), ( )2 2 2 2 2 217 1sin 0 cos 2 cos8 4m m m m mma ma mgaφ θ φ θ θ+ + −! !

( )2 2 2 2 2 20 0 0

17 1sin 0 cos 90 2 cos908 4

ma ma mgaφ θ φ= + + ° − °! !

( ) ( )2 22 2 2 20

17 1 172 sin 2 cos 2 cos8 4 8m m mma mgaθ θ φ θ + − =

!

2 202.1250 1.5055ma mgaφ =!

20

32.20.70849 0.70849 30.4180.75

ga

φ = = =! 0 5.52 rad/sφ =! !



Chapter 18, Solution 133. Let the Z axis be vertical.

For principal axes xyz with origin at A, the principal moments of inertia are

( )22 21 1724 4x yI I I m a a ma ′ = = = + =

212zI I ma= =







( )2 2 2 21 1sin2 2 zT I Iφ θ θ ω′= + +! !

Potential energy: 2 cosV mga θ= −

Conservation of angular momentum about fixed Z axis:









Constraint of clevis: 0 coszψ ω φ θ= = !!

(a) At 30 and at 90 , 0.m oθ θ θ θ θ= = ° = = ° =!

From (1), 2 2 2 2 2 20 0

17 1 17sin cos sin 04 2 4m m m mma ma maφ θ φ θ φ θ+ = +! ! !

22

2 20 0

17 1 1 3 17 68 4 2 2 2 4 23m mma maφ φ φ φ

+ = =

! ! ! !

From (2), 2 2

2 2 2 20 0

17 68 1 68sin 30 cos 30 2 cos308 23 4 23

ma ma mgaφ φ ° + ° − ° ! !

2 2 2 2 2 20 0

17 1sin 90 cos 90 2 cos908 4

ma ma mgaφ φ= ° + ° − °! !

22 2

017 3 68 17 2 cos3032 16 23 8

ma mgaφ + − = °

!

20

32.20.41660 0.41660 17.8860.75

ga

φ = = =!

0 4.2292 rad/sφ =! 0 4.23 rad/sφ =! !

(b) ( )68 4.229223mφ =! 12.50 rad/smϕ =! !



Chapter 18, Solution 134. Let the Z axis be vertical with positive direction up.

For principal axes with the origin at A, the principal moments of inertia are

225zI ma=

( )22 22 2225 5x yI I ma m a ma= = + =

Unit vector along the Z axis: cos sinβ β= +k i k

Angular velocity: ω φ β= +k j!!

cos sinφ β β φ β= + +i j k!! ! Angular momentum about fixed point A:


2 2 222 22 2cos sin5 5 5

ma ma maφ β β βφ= + +i j k!! !

Kinetic energy:

2 2 21 1 12 2 2x x y y z zT I I Iω ω ω= + +

( )2 2 2 2 2 21 111cos sin5 5

ma maβ β φ β= + + !!

Potential energy: ( )2 sinV mg a β= −

Conservation of angular momentum about the fixed Z axis: ( ) ( )0A A⋅ = ⋅H K H K

2 2 2 20

22 2 22cos sin5 5 5

ma maφ β β φ + = ! !

02 211

11cos sinφφ

β β=

+

!!



When max 1 02230 ,17

β β φ φ φ= = ° = =! ! ! (1)

Conservation of energy: 1 1 0 0T V T V+ = +

State 0: 0, 0, 0φ φ β β= = =!! !

State 1: 1, 30 , 0φ φ β β= = ° =!! !

( )2 2 2 2 2 21 0

1 1111cos 30 sin 30 2 sin 30 05 5

ma mga maφ φ° + ° − ° = +! !

2

2 2 20 0

17 22 1110 17 5

ma mga maφ φ − = ! !

2 20

1117

ma mgaφ =!

(a) 01711

ga

φ =! !

(b) From (1), 122 1717 11

ga

φ =! 14417

ga

φ =! !




Let the Z axis be vertical.


( )22 21 1724 4x yI I I m a a ma ′ = = = + =

212zI I ma= =







( )2 2 2 21 1sin2 2 zT I Iφ θ θ ω′= + +! !











Let the reference frame Axyz be turning with angular velocity

sin cosφ θ θ φ θΩ = − +i j k! ! !

Then ( )A A A AAxyzΣ = = + ×M H H H! ! Ω

( )sin sin zA

d dmga I I Idt dt

ωθ φ θ θ′ ′= − + + ×j i j k Η! !! Ω

z-component: 0 zΑ

dIdtω= + ⋅ ×k ΗΩ

0 0 10 sin cos 0

sin

z z z

z

d d dI I Idt dt dt

I I

ω ω ωφ θ θ φ θφ θ θ ω

= + − = + =′ ′−

! ! !! !

Integrating, zIω β= (3)

where β is a constant.

Data: 0 0 090 , 0, 0,θ φ θ= ° = =! !

From (3), 0zω ψ= !

From (1), 2 2 2 2 2 20

17 1 17 1sin cos sin 90 cos904 2 4 2z zma ma ma maφ θ ω θ φ ω+ = ° + °! !

22 cos

17 sinzω θφ

θ= −! (4)

From (2), ( )2 2 2 2 217 1sin 2 cos8 4 zma m mgaφ θ θ ω θ+ + −! !

( )2 2 2 2 20 0 0

17 1sin 2 cos908 4 zma m gmaφ θ θ ω= + + − °! !

or ( )2 2 2 217 sin 2 cos8

ma mgaφ θ θ θ+ =! !

(a) At ,mθ θ= 0θ =!

2 2 217 sin 2 cos8 m m mma mgaφ θ θ=!

22 2

217 2 cos sin 2 cos8 17 sin

z mm m

mma mgaω θ θ θ

θ

− =



2 2 2

21 cos 2 cos34 sin

z mm

m

ma mgaω θ θθ

=

( )( )( )( )

22 2 2 2 0.75 50sin 0.85632cos 68 68 68 32.2

m z z

m

ma amga g

θ ω ωθ

= = = =

21 cos 0.85632cosm mθ θ− =

2cos 0.85632cos 1 0m mθ θ+ − =

[ ]1cos 0.85632 2.17561 0.65965, 1.5142mθ = − ± = −

( )1cos 0.65965 48.727mθ −= = ° 48.7mθ = °!

(b) From (4), 22 50cos 48.727 6.8694

17 sin 48.727mφ °= − = −! 6.87 rad/smφ = −! !

( )( )cos 50 6.8694 0.65965m z m mψ ω φ θ= − = − −!!

54.5 rad/smψ =! !




Let the Z axis be vertical.


( )22 21 172

4 4x yI I I m a a ma ′ = = = + =

21

2zI I ma= =


sin , , cosx y zω φ θ ω θ ω ψ φ θ= = − = +& & &&



sin zI I Iφ θ θ ω′ ′= − +i j k& &

Kinetic energy: 2 2 21 1 1

2 2 2x x y y z zT I I Iω ω ω= + +

( )2 2 2 21 1sin

2 2 zT I Iφ θ θ ω′= + +& &




2sin coszI Iφ θ ω θ′= +&

2 2 217 1sin cos

4 2 zma maφ θ ω θ α= + =& (1)



( )2 2 2 2 2 217 1sin 2 cos

8 2 zma ma mga Eφ θ θ ω θ+ + − =& & (2)



Let the reference frame Axyz be turning with angular velocity

sin cosφ θ θ φ θΩ = − +i j k& & &

Then ( )A A A AAxyzΣ = = + ×M H H H& & ΩΩΩΩ

( )sin sin zA

d dmga I I I

dt dt

ωθ φ θ θ′ ′= − + + ×j i j k Η& && ΩΩΩΩ

z-component: 0 zΑ

dI

dt

ω= + ⋅ ×k ΗΩΩΩΩ

0 0 1

0 sin cos 0

sin

z z z

z

d d dI I I

dt dt dtI I

ω ω ωφ θ θ φ θφ θ θ ω

= + − = + =′ ′−

& & &

& &

Integrating, zIω β= (3)


Data: 0 0 090 , 0, 0,θ φ θ= ° = =& &

From (3), 0zω ψ= &

From (1), 2 2 2 2 2 20

17 1 17 1sin cos sin 90 cos90

4 2 4 2z zma ma ma maφ θ ω θ φ ω+ = ° + °& &

2

2 cos

17 sinzω θφ

θ= −& (4)

From (2), ( )2 2 2 2 217 1sin 2 cos

8 4 zma m mgaφ θ θ ω θ+ + −& &

( )2 2 2 2 20 0 0

17 1sin 2 cos90

8 4 zma m mgaφ θ θ ω= + + − °& &

or ( )2 2 2 217sin 2 cos

8ma mgaφ θ θ θ+ =& &

(a) At ,mθ θ= 0θ =&

2 2 217sin 2 cos

8 m m mma mgaφ θ θ=&

2

2 22

17 2 cossin 2 cos

8 17 sinz m

m mm

ma mgaω θ θ θ

θ

− =

2 2 2

2

1 cos2 cos

34 sinz m

mm

mamga

ω θ θθ

=



( )2 2

2 sin 32.2 sin 3068 68 842.78

cos 0.75 cos30m

zm

g

a

θωθ

° = = = °

29.031 rad/szω =

From (3), 0 zψ ω=& 0 29.0 rad/sψ =& !

(b) From (4), ( )

2

29.031 cos302

17 sin 30mφ

°= −

°& 11.83 rad/smφ = −& !

( )cos 29.031 11.83 cos30m z m mψ ω φ θ= − = − − °&& 39.3 rad/smψ =& !




Use a rotating frame of reference with the y axis pointing into the paper.

Angular velocity of the frame:

sin cosφ θ θ φ θ+ +i j k& & &Ω = −Ω = −Ω = −Ω = −

Angular velocity of the top:

( )sin cosφ θ θ ψ φ θ= − + + +ω i j k& & &&

Its angular momentum about O.

O x x y y z zI I Iω ω ω= + +H i j k

( )sin cosI I Iφ θ θ ψ φ θ′ ′= − + + +i j k& & &&

where and .x y zI I I I I′= = =

The moment OM about O is due to the weight mg.

0 sinM mgc θ= j

(a) Since the fixed Z axis refers to a Newtonian frame of reference and

( ) 0,O ZM = it follows that ( )O Z

H is constant. Thus,

( ) ( )sin cosO O OZH θ θ= ⋅ = ⋅ − +H K H i k

( )2sin cos cosI Iφ θ ψ φ θ θ α′= + + =& && (1) !


continued



( )O O O OOxyz= = + ×M H H H& & ΩΩΩΩ

( ) ( )sin sin cosd d

mgc I I Idt dt

θ φ θ θ ψ φ θ′ ′= − + + +j i j k& && &&

( )cos cosI Iψ φ θ θ φθ θ ′+ + − i& & & &&

( ) 2cos sin sin cosI Iψ φ θ φ θ φ θ θ ′+ + − j& & &&

z-components: ( )0 cosd

dtψ φ θ= + &&

Integrating, ( )cosI ψ φ θ β+ =&& (2) !


(b) Since cos ,zω ψ φ θ= + && constantz I

βω = = (3) !

From (1) and (2), 2sin cosI φ θ β θ α′ + =&

2

cos

sinI

α β θφθ

−=′

& (4) !

which is a function of .θ




(a) Angular velocity of the top:

( )sin cosφ θ θ ψ φ θ= − + + +ω i j k& & &&

Kinetic energy:

2 2 21 1 1

2 2 2x x y y z zT I I Iω ω ω= + +

( )2 2 21 1 1sin

2 2 2 zI I Iφ θ θ ω′ ′= + +& &

Potential energy: cosV mgc θ=

Principle of conservation of energy: T V E+ =

( )2 2 21 1 1sin cos

2 2 2 zI I I mgc Eφ θ θ ω θ′ ′+ + + =& & !

(b) Solving for 2,θ&

( ) ( )22 212 2 cos sinzE I mgc

Iθ ω θ φ θ= − − −

′& & (A)

Equation (2) of Problem 18.137, with coszω ψ φ θ= + && gives

2

2zI

I

βω = (B)

Equation (1) of Problem 18.137 gives

2sin cosI φ θ β θ α′ + =& (C)

2

cos

sinI

α β θφθ

−=′

& (D)



Substituting Equations (D) and (B) into Equation (A),

( )2 ,fθ θ=&

where ( )221 cos

2 2 cossin

f E mgcI I I

β α β θθ θθ

− = − − − ′ ′ (1) !

(c) Maximum and minimum values of θ occur when ( ) 0.f θ = Setting cos xθ = and 2 2sin 1 xθ = − in

Equation (1), and letting ( ) 0f θ = gives

( )( ) ( )

22

2 2

12 2 0

1

xE mgcx

I I I x

α ββ −− + − = ′ ′ −

Multiplying by ( )21I x′ − gives the cubic equation ( ) 0:F x =

( ) ( )2

22 12 2 1 0E mgcx x x

I I

β α β

− − − − − = ′ (2) !

Solving this equation will yield three values of x. The two values lying between 1− and 1+ correspond to the maximum and minimum values of .θ




Data: 360 mm 0.06 m, 180 mm 0.18 m, 0.135 m,4

r h c h= = = = = =

0 0 0 030 , 300 rad/s, 20 rad/s, 0θ ψ φ θ= ° = = =! !!

Calculate the following:

( )22 3 23 3 0.06 1.08 10 m10 10

I rm

−= = = ×

( ) ( )2 22 2 3 23 1 3 1 0.06 0.18 19.98 10 m5 4 5 4

I r hm

−′ = + = + = ×

Initially, 0 0 0sin 20sin 30 10 rad/s, 0x yω φ θ ω θ= − = − ° = − = =! !

0 0 0cos 300 20cos30 317.32 rad/szω ψ φ θ= + = + ° =!!

( )2 2 21 12 2x y z

T I Im m m

ω ω ω′= + +

( )( ) ( )( )23 2 3 2 21 119.98 10 10 0 1.08 10 317.32 55.3728 m /s2 2

− −= × + + × =

( )( ) 2 20cos 9.81 0.135 cos30 1.1469 m /sV gc

mθ= = ° =

( ) 2 22 2 55.3728 1.1469 113.0394 m /sEm

= + =

( )( )3 21.08 10 317.32 0.342706 m /szI

m mβ ω −= = × =

( )( )2 3 20 0 0sin cos 19.98 10 20 sin 30 0.342706cos30I

m mα φ θ β θ −′

= + = × ° + °!

20.396692 m /s=



After dividing by m, Equation (2) of Problem 18.138 becomes

( ) ( )2

22

22 2 1 0mIm

E m xF x gcx xm I m m

βα β

= − − − − − = ′

( ) ( )( )( ) ( ) ( )2 22

3 30.342706 0.396692 0.342706

113.0394 2 9.81 0.135 1 01.08 10 19.98 10

xx x− −

− − − − − =

× ×

( )( ) ( )224.29198 2.6487 1 5.87825 1.157529 0x x x− − − − =

(a) Roots are: cos 0.68170, 0.86603, 2.2919x θ= =

( )1max cos 0.68170 47.023θ −= = ° max 47.0θ = °!

(b) By Equation (4) of Problem 18.137,

( )( )( )2 3 2

0.396692 0.342706 0.68170cos 15.2474sin 19.98 10 sin 47.023I

α β θφθ −

−−= = =′ × °

!

317.32 rad/szω =

( )( )cos 317.32 15.2474 0.68170zψ ω φ θ= − = −!! spin: 307 rad/sψ =! !

precession: 15.25 rad/sφ =! !




Data: 360 mm 0.06 m, 180 mm 0.18 m, 0.135 m4

r h c h= = = = = =

0 0 0 030 , 300 rad/s, 4 rad/s, 0θ ψ φ θ= ° = = − =! !!

Calculate the following:

( )22 3 23 3 0.06 1.08 10 m10 10

I rm

−= = = ×

( ) ( )2 22 2 3 23 1 3 1 0.06 0.18 19.98 10 m5 4 5 4

I r hm

−′ = + = + = ×

Initially, ( )0 0 0sin 4 sin 30 2 rad/s, 0x yω φ θ ω θ= − = − − ° = = =! !

( )300 4 cos30 296.536 rad/szω = + − ° =

( )2 2 21 12 2x y z

T I Im m m

ω ω ω′= + +

( ) ( )( ) ( )( )2 23 3 2 21 119.98 10 2 0 1.08 10 296.536 47.5241 m /s2 2

− −= × + + × =

( )( ) 2 2cos 9.81 0.135 cos30 1.1469 m /sV gcm

θ= = ° =

( ) 2 22 2 47.5241 1.1469 97.3419 m /sEm

= + =

( )( )3 21.08 10 296.536 0.320259 m /szI

m mβ ω −= = × =

( )( )2 3 20 0 0sin cos 19.98 10 4 sin 30 0.320259cos30I

m mα φ θ β θ −′

= + = × − ° + °!

20.257372 m /s=



After dividing by m, Equation (2) of Problem 18.138 becomes

( ) ( )2

22

22 2 1 0mIm

E m xF x gcx xm I m m

βα β

= − − − − − = ′

( ) ( )( )( ) ( ) ( )2 22

3 30.320259 0.257372 0.320259

97.3419 2 9.81 0.135 1 01.08 10 19.98 10

xx− −

− = − − − − =

× ×

( )( ) ( )222.37324 2.6487 1 5.13342 0.80364 0x x x− − − − =

(a) Roots are: cos 0.23732, 0.86603, 1.73x θ= =

( )1max cos 0.23732 76.272θ −= = ° max 76.3θ = °!

(b) By Equation (4) of Problem 18.137,

( )( )( )2 3 2

0.257372 0.320259 0.23732cos 9.6192 rad/ssin 19.98 10 sin 76.272I

α β θφθ −

−−= = =′ × °

!

296.536 rad/szω =

( )( )cos 296.536 9.6192 0.23732zψ ω φ θ= − = −!! spin: 294 rad/sψ =! !

precession: 9.62 rad/sφ =! !

(c) 2cos 0

sinIα β θφ

θ−= =′

!

0.257372cos0.320259

αθβ

= = 36.5θ = °!




(a) From Equation (18.27), O OΣ =M H!

Since 0, 0.O OΣ = =M H! constant (1)O =H !

Conservation of energy: T V+ = constant

Since 0,V = constant (2)T = !

For a rigid body rotating about point O, ( )2 2 212 xx y y z zT I I Iω ω ω= + +

, O x x y y z z x y zH I I Iω ω ω ω ω ω= + + = + +i j k i j kω

2 2 2 2x y zO x y zI I I Tω ω ω ω⋅ = + + =H

Let β be the angle between the vectors OH and .ω

cosO OH ω β⋅ =H ω

The projection of ω along is cosO ω βH

2cos constantO

TH

ω β = = cos constant (3)ω β = !

(b) cosω β is the perpendicular distance from the invariable plane. This distance is equal to 2 .O

TH



(c) For a frame of reference attached to the body, the moments of inertia with respect of orthogonal axes of the frame do not change.

2 2 2 2x x y y z zI I I Tω ω ω+ + = (4)

Let 1 1 12 2 2, ,

x y z

T T Ta b cI I I

= = = (5)

Then, 1 1 1

22 2

2 2 2 1yx z

a b cωω ω+ + = (6)

which is the equation of an ellipsoid.




(a) From Problem 18.141, the equation of the Poinsot ellipsoid is

2 2 2 2 constantx y zx y zI I I Tω ω ω+ + = =

Let ( ) 2 2 2, ,x y zx y z x y zF I I Iω ω ω ω ω ω= + +

grad x y z

F F FF

ω ω ω∂ ∂ ∂= + +∂ ∂ ∂

i j k

2 2 2 2x x y y z z OI I Iω ω ω= + + =i j k H

The direction of the normal at a point on the surface of the ellipsoid is parallel to grad F, which, in turn is parallel to .OH Since OH is normal also to the invariable plane, it follows that the Poinsot ellipsoid is tangent to the invariable plane at the point common to the plane and the ellipsoid.

(b) The Poinsot elipsoid moves with the body. Thus, its angular velocity is ,ωωωω the angular velocity of the body. Since point O is regarded as fixed, the angular velocity vector lies along the axis of rotation, or the locus of points of zero velocity. Thus, the velocity of the point of contact of the Poinsot ellipsoid with the invariable plane is zero. The Poinsot ellipsoid rolls without slipping on the invariable plane.




Let and x y zI I I I I′= = = so that the z axis is the symmetry axis. Then, the equation of the Poinsot ellipsoid

(Equation (4) of Problem 18.141) becomes

( )2 2 2 2 constantx y zI I Tω ω ω′ + + = =

which is the equation of an ellipsoid of revolution. It follows that the tip of ωωωω describes circles on both the Poinsot ellipsoid and on the invariable plane, and that the vector ωωωω itself describes circular body and space cones. The Poinsot ellipsoid, the invariable plane, and the body and space cones are shown below for cases a and b.

(a) I I ′< (b) I I ′>

Direct Precession Retrograde Precession




(a) Equation (1) expresses conservation of energy as shown in the solution to Problem 18.141. It is the

equation of the Poinsot ellipsoid.

Let 1 1 12 2 2, ,

x y z

T T Ta b cI I I

= = =

Then, 22 2

2 2 21 1 1

1yx z

a b cωω ω+ + = (3)

which is the equation of an ellipsoid.

Equation (2) expresses the constancy of ( ) ( ) ( )2 2 22O O O Ox y zH H H H= + + , the square of the magnitude

of the angular momentum vector.

Let 2 2 2, , O O O

x y z

H H Ha b cI I I

= = =

Then, 22 2

2 2 22 2 2

1yx z

a b cωω ω+ + = (4)

which is the equation of a second ellipsoid. Since the coordinates , , x y zω ω ω of the tip of the vector ω must satisfy both Equations (1) and (2), the curve described by the tip of ω is the intersection of the two ellipsoids.

(b) Assume .x y zI I I> >

Then, 1 1 1 2 2 2 and .a b c a b c< < < <



Thus, for both ellipsoids, the minor axis is directed along the x axis, the intermediate axis along the y axis, and the major axis along the z axis. However, because the ratio of the major to minor

semi axis is x

z

II

for the Poinsot ellipsoid and is x

z

II

for the second

ellipsoid, the deviation from a spherical shape is more pronounced in the second ellipsoid.

The largest ellipsoid of the second type to be in contact with the Poinsot ellipsoid will lie outside that ellipsoid and touch it at its points of intersection with the x axis, and the smallest will lie inside the Poinsot ellipsoid and touch it at its points of intersection with the z axis (see left-hand sketch). All ellipsoids of the second type comprised between these two will intersect the Poinsot ellipsoid along the curves called polhodes, as shown in the right-hand figure.

Note that the ellipsoid of the second type, which has the same intermediate axis as the Poinsot ellipsoid, intersects that ellipsoid along two ellipses whose planes contain the y axis. These curves are not polhodes, since the tip of ω will not describe them, but they separate the polhodes into four groups. Two groups loop around the minor axis (x axis) and the other two around the major axis (z axis).

(c) If the body is set to spin about one of the principal axes, the Poinsot ellipsoid will remain in contact with the invariable plane at the same point (on the x, y, or z axis); the rotation is steady. In any other case, the point of contact will be located on one of the polhodes, and the tip of ω will start describing that polhode, while the Poinsot ellipsoid rolls on the invariable plane.

A rotation about the minor or the major axis (x or z axis) is stable. If that motion is disturbed, the tip of ω will move to a very small polhode surrounding that axis and stay close to its original position.

On the other hand, a rotation about the intermediate axis (y axis) is unstable. If that motion is disturbed, the tip of ω will move to one of the polhodes located near that axis and start describing it, departing completely from its original position and causing the body to tumble.




230.093168 lb s /ft, 24 in. 2 ft,

32.2

Wm l

g= = = ⋅ = =

( )12 rad/s=ω i

For rod ADB, 0, since 0.D x xI Iω= ≈ ≈H i

For rod CDE, use principal axes , x y′ ′ as shown.

9cos , 41.410

12θ θ= = °

2cos 9 rad/sxω ω θ′ = =

2sin 7.93725 rad/syω ω θ′ = =

0zω ′ =

0′ ≈xI

( )( )221 10.093168 2

12 12yI ml′ = =

20.0310559 lb s ft= ⋅ ⋅

D x x y y z zI I Iω ω ω′ ′ ′ ′ ′ ′′ ′ ′= + +H i j k

( )( )0 0.0310559 7.93725 0′= + +j

0.246498 ′= j

0.246 lb s ftDH = ⋅ ⋅ !

( )0.246498 sin cos 0.163045 0.184874D θ θ= + = +H i j i j

0.163045

cos0.246498xθ = 48.6θ = °x !

0.184874

cos0.246498

θ =y 41.4θ = °y !

cos 0zθ = 90θ = °z !




230.093168 lb s /ft, 24 in. 2 ft,

32.2

Wm l

g= = = ⋅ = =

( )12 rad/s=ω i


For rod CDE, use principal axes , x y′ ′ as shown.

9cos , 41.410

12θ θ= = °

2cos 9 rad/sxω ω θ′ = = 2sin 7.93725 rad/syω ω θ′ = =

0zω ′ =

0xI ′ ≈

( )( )221 10.093168 2

12 12yI ml′ = =

20.0310559 lb s ft= ⋅ ⋅

2 2 2 21 1 1 1

2 2 2 2x y zx y zT mv I I Iω ω ω′ ′ ′′ ′ ′= + + +

( )( )10 0 0.0310559 7.93725 0

22= + + +

0.97826 ft lb= ⋅ 0.978 ft lbT = ⋅ !




Let 1

6m m′ = be the mass of one side of the cube. Choose x, y, and z axes perpendicular to the faces of the

cube. Let a be the side of the cube.

For a side perpendicular to the x axis, ( ) 21

1.

6xI m a′=

For a side perpendicular to the y or z axis, ( ) 2 22

1 1 1

12 4 3xI m a m a ′ ′= + =

Total moment of inertia: ( ) ( ) 2 21 2

5 52 4

3 18x x xI I I m a ma′= + = =

By symmetry, y xI I= and .z xI I=

Since all three moments of inertia are equal the ellipsoid of inertia is a sphere. All centroidal axes are principal axes.

Moment of inertia about the vertical axis: 25

18vI ma=

Let 2

3b a= be the moment arm of the impulse applied to the corner.

Using the impulse-momentum principle and taking moments about the vertical axis,

( ) 25

18v vbF t H I maω ω∆ = = = (1)

Data: ( )21.5 m, 1.5 1.22474 m

3a b= = =

21.25664 rad/s, 50 N, 1.2 s.

5F t

πω = = = ∆ =

Solving Equation (1) for m, ( ) ( )( )( )

( ) ( )2 2

1.22474 50 1.218 1893.563 kg

5 5 1.5 1.25664

bF tm

a ω∆

= = = 93.6 kgm = !




( )2 2 2 22G x x y y z z x x y y zI I I m k k kω ω ω ω ω ω= + + = + +H i j k i j k

( ) ( ) ( ) ( ) ( ) ( )2 2 250002.45 0.040 2.65 0 2.55 0.060

32.2 = + +

i j k

( ) ( )37.283 lb s ft 60.582 lb s ft= ⋅ ⋅ + ⋅ ⋅i k

Let , , ,A B C− − −j j j and D− j be the impulses provided by the 5-lb thrusters at A, B, C, and D respectively. Let Ej be that provided by the 125-lb main thruster.

Position vectors for intersections of the lines of action of the truster impulses with the xz plane.

( )14 2 ft, 2 2 2 4.8284 ft

2a b= = = + =

, A Ba b a b= − + = +r i k r i k

, C Da b a b= − = − −r i k r i k

The final linear and angular momenta are zero.

Principle of impulse-momentum. Moments about G:

( ) ( ) ( ) ( ) 0G A B C DA B C D+ × − + × − + × − + × − =H r j r j r j r j

( ) ( ) 0G b A B C D a A B C D+ + − − + − − + =H i k


( ) 37.283

: 7.7216 lb s4.8284

G xH

A B C Db

+ − − = − = − = − ⋅i (1)

( ) 60.582

: 30.291 lb s2

G zH

A B C Da

− − + = − = − = − ⋅k (2)

Of A, B, C, and D, two must be zero or positive, the other two zero.

Set 0A = and .B D N− = Solve the simultaneous equations (1) and (2).



19.006 lb sC = ⋅ and 11.285 lb s.N = ⋅ Set 0D = and 11.285 lb sB = ⋅

(a) Use thrusters C and B. !

(b) ( ) 19.006,

5C C CC

CF t C t

F∆ = ∆ = = 3.80 sCt∆ = !

( ) 11.285,

5B B BB

BF t B t

F∆ = ∆ = = 2.26 sBt∆ = !

(c) Linear momentum: 0, 30.291 lb sE B C E− − = = ⋅j j j

( ) 30.291

125E E EE

EF t E t

F∆ = ∆ = = 0.242 sEt∆ = !




For the ′x and ′y axes shown, the initial angular velocity 0ω j has components

0 02 2

, ,2 2

ω ω ω ω′ ′= =x y

Initial angular momentum about the mass center:

( ) ( )200

1 2

12 2ω ω ω′ ′ ′ ′′ ′ ′ ′= + = +G x x y yI I maH i j i j

Initial velocity of the mass center: 0 0=v

Let ωωωω be the angular velocity and v be the velocity of the mass center immediately after impact. Let

( )F t∆ k be the impulse at B.

Kinematics: ( ) ( )/ ω ω ω′ ′ ′′ ′ ′ ′= × = + + × −B B A x y z av r i j k jωωωω

( )B z xa ω ω′ ′ ′= −v i k

Since the corner B does not rebound, ( ) 0 or 0ω ′′ = =B xzv

( ) ( ) ( )/1 1

2 2G A y z z z ya aω ω ω ω ω′ ′ ′ ′ ′ ′ ′ ′ ′ ′= × = + × − − = − +

v r j k i j i j kωωωω

Also, ( )2/

12

4ω ω ω′ ′ ′′ ′ ′× = − + +G A y y zm mar v i j k

and

2 21 1

12 6ω ω ω ω ω′ ′ ′ ′ ′ ′ ′ ′′ ′ ′ ′ ′= + + = +G x x y y z z y zI I I ma maH i j k j k




Moments about A: ( ) ( ) ( )0A Aa F t+ − × ∆ =H j k H

( ) ( )/ 0 /0G G A G G Am aF t m+ × − ∆ = + ×H r v i H r v


( )2 20

1 1: 2

24 4 yma aF t maω ω ′′ − ∆ = −i

2 2 20 0

1 1 1 2: 2

24 12 4 8y y yma ma maω ω ω ω ω′ ′ ′′ = + =j

2 21 1: 0 0

6 2z z zma maω ω ω′ ′ ′′ = + =k

(a) ( )0 02 1 2

28 8 2

ω ω′= = −j j iωωωω ( )01

8ω= − +i jωωωω !

(b) 01 2

2 16ω ω′= =ya av k k 00.0884 ω= av k!




For the simpler x′ and y′ axes, the initial angular velocity 0ω j has components

0 02 2

, ,2 2

ω ω ω ω′ ′= =x y

Initial angular momentum about the mass center:

( ) ( )200

1 2

12 2ω ω ω′ ′ ′ ′′ ′ ′ ′= + = +G x x y yI I maH i j i j

Initial velocity of the mass center: 0 0=v

Let ωωωω be the angular velocity and v be the velocity of the mass center immediately after impact. Let

( )F t∆ k be the impulse at B.

Kinematics: ( ) ( )/ ω ω ω′ ′ ′′ ′ ′ ′= × = + + × −B B A x y z av r i j k jωωωω

( )B z xa ω ω′ ′ ′= −v i k

Since the corner B does not rebound, ( ) 0 or 0ω ′′ = =B xzv

( ) ( ) ( )/1 1

2 2G A y z z z ya aω ω ω ω ω′ ′ ′ ′ ′ ′ ′ ′ ′ ′= × = + × − − = − +

v r j k i j i j kωωωω

Also, ( )2/

12

4ω ω ω′ ′ ′′ ′ ′× = − + +G A y y zm mar v i j k

and

2 21 1

12 6ω ω ω ω ω′ ′ ′ ′ ′ ′ ′ ′′ ′ ′ ′ ′= + + = +G x x y y z z y zI I I ma maH i j k j k




Moments about A: ( ) ( ) ( )0A Aa F t′+ − × ∆ =H j k H

( ) / 0 /0G G A G G Am aF t m′+ × + ∆ = + ×H r v i H r v


( )2 20

1 1: 2

24 4 yma aF t maω ω ′′ − ∆ = −i (1)

2 2 20 0

1 1 1 1: 2 2

24 12 4 8y y yma ma maω ω ω ω ω′ ′ ′′ = + =j

2 21 1: 0 0

6 2z z zma maω ω ω′ ′ ′′ = + =k

(a) From (1), 0 0 01 1 7

2 2 224 32 96

F t ma ma maω ω ω∆ = + =

( ) 00.1031F t maω∆ =k k!

01 1

22 16

ω ω′ ′= =ya av k k

Linear momentum: 0m t F t m+ ∆ + ∆ =v A k v

0 07 1

0 2 296 16

ω ω′ ′+ ∆ + =t ma maA k k

(b) 01

296

ω∆ = −t maA 00.01473 ω∆ = −t maA k!




( )23 0.093168 lb s /ft, 24 in. 2 ft, 12 rad/s32.2

= = = ⋅ = = =wm lg

iω


For rod CDE, use principal axes , ′ ′x y as shown.

9cos , 41.41012

θ θ= = °

2cos 9 rad/sω ω θ′ = =x

2sin 7.93725 rad/sω ω θ′ = =y

0ω ′ =z

( )296 rad/sα = − i

cos 72 rad/sα α θ′ = = −x

2sin 63.498 rad/sα α θ′ = = −y

0′ ≈xI

( )( )22 21 1 0.093168 2 0.0310559 lb s ft12 12yI ml′ = = = ⋅ ⋅

′ ′=z yI I

ω ω ω′ ′ ′ ′ ′ ′′ ′ ′= + +D x x y y z zI I IH i j k

( )( ) ( )0 0.0310559 7.93725 0 0.246498 lb s ft ′= + + = ⋅ ⋅j j

Let the reference frame D x′y′z′ be rotating with angular velocity

( ) ( )9 rad/s 7.93725 rad/sx yω ω ω′ ′′ ′ ′ ′= = + = +i i j i jΩ



( ) ′ ′ ′= + ×! !

D D DDx y zH H HΩ

α α α′ ′ ′ ′ ′ ′′ ′ ′= + + + ×x x y y z z DI I Ii j k HΩ

( )( ) ( ) ( )0 0.0310559 63.498 0 9 7.93725 0.246498′ ′ ′= + − + + + ×j i j j

1.97199 2.21848′= − +j k

( )( ) ( )1.97199 lb ft sin cos 2.21848 lb ftθ θ= − ⋅ + + ⋅i j k

( ) ( ) ( )1.304 lb ft 1.479 lb ft 2.22 lb ftD = − ⋅ − ⋅ + ⋅H i j k! !




( ) ( )2 2 75040 1.24224 lb s /ft, 78.54 rad/s32.2 60

π= = ⋅ = =m i iω

(a) The force-couple system acting on the wheel is ( ) ( )36.2 lb , 10.85 lb ft′ ′= − = − ⋅CF j M k

22 nm m y z

mω

ω′′ = = − = + = FF a r r j k

( )( )3

2 236.2 4.7241 10 ft 0.0567 in.

1.24224 78.54ω−′

= = = × =yFy y

m

2 0ω

′= =zFz

m 0.0567 in.=r !

ω ω ω= − −C x xy xzI I IH i j k

( )ω ω ω ω′ = = × = × − −!C C C x xy xzI I IM H H i i j kω

( ) ( ) 2 2ω ω′ ′+ = −C C xz xyy zM M I Ij k j k

( ) ( )( )

3 22 2

10.851.75893 10 lb s /ft

78.54ω−′ − −

= − = = × ⋅C zxy

MI

3 21.759 10 lb s ft−= × ⋅ ⋅xyI !

( )

2 0ω

′= =

C yxz

MI 0=xzI !

(b) Positions for the balance masses. 7.25 in. 0.60417 ft= = − = − = =A B E Dy y y y

3 in. 0.25 ft− = = = − = =A B E Dx x x x



Balance masses must be added to move the mass center to point C and to reduce the product of inertia to zero.

0+ + + + =A A B B E E D Dm y m y m y m y my

( )( ) ( )( )30.60417 1.24224 4.7241 10−+ − − + ×A B E Dm m m m

39.7133 10−+ − − = − ×A B E Dm m m m (1)

0+ + + + =A A A B B B E E E D D D xym x y m x y m x y m x y I

( )( )( ) 30.25 0.60417 1.75893 10 0−− + − + + × =A B E Dm m m m

11.6452− + − + = −A B E Dm m m m (2)

To solve (1) and (2), set 0=Bm and let = −AD A Dm m m .

Then, 39.7133 10−− = − ×AD Em m

and 311.6452 10−− − = − ×AD Em m

Solving, 3 2 3 20.9660 10 lb s /ft, 10.6792 10 lb s /ftAD Em m− −= × ⋅ = × ⋅

Set 0,=Dm so that 3 20.9660 10 lb s /ft.Am −= × ⋅

Then, ( )( )310.6792 10 32.2 0.3439 lb,−= = × =E EW m g 5.50 oz=EW !

( )( )30.9660 10 32.2 0.0311 lb,A AW m g −= = × = 0.498 ozAW = !




( )( )1

2 2500261.8 rad/s

60

π= =ωωωω 1 1ω= kωωωω

Angular velocity: 2 1ω ω= +j kωωωω

Angular momentum of rotor: 2 1G y zI Iω ω= +H j k

Let the reference frame Gxyz be rotating with angular velocity 2 .ω= jΩΩΩΩ

( ) ( )2 2 1 1 20G G G y z zGxyzI I Iω ω ω ω ω= + × = + × + =H H H j j k i& & ΩΩΩΩ

Couple exerted on the saw: 21 2 1 2G z zI mkω ω ω ω= = =M H i i&

( )( ) ( )( )231.75 30 10 261.8 2.4−= × −M i ( )0.990 N m= − ⋅M i!




Angular velocity of shaft DE and arm CBA: 2ω= iΩ

Angular velocity of disk A: 2 1ω ω= +i jω

Angular momentum about its mass center A:

2 22 1 2 1

1 14 2A x x y y z z x yI I I I I mr mrω ω ω ω ω ω ω= + + = + = +H i j k i j i j

Let the reference frame Oxyz be rotating with angular velocity 2 .ω= iΩ

( ) 2 22 1 2

1 14 2A A A AOxyz

mr mrω ω ω= + × = + + ×H H H i j i H! ! ! !Ω

2 2 22 1 1 2

1 1 14 2 2

mr mr mrω ω ω ω= + +i j k! !

Velocity of point A: ( )2 / 2 2 2A A O b c b cω ω ω ω= × = × − + = +v i r i k j j k

Acceleration of point A: 2 / 2A A O Aω ω= × + ×a j r j v!

( ) ( )2 22 2 2 2A b c c bω ω ω ω= − + +a j k! !

Consider the system of particles consisting of the shaft, the arm, and the disk. Neglect the mass of the arm.

AmΣ =F a

y z y z AD D E E m+ + + =j k j k a


( )22 2y yD E m b cω ω+ = −!

( )22 2z zD E m c bω ω+ = +!

/D D A A D AmΣ = = + ×M H H r a! !



( ) ( ) ( )0 2 y z A AM l E E l c b m+ × + = + + − ×i i j k H i j k a!

( ) 2 2 2 2 20 2 1 2 2

1 12 24 2z yM lE lE m r b c m r lc lbω ω ω ω − + = + + + − −

i j k i j! ! !

2 21 2 2 2

12

m r lb lcω ω ω ω + + −

k!

2 2 20 2

1: 4

M m r b c ω = + +

i !

2 2 2 21 2 2 2 1 2 2 2

1 1: , 2 2 2 2y ym mE r lb lc D r lb lcl l

ω ω ω ω ω ω ω ω = + − = − + −

k ! !

2 2 2 22 2 1 2 2 1

1 1: , 2 2 2 2z zm mE lc lb r D lc lb rl l

ω ω ω ω ω ω = + − = + +

j ! ! ! !

Data: 3 kg, 0.2 m, 0.225 m, 0.3 m= = = = =m r b c l

1 1 2 216 rad/s, 0, 8 rad/s, 0ω ω ω ω= = = =! !

( )( ) ( ) ( )( ) ( )( )( )2 23 1 0.2 16 8 0 0.3 0.225 8 34.4 N2 0.3 2yD = − + − = −

( )( ) ( )( )( )23 0 0.3 0.225 8 0 21.6 N2 0.3zD = + + =

( ) ( )34.4 N 21.6 N= − +D j k!

( )( ) ( ) ( )( ) ( )( )( )2 23 1 0.2 16 8 0 0.3 0.225 8 8.80 N2 0.3 2yE = + − = −

( )( ) ( )( )( )23 0 0.3 0.225 8 0 21.6 N2 0.3zE = + + =

( ) ( )8.80 N 21.6 N= − +E j k!




Angular velocity of shaft DE and arm CBA: 2ω= iΩ

Angular velocity of disk A: 2 1ω ω= +i jω

Angular velocity about its mass center A:

2 22 1 2 1

1 14 2A x x y y z z x yI I I I I mr mrω ω ω ω ω ω ω= + + = + = +H i j k i j i j

Let the reference frame Oxyz be rotating with angular velocity 2 .ω= iΩ

( ) 2 22 1 2

1 14 2A A A AOxyz

mr mrω ω ω= + × = + + ×H H H i j i H! ! ! !Ω

2 2 22 1 1 2

1 1 14 2 2

mr mr mrω ω ω ω= + +i j k! !

Velocity of point A: ( )2 / 2 2 2A A O b c b cω ω ω ω= × = × − + = +v i r i k j j k

Acceleration of point A: 2 / 2A A O Aω ω= × + ×a j r j v!

( ) ( )22 2 2 2A b c c bω ω ω ω= − + +a j k! !

Consider the system of particles consisting of the shaft, the arm, and the disk. Neglect the mass of the arm.

AmΣ =F a

y z y z AD D E E m+ + + =j k j k a




( )22 2y yD E m b cω ω+ = −!

( )22 2z zD E m c bω ω+ = +!

/D D A A D AmΣ = = + ×M H H r a! !

( ) ( ) ( )2O y z A AM l E E l c b m+ × + = + + − ×i i j k H i j k a!

( ) 2 2 2 2 22 1 2 2

1 12 24 2O z yM lE lE m r b c m r lc lbω ω ω ω − + = + + + − −

i j k i j! ! !

2 21 2 2 2

12

m r lb lcω ω ω ω + + −

k!

2 2 22

1: 4OM m r b c ω = + +

i !

2 2 2 21 2 2 2 1 2 2 2

1 1: , 2 2 2 2y ym mE r lb lc D r lb lcl l

ω ω ω ω ω ω ω ω = + − = − + −

k ! !

2 2 2 22 2 1 2 2 1

1 1: , 2 2 2 2z zm mE lc lb r D lc lb rl l

ω ω ω ω ω ω = + − = + +

j ! ! ! !

Data: 3 kg, 0.2 m, 0.225 m, 0.3 mm r b c l= = = = = 2

1 1 2 216 rad/s, 0, 8 rad/s, 6 rad/sω ω ω ω= = = =! !

(a) ( ) ( ) ( ) ( ) ( )2 2 213 0.2 0.225 0.225 64OM = + +

( )2.00 N mO = ⋅M i!

(b) ( )( ) ( ) ( )( ) ( )( )( ) ( )( )( )2 23 1 0.2 16 8 0.3 0.225 6 0.3 0.225 8 32.4 N2 0.3 2yD = − + − = −

( )( ) ( )( )( ) ( )( )( )23 0.3 0.225 6 0.3 0.225 8 0 23.6 N2 0.3zD = + + =

( ) ( )32.4 N 23.6 N= − +D j k!

( )( ) ( ) ( )( ) ( )( )( ) ( )( )( )2 23 1 0.2 16 8 0.3 0.225 6 0.3 0.225 8 6.78 N2 0.3 2yE = + − = −

( )( ) ( )( )( ) ( )( )( )23 0.3 0.225 6 0.3 0.225 8 0 23.6 N2 0.3zE = + + =

( ) ( )6.78 N 23.6 N= − +E j k !




( )02 rad 1 h36 rev/h 0.062832 rad/s

rev 3600 sπω = =

25000 155.28 lb s /ft32.2

= = ⋅m

( )( )22 2155.28 2.7 1131.99 lb s ft= = = ⋅ ⋅x xI mk

21131.99 lb s ft= = ⋅ ⋅z xI I

( )( )22 2155.28 2.94 1342.17 lb s ft= = = ⋅ ⋅y yI mk

( ) ( )( ) ( )00 1342.18 0.062832 84.331 lb s ftω= = = ⋅ ⋅H j j jG yI

1.8 ft, 1.8 3.6cos 45 4.3456 ft= = + ° =a b

When thrusters A and B are activated,

( ) ( )( ) ( )4.3456 10 43.456 lb ftG A Bb F F= − + = − = − ⋅M i i i

Angular momentum after 2 s:

( ) ( ) ( )( )0 84.331 43.456 2G G G t= + ∆ = + −H H M j i

( ) ( )86.912 lb s ft 84.331 lb s ft= − ⋅ ⋅ + ⋅ ⋅i j

86.912 0.076778 rad/s 43.991 rev/h1131.99

ω = = − = − = −xx

x

HI

36 rev/h 0ω ω= = = =y zy z

y z

H HI I



(a) Precession axis:

86.912tan 45.86484.331

θ θ= − = = °x

y

HH

44.1 , 45.9 , 90θ θ θ= ° = ° = °x y z "

43.992tan 50.70536

ωγ γω

= − = = °x

y

4.841γ θ− = °

2 2 56.844 rev/hx yω ω ω= + =

Law of sines

( )sin sin sinφ ψ ω

γ γ θ θ= =

−

! !

(b) 56.844sin 50.705sin 45.864

φ °=°

!

61.3 rev/hφ =! "

(c) 56.844sin 4.841sin 45.864

ψ °=°

!

6.68 rev/hψ =! "

solucionario beer, johnton - octava edicion parcial ultimo

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