review exercise 1€¦ · (1 sin ) substituting for from equation cos sin (1 sin ) tan cos cos t t...

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Review exercise 1

1 Changing the units:

diameter 7cm radius ( ) 3.5cm 0.035m

1000 2angular speed ( ) 1000 revolutions per minute radians per second

60

r

Using v r gives:

v 0.035

1000 260

3.67 ms-1 (3 s.f.)

So the speed is 3.67 m s–1 (3 s.f.)

2 a Let the tension in the string be T and let the string make an angle with the vertical.

Let OP r, then r 1.5sin

Acceleration towards the centre of the circle = r

2 1.5sin 2.72

Force towards the centre of the circle = T sin

So using F ma gives:

T sin 0.51.5sin 2.72

T 0.51.5 2.72 5.4675 5.5N (2 s.f.)

b Resolving the forces vertically R() and substituting for T gives:

T cos - 0.5g 0

T cos 0.5g

cos 0.5g

5.4675 0.8962

So cos-1 0.8962 26° (to the nearest degree)


3 a These are the forces acting on P.

R( ) : T cos60° - mg 0

T mg

cos60° 2mg

b R( ¬) : T sin60° ma mr 2 using F ma and r is the radius of the circular motion

As r Lsin60° this gives:

2

2

sin 60 sin 60

2 2 substituting result for from part

2So

T mL

T mg gT

mL mL L

g

L

° °

a

4 The forces acting on the car are its weight mg and the normal reaction R.

R( ) : Rcos10° - mg 0

R mg

cos10°

Using F ma horizontally gives:

2 2

2

2

18sin10

18sin10 substituting for

cos10

18187.4995 190 m (2 s.f.)

tan10

mv mR

r r

mg mR

r

rg

°

°

°

°

…


5 The forces acting on the cyclist and bicycle are their weight mg, the normal reaction R and the

friction acting down the slope.

R( ) : Rcos25° - F sin25° - mg 0

If m is the coefficient of friction between the cycle’s tyres and the track, then the maximum friction

for which the tyres do not slip is F mR 0.6R . Substituting for F gives:

cos25 0.6 sin 25 0

cos25 0.6sin 25

R R mg

mgR

° - ° -

°- °

(1)

2 2

2

2

R( ) : sin 25 cos25 using and

sin 25 0.6 cos25 as 40 and 0.6 at maximum speed40

40(sin 25 0.6cos25 )

mv vR F F ma a

r r

mvR R r F R

mvR

¬ ° °

° °

° °

(2)

Using equations (1) and (2) gives:

mv2

40(sin25° 0.6cos25°)

mg

cos25° - 0.6sin25°

v2 40g(sin25° 0.6cos 25°)

cos25° - 0.6sin25° 580.37 (2 d.p.)

v 24ms-1

(2 s.f.)


6 a The forces acting on the ring are the the weight of the ring and tensions of equal magnitude T

along each section of string.

Note that is an an angle in a 3-4-5 triangle, with cos

3

5, sin

4

5

R( ) : T sin - mg 0

T mg

sin

5mg

4

b R( ¬) : T T cos

mv2

r using F ma and a

v2

r

T 13

5

æ

èçö

ø÷

mv2

3l as r 3l and cos

3

5

5mg

4

8

5

mv2

3l substituting for T

v2 6gl

v 6gl

c If the ring is firmly attached to the string, then the tensions in each section of the string cannot

be assumed to have the same magnitude. So the approach adopted for parts a and b would not

be valid.

7 a The forces acting on the metal ball are the weight of the ball and tension along the string.

R( ) : 3mg cosa - mg 0

cosa mg

3mg

1

3

a 70.5° (3 s.f.)


7 b R( ¬) : 3mg sina mr2gk using F ma and a r 2

Let the length of the string be l, then from DAOB is is clear that sina

r

l

So 3mgr

l mr2gk

l 3

2k

8 a The forces acting on the particle are its weight and the normal reaction.

Let be the angle between the normal reaction and the horizontal.

Then sin

12r

r

1

2 30°

R( ) : Rsin30° - mg 0

R mg

sin30° 2mg

b R(®) : Rcos30° mx 2 using F ma and a r 2

Using the result from part a and as x r cos30° this gives:

2mg mr 2

2g

r

Time to complete one revolution

2

2r

2g


9 a Let T1 be the tension in AP and T2 be the tension in BP. The forces acting on P are the tensions in

the two strings and its weight.

From the diagram, it can be seen that the equilateral triangle APB can be divided into two

right-angled triangles, where:

tan 60

2

3tan 60

2 2

r

h

h hr

°

°

b Resolving the forces from the diagram in part a

1 2

1 2

R( ) : cos 60 cos 60 0

2

T T mg

T T mg

° - ° -

- (1)

2 2

1 2

2

1 2

2

1 2

R( ) : sin 60 sin 60 using and

3 3 3 3 as from part

2 2 2 2

T T mr F ma a r

T T m h r h

T T mh

¬ ° °

a

(2)

Adding equations (1) and (2) gives: 2T

1 2mg mh 2 T

1 mg

1

2mh 2

Substituting for T1 in equation (2) gives: T

2

1

2mh 2 - mg


9 c Both strings are taut, therefore T1 > 0 and T2 > 0. From part b, T1 > 0 for all values of

From the formula for T2 in part b, the condition that T

2> 0 >

2g

h

As T 2

, 2T

So >2g

h

2T

>2g

hT < 2

h

2g as T > 0 and

h

2g< 0

T < 2h

g as required

10 a The forces acting on the ring are the the weight of the ring and, as the ring is threaded on the

strong, tensions of equal magnitude T along each section of string.

R( ) : cos 0

cos

T mg

mgT

-

(1)

2 2

2

2

R( ) : sin using and

(1 sin ) substituting for from equation cos

sin(1 sin ) tan

cos cos

T T mr F ma a r

mgmr T

mgmh mh

¬

(1)

2

2

as tan by basic trigonometry

1 sin

sin

r h

g

h

æ ö ç ÷è ø

b From part a,

2 g

h

1 sinsin

æ

èçö

ø÷

g

h

1

sin1

æ

èçö

ø÷

As sin <1 it follows that

1

sin>1

Therefore 2 >g

h 2

>2g

h


10 c Given 3g

h , then

1 sin 3

sin

g g

h h

æ ö ç ÷è ø

1 31 sin 3sin sin and so cos

2 2

2 2 3From equation ,

cos 33

mgT mg mg

(1)

11 Let a be the angle between the slant side and the axis of the cone, r the radius of the horizontal circle

with centre C and h the height of C above V. The forces acting on the particle are its weight and the

normal reaction.

From the diagram, tana

3a

4a

3

4 and tana

r

h

R( ) : sin 0

sin

R mg

R mg

aa

-

(1)

2 2R( ) : cos using and

8 8cos using

9 9

R mr F ma a r

mr g gR

a a

a

a

¬

(2)

Dividing equation (1) by equation (2) gives:

8 9tan

9 8

3 9 3 using tan

4 8 4

3

2

3 4As tan ,

4 3

4 3So 2

3 2

mrg amg

a r

a

r

ar

rh r

h

ah a

a

a

a

The height of C above V is 2a.


12 a Let the tension in the string be T.

2 2R( ) : cos30 using and

cos30 2 cos30 using and 2 cos30 by trigonometry3 3

2 as required

3

T mr F ma a r

kg kgT m a r a

a a

kmgT

¬ °

° ° °

b Let the normal reaction be R and resolve vertically.

R( ) : R T sin30° - mg 0

R mg -2kmg

3

1

2 mg 1-

k

3

æ

èçö

ø÷

c Using the result from part b, as R > 0,1-

k

3> 0 k < 3

d Let the new tension be ¢T and let AP make an angle with the horizontal. So PX 2acos .

R( ¬) : ¢T cos mr 2 us ing F ma and a r 2

¢T cos m2acos2g

a using

2g

a and r 2acos

¢T 4mg

R( ) : ¢T sin - mg 0

sin mg

¢T

mg

4mg

1

4

Hence AX 2asin 2a 1

4

a

2

AO 2asin30° 2a 1

2 a

Therefore AX AO

2

So the point X is the midpoint of AX.


13 a Let the tension in AP be T and in BP be S. Let ÐBAP . As the triangle APB is an equilateral

triangle ÐBAP ÐPBA .

AO

3

2l

1

2

3l

4, so from the right-angled triangle APO, cos

3l

4l

3

4

R( ) : cos cos 0

4

cos 3

T S mg

mg mgT S

- -

- (1)

2 2

22

R( ) : sin sin using and

as sin sin

T S mr F ma a r

mr rT S ml

l

¬

(2)

Adding equations (1) and (2) gives:

2T

4mg

3 ml 2 T

1

6m(4g 3l 2 )

1

6m(3l 2 4g) as required

b Subtracting equation (1) from equation (2) gives:

2S ml 2 -

4mg

3 S

1

6m(3l 2 - 4g)


13 c From part b), 23 4

6BP

mT l g -

Since BP is taut, 0BPT �

2

3 4 06

ml g - �

23 4 0l g - �

23 4l g �

2 4

3

g

l �

14 a The forces acting on the particle are its weight, the normal reaction and friction.

R( ) : R- mg 0 R mg

R( ¬) : F mr 2 us ing F ma and a r 2

F 4ma 2

3 a s r

4

3a

As P remains at rest F Rm�

2

2

2

4 3So

3 5

4 3

3 5

9 as required

20

maR

ma mg

g

a

�

�

�


14 b The forces now acting on the particle are its weight, the normal reaction, friction and the tension in

the elastic string.

Find T using Hooke’s law (Further Mechanics 1, Chapter 3): T

lx

l,

where l is the modulus of elesticity, x is the extension of the string and l is its natural length

So in this case, T

2mg

a

a

3

2mg

3

R( ¬) : T F mr 2 using F ma and a r 2

2 3

4ma

2mg

3 F

æ

èçö

ø÷

Note that the frictional force can act away from O against the pull of the elastic string, or

towards O through the force of the acceleration of the particle generated by the circular motion. As

P remains at rest R F Rm m- � � (where 0.6R mgm from part a).

So, from the equation for 2 , it is maximum when F mR 3mg

5

So max

2 3

4ma

2mg

3

3mg

5

æ

èçö

ø÷

3

4ma

19mg

15

19g

20a

Similarly 2 is minimum when F -mR -3mg

5

So min

2 3

4ma

2mg

3-

3mg

5

æ

èçö

ø÷

3

4ma

mg

15

g

20a

15 a i AP

x

cos, so extension in the string

x

cos- x

Using Hooke’s law: 10 10

10 cos cos

g x gT x g

x æ ö - -ç ÷è ø

(1)

2R( ) : cos 2 0 cos

Substituting for cos into equation gives:

5 10

102.5 N

4

gT g

T

T T g

gT g

-

-

(1)

ii

cos 2g

T

2g

2.5g 0.8

So arccos0.8 36.9° (3 s.f.)


15 b

2 22R( ) : sin using and

v vT F ma a

r r¬

As cos

4

5, APO is a 3-4-5 right-angled triangle, with

sin

3

5 and tan

r

x

3

4 r

3

4x

So T sin 2v2

r

2.5g 3

5

4

3x 2v2

v2 9gx

16

v 3

4gx

16 a Let u be the speed of P when the string is horizontal.

Applying the work-energy principle, the sum of the particle’s kinetic and gravitational potential

energy remains constant, so when it reaches the horizontal the loss of kinetic energy = the gain in

potential energy.

Take the starting point as the zero level for potential energy.

So 1

2m

5gl

2-

1

2mu

2 mgl

u2 5gl

2- 2gl

gl

2

u gl

2


16 b Let the particle move in a semi-circle about B with radius r.

Taking the line AB as the zero level for potential energy, then applying conservation of energy at

the highest point of the semi-circle:

1

2m(u

2 - v2) mgr

v2 u2 - 2gr (1)

Resolving the vertical forces at the highest point:

R( ¯) : T mg mv2

r us ing F ma and a

v2

r

T m(u

2 - 2gr )

r- mg substituting for v

2 using equation (1)

T mu

2

r- 3mg

T mgl

2r- 3mg s ubstituting for u2 using result from part a

As the string does not go slack T > 0, so

mgl

2r- 3mg > 0

mgl > 6mgr

r <l

6

As AB = l – r, this shows that AB >

5l

6


17 a Let the speed of the particle be v when it makes an angle with the downward vertical.

Take the horizontal through B as the zero level for potential energy. Using conservation of energy:

2

2

2

1(3 ) (1 cos )

2

3 2 (1 cos )

2 cos

m gl v mgl

v gl gl

v gl gl

- -

- -

(1)

Resolving along the string:

2 2

2

R( ) : cos using and and

2 coscos substituting for using equation

3 cos (1 3cos )

mv vT mg F ma a r l

l r

mgl glT mg v

l

T mg mg mg

-

(1)

տ

as required

b The instant the string becomes slack, T = 0.

Using the expression for T from part a, this occurs when:

1 3cos 0 cos -

1

3 Substituting for cos in equation (1) gives:

v2 gl 2gl -1

3

gl

3

v gl

3


17 c The particle now moves as a projectile under gravity. The maximum height is achieved when the

vertical component of the velocity is zero.

vy vsin(180°- )

gl

3sin

If 2< < and cos -

1

3, then sin

2 2

3

So vy

2 2

3

gl

3

Considering the particle’s vertical motion, u v

y, v 0, a -g, s h, where h is the height

above the point the string becomes slack.

Using v2 u2 - 2gh

h v

y

2

2g

8

9

gl

3

1

2g

4l

27

The height at which string becomes slack l l cos(180° - ) l(1- cos )

4l

3

So if H is the maximum height above the level of B reached by P

H

4l

3

4l

27

40l

27

18 a Take the horizontal through l as the zero level for potential energy. When angle a , the particle

is momentarily at rest. Using conservation of energy, with the loss of kinetic energy equal to the

gain in potential energy:

1

2mu

2 mgl(1- cosa ) mgl

3 a s cosa=

2

3

u2 2

3gl

u 2gl

3


18 b Let the speed of the particle at be v.

Resolving along the string gives:

2 2

2

R( ) : cos using and and

cos

mv vT mg F ma a r l

l r

mvT mg

l

-

տ

Applying conservation of energy:

1

2mu2 -

1

2mv2 mgl(1- cos )

v2 u

2 - 2gl(1- cos ) 2gl

3- 2gl(1- cos ) using the result from part a

v2 2glcos -4gl

3

Substituting for v2 in the equation for T gives:

T mg cos 2mg cos -4mg

3

3mg cos -4mg

3

mg

3(9cos - 4)

c Maximum value of T is when cos 1

So T

max

5mg

3

Minimum value of T is when cos

2

3

min

2So

3

2 5Hence

3 3

mgT

mg mgT

� �

19 a The lengths on the diagram can be deduced from the question:

OA OB 0.8m AN 0.2m ON OA- AN 0.6m

cos

ON

OB

0.6

0.8

3

4


19 b As the particle leaves the hemisphere at B the normal reaction R = 0.

So resolving along the radius OB gives:

2 2

2

3R( ) : cos using and and 0.8

4 0.8

30.8 0.6 5.88

4

mg mv vmg F ma a r

r

gv g

ւ

c Take the horizontal through A as the zero level for potential energy. Using conservation of energy,

the gain of kinetic energy at B equals the loss in potential energy:

1

2m5.88-

1

2mu

2 mg 0.2

u2 5.88- 0.4g 0.2g

So u 1.4

20 a This is a diagram of the problem.

Take the horizontal through A as the zero level for potential energy. Using conservation of energy,

the gain of kinetic energy at P equals the loss in potential energy:

1

2mv

2 mg(acosa - acos )

v2 2ga(cosa - cos )

b Resolving along the radius OP:

2

22

R( ) : cos

At the point when loses contact with the sphere 0

cos 2 (cos cos ) substituting for from part

33cos 2cos

2

1cos , so 60 or r

2 3

mvmg R

a

P R

mvmg gm v

a

a

a

-

-

°

a

ւ

adians


20 c Let the speed of P when it hits the table be w. Then taking the horizontal through A as the zero

level for potential energy and using conservation of energy, the gain of kinetic energy at B equals

the loss in potential energy:

1

2mw

2 mg(a acosa )

w2 2ga 1

3

4

æ

èçö

ø÷

7ga

2

So w 7ga

2ms-1

There are alternative ways to calculate w that use projectiles, but the approach shown above is the

shortest method.

21 a Let the speed of P at point B be v. Then taking the horizontal through A as the zero level for

potential energy and using conservation of energy, the loss of kinetic energy at B equals the


1

2mu2 -

1

2mv2 mga

1

2m

7ag

2- mga

1

2mv

2

v2 3

2ga

R( ¬) : R mv2

a

R m

a

3

2ga

3

2mg

b Let the speed of P at point C be w. Then taking the horizontal through A as the zero level for

potential energy and using conservation of energy, the loss of kinetic energy at C equals the


21 7 1(1 cos )

2 2 2

gam mw mga - (1)

Resolving along the radius OP

2

2

2

R( ) : cos

At the point when loses contact with the sphere 0

So cos

Substituting for in equation gives:

7 cos(1 cos )

4 2

3 3cos

2 4

1cos , hence 60

2

mwmg R

a

P R

ga w

w

mga mgamga

-

°

(1)

ւ


21 c Consider the motion of the projectile P as it moves from C to A in the horizontal direction.

The horizontal distance from C to A asin60°

The horizontal component of the speed at C, wx wcos60°

So the time t that P takes to travel from C to A is given by t asin60°wcos60°

From part b, w2 ag cos60° w ag

2

So t asin60°cos60°

ag

2 a 3

2

ag

6a

g

22 a Let the speed of the trapeze artist at the lowest point of her path be v. Then taking the horizontal

through A as the zero level for potential energy and using conservation of energy, the gain in

kinetic energy at this point equals the loss in potential energy:

1

2mv2 -

1

2m( 15)2 mg 5(1- cos60°)

v2 155g 64

v 8ms-1


22 b Let the velocity after catching the ball be w. Then the loss in kinetic energy at point Q where the

trapeze artist becomes momentarily stationery equals the gain in potential energy from the lowest

point of the trapeze artist’s path.

1

2(60 m)w2 - 0 (60 m)g5(1- cos60°)

w2 5g 49, so w 7ms-1

So at the instant just prior to catching the ball, the trapeze artist is travelling at 8 m s–1 (part a) and

the ball is travelling at 3 m s–1 in the opposite direction, and immmediately after catching the ball

she is travelling at 7 m s–1. Using conservation of linear momentum at the instant when she catches

the ball, this gives:

60 8 ( 3 ) (60 ) 7

480 3 420 7

10 60

So 6 kg

m m

m m

m

m

-

-

c R( ) : T - 66g

66w2

r us ing F ma and a

v2

r

T 66g 66

72

5 1293.6 1300 N (2 s.f.)

23 a Let the speed at C be v m s–1. Then taking the horizontal through B as the zero level for potential

energy and using conservation of energy, the gain in kinetic energy at point C equals the loss in

potential energy:

1

2mv2 -

1

2m 202 mg 50(1- cos60°)

v2 202 50g v2 890

So v 30ms-1 (2 s.f.)

b At C, R( ) : R- 70g

70v2

50 us ing F ma and a

v2

r

R 70g

70890

50 686 1246 1900 N (2 s.f.)


23 c Consider motion C to D. Let the speed at D be w m s–1. Then taking the horizontal through C as the

zero level for potential energy and using conservation of energy, the loss in kinetic energy at point

C equals the gain in potential energy:

1

2m890-

1

2mw2 mg 50(1- cos30°)

w2 890-100g(1- cos30°) 759 (3 s.f.)

w 28ms-1 (2 s.f.)

d Resolving perpendicular to the slope at B just before the circular motion and just after circular

motion begins:

Before: R mg cos60° 35g

After: R - mg cos60°

m 202

50 R 35g 560

So change in R = 560 N

e Allowing for the influence of friction would mean that the skier would arrive at C with lower

speed. From the equation used in part b, this would result in a lower normal reaction.

24 a Taking the horizontal through A as the zero level for potential energy and using conservation of

energy, the loss in kinetic energy when OP makes an angle equals the gain in potential energy:

1

2m3ag -

1

2mv

2 mga(1 cos )

v2 ag(1- 2cos )

b Resolving along the radius when OP makes an angle

2

R( ) : cos

(1 2cos )cos (1 3cos )

mvT mg

a

magT mg mg

a

- - -

ւ

c The string becomes slack when T = 0, i.e. from part b when cos

1

3

Height above A a acos a

1

3a

4

3a


24 d Using result for v2 from part a, and that cos

1

3 at point B, then 2

3

lgv at point B

Consider the vertical component of the motion of the particle under gravity

Using v2 u2 2as , where u is the vertical component of motion at B = sin ,

3

lg v is the

vertical component of motion at C = 0, a = –g and s is the vertical height of C above B

2

2 2

So 0 sin 23

8 4sin (1 cos )

6 6 6 9 27

lggs

l l l ls

-

-

The solution can also be found using conservation of energy as follows.

At C the particle has speed cos ,3

lg the horizontal component of the speed at B

Let the vertical height of C above B be h. Taking the horizontal through B as the zero level for

potential energy and using conservation of energy, the loss in kinetic energy at C equals the gain in potential energy:

21 1cos

2 3 2 3

1 8 41

6 9 54 27

lg lgm m mgh

l l lh

-

æ ö - ç ÷è ø

25 a Taking the horizontal through O as the zero level for potential energy and using conservation of

energy, the gain in kinetic energy equals the loss in potential energy:

1

2mv2 -

1

2mu2 mga sin

v2 u

2 2gasin

v2 3

2ga 2gasin

b Resolving along the radius when OP makes an angle with the horizontal:

2

R( ) : sin

3 3 3sin 2 sin 3 sin (1 2sin )

2 2 2

mvT mg

a

m ga mg mgT mg ga mg

a

-

æ ö ç ÷è ø

տ

c The string becomes slack when T = 0, i.e. from part b when 2sin -1

So sin -

1

2 210°

d From part a, v2

3

2ga 2gasin

To complete the circle, v ¹ 0 before the rod reaches the top point, i.e. 0 for 0 270v > ° °� �

But v = 0 when 3

sin 229 (3 s.f.)4

- ° , so P would not complete a vertical circle.


25 e Using conservation of energy, as there is no change in potential energy when the particle reaches A, as A and O are on the same level, there can be no change in its kinetic energy:

So v u

3ga

2

f Using the results from parts a and c, when the string becomes slack 210° and

v2 3

2ga 2ga -

1

2

æ

èçö

ø÷

1

2ga

Its horizontal component of velocity is

1

2ga cos60°

From part e, when it reaches point A, its horizontal component of velocity is

3

2ga cosf

When P moves freely under gravity, its horizontal component of velocity will remain constant, so

3ga

2cosf

1

2ga cos60°

cosf cos60°

3

1

2 3

f 73.2° (3 s.f.)

26 a Let the velocity of the particle at C be v. Taking the horizontal through A as the zero level for potential energy and using conservation of energy, the gain in kinetic energy at C equals the loss in

potential energy:

2 2

2 2

1 1(1 cos )

2 2

2 (1 cos )

mv mu mga

v u ga

- -

- (1)

Resolving along the radius through C:

2

R ( ) : cosmv

R mga

- ւ

As the particle leaves the sphere at C, 0R so v

2 ag cos

Substituting for v2 in equation (1) gives:

ag cos u2 2ga(1- cos )

3cos u2

ag 2

cos 2

3

u2

3ag

b Using conservation of energy, the gain in kinetic energy from C to when the particle hits the

ground equals the loss in potential energy:

1

2m

9ag

2

æ

èçö

ø÷-

1

2m(ag cos ) mga(1 cos )

3

2cos

9

4-1

5

4 cos

5

6

So 34° (2 s.f.)


27 a Taking the horizontal through A as the zero level for potential energy and using conservation of energy, the loss in kinetic energy at B equals the gain in potential energy:

2 2

2 2

1 1(1 cos 60 )

2 2

3

mu mv mga

v u ga

- °

- (1)

b Resolving along the radius through B:

2

R ( ) : cos 60 mv

R mga

° (2)ւ

When u

2 6ga, v2 3ga from equation (1)

So R

mv2

a- mg cos60° 3mg -

mg

2

5mg

2

c The least value for u will be that which enables the particle to reach B, but to leave the surface at B

so that R = 0 at point B.

22

2 2

If 0 at , then from equation 2 2

7So from equation 3

2 2

7Hence

2

mg mv agR B v

a

ag agu ga u

agu

-

(2)

(1)

d The distance BC = 2 asin60° 3a

The horizontal component of the speed at B = vcos60°

For motion under gravity the horizontal velocity of the particle in travelling from B to C is

constant and using s vt gives:

3a vcos60°t t 2a 3

v

The vertical component of the speed at B = vsin60°

The vertical component of the distance the particle travels is 0 (as the particle is now at C, which is level with B)

So using s ut

1

2at2

gives:

0 vsin60°t -1

2gt

2

t 2vsin60°

g

v 3

g

So when the particle leaves the bowl at B and meets it at C

v 3

g

2a 3

v v2 2ag

So using equation (1) from part a, 2ga u2 - 3ga u2 5ga

Hence u 5ga


28 a The total mass is 3m5m lm (8 l )m

Taking moments about the y-axis:

3m 4 5m 0 lm 4 (8 l)m 2

12 4l 16 2l2l 4

l 2

b Taking moments about the x-axis:

3m 0 5m -3 lm 2 (8 l)m k

-15 4 10k substituting for l 2

-11 10k

So k = 1.1

29 The total mass is 2M xM yM (2 x y)M

Taking moments about the y-axis:

(2 ) 2 2 2 1 3

4 2 2 4 3

0

x y M M xM yM

x y x y

x y

- (1)

Taking moments about the x-axis:

(2 ) 4 2 5 3 1

8 4 4 10 3

3 2

x y M M xM yM

x y x y

x y

(2)

Subtracting equation (1) from equation (2) gives:

14 2

2

1And so from equation

2

y y

x

(2)

30 Using

mir

i r m

iåå

2 2 40.1 0.2 0.3 (0.1 0.2 0.3)

1 5 2

0.2 0.4 1.2(0.6)

0.1 1 0.6

1.8(0.6)

1.5

3

2.5

x

y

x

y

x

y

x

y

æ ö æ ö æ ö æ ö ç ÷ ç ÷ ç ÷ ç ÷-è ø è ø è ø è ø


æ ö æ öç ÷ ç ÷

è ø è ø


è ø è ø

So the centre of mass is (3i + 2.5j) m


31 a Using

mir

i r m

iåå

6 0 2 32 (3 )

0 4 2

12 0 2 9 3

0 4 2 3

M M kM k Mc

k k

k c ck


æ ö æ ö æ ö æ ö ç ÷ ç ÷ ç ÷ ç ÷- è ø è ø è ø è ø

Equating i component gives 12 2 9 3 3 as required.k k k

b Equating j component and substituting for k gives:

4- 2k 3c ck -2 6c c -

1

3

32 a The area of the lamina is 2010 200 cm2

The area of the circle is 32 9 cm2

The area of the plate is (200 - 9) cm2

Let the distance of the centre of mass of the plate from AD be x cm

As the lamina is uniform, masses are proportional to areas. By symmetry, the centre of mass of the rectangular lamina is 10 cm from AD and the centre of mass of the circle is 6 cm from AD.

Lamina Circle Plate

Area 200 9 200- 9

Distance of centre of mass from AD 10 6 x

The moment of the plate about AD equals the moment of the complete lamina less the moment of the circle that has been removed:

(200 9π) 200 10 9 6

2000 5410.658 10.7 cm (3 s.f.)

200 9

x

x

- -

-

- …


32 b Let the angle between AB and the vertical be a. When the plate is suspended freely from A, its

centre of mass G is vertically below the point of suspension A. The distance of G from AD was

found in part a and the distance of G from AB is 5 cm by symmetry.

tana 5

x

5

10.658 0.469 (3 s.f.)

a 25° (to the nearest degree)

33 a The area of rectangle ABDE is 26 8 48a a a

The centre of mass of rectangle is 4a from X

The area of ∆BCD is

1

2 6a 4a 12a2

The centre of mass of the triangle is

1

3h

1

3 4a from the base BD of the triangle.

The area of lamina ABCDE is 48a2 12a2 60a2

Lamina Rectangle Triangle

Area 60a2 48a2 12a2

Displacements from X GX 4a 4

3a-

Taking moments about X, the rectangle is on one side of X, taken as positive, and the triangle is on

the other side of X, taken as negative:

2 2 2 3 3 3

3

2

460 48 4 12 192 16 176

3

176 44 as required

60 15

a GX a a a a a a a

aGX a

a

æ ö - - ç ÷è ø


33 b With the particle at C, taking moments about X:

Mg GX lMg 4a

M g 44

15a l M g 4 a substituting for GX from part a

l 44

15 4

11

15

There is a vertical force at B but, as the line of action of this force passes through X, its moment

about X is zero.

34 a Let the distance of the centre of mass, G, of the loaded plate from AB and BC be x cm and y cm

respectively.

Taking B as the origin and using

mir

i r m

iåå

0 0 2 22 3 4 10 (1 2 3 4 10)

2 0 0 2

2420

20

l l l xM M M M M M

l l l y

l x

l y

æ ö æ ö æ ö æ ö æ ö æ ö ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷

è ø è ø è ø è ø è ø è ø


è ø è ø

So distance from AB x

24l

20

6l

5

b From working in part a, distance from BC y

20l

20 l

c When the framework hangs freely from D, its centre of mass G is vertically below D. Let the angle

made by DA with the downward vertical be a. Then from the diagram in part a:

tana 2l - y

2l - x

2l - l

2l - 6l5

l4l5

5

4

a 51° (to the nearestdegree)


35 a The area of rectangle ABCD is 3a 2a 6a2

The area of triangle BCE is

1

2a 2a a2

The area of lamina ABCD is 6a2 - a2 5a2

Let the distance of the centre of mass of the lamina, G, from AD be x cm. The distance of the

centre of mass of the rectangle from AD is 1.5a. Consider the centre of mass of the triangle, .G ¢

If Y is the midpoint of BE then CG¢ : G¢Y 2 :1.

Using similar triangles,

CX

XE

CG¢G¢Y

2

1

As CE a, XE

1

3a , so ¢G is

1

3a from BE and

3a-

1

3a

8

3a from AD.

Rectangle Lamina Triangle

Area 6a2 5a2 a2

Distance from AD

3

2a

x

8

3a

The moment of the lamina about AD is the moment of the complete rectangle less the moment of

the triangle which has been removed from the rectangle:

5a2 x 6a2 3

2a - a2

8

3a

5a2x 9a

3 -8

3a

3 19

3a

3

x 19

15a

b Let N be the mid-point of AB. Taking moments about N:

Mg3

2a - x

æ

èçö

ø÷ mg

3

2a

m 2

3aM

3

2a - x

æ

èçö

ø÷

2

3aM

3

2a -

19

15a

æ

èçö

ø÷

2

3

7

30M

7

45M


36 a Let the distance of the centre of mass of the decoration, G, from B be x cm.

The area of the smaller circle is 102 100

The area of the larger circle is 202 400

The area of the decoration is 100 400 500

As the card is uniform, the masses of the decoration and circles are proportional to their areas.

Decoration Small circle Large circle

Area 500 100 400

Distance from B (cm) x 30 0

Taking moments about B:

500 x 100 30 400 0

x 3000

500 6cm

b When the decoration is freely suspended from C, its centre of mass G is vertically below C.

Let the angle between AB and the vertical be a. Then from the diagram drawn for part a:

tana AC

GA

10

10 (20- x )

10

24

5

12

a 22.6° (1 d.p.)


37 a Let the distance of the centre of mass, G, of the lamina from the midpoint of AB be x cm.

The area of the semicircle is 21 25

52 2

So the area of the lamina L is 25

1002

-

From the question, the distance of the centre of mass of the semicircle from the midpoint of AB

is 4 20

as the radius is 5cm in this case.3 3

aa

Square Semicircle Lamina L

Area 100 25

π2

25

100 π2

-

Distance from midpoint of AB (cm) 5 20

3π x

Taking moments about the midpoint of AB:

100 -25

2

æ

èçö

ø÷x 100 5-

20

3

25

2 500 -

250

3

1250

3

x 2500

3(200 - 25 6.86 (2 d.p.)

b When L is suspended freely from D, the centre of mass G hangs vertically below D.

The downward vertical is drawn in the diagram. Let the angle between CD and the vertical be a.

tana GN

DN

(10- x )

5

10 - 6.86

5 0.628

a 32.1° (1 d.p.)


38 a Let the distance of the centre of mass of T from A be x cm, i.e. AG x .

The area of the smaller circle is 2 64

The area of the larger circle is 2 5762 The area of T is 576 64 512 -

As the lamina is uniform, the masses of T and the large circle are proportional to their areas.

T Small circle Large circle

Area 512 64 576

Distance from A (cm) x 16 24

Taking moments about A:

512 576 24 64 16

576 24 64 1625cm

512

x

x

-

-

b Let C be the midpoint of OB and the mass of T be M.

12cm, 36 25 11cmBC CG AC AG - -

Taking moments about C

Mg 111

4mg12

M 3m

11

39 a Using

mir

i r m

iåå

0 9 04 6 2 (4 6 2)

4 0 4

5412

8

xm m m m

y

x

y



è ø è ø

So x

54

12

9

2 and y

8

12

2

3

So the cooordinates of the centre of mass of the particles without the lamina are 9 2

,2 3

æ öç ÷è ø


39 b The uniform lamina is a triangle so its centre of mass is (3, 0). Its mass is km.

The combined system of the lamina and the three particles has centre of mass (4,l).

Its mass 4m 6m 2m km (12 k)m

Using

mir

i r m

iåå and the result from part a to treat the three particles as a single entity:

92

23

3 412 (12 )

0

54 3 4(12 )

8

m km k m

kk

l

l

æ ö æ ö æ ö ç ÷ ç ÷ ç ÷

è ø è øè øæ ö æ ö

ç ÷ ç ÷è ø è ø

So equating the i components gives:

54 3k 48 4k

k 6

c Equating the j components from the vector equation in part b and substituting for k gives:

8 (12 k)l 18l

l 8

18

4

9

d Let the angle between AC and the vertical be a.

tana 4

l 4

4

9 9

a 83.7° (1 d.p.)


40 a Let the distances of the centre of mass of L, say G, from AD and AB be x and y respectively.

The mass of L is 3 4 2 10 .m m m m m

Plate Rectangle Particles

L ABCD A B C

Mass 10m 3m 4m m 2m

Distance from AD x 2.5a 0 5a 5a

Distance from AB y a 0 0 2a

Taking moments about AD:

10m x 3m 2.5a m 5a 2m5a 22.5ma

x 22.5a

10 2.25a as required

b Taking moments about AB:

10m y 3m a 2m 2a 7ma

y 7a

10 0.7a

c When L is freely suspended from O, the centre of mass G of the complete system hangs vertically

below O. Let a be the angle between OA and the vertical.

tana y

2.5a - x

0.7a

0.25a 2.8

a 70° (to the nearestdegree)

So the angle that AB makes with the horizontal is (90- 70)° 20° (to the nearest degree)

d The total weight 10mg of the loaded plate L acts vertically through the centre of mass G, so from

part a the force has a perpendicular distance from O of 2.5a - x 2.5a - 2.25a 0.25a

The force P acts horizontally through C, so this force has a perpendicular distance from O of 2a

The loaded plate is in equilibrium, so taking moments about O:

P 2a 10mg 0.25a

P 2.5mg

2

5

4mg as required


40 e The forces acting on the system are the horizontal force at C, the weight of the loaded plate

through G and the reaction force at O. Let the horizontal and vertical components of the force

acting on L at O be X and Y respectively.

R(®) X P 5

4mg

R() Y 10mg

Let the magnitude of the force acting on L at O be R.

2

2 2 2 2 2 25 1625(10 )

4 16

1625 5 65

4 4

R X Y mg mg m g

R mg mg

æ ö ç ÷è ø

41 a As the wire is uniform, each side has a mass proportional to its length and the centre of mass

of each side is at the middle of the side. The triangle has sides in the ratio of 3 : 4 : 5 so it is a

right-angled triangle.

Taking B as the origin and axes along AB and BC:

0 4 46 8 10 (6 8 10)

3 0 3

7224

48

x

y

x

y

æ ö æ ö æ ö æ ö ç ÷ ç ÷ ç ÷ ç ÷

è ø è ø è ø è ø


è ø è ø

So distance from AB x

72

24 3cm

b 48

Using vector calculation in part , distance from 2cm24

BC y a


41 c When the frame hangs freely from A, its centre of mass G is vertically below A.

Let the acute angle made by AB with the downward vertical be a.

tana x

6- y

3

6 - 2

3

4


42 a Let the distance of the centre of mass of the loaded framework, say G, from AB and AD be x and

y respectively. As the rods that make up the framework are uniform, the centre of mass of each

rod is at its midpoint.

Taking A as the origin and axes along AB and AD:

0 1.5 1.5 3 3 36 2 (1 1 1 1 6 2)

0 2 2

3012

16

a a a a a xm m m m m m m

a a a a a y

a x

a y

æ ö æ ö æ ö æ ö æ ö æ ö æ ö ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷

è ø è ø è ø è ø è ø è ø è ø


è ø è ø

i Distance of the centre of mass of the loaded framework from AB x

30

12a

5

2a

ii Distance of the centre of mass of the loaded framework from AD y

16

12a

4

3a


42 b Let a be the angle BC makes with the vertical (see diagram for part a).

43

52

22 2 2 4tan

3 5 15

15 (to the nearest degree)

a aa y

x aa

a

--

°

43 a Let the distance of the centre of mass of the loaded framework, say G, from BC and AB be x and

y respectively. As l

2 l 2 (l 2)2 , the angle at B is a right angle. As each rod is uniform, the

centre of mass of each rod is at its midpoint.

Taking B as the origin and axes along AB and BC:

0 0.5 0.52 3 (1 2 3)

0.5 0 0.5

2.56

2

l l xm m m m

l l y

l x

l y

æ ö æ ö æ ö æ ö ç ÷ ç ÷ ç ÷ ç ÷

è ø è ø è ø è ø


è ø è ø

i Distance of the centre of mass of the framework from BC x

2.5l

6

5l

12

ii Distance of the centre of mass of the loaded framework from AB y

2l

6

l

3

b When the framework changes freely from A, its centre of mass G is vertically below A. The

vertical line has been drawn in the diagram (see part a). Let the angle that BC makes with the

vertical be a.

3

512

1 12 4tan 0.8

3 5 5


l

l

y

xa

a

°


44 a Let G be the origin so that GE lies on the x-axis and AG lies on the y-axis.

The coordinates of the centre of mass of the rectangle ABGF are 3

, .2 2

d dæ öç ÷è ø

The coordinates of the centre of mass of the square CDFE are 3

, .2 2

d dæ öç ÷è ø

Rectangle ABGF has area 3d2, square CDFE has area d2, but the square has density three times

that of the rectangle. So the mass ratios of the rectangle, square and lamina are 3d2, 3 × d2 = 3d2 and 6d2 respectively.

2 2 2

3

2

3

3

2 23 3 6

3

2 2

66

6

d d

xd d d

d d y

xdd

yd

æ ö æ öç ÷ ç ÷ æ öç ÷ ç ÷

æ öç ÷

ç ÷ç ÷ ç ÷ è øç ÷ ç ÷è ø è ø

æ

è ø

öç ÷è ø

3

2

6Distance of the centre of mass of the framework from .

6

dAG x d

d

b

3

2

6Distance of the centre of mass of the framework from .

6

dGE y d

d

c The centre of mass of the lamina lies at (d, d), which are the coordinates of point C. If the lamina

is freely suspended from point A, the centre of mass will align itself vertically below point A, so that AC will be the new vertical.

Let the angle that AB makes with the vertical be , then .CAB Ð

2

tan 2


CB d

AB d

°


45 a Let N be the foot of the perpendicular from D to BC and NC = p cm.

The total length of the frame is AB BC CD AD 20 4 BC 5 4 20 BC 7cm

Hence NC = BC – BN = 7 – 4 = 3

Let the distance of the centre of mass of the frame from AB be x cm

Calculating the mass of each section of wire from the densities given in the question:

AB BC CD DA

Mass (kg) 0.04M 0.07M 0.075M 0.04M

Distance from AB (cm) 0 3.5 5.5 2

Taking moments about AB:

(0.04 0.07 0.075 0.04) 0.04 0 0.07 3.5 0.075 5.5 0.04 2

0.73753.2777 3.28cm (3 s.f.)

0.225

M x M M M M

x

…

b The weight of the framework acts through the centre of mass, which is (3.5- x )cm from the

vertical through the midpoint of BC.

Taking moments about the midpoint of BC:

(3.5 ) 3.5

3.5 3.5 3.27770.0635 (3 s.f.)

3.5 3.5

Mg x kMg

xk

-

- -


46 a Let the midpoint of BC be G and the midpoint of EG be F. Divide the lamina into three sections, a rectangle ABEG, a square CDFG and a triangle EDF.

Let A be the origin so that AB lies on the x-axis and AE lies on the y-axis.

The coordinates of the centre of mass of the rectangle ABEG are (40, 20).-

The coordinates of the centre of mass of the square CDFG are (60, 60).-

The coordinates of the centre of mass of the triangle EDF are found by taking the average of the

coordinates of its three vertices, in this case (0,-40) , (40,-40) and (40,-80) , hence it is:

0 40 40

3,-40 -80 - 40

3

æ

èçö

ø÷

80

3,-

160

3

æ

èçö

ø÷

Rectangle ABEG has area 3200 cm2, square CDFG has area 1600 cm2, and triangle EDF has

area 800 cm2.

Using

mir

i r m

iåå

80

40 60 33200 1600 800 (3200 1600 800)

20 60 160

3

5600736 000

3

608 000

3

x

y

x

y

æ öç ÷æ ö æ ö æ ö

ç ÷ç ÷ ç ÷ ç ÷- - ç ÷è ø è ø è ø-ç ÷è ø

æ öæ ö ç ÷

ç ÷ è øç ÷ç ÷ç ÷-ç ÷è ø

The distance of the centre of mass of the lamina from AE 736 000

43.8cm (3 s.f.)3 5600

x

b 1 2

R( ) : T T W (1)


2

2

2

80 0

736 00080

16800

736 230.548 N (3 s.f.)

1344 42

W x T

T W

T W W W

-

1 2

19From equation 0.452 N (3 s.f.)

42T W T W W - (1)


46 c 1 2

R( ) : T T W kW (2)


2

2

2

80 80 0

736 00080 80

16 800

23

42

W x kW T

T W kW

T W kW

-

2

23So if 8 8

42

238

42

7.45 (3 s.f.)

T W W kW W

k

k

-

� �

�

�

1 2

1

19From equation

42

10

T W kW T W

T W

-

(2)

�

So the largest value of k satisfying T1 ⩽ 10W and T2 ⩽ 8W is k = 7.45 (3 s.f.)

47 The centre of mass lies on the axis of symmetry, y = 0.

To find the x coordinate, using

x y2xdxòy2 dxò

, and substituting y x gives:

4 42 3130 0

4 421

2 00

d 64 88

3 3d

x x xx

xx x

òò

So the coordinates of the centre of mass of the solid are 8

, 03

æ öç ÷è ø

, a distance of 8

3 from O.


48 a Let the straight edge lie along the y-axis. Then the centre of mass lies on the x-axis from

symmetry.

If P has coordinates (x, y) and the elemental strip PQRS has width d x then its area is 2 .y xd

The mass M of the lamina = 21,

2a where is the mass per unit area of the lamina.

Let x be the distance of the centre of mass from O, then M x 2xydx

0

a

ò

As the boundary of the semicircle has the equation x

2 y2 a

2, then y (a

2 - x2)

12

312 22 2 2 2 3

00

323

212

2 2So 2 ( ) d ( )

3 3

4 as required

3

aa

M x x a x x a x a

a ax

a

- - -

ò

b Using the result from part a and letting x be the distance of the centre of mass of the resulting

lamina from O:

Shape Mass Distance of centre

of mass from O

Semicircle radius a

1

2a2

4a

3

Semicircle radius b

1

2b2

4b

3

Resulting

1

2(a2 - b2) x

Taking moments about O:

2 2 2 2

3 3 2 2 2 2

2 2

1 1 4 1 4( )

2 2 3π 2 3π

4 ( ) 4 ( )( ) 4 ( ) as required

3 ( ) 3 ( )( ) 3 ( )

a ba b x a b

a b a b a ab b a ab bx

a b a b a b a b

- -

- -

- -

c As b ® a, the area becomes a circular arc and from the equation found in part b

x ®4

3

(a2 a2 a2 )

(a a)

4

3

3a2

2a

2a


49 a Let B be the origin so that BC lies on the x-axis and AB lies on the y-axis. Let the equation of the

line AC be y c- mx , so that the coordinates of the triangle are (0, 0), (0, c) and ,0c

m.

If a point on the line has coordinates (x, y), then an elemental horizontal strip of width d y at that

point has area is xd y

The mass M of the triangular lamina =

1

2c

c

m

c22m

, where is the mass per unit area

Let y be the distance of the centre of mass from BC, then M y xyd y

0

c

ò

2

2 2 3

0 0 00

3 3

2

( )So given

1 1d ( ) d ( )d

2 2 3

2 1 1

2 3 3

cc c c

c yx

m

cy xy y c y y y cy y y cy y

m m m m

cy c c

c

-

æ ö - - -ç ÷ è ø

æ ö - ç ÷è ø

ò ò ò

b Divide the lamina into two sections, a rectangle and a triangle. Let be the mass per unit area of

the lamina. Let the point P be the origin, so that PQ is the x-axis and SP is the y-axis, and the

coordinates of G, the lamina’s centre of mass be (x , y ) .

The mass of the rectangle is 2a2 and its centre of mass is , .

2

aa

æ öç ÷è ø

The mass of the triangle is a

2 and its centre of mass is

4a

3,

2a

3

æ

èçö

ø÷, using the result from part a.

Using

mir

i r m

iåå

2 2 2 2

4

32 (2 )2

2

3

7

33

8

3

aa

xa a a a

a ya

a

x

a y

æ öæ ö ç ÷ æ öç ÷ ç ÷ ç ÷ç ÷ç ÷ ç ÷ è øç ÷è ø

è ø

æ öç ÷ æ ö

ç ÷ ç ÷ç ÷ è øç ÷è ø

The distance of the centre of mass G of the lamina from PS 7

9

ax

c The distance of the centre of mass G of the lamina from PS 8

9

ay


50 a Use

x y2xdxòy2 dxò

, and substitute 2

1

2y

x

x xy2 dx

1

2

òy2 dx

1

2

ò

x 14

x-4 dx1

2

ò14

x-4 dx1

2

ò

x-3 dx1

2

òx-4 dx

1

2

ò

-1

2x-2

1

2

-1

3x-3

1

2

3 1-2 - 2-2 2 1

-3 - 2-3

3

2

3

4

8

7

9

7

The centre of mass is 9

m7

from the y-axis, hence 9 2

1 m7 7

æ ö- ç ÷è ø

from the larger plane face.

b Let be the mass per unit volume of the hemisphere H and the solid S.

Then the mass of the S = 2

2 22 4 3

1 11

1 7d d 1

4 12 12 8 96y x x x x

- - æ ö - - ç ÷ è øò ò

The mass of the hemisphere and its distance of its centre of mass from its plane face can be found

from the standard formulae, using r = 0.5.

Shape Mass Distance of centre of mass from

trophy’s plane face

Solid, S

796

1-

2

7

5

7

Hemisphere, H

23

1

2

æ

èçö

ø÷

3

12

1

3

8

1

2

19

16

Trophy, T

7

96

1

12

æ

èçö

ø÷

532

x

Taking moments around the trophy’s plane face (the bottom of the trophy as shown):

532

x 12

19

16

796

5

7

x 32

5

19

192

5

96

æ

èçö

ø÷

32

5

29

192

29

30 0.967 m (3 s.f.)


51 a A hemisphere is generated when a semicircle is notated through 180° about the x-axis.

Divide the hemisphere into circular discs, with each disc having mass y2d x and centre of mass

at a distance x from O.

2 2 2

0 0

2 2 2

0 0

2 2 4 4 4

0

3 32 3

0

d ( )dSo

d ( )d

1 1 1 13 1 32 4 2 4

1 2 4 8133

R R

R R

R

R

xy x x R x xx

y x R x x

R x x R RR R

R RR x x

-

-

- - --

ò òò ò

b Using the standard formulae for a cone:

Shape Mass Distance of centre of mass

from V

Cone

1

3a2ka

3

4ka

Hemisphere

2

3a3

ka

3

8a

Top

1

3a3(k 2) x

Taking moments about V:

1

3a3(k 2)x

1

3a3k

3

4ka

æ

èçö

ø÷

2

3a3 ka

3a

8

æ

èçö

ø÷

(k 2)x 3

4k 2a 2ka

3a

4

x (3k 2 8k 3)a

4(k 2)


51 c The manufacturer’s requirement is that x ka

Hence from part b 3k

2 8k 3

4(k 2) k

3k 2 8k 3 4k 2 8k

k 2 3, so k 3

52 a Using the standard results for a hemisphere:

Shape Mass Ratio of masses Distance of centre

of mass from O

Large hemisphere

2

3a

3 8

3

8a

Small hemisphere

2

3

a

2

æ

èçö

ø÷

3

1

3

16a

Remainder

2

3

7a3

8 7 x


7x 83

8a -1

3

16a

x 1

7

45

16a

45a

112

b Using the result from part a:

Mass ratios Distance of centre

of mass from O

Bowl M 45

112a

Liquid kM 3

16a

Bowl + liquid (k + 1)M

17

48a


(k 1) M 17

48a M

45

112a kM

3

16a

k17

48-

3

16

æ

èçö

ø÷

45

112-

17

48

8

48k

45

112-

17

48

k 645

112-

17

48

æ

èçö

ø÷

6(2160 -1906)

5376

254

896

2

7


53 a V y2

0

1

ò dx 4

(x - 2)4

0

1

ò dx substituting y 1

2(x - 2)2

2

5 5 3

0

32 8( 2) ( 2) cm

20 20 20 5V x

- - -

b Let x be the distance of the centre of mass of S from its plane face

To find, using M x y2xdxò , and substituting 2 21 8

( 2) and from part 4 5

y x M

- a

gives:

x

5

32(x - 2)4 xdx

0

2

ò

Integrate using the substitution u x - 2

x 5

32u4(u 2)du

-2

0

ò 5

32u5 2u4 du

-2

0

ò

5

32

1

6u6

2

5u5

-2

0

5

32

2

5 25 -

1

626æ

èçö

ø÷

5

32

64

5-

64

6

æ

èçö

ø÷

101

5-

1

6

æ

èçö

ø÷

10

30

1

3

The centre of mass lies on the axis of symmetry at a distance of 1

cm3

from the plane base.

c Let the reaction force at A be A, and at B be B.


18 2 10 4

3

240

118 593

8 24 12

A W W

WW W

A

æ ö-ç ÷è ø


54 a Using the formulae for standard uniform bodies:

Shape Mass Mass ratios Distance of centre

of mass from base

Cylinder r 2h 1 2

h

Cone

1

3r 2 h

2

1

6

h-1

4

h

2

æ

èçö

ø÷

7h

8

Ornament

5

6r 2h

5

6 x

Take moments about O, the centre of plane base:

5

6x 1

h

2-

1

6

7h

8

5

6x

h

2-

7h

48

17h

48

x 17h

48

6

5

17h

40

b The centre of mass G of the ornament will be directly below the point of suspension.

tana h-

17h

40r

23h

40r

As h 4r, this gives tana 23r

10r 2.3

a 66.5° (1 d.p.)


55 a Let O be the midpoint of YZ. Let OZ be the x-axis and OX the y-axis. The triangle is isosceles and

the y-axis is the line of symmetry.

The equation of the line XZ is y h- mx x

h- y

m

The area of the elemental strip of width d y is 2xd y

Let y be the distance of the centre of mass from O, then:

20

0 0

0 00

2 3 3

0

22

0

d2 d d

2 d dd

1 1

12 3 61 3122

hh h

h hh

h

h

h yy yyx y yh y ym

yh yx y h y yy

m

hy y h

h

hhy y

-æ öç ÷ -è ø -æ ö -ç ÷

è ø

- -

òò òò òò

So distance from 2

3 3

h hX h -

b Let E be the midpoint of BC, then BE = 4a and AE = 3a, as DABE is a 3 : 4 : 5 triangle,

and DE = a, from part a.

Let be the mass per unit area of the lamina.


of mass from A

∆ABC 212 a 2a

∆BCD 4a2

22

3

aa

Plate ABDC 8a2 x


8a2x 12a2 2a - 4a2 8a

3

æ

èçö

ø÷

24a2 -32a

3

3

40a3

3

So x

40a

8 3

5a

3


55 c Let be the angle between CB and the vertical.

From part b, the distance GE is 5 4

33 3

a aa -

Taking moments about C:

Mg 8asin Mg4a

3cos - 4asin

æ

èçö

ø÷

12 Mgasin 4

3Mgacos

sincos

tan 4

36

1

9

So arctan1

9

So CB makes an angle1

arctan9

æ öç ÷è ø

with the vertical.

56 a Let be the mass per unit area of the material and x the distance of the centre of mass of the

closed container from O. Using the formulae for standard uniform bodies:

Shape Mass Distance of centre of

mass from O

Circular disc a2 0

Hemispherical bowl 2a2

1

2a

Closed container 3a2 x


2 23π 0 2π2

3

aa x a

ax


56 b The container is resting in equilibrium, so the weight of P acts to one side and the weight of C

balances on the other side.


Mg a

3sin

1

2Mg acos

So tan 3

2

56° (to the nearest degree)

57 Let V be the vertex of the cone and O be the centre of its base.

Let G be the position of its centre of mass.

From ∆VAO, tana

OA

OV

a

h, where h is the height of the cone.

So tana 1

3

a

h h 3a

So from the standard result for a uniform solid right circular cone OG 1

4h

3a

4

Then from ∆GAO

tan AO

GO

a

3a

4

4

3

53.1° (1 d.p.)


58 a The hemisphere K has mass M. The hemisphere H has double the radius, so it has 8 times the

volume. Its mass is 8M. Therefore the composite body S has mass M 8M 9M .

Shape Mass Distance of centre of

mass from C

H 8M 3

8

a

K M 3

16

a-

S 8M M x

Taking moments about C:

9M x 8M 3a

8- M

3a

16 M

45a

16

x 45a

916

5a

16

b The composite body with particle P attached rests in equilibrium, with C above the point of

contact with the plane, X.

Mg acosa 9Mgx sina 95

16Mg sina substituting result for x from part a

tana 16

45


59 a A line R

y R xh

- rotated around the x-axis with generate a solid cone with its base on the y-axis.

The volume of an elemental strip of width d x is 2 .x xd

Let x be the distance of the centre of mass of the solid come from its base, then:

22 2

2 2 3220 00

2 2 22

2 2

0 200

2 2 2 3 2 4

2 22

0

2 2 2 32

2

2

0

d 2 dd

d 2 dd

2 1 2 1

2 3 4 2 3 4

11 1

33

13

12 4

h hh

hhh

h

h

R R Rx R x x R x x x xxy x h h hxR RRy x R x x xR x xh hh

R x R x R xR h

h h

R x R x R hR xh h

hh

æ ö - - ç ÷ è ø æ ö - -ç ÷è ø

æ ö- - ç ÷ è ø æ ö - ç ÷- è ø

ò òòò òò


59 b Using the result from part a:

Shape Mass Mass ratio Distance from base of

centre of mass

Large cone

1

3a2 H H

H

4

Small cone

1

3a2h h

h

4

Solid S

1

3a2(H - h) H – h x

Taking moments about the base:

(H - h)x H H

4- h

h

4

x 1

4

(H 2 - h2 )

(H - h)

1

4

(H - h)(H h)

(H - h)

1

4(H h)

So the distance from the vertex is H -

1

4(H h)

3

4H -

1

4h

1

4(3H - h) as required

c Let the tension in the string attached to the vertex be T1 and tension in the other string be T2.

Taking moments about the centre of mass G:

2 1

11 4

12 4

1( ) where (3 ) from part

4

( )

(3 ) 3

T x T H x H x H h

H hT x H h

T H x H h H h

- - -

- - -

b


60 a Let be the mass per unit volume of the material of the cylinder and x the distance of the centre

of mass of the toy from O. Using the formulae for standard uniform bodies:

Shape Mass Mass ratio Distance of centre

of mass from O

Hemisphere 32

π 63

r 4r 5

8

r

Cylinder 2πr h h 2

hr

Toy 2π (4 )r r h 4r h x


(4r h)x 4r 5r

8 h

h

2 r

æ

èçö

ø÷

x 5r 2 h2 2rh

2(4r h)

h2 2hr 5r 2

2(h 4r) as required

b If the toy remains in equilibrium when resting on any point of the curved surface, its centre of

mass must be in the centre of the hemisphere’s flat surface, so x r.

Hence from part a h2 2hr 5r 2

2(h 4r) r

h2 2hr 5r 2 2rh8r 2

h2 3r 2

h 3r

61 a Let x be distance of the centre of mass from AB on the axis of symmetry in the direction away

from O. Using the formulae for standard uniform bodies:


of mass from AB

Hemisphere M 3

8r

Cone m 3

4r-

Toy m + M x

Taking moments about AB:

(m M )x 3

8Mr -

3

4mr

3r

8( M - 2m)

x 3( M - 2m)

8( M m)r as required


61 b Let D be the point on the axis of symmetry vertically above B when the toy is placed on a

horizontal surface. The toy will not remain in equilibrium if its centre of mass is not on the line

segment OD, i.e if x > CD .

From the diagram: tana

r

3r

CD

rCD

1

3r

So x >CD 3( M - 2m)

8( M m)r >

1

3r

9( M - 2m) > 8( M m)

M > 26m as required

62 a The centre of mass lies on the axis of symmetry OX. Let the distance of the centre of mass of S

from O be x and let it be at point G.

Using

x y2xò dx

y2 dxò

and substituting y

2 rx gives:

x rx

2dx

0

y

òrxdx

0

y

ò

13rx3 0

y

12rx

2 0

y

1

3r 4

1

2r3

2

3r

b The solid will not topple providing G is vertically above a point on the plane face in contact with

the inclined plane. The maximum value of a is when G is directly above N, the edge of the solid.

tana r

r -2

3r

r 1

3r 3



63 a Using

y 12

y2 dxòydxò

with y sin x gives:

y 12sin2 xdx

0

òsin xdx

0

ò

14

1- cos2xdx0

òsin xdx

0

ò

1

4

x - 12sin2x 0

-cos x 0

1

4

(11)

8

b When S is on the point of toppling, G is above O. Let A be the point midway along the base of S.

2 2tan 4

2


y

°


64 a Let be the mass per unit volume of the material and x be the distance of the centre of mass of

the solid S from O. Using the formulae for standard uniform bodies:

Shape Mass Mass ratios Distance of centre

of mass from O

Cylinder

(2a)2 3

2a

æ

èçö

ø÷ 6

3

4a

Hemisphere

2

3a

3 2

3

3

8a

Solid, S

6a3 -2

3a3æ

èçö

ø÷

16

3 x


16

3x 6

3

4a -

2

3

3

8a

9

2a -

1

4a

17

4a

x 51a

64 0.797a (3 s.f.)

b On the point of toppling: G is above S – the lowest point on the bottom circular face.

Let X be the centre of the base of the cylinder.

2 64 2 128tan substituting value for from part

3 96 51 45

2

70.6 (3 s.f.)

SX ax

XGa x

a

a

--

°

a


64 c There are three forces acting on S, its weight, the reaction force and friction.

When S is on the point of sliding F mR 0.8R

Resolving perpendicular to the plane

R( ) : cos 0 cos

Resolving along the plane

R( ) : sin 0 sin

Using 0.8 gives

sin 0.8 cos tan 0.8

So 38.7 (3 s.f.)

R mg R mg

F mg F mg

F R

mg mg

-

-

°

տ

ր

65 a Let be the mass per unit volume of the material and x be distance of the centre of mass of the

bowl B from O. Using the formulae for standard uniform bodies:


of mass from O

Large hemisphere

2

3(6a)

3 144a3 27

3

8 6a

Small hemisphere

2

3(2a)

3 16

3a

3 1

3

8 2a

Bowl B

2

3(6

3 - 23)a

3 416

3a

3 26 x


26x 27 3

8 6a -1

3

8 2a

243a

4-

3a

4 60a

x 30a

13


65 b Let y be distance of the centre of mass of the bowl B from O.


of mass from O

Bowl, B

416

3a3

52 30

13

a

Cylinder 24a3 9 6a + 3a

Combined solid, S

488

3a3

61 y


61y 52 30a

13 9 9a 120a 81a

y 201

61a

c Let N be the point on the plane vertically below the centre of mass of the combined solid G, and

let X be the centre of the cylindrical base of S.

NX GX tan12° 12a-201

61a

æ

èçö

ø÷tan12°

531

61a tan12° 1.85a (3 s.f.)

The solid will topple when point N is outside the base circle; as NX < 2a, S will not topple.



the solid from O. Using the formulae for standard uniform bodies:

Shape Mass Mass ratio Distance from O of

centre of mass

Hemisphere

2

3a3 2

3a

8

Cone

1

3a

3 1

a

4

Solid

1

3a

3 1 x


1 x 23

8a-1

a

4

x a

2

b As the solid is resting in equilibrium, its centre of mass G is above the point of contact N between

the solid and the plane.

From DOGN , and using the result from part a:

sina

a

2a

1

2a

6


66 c There are three forces acting on the solid, its weight, the reaction force and friction.

For limiting equilibrium, when the solid is about to slip .F Rm

Resolving the forces perpendicular and parallel to the inclined plane:

R( ) : sin

R( ) cos

As , sin cos

sintan

cos

1tan

6 3

1 1So , hence is thesmallest valueof

3 3

F mg

R mg

F R mg mg

a

a

m a m aa

m m aa

a a

m m

ր

տ

� �

�

67 a Let be the mass per unit volume of the material and x be distance of the centre of mass of the

bollard from O. Using the formulae for standard uniform bodies:

Shape Mass Mass ratio Distance from O of

centre of mass

Cone

1

3(3r)2 h 3h

3

4h

Cylinder (4r)2 r 1 6 r

h

r

2

Bollard (16r 3h)r 2 16 3r h x


(16r 3h)x 3h3h

416r h

r

2

æ

èçö

ø÷

9h2

416rh 8r 2

x 32r 2 64rh 9h2

4(16r 3h)


67 b When the bollard is on the point of toppling, its centre of mass G lies above N a point on the plane

which is at the lowest point on its base.

Let X be the centre point of the base, so GX OX - x

As h 4r, OX h r 5r

And from part a: x 32r 2 256r 2 144r 2

4(28r)

432r

112

27r

7

So GX 5r -27r

7

8r

7

From DGNX : tana NX

GX

4r

8r

7

28

8 3.5



the tree from O. Using the formulae for standard uniform bodies:

Shape Mass Mass ratio Position of centre of

mass from O

Cylinder r 2h 3 2

h

Cone

1

3(2r)2 h 4

4

h-

Tree

r 2h 14

3

æ

èçö

ø÷ 7 x


7x 3h

2- 4

h

4

h

2

x h

14


68 b When the tree is on the point of toppling, its centre of mass G lies above S, a point on the desk

which is at the lowest point on its cylindrical base. GX can be found using the result in part a.

tana r

h- h14

7

26

r 7

26

13h

14

æ

èçö

ø÷

1

4h

69 a Using the formulae for standard uniform bodies:

Shape Mass Distance of centre of mass from O

Hemisphere, H 2M 3

8h r

Cylinder, C 3M

h

2

Body 5M x


5M x 2M h3

8r

æ

èçö

ø÷ 3M

h

2 2h

3

4r

3h

2

7h

2

3r

4

x 14h 3r

20


69 b When the body is on the point of toppling, its centre of mass G lies directly above a point on the

plane which is at the lowest point on its cylindrical base.

From the diagram: tana

r

x

20r

14h 3r

As tana 4

3, this gives

20r

14h 3r

4

3

60r 56h12r

48r 56h

h 48

56r

6

7r

70 a Let be the mass per unit volume of the material in the cylinder and x be the distance of the

centre of mass of the top from O. Using the formulae for standard uniform bodies:

Shape Mass Mass ratio Position of centre of mass

from O

Cylinder 36r3

1 2r

Cone 36r3 1 –r

Toy 72r3 2 x


2x 1 2r 1 (-r) r

x r

2

b In ∆AGX, AX is the radius 3r of the cylinder and X is the centre of the base of the cylinder.

tan AX

GX

AX

OX -OG

3r

4r - x

3r

4r - r2

6

7

47.5° (1 d.p.)


70 c The toy will not topple if G (the centre of mass) is vertically above a point between B and V.

Let Y be the point directly above B on the axis of symmetry of the toy.

tana 3r

4rfrom DVOB and tana

OY

OB

OY

3rfrom DBOY

So OY

3r

3

4OY

9r

4

But OG r

2so OG < OY

This means that the centre of mass is above the face of the body in contact with the place, so the

toy will not topple.

71 a Total mass 1.5

0

3d

(1 )(2 )h

h h

ò

Rewrite

3

(1 h)(2 h) as

1

1 h-

1

2 h

æ

èçö

ø÷ so:

1.5 1.5

00

1 13 d 3 ln(1 ) ln(2 )

1 2

53(ln 2.5 ln 3.5 ln 2) 3ln 1.07kg (2 d.p.)

3.5

m h h hh h

æ ö - - ç ÷ è ø

-

ò

b Let h be the distance of the centre of mass of the rod from its base.

Taking moments about the base of the rod:

1.5 1.5

0 0

1.52

1.5

0

0

5 3 2 13ln d 3 d

3.5 (1 )(2 ) 2 1

(2 )3 2 ln(2 ) ln(1 ) 3 ln

1

493(ln 4.9 ln 4) 3ln

40

hh h h

h h h h

hh h

h

æ ö -ç ÷ è ø

-

-

ò ò

So 4940

53.5

ln0.57 m (2 d.p.)

lnh


72 a Total mass m

5

1 4x2dx

0

2.4

ò 51

1 4x2dx

0

2.4

ò 51

2arctan 2x

0

2.4

So m 2.5arctan4.8 3.41kg (2 d.p.)

b Let x be the distance of the centre of mass of the oar from its end.

Taking moments about the end of the oar:

2.5arctan4.8 x 5x

1 4x2dx

0

2.4

ò 5

8ln(1 4x2 )

0

2.4

5

8ln24.04

So 2 ln 24.04

0.58m (2 d.p.)8 arctan 4.8

x

73 a Since the mass per unit length of the rod increases as the distance from A increases, the centre of mass of the rod will be closer to B than to A.

b Total mass m

(10 kx)dx0

20

ò 10x k

2x2

0

20

200 200k

As m = 750, this gives 200 200 750k

k

11

4

c Let x be the distance of the centre of mass of the oar from A.


2020

2 3

00

11 11 22 000 28 000750 10 d 5 2000

4 12 3 3

28 000 112 112m

750 3 3 3 9

x x x x x x

x

æ ö ç ÷ è ø

ò


74 Let the length of the rod be L and its mass be M. Let x be the distance of the centre of mass of the oar from A.

M (8 x2 )dx0

L

ò 8x x3

3

0

L

8LL3

3

M x (8 x2 )xdx0

L

ò 4x2 x

4

4

0

L

4L2 L

4

4

If the tension in the string at A is T then the tension in the string at B is 2T.

3

3

R( ) : 2 83

8

3 9

LT T Mg L g

L LT g

æ ö ç ÷

è ø

æ ö ç ÷

è ø


4

2

3

2

4 24

28

Mg x LT

LL g LT

LT L g

æ ö ç ÷

è ø

æ ö ç ÷

è ø

Equating the two expressions for T gives:

8L

3

L3

9

æ

èçö

ø÷g 2L

L3

8

æ

èçö

ø÷g

2

3

L2

72 L2 48

L 48 4 3 m


Challenge

1 The two particles have equal but opposite velocities just before collision at point A. Let their velocities at point A be u and –u respectively.

After collision, P1 travels three times the distance travelled by P2 before colliding at point B. Therefore, the velocities after collision are –3v and v respectively.

Using conservation of momentum:

mu 2m(-u) m(-3v) 2mv

u v

Coefficient of restitution:

1

3 2

u ue

u u

- -

- -

2 a Let the instantaneous speed at any point be w and the normal reaction between the ball and the ring

is N. Then resolving towards the centre of the circle:

N

mw2

R using Newton’s second law

So the frictional force opposite to the direction of motion is given by:

F m

mw2

R

Resolving in the direction of motion:

m

dw

dt -m

mw2

R

Let v be the velocity at time t, so integrating:

2 0

0

1d d

1

1 1

1 1

v t

u

vt

u

w tw R

tw R

tv u R

R u tt

v u R uR

uRv

R u t

m

m

m

m m

m

-

- -

- -

ò ò


Challenge

2 b From part a

v dx

dt

uR

R umt

The ball completes one revolution in time t when x 2R

1dx0

ò uR

R umtdt

0

t

ò 20

0.510tdt

0

t

ò

2ln(0.510t) 0

t

2(ln(0.510t)- ln0.5)

2ln(1 20t)

20t e2 -1

t e2 -1

20 0.191s (3 s.f.)

3 a R() : T

1T

2T

3cos45° 4Mg

3R( ) : cos 45 0T® °

Therefore T1 + T2 = 4Mg (1)


2 1

5 10 20T T Mg

2 12 4T T Mg (2)

Subtracting (1) from (2) gives T1 = 0 and T2 = 4Mg


Challenge

3 b First find the centre of mass of the lamina. Let A be the origin. Using geometry, the coordinates of the points on the lamina are:

A(0,0), B(10,0), C(5,0), D15

2,-5 2

æ

èçö

ø÷, E

5

2,-5 2

æ

èçö

ø÷

The x-coordinate of the centre of mass of the lamina is x 5 by symmetry.

The lamina is composed of three equivalent triangles ACE, CED, BCD, each with height 5 2 and

triangle CED has twice the density as the other two triangles.

So the y-coordinate of the centre of mass of the lamina is given by:

5 2 10 2 5 24 2

3 3 3

30 2 5 2

12 2

M y M M M

y

æ ö æ ö æ ö - - -ç ÷ ç ÷ ç ÷ç ÷ ç ÷ ç ÷

è ø è ø è ø

- -

If a mass of 10M is attached to the lamina at B, while AB remains horizontal the position of the

new centre of mass is:

510

14 4 105 20

2

60

7

5 2

7

xM M M

y

x

y

æ öæ ö æ öç ÷ ç ÷ ç ÷ç ÷-è ø è øç ÷

è ø

æ öç ÷æ öç ÷ç ÷ç ÷è ø -ç ÷è ø

The new centre of mass will align itself so that it is vertically below point A.

So angle made by AB with the vertical in this new position is 5 2

arctan 6.72 (3 s.f.)60

æ ö

°ç ÷ç ÷è ø


Challenge

4 Consider a cylindrical disc at a distance x cm from the base of the hemisphere. The thickness of the

disc is d x and the radius of the disc is 122 2)(r x- so that the volume of the disc is 2 2( ) .r x xd -

Let m be the total mass of the hemisphere:

2 2 2 2 2 3 4

00

3 4

5 2 5( )(5 2)d 2

2 3 4

4 5

3 4

rr

m r x x x r x r x x x

r r

- - -

ò

Let x be the distance of the centre of mass of the hemisphere from its plane face. Taking moments about the face:

3 4 2 2 2 2 2 3 4 5

00

4 5

2

2

4 5 5 1( )(5 2) d

3 4 3 2

2

2 3

26 82 3

4 5 16 15

3 4

rr

r r x r x x x x r x r x x x

r r

r rr r

xr r

æ ö - - -ç ÷ è ø

ò

When the hemisphere is on the point of tipping, tan(arctan 2)r

x

26 8So 2

16 15

12 16 16 15

4 cm

r rr

r

r r

r

æ öç ÷è ø

review exercise 1€¦ · (1 sin ) substituting for from equation cos sin (1 sin ) tan cos cos t t...

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