tenth edition - welcome to apl100 -...

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS

Tenth Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

Phillip J. Cornwell

Lecture Notes:

Brian P. Self California Polytechnic State University

CHAPTER

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

15 Kinematics of

Rigid Bodies


Vector Mechanics for Engineers: Dynamics

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2 - 2

Quiz Date 17 August

Time: 6:15 p.m.



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Contents

15 - 3

Introduction

Translation

Rotation About a Fixed Axis: Velocity

Rotation About a Fixed Axis:

Acceleration

Rotation About a Fixed Axis:

Representative Slab

Equations Defining the Rotation of a

Rigid Body About a Fixed Axis

Sample Problem 5.1

General Plane Motion

Absolute and Relative Velocity in Plane

Motion

Sample Problem 15.2

Sample Problem 15.3

Instantaneous Center of Rotation in

Plane Motion

Sample Problem 15.4

Sample Problem 15.5

Absolute and Relative Acceleration in

Plane Motion

Analysis of Plane Motion in Terms of a

Parameter

Sample Problem 15.6

Sample Problem 15.7

Sample Problem 15.8

Rate of Change With Respect to a

Rotating Frame

Coriolis Acceleration

Sample Problem 15.9

Sample Problem 15.10

Motion About a Fixed Point

General Motion


Three Dimensional Motion. Coriolis

Acceleration

Frame of Reference in General Motion




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Applications

15 - 4

How can we determine the velocity of the tip of a turbine blade?



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Applications

2 - 5

Planetary gear systems are used to get high reduction ratios

with minimum weight and space. How can we design the

correct gear ratios?



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Applications

2 - 6

Biomedical engineers must determine the velocities and

accelerations of the leg in order to design prostheses.



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15 - 7

• A rigid body is made of infinite points and

distance between any two points does not

change during motion.

• Kinematics of rigid bodies: relations

between time the positions, velocities, and

accelerations of the particles forming a rigid

body.



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Introduction

15 - 8

• Classification of rigid body

motions:

- general motion

- motion about a fixed point

- general plane motion

- rotation about a fixed axis

• curvilinear translation

• rectilinear translation

- translation:



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The rate of change of a scalar A is

independent of frame of reference.

If F is a fixed frame of reference and m

is a moving coordinate system

ȦǀF = Ȧǀm

Scalar



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1. Rate of change of vector in

translating frame

2. .Rate of change of vector in a

rotating frame

3. Relation between velocities and

accelerations of points on a rigid

body.

10

Topics



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Translating frames

•Let F & T be the two reference frames

such that frame T translates w.r.t F i.e

no rotations are involved.

•During translatory motion, the

coordinate axes in the two frames

retain same orientation w.r.t each

other.



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xyz translate relative to XYZ.

i,j,k as seen from XYZ remain constant in

magnitude and direction.



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Velocity of a point moving in a translating

frame



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Acceleration of a point in translating frame



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Rate of change of vector in rotating frame



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XYZ fixed frame F

X’Y’Z’ fixed frame F’ with Z’ oriented in

direction of w

xyz moving frame with angular velocity w. z

is also oriented in direction of w



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Rate of change of vector in fixed frame

=

Rate of change of vector in moving

frame + w x A



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• For a vector fixed in m, i.e.,



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2nd derivative



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• Relation between velocity of two points

on a rigid body in fixed frame of

reference.



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body in fixed frame of reference.



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2 - 31

While Solving problems se will use vectors.

We will not use graphical method,

i.e. velocity or acceleration diagrams



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Concept Quiz

15 - 32

What is the direction of the velocity

of point A on the turbine blade?

A a) →

b) ←

c) ↑

d) ↓

Av rw

x

y

ˆ ˆAv k Liw

ˆAv L jw

w

L



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Concept Quiz

15 - 33

What is the direction of the normal

acceleration of point A on the turbine

blade?

A a) →

b) ←

c) ↑

d) ↓ 2

na rw

x

y

2 ˆ( )na Liw 2ˆ

na L iw

w

L



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Group Problem Solving

15 - 34

A series of small machine components being moved by a

conveyor belt pass over a 150 mm-radius idler pulley. At

the instant shown, the velocity of point A is 375 mm/s to the

left and its acceleration is 225 mm/s2 to the right. Determine

(a) the angular velocity and angular acceleration of the idler

pulley, (b) the total acceleration of the machine

component at B.



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15 - 35

SOLUTION:

• Using the linear velocity and

accelerations, calculate the angular

velocity and acceleration.

• Using the angular velocity,

determine the normal acceleration.

• Determine the total acceleration

using the tangential and normal

acceleration components of B.



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15 - 36

v= 375 mm/s at= 225 mm/s2

Find the angular velocity

of the idler pulley using

the linear velocity at B.

375 mm/s (150 mm)

v r

2.50 rad/sw

B



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15 - 37

v= 375 mm/s at= 225 mm/s2

2225 mm/s (150 mm)

a r

21.500 rad/s

B Find the angular

acceleration of the idler

pulley.



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15 - 38

v= 375 mm/s at= 225 mm/s2 B Find the normal

acceleration of point B.

2

2(150 mm)(2.5 rad/s)

na r

2937.5 mm/san

What is the direction of the normal

acceleration of point B?

Downwards, towards the center



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2 - 39



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2 - 40

Video

newfilefinal.avi



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Knee joint

2 - 41

video

Superimposed four-bar linkage to follow joint flexion there are no fixed link lengths.flv



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2 - 42

In the position shown, bar AB

has an angular velocity of 4 rad/s

clockwise. Determine the angular

velocity of bars BD and DE.

Which of the following is true?

a) The direction of vB is ↑

b) The direction of vD is →

c) Both a) and b) are correct



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2 - 43

In the position shown, bar AB has an angular

velocity of 4 rad/s clockwise. Determine the angular

velocity of bars BD and DE.



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2 - 44

How should you proceed?

Determine vB with respect to A, then work

your way along the linkage to point E.

Determine the angular velocity of bars

BD and DE.

wAB= 4 rad/s

(4 rad/s)AB kw

/ /–(175 mm) (–4 ) (–175 )

(700 mm/s)

r i v k i

v j

B A B AB B A

B

r

Does it make sense that vB is in the +j direction?

x

y

/AB A AB B v v rw

Write vB in terms of point A, calculate vB.



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2 - 45

wAB= 4 rad/s

(4 rad/s)AB kw

x

y

/AB A AB B v v rw

Write vB in terms of point

A, calculate vB.



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2 - 46

wAB= 4 rad/s

Does it make sense that vB is in the +j direction?

x

y

/ –(175 mm)r iB A

/ (–4 ) (–175 )

(700 mm/s)

v k i

v j

B AB B A

B

r

w



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2 - 47

wAB= 4 rad/s

x

y

Determine vD with respect to

E, then equate it to equation

above.

,Dv

kwDE DEw

/ (275 mm) (75 mm)r i jD E



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2 - 48

wAB= 4 rad/s

x

y


E, then equate it to equation

above.

,Dv

/

( ) ( 275 75 )

275 75

v r

k i j

v j i

wD DE D E

DE

D DE DE

w

w w



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2 - 49


B. wAB= 4 rad/s

x

y

,Dv

kwBD BDw

/ (200 mm)r jD B



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2 - 50


B. wAB= 4 rad/s

x

y

,Dv

/

700 ( ) ( 200 )

700 200

v v r

j k j

v j i

wD B BD D B

BD

D BD

w

w



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2 - 51

Equating components of the two expressions

for vD

,Dv

j: 700 275 2.5455 rad/sDE DEw w

3: 200 75

8i BD DE BD DEw w w w

0.955 rad/sBD w

2.55 rad/sDE w



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2 - 52

Knowing that at the instant

shown bar AB has a constant

angular velocity of 4 rad/s

clockwise, determine the angular

acceleration of bars BD and DE.

Which of the following is true?

a) The direction of aD is

b) The angular acceleration of BD must also be constant

c) The direction of the linear acceleration of B is →



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2 - 53

Knowing that at the instant

shown bar AB has a constant

angular velocity of 4 rad/s

clockwise, determine the

angular acceleration of bars

BD and DE.

SOLUTION:

• The angular velocities were determined

in a previous problem by simultaneously

solving the component equations for

BDBD vvv

• The angular accelerations are now

determined by simultaneously solving

the component equations for the relative

acceleration equation.



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2 - 54

From our previous problem, we used the relative

velocity equations to find that: wAB= 4 rad/s

0.955 rad/sBD w2.55 rad/sDE w

0AB

We can now apply the relative

acceleration equation with



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2 - 55

wAB= 4 rad/s

2/A /AB A AB B AB Bw a a r r

2 2 2/ (4) ( 0.175 ) 2.8 m/sa r i iB AB B Aw

Analyze Bar AB



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2 - 56

wAB= 4 rad/s

Analyze Bar BD

2/ /

22.8 ( 0.2 ) (0.95455) ( 0.2 )

D B BD D B BD D B

BD

w

a a r r

i k j j



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2 - 57

(2.8 0.2 ) 0.18223a i jD BD



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2 - 58

wAB= 4 rad/s

Analyze Bar DE

2/ /

2

( 0.275 0.075 )

(2.5455) ( 0.275 0.075 )

a r

k i j

i j

D DE D E DE D E

DE

rw



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2 - 59

( 0.075 1.7819)

(0.275 0.486)

a i

j

D DE

DE



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2 - 60

(2.8 0.2 ) 0.18223a i jD BD

( 0.075 1.7819)

(0.275 0.486)

a i

j

D DE

DE



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2 - 61

wAB= 4 rad/s Equate like components of aD

j: 0.18223 (0.275 0.486)DE

22.43 rad/sDE

i: 2.8 0.2 [ (0.075)( 2.43) 1.7819]BD

24.18 rad/sBD



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Planetary gears

2 - 62

Video

Planetary Gear Set.mp4



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15 - 63

• General plane motion is neither a translation nor a

rotation.

• General plane motion can be considered as the sum

of a translation and rotation.



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15 - 64





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15 - 65



• Displacement of particles A and B to A2 and

B2 can be divided into two parts:

- translation to A2 and

- rotation of about A2 to B2 1B

1B



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Instantaneous Center of Rotation in Plane Motion

15 - 66

• The same translational and

rotational velocities at A are

obtained by allowing the slab to

rotate with the same angular

velocity about the point C on a

perpendicular to the velocity at A.



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15 - 67

• Plane motion of all particles in a

slab can always be replaced by the

translation of an arbitrary point A

and a rotation about A with an

angular velocity that is independent

of the choice of A.



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15 - 68

• The velocity of all other particles

in the slab are the same as

originally defined since the angular

velocity and translational velocity

at A are equivalent.

• As far as the velocities are

concerned, the slab seems to rotate

about the instantaneous center of

rotation C.



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15 - 69

• If the velocity at two

points A and B are

known, the

instantaneous center of

rotation lies at the

intersection of the

perpendiculars to the

velocity vectors through

A and B .



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15 - 70

• If the velocity vectors at A

and B are perpendicular to

the line AB, the

instantaneous center of

rotation lies at the

intersection of the line AB

with the line joining the

extremities of the velocity

vectors at A and B.



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15 - 71

• If the velocity vectors are parallel and

equal ,

• the instantaneous center of rotation is

at infinity and

• the angular velocity is zero.



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15 - 72

• The instantaneous center of rotation lies at

the intersection of the perpendiculars to the

velocity vectors through A and B .

w

cosl

v

AC

v AA

w

tan

cossin

A

AB

v

l

vlBCv

• The velocities of all particles on the rod are as if

they were rotated about C.

• The particle at the center of rotation has zero

velocity.



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15 - 73

• The particle coinciding with the center of rotation

changes with time and the acceleration of the particle

at the instantaneous center of rotation is not zero.

• The acceleration of the particles in the slab cannot be

determined as if the slab were simply rotating about

C.



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15 - 74

At the instant shown, what is the

approximate direction of the velocity

of point G, the center of bar AB?

a)

b)

c)

d)

G



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2 - 75

vD

What direction is the velocity of B?

vB

What direction is the velocity of D?

wAB= 4 rad/s

( ) (0.25 m)(4 rad/s) 1m/sB ABAB w vWhat is the velocity of B?

1 0.06 mtan 21.8

0.15 m

Find



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2 - 76

vD

vB

B

D

Locate instantaneous center C at intersection of lines drawn

perpendicular to vB and vD.

C

100 mm

1m/s (0.25 m) BDw

4 rad/sBD w

( )B BDv BC w

Calculate wBD



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2 - 77

vB

B

vD

D

C

100 mm

Find wDE

0.25 m( ) (4 rad/s)

cosD BDv DC w

1 m/s 0.15 m( ) ; ;

cos cosD DE DEv DE w w

6.67 rad/sDE w



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Absolute and Relative Velocity in Plane Motion

15 - 78

• Any plane motion can be replaced by a translation of an

arbitrary reference point A and a simultaneous rotation

about A.

ABAB vvv

ww rvrkv ABABAB

ABAB rkvv

w



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Sample Problem 15.2

15 - 79

The double gear rolls on the stationary lower rack: the

velocity of its center is 1.2 m/s.

Determine (a) the angular velocity of the gear, and (b) the

velocities of the upper rack R and point D of the gear.



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Sample Problem 15.2

15 - 80



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Sample Problem 15.2

15 - 81

The double gear rolls on the stationary lower

rack: the velocity of its center is 1.2 m/s.

Determine (a) the angular velocity of the gear,

and (b) the velocities of the upper rack R and

point D of the gear.



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Sample Problem 15.2

15 - 82

SOLUTION:

• The displacement of the gear center in one

revolution is equal to the outer circumference.

Relate the translational and angular

displacements. Differentiate to relate the

translational and angular velocities.



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Sample Problem 15.2

15 - 83

• The velocity for any point P on the gear

may be written as

Evaluate the velocities of points B and D.

APAAPAP rkvvvv

w



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Sample Problem 15.2

15 - 84

x

y

SOLUTION:

w < 0 (rotates clockwise).

m0.150

sm2.1

1

1

r

v

rv

A

A

w

w

kk

srad8 ww



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Sample Problem 15.2

15 - 85

• For any point P on the gear,

APAAPAP rkvvvv

w



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Sample Problem 15.2

15 - 86

Velocity of the upper rack is equal to

velocity of point B:

ii

jki

rkvvv ABABR

sm8.0sm2.1

m 10.0srad8sm2.1

w

ivR

sm2



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Sample Problem 15.2

15 - 87

Velocity of the point D:

iki

rkvv ADAD

m 150.0srad8sm2.1

w

sm697.1

sm2.1sm2.1

D

D

v

jiv



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Sample Problem 15.2

15 - 88



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Sample Problem 15.6

15 - 89



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Sample Problem 15.6

15 - 90

The center of the double gear has a velocity

and acceleration to the right of 1.2 m/s and 3

m/s2, respectively. The lower rack is

stationary.

Determine (a) the angular acceleration of the

gear, and (b) the acceleration of points B, C,

and D.



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Sample Problem 15.6

15 - 91

SOLUTION:

• The expression of the gear position as a

function of is differentiated twice to define

the relationship between the translational

and angular accelerations. • The acceleration of each point on the gear is

obtained by adding the acceleration of the gear

center and the relative accelerations with

respect to the center. The latter includes

normal and tangential acceleration

components.



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Sample Problem 15.6

15 - 92

11 rraA

m 150.0

sm3 2

1

r

aA

kk 2srad20



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Sample Problem 15.6

15 - 93

• The acceleration of each point is obtained by

adding the acceleration of the gear center and

the relative accelerations with respect to the

center.

The latter includes normal and tangential

acceleration components.



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Sample Problem 15.6

15 - 94

jii

jjki

rrka

aaaaaa

ABABA

nABtABAABAB

222

222

2

sm40.6sm2sm3

m100.0srad8m100.0srad20sm3

w

222 sm12.8sm40.6m5 BB ajisa



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Sample Problem 15.6

15 - 95

jii

jjki

rrkaaaa ACACAACAC

222

222

2

sm60.9sm3sm3


w

jac

2sm60.9

iji

iiki

rrkaaaa ADADAADAD

222

222

2

sm60.9sm3sm3


w

222 sm95.12sm3m6.12 DD ajisa



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12 - 96

Figure shows a cyclist on a bicycle. Assume

that he is pedaling at 90 rpm and

chain runs between the 12 cm radius front

chain ring and the 3 cm radius rear cog. The

wheels have a radius of 25 cm. Determine how

fast the cyclist is travelling? Assume no

slipping between the wheel and the road.



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12 - 97



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12 - 98

Since, O only moves horizontally, b1

component must be 0.



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12 - 99

To keep in shape during the winter, cyclists often put their bikes on

rollers which allows them to pedal as hard as they wish without

going anywhere. Assume that the cyclist is pedaling at 4 rad/s and

then begins to accelerate so that the angular acceleration of the main

crank assembly is equal to 1.2rad/s2. After accelerating at this rate

for 2 s. what is the acceleration of the pedal spindle P? The

distance from O, the center of

the bottom bracket, to the pedal spindle is 17 cm, and the bottom

bracket center O is fixed in space.

Find the acceleration of a bike's pedal spindle.



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12 - 100

First we will need to determine the rotation rate

after 2 seconds by integrating with respect to

time.



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Example – General Plane Motion

2 - 101

The knee has linear velocity and acceleration from both

translation (the runner moving forward) as well as rotation

(the leg rotating about the hip).



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Sample Problem 15.4

15 - 102

The double gear rolls on the

stationary lower rack: the velocity

of its center is 1.2 m/s.

Determine (a) the angular velocity

of the gear, and (b) the velocities of

the upper rack R and point D of the

gear.

SOLUTION:

• The point C is in contact with the stationary

lower rack and, instantaneously, has zero

velocity. It must be the location of the

instantaneous center of rotation.

• Determine the angular velocity about C

based on the given velocity at A.

• Evaluate the velocities at B and D based on

their rotation about C.



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Sample Problem 15.4

15 - 103

SOLUTION:

• The point C is in contact with the stationary lower rack

and, instantaneously, has zero velocity. It must be the

location of the instantaneous center of rotation.

• Determine the angular velocity about C based on the

given velocity at A.

srad8m 0.15

sm2.1

A

AAA

r

vrv ww

• Evaluate the velocities at B and D based on their rotation

about C.

srad8m 25.0 wBBR rvv

ivR

sm2

srad8m 2121.0

m 2121.02m 15.0

wDD

D

rv

r

sm2.12.1

sm697.1

jiv

v

D

D



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Golf Robot

2 - 104

w,

If the arm is shortened to ¾ of its

original length, what happens to

the tangential acceleration of the

club head?

A golf robot is used to test new

equipment. If the angular

velocity of the arm is doubled,

what happens to the normal

acceleration of the club head?

If the arm is shortened to ¾ of its

original length, what happens to

the tangential acceleration of the

club head?

If the speed of the club head is

constant, does the club head have

any linear accelertion ?

Not ours – maybe Tom Mase has pic?



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Instantaneous Center of Zero Velocity

2 - 105

What happens to the location of the instantaneous center of

velocity if the crankshaft angular velocity increases from

2000 rpm in the previous problem to 3000 rpm?

What happens to the location of the instantaneous center of

velocity if the angle is 0?



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12 - 106



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12 - 107

A disc C is mounted on a shaft AB in Figure.

The shaft and disc rotate with a constant angular

speed ω2 of 10 rad/sec relative to the platform to

which bearings A and B are attached. Meanwhile,

the platform rotates at a constant angular speed ω1

of 5 rad/sec relative to the ground in a direction

parallel to the Z axis of the ground reference XYZ.

What is the angular

velocity vector ω for the disc C relative to XYZ?

What are (dω/dt)xyz and (d2ω /dt2)xyz ?

Consider the vector w2. Note that this

vector is constrained in direction to be

always collinear with the axis AB of the

bearings of the shaft. This clearly is a

physical requirement.

Also, since w2 is of constant value, we may

think of the vector w2 as fixed to the

platform along AB.

121 www

121 www

12112 www

Example 15.1:

3

1211

1211

2

121

sec/250)105(5

)(0

sec/50105

jradjkk

radijk

w

www

wwww

w

www

Consider a position vector ρ between two points on

the rotating disc. The length of ρ is 100 mm and at

the instant of interest is the vertical direction. What

are the first and second time derivatives of ρ at this

instant as seen from ground reference?

r is fixed in a frame rotating at w1 + w2|1

rwwr )(121

sec/1000100)105( mmikjk r

rwwrwwr )()(121121

rwwrwwr )()0(121121

rwwrwwr )()(121121

rwwrwwr )()0(121121

2

2

sec/1010

sec/1000)105(100)105(

mmkj

mmijkkjk

r

r



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12 - 114

A tank is maneuvering its gun into position.

At the instant of interest, the turrent A, is

rotating at an angular speed of 2 rad/sec

relative to the tank and is in position . Also

at this instant , the gun is rotating at an

angular speed of 1 rad/sec relative to the

turret and forms an angle =300 with the

horizontal plane . What are and of the

gun relative to the ground?

ww , w



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12 - 115



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12wConsider Fixed in turret.

121 www

121121 0 wwwww

0 0

2 1(1)cos 20 i (1)sin 20 j

0.940 i 0.342 j

w

2

2k (0.940 i 0.342 j)

1.880 j 0.684 irad / sec

w

)()(01211

1211

www

wwww

At instant of interest:

3sec/368.176.3

)]342.0940.0()2[(2

radji

jikk

w



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Concept Question

15 - 119

The figure depicts a model

of a coaster wheel. If both

w1 and w2 are constant, what

is true about the angular

acceleration of the wheel?

a)It is zero.

b)It is in the +x direction

c)It is in the +z direction

d)It is in the -x direction

e)It is in the -z direction



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15 - 120



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2 - 121

The crane rotates with a constant

angular velocity w1 = 0.30 rad/s

and the boom is being raised with a

constant angular velocity w2 = 0.50

rad/s. The length of the boom is l

= 12 m.



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15 - 122

Determine:

• angular velocity of the boom,

• angular acceleration of the boom,

• velocity of the boom tip, and

• acceleration of the boom tip.



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15 - 123

SOLUTION:

With

• Angular velocity of the boom,

1 2w w w

ji

jir

kj

639.10

30sin30cos12

50.030.0 21

ww



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15 - 124

•Angular acceleration of the boom,

21

22221

ww

wwwww

Oxyz

•Velocity of boom tip,

rv

w



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125

•Acceleration of boom tip,

vrrra

www



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15 - 126 jir

kj

639.10

50.030.0 21

ww



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15 - 127

SOLUTION:

•Angular velocity of the

boom,

21 www

kj

srad50.0srad30.0 w



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2 - 128

•Angular acceleration of the

boom,

kj

Oxyz

srad50.0srad30.021

22221

ww

wwwww

i 2srad15.0



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15 - 129

•Velocity of boom tip,

0639.10

5.03.00

kji

rv

w

kjiv

sm12.3sm20.5sm54.3



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15 - 130

• Acceleration of boom tip,

kjiik

kjikji

a

vrrra

90.050.160.294.090.0

12.320.53

50.030.00

0639.10

0015.0

www

kjia 222 sm80.1sm50.1sm54.3



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Absolute and Relative Acceleration in Plane Motion

2 - 131

As the bicycle accelerates, a point on the top of the wheel will

have acceleration due to the acceleration from the axle (the

overall linear acceleration of the bike), the tangential

acceleration of the wheel from the angular acceleration, and

the normal acceleration due to the angular velocity.



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Concept Question

2 - 132

A) Energy will not be conserved when I kick this ball

B) In general, the linear acceleration of my knee is equal to

the linear acceleration of my foot

C) Throughout the kick, my foot will only have tangential

acceleration.

D) In general, the angular velocity of the upper leg (thigh)

will be the same as the angular velocity of the lower leg

You have made it to the kickball

championship game. As you try to kick

home the winning run, your mind

naturally drifts towards dynamics.

Which of your following thoughts is

TRUE, and causes you to shank the ball

horribly straight to the pitcher?



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Applications

2 - 133

Rotating coordinate systems are often used to analyze mechanisms

(such as amusement park rides) as well as weather patterns.



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• Vector moving in moving frame of

reference

139



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15 - 147

• Consider a collar P which is

made to slide at constant relative

velocity u along rod OB. The

rod is rotating at a constant

angular velocity w. The point A

on the rod corresponds to the

instantaneous position of P.

cPAP aaaa

F

• Absolute acceleration of the collar is



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15 - 148

cPAP aaaa

F

0 OxyP ra F

uava cPc w22 F

• The absolute acceleration consists of the

radial and tangential vectors shown

2wrarra AA

where



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15 - 149

uvvtt

uvvt

A

A

,at

,at



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15 - 150

uvvtt

uvvt

A

A

,at

,at

• Change in velocity over t is

represented by the sum of three

vectors TTTTRRv

2wrarra AA

recall,

• is due to change in

direction of the velocity of

point A on the rod,

AAtt

arrt

vt

TT

2

00limlim www

TT



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15 - 151

uvvtt

uvvt

A

A

,at

,at

• result from

combined effects of relative

motion of P and rotation of the

rod

TTRR and

uuu

t

r

tu

t

TT

t

RR

tt

www

w

2

limlim00

uava cPc w22 F

recall,



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Concept Question

2 - 152

v w

You are walking with a

constant velocity with

respect to the platform,

which rotates with a

constant angular

velocity w. At the

instant shown, in which

direction(s) will you

experience an

acceleration (choose all

that apply)?

x

y

OxyOxyP rrrra

2



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Concept Question

2 - 153

v w

a) +x

b) -x

c) +y

d) -y

e) Acceleration = 0

x

y

OxyOxyP rrrra

2



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15 - 154

•Determine the

acceleration of the

virtual point D’.

•Calculate the Coriolis

acceleration.

•Add the different components to get the

overall acceleration of point D.



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155



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15 - 156

The sleeve BC is welded to an arm that

rotates about stationary point A with a

constant angular velocity w = (3 rad/s) j.

In the position shown rod DF is being

moved to the left at a

constant speed u =

400 mm/s relative to

the sleeve. Determine

the acceleration of

Point D.



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2 - 157

SOLUTION:

•The absolute

acceleration of point D

may be expressed as

'D D D BC ca a a a



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2 - 158

Given: u= 400 mm/s, w = (3 rad/s) j.

Find: aD

Write overall expression for aD

2DOxy Oxy

a r r r r



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159

Do any of the terms go to zero?

2

D

Oxy Oxy

a r r

r r



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2 - 160

Determine the normal

acceleration term of

the virtual point D’

2

(3 rad/s) (3 rad/s) [ (100 mm) (300 mm) ]

(2700 mm/s )

a

j j j k

k

D r

where r is from A to D



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2 - 161

Determine the Coriolis

acceleration of point D

2DOxy Oxy

a r r r r

/

2

2

2(3 rad/s) (400 mm/s)

(2400 mm/s )

a v

j k

i

C D F

w



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2 - 162

/

2 2(2700 mm/s ) 0 (2400 mm/s )

a a a a

k i

D D D F C

Add the different

components to obtain the

total acceleration of point

D

2 2(2400 mm/s ) (2700 mm/s )a i kD



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15 - 163

In the previous problem, u

and w were both constant.

What would happen if

u was increasing?

a)The x-component of aD would increase

b)The y-component of aD would increase

c)The z-component of aD would increase

d)The acceleration of aD would stay the same

w



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15 - 164

What would happen if w was

increasing?

a)The x-component of aD would increase

b)The y-component of aD would increase

c)The z-component of aD would increase

d)The acceleration of aD would stay the same

w



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2 - 165

Geneva

mechanism

GENEVA_MECHANISM.mp4



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Sample Problem 15.9

15 - 166

Disk D of the Geneva

mechanism rotates

with constant

counterclockwise

angular velocity wD =

10 rad/s.

At the instant when = 150o, determine (a)

the angular velocity of disk S, and (b) the

velocity of pin P relative to disk S.



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Sample Problem 15.9

15 - 167

SOLUTION:

• The absolute velocity of the point P

may be written as

sPPP vvv

• Magnitude and direction of velocity

of pin P are calculated from the

radius and angular velocity of disk D. Pv

• Direction of velocity of point P’ on

S coinciding with P is perpendicular to

radius OP.

Pv

• Direction of velocity of P with

respect to S is parallel to the slot. sPv

• Solve the vector triangle for the

angular velocity of S and relative

velocity of P.



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Sample Problem 15.9

15 - 168

SOLUTION:

• The absolute velocity of the

point P may be written as

sPPP vvv

• Magnitude and direction of

absolute velocity of pin P are

calculated from radius and

angular velocity of disk D.

smm500srad 10mm 50 DP Rv w



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Sample Problem 15.9

15 - 169

• Direction of velocity of P

with respect to S is parallel to

slot. From the law of cosines,

mm 1.37551.030cos2 2222 rRRllRr

4.42742.0

30sinsin

30sin

R

sin

r

6.17304.4290

The interior angle of the vector

triangle is



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Sample Problem 15.9

15 - 170

• Direction of velocity of point

P’ on S coinciding with P is

perpendicular to radius OP.

mm 1.37

smm2.151

smm2.1516.17sinsmm500sin

ss

PP

r

vv

ww

ks

srad08.4w

6.17cossm500cosPsP vv

jiv sP

4.42sin4.42cossm477

smm 500Pv



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15 - 171

In the Geneva

mechanism, disk D

rotates with a constant

counter-clockwise

angular velocity of 10

rad/s. At the instant

when j = 150o,

determine angular

acceleration of disk S.



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15 - 172

SOLUTION:

• The absolute acceleration of the pin P may

be expressed as

csPPP aaaa

• The instantaneous angular velocity of Disk

S is determined as in Sample Problem 15.9.

• The only unknown involved in the

acceleration equation is the instantaneous

angular acceleration of Disk S.

• Resolve each acceleration term into the

component parallel to the slot. Solve for

the angular acceleration of Disk S.



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15 - 173

SOLUTION:

• Absolute acceleration of

the pin P may be expressed

as csPPP aaaa

• From Sample Problem

15.9.

jiv

k

sP

S

4.42sin4.42cossmm477

srad08.44.42 w



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2 - 174



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15 - 175

• Considering each term in the

acceleration equation,

jia

Ra

P

DP

30sin30cossmm5000

smm5000srad10mm500

2

222w



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15 - 176

jia

jira

jira

aaa

StP

StP

SnP

tPnPP

4.42cos4.42sinmm1.37

4.42cos4.42sin

4.42sin4.42cos2

w

note: S may be positive or negative



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15 - 177

sPv

• The direction of the Coriolis

acceleration is obtained by

rotating the direction of the

relative velocity

by 90o in the sense of wS.

ji

ji

jiva sPSc

4.42cos4.42sinsmm3890

4.42cos4.42sinsmm477srad08.42

4.42cos4.42sin2

2

w



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15 - 178

• The relative acceleration

must be parallel to the

slot.

sPa

• Equating components of

the acceleration terms

perpendicular to the slot,

srad233

07.17cos500038901.37

S

S

kS

srad233



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15 - 179

For the disk mounted

on the arm, the

indicated angular

rotation rates are

constant.

Determine:

• the velocity of the point P,

• the acceleration of P, and

• angular velocity and angular acceleration of the

disk.



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15 - 180



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15 - 181

SOLUTION:

• Define a fixed reference frame OXYZ at O and a moving reference frame Axyz or F attached to the

arm at A.

j

jRiLr

1w

k

jRr

D

AP

2ww

F



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15 - 182

• With P’ of the moving reference frame coinciding

with P, the velocity of the point P is found from

iRjRkrv

kLjRiLjrv

vvv

APDP

P

PPP

22

11

www

ww

FF

F

kLiRvP

12 ww



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15 - 183



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15 - 184

• The acceleration of P is found from

cPPP aaaa

F

iLkLjraP

2111 www

jRiRk

ra APDDP

2

222 www

ww

FFF

kRiRj

va Pc

2121 22

2

wwww

F

kRjRiLaP

21

22

21 2 wwww



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15 - 185

• Angular velocity and acceleration of the disk,

FDww

kj

21 www

kjj

211 www

ww

F

i

21ww



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15 - 186

Knowing that at the instant shown the length of

the boom is 6 m and is increasing at the constant

rate u= 0.5 m/s determine the acceleration of Point

B.

The crane shown

rotates at the constant

rate w1= 0.25 rad/s;

simultaneously, the

telescoping boom is

being lowered at the

constant rate w2= 0.40

rad/s.



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15 - 187

SOLUTION:

• Define a moving reference frame Axyz or F

attached to the arm at A.

• The acceleration of P is found from

'B B B ca a a a F

• The angular velocity and angular acceleration of

the disk are

Bw w

w w

F

F



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15 - 188

Given: w1= 0.25 rad/s,

w2= -0.40 rad/s. L= 6 m,

u= 0.5 m/s. Find: aB.

Equation of overall

acceleration of B

2DOxy Oxy

a r r r r

Do any of the terms go to zero?

2DOxy Oxy

a r r r r



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15 - 189

Let the unextending

portion of the boom AB

be a rotating frame of

reference.

What are and ?

2 1

(0.40 rad/s) (0.25 rad/s) .

w w

i j

i j

1 2

1 2

2(0.10 rad/s ) .

w w

w w

j i

k

k



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15 - 190

Find

2DOxy Oxy

a r r r r

20 0 0.10 (0.3 m/s )

0 3 3 3

i j k

r iB

r

/

(6 m)(sin30 cos30 )

(3 m) (3 3 m)

r r

j k

j k

B A B

Determine the position vector rB/A



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15 - 191

Find r

2 2 2

(0.40 0.25 ) (0.40 0.25 ) (3 3 3 )

(0.3 m/s ) (0.48 m/s ) (1.1561 m/s )

r i j i j j k

i j k

B



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15 - 192

2DOxy Oxy

a r r r r

Determine the Coriolis

acceleration – first define

the relative velocity term

/ (sin30 cos30 )

(0.5 m/s)sin30 (0.5 m/s)cos30

v j k

j k

B F u



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15 - 193

/

2 2 2

2 (2)(0.40 0.25 ) (0.5sin30 0.5cos30 )

(0.2165 m/s ) (0.3464 m/s ) (0.2 m/s )

Ω v i j j k

i j k

B F

Calculate the Coriolis

acceleration

Add the terms together

2 2 2(0.817 m/s ) (0.826 m/s ) (0.956 m/s )a i j kB

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