Ch11 Kinematics of Particles

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS

Tenth Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

Phillip J. Cornwell

Lecture Notes:

Brian P. SelfCalifornia Polytechnic State University

CHAPTER

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

11Kinematics of Particles


Vector Mechanics for Engineers: Dynamics

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Contents

11 - 2

Introduction

Rectilinear Motion: Position,

Velocity & Acceleration

Determination of the Motion of a

Particle

Sample Problem 11.2

Sample Problem 11.3

Uniform Rectilinear-Motion

Uniformly Accelerated Rectilinear-

Motion

Motion of Several Particles:

Relative Motion

Sample Problem 11.4

Motion of Several Particles:

Dependent Motion

Sample Problem 11.5

Graphical Solution of Rectilinear-

Motion Problems

Other Graphical Methods

Curvilinear Motion: Position, Velocity

& Acceleration

Derivatives of Vector Functions

Rectangular Components of Velocity

and Acceleration

Motion Relative to a Frame in

Translation

Tangential and Normal Components

Radial and Transverse Components

Sample Problem 11.10




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2 - 3

Kinematic relationships are used to

help us determine the trajectory of a

golf ball, the orbital speed of a

satellite, and the accelerations

during acrobatic flying.



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Introduction

11 - 4

• Dynamics includes:

Kinetics: study of the relations existing between the forces acting on

a body, the mass of the body, and the motion of the body. Kinetics is

used to predict the motion caused by given forces or to determine the

forces required to produce a given motion.

Kinematics: study of the geometry of motion.

Relates displacement, velocity, acceleration, and time without reference

to the cause of motion.Fthrust

Flift

Fdrag



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Introduction

11 - 5

• Particle kinetics includes:

• Rectilinear motion: position, velocity, and acceleration of a

particle as it moves along a straight line.

• Curvilinear motion: position, velocity, and acceleration of a

particle as it moves along a curved line in two or three

dimensions.



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Rectilinear Motion: Position, Velocity & Acceleration

11 - 6

• Rectilinear motion: particle moving

along a straight line

• Position coordinate: defined by

positive or negative distance from a

fixed origin on the line.

• The motion of a particle is known if

the position coordinate for particle is

known for every value of time t.

• May be expressed in the form of a

function, e.g., 326 ttx

or in the form of a graph x vs. t.



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11 - 7

• Instantaneous velocity may be positive or

negative. Magnitude of velocity is referred

to as particle speed.

• Consider particle which occupies position P

at time t and P’ at t+Dt,

t

xv

t

x

t D

D

D

D

D 0lim

Average velocity

Instantaneous velocity

• From the definition of a derivative,

dt

dx

t

xv

t

D

D

D 0lim

e.g.,

2

32

312

6

ttdt

dxv

ttx



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11 - 8

• Consider particle with velocity v at time t and

v’ at t+Dt,

Instantaneous accelerationt

va

t D

D

D 0lim

tdt

dva

ttv

dt

xd

dt

dv

t

va

t

612

312e.g.

lim

2

2

2

0

D

D

D

• From the definition of a derivative,

• Instantaneous acceleration may be:

- positive: increasing positive velocity

or decreasing negative velocity

- negative: decreasing positive velocity

or increasing negative velocity.



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11 - 9

• From our example,

326 ttx

2312 ttdt

dxv

tdt

xd

dt

dva 612

2

2

- at t = 2 s, x = 16 m, v = vmax = 12 m/s, a = 0

- at t = 4 s, x = xmax = 32 m, v = 0, a = -12 m/s2

• What are x, v, and a at t = 2 s ?

• Note that vmax occurs when a=0, and that the

slope of the velocity curve is zero at this point.

• What are x, v, and a at t = 4 s ?



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Concept Quiz

2 - 10

What is true about the kinematics of a particle?

a) The velocity of a particle is always positive

b) The velocity of a particle is equal to the slope of

the position-time graph

c) If the position of a particle is zero, then the

velocity must zero

d) If the velocity of a particle is zero, then its

acceleration must be zero



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Determination of the Motion of a Particle

11 - 11

• We often determine accelerations from the forces applied

(kinetics will be covered later)

• Generally have three classes of motion

- acceleration given as a function of time, a = f(t)

- acceleration given as a function of position, a = f(x)

- acceleration given as a function of velocity, a = f(v)

• Can you think of a physical example of when force is a

function of position? When force is a function of velocity?

a spring drag



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Acceleration as a function of time, position, or velocity

2 - 12

a a t 0 0

v t

v

dv a t dt ( )dv

a tdt

vdv a x dx

0 0

v x

v x

v dv a x dx a a x

and dx dv

dt av dt

dv

v a vdx

0 0

v t

v

dvdt

a v

0 0

x v

x v

v dvdx

a v

a a v

( )dv

a vdt

If…. Kinematic relationship Integrate



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Sample Problem 11.2

11 - 13

Determine:

• velocity and elevation above ground at

time t,

• highest elevation reached by ball and

corresponding time, and

• time when ball will hit the ground and

corresponding velocity.

Ball tossed with 10 m/s vertical velocity

from window 20 m above ground.

SOLUTION:

• Integrate twice to find v(t) and y(t).

• Solve for t when velocity equals zero

(time for maximum elevation) and

evaluate corresponding altitude.

• Solve for t when altitude equals zero

(time for ground impact) and evaluate




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Sample Problem 11.2

11 - 14

tvtvdtdv

adt

dv

ttv

v

81.981.9

sm81.9

00

2

0

ttv

2s

m81.9

s

m10

2

21

00

81.91081.910

81.910

0

ttytydttdy

tvdt

dy

tty

y

2

2s

m905.4

s

m10m20 ttty

SOLUTION:

• Integrate twice to find v(t) and y(t).



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Sample Problem 11.2

11 - 15

• Solve for t when velocity equals zero and evaluate

corresponding altitude.

0s

m81.9

s

m10

2

ttv

s019.1t

• Solve for t when altitude equals zero and evaluate


22

2

2

s019.1s

m905.4s019.1

s

m10m20

s

m905.4

s

m10m20

y

ttty

m1.25y



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Sample Problem 11.2

11 - 16

• Solve for t when altitude equals zero and evaluate


0s

m905.4

s

m10m20 2

2

ttty

s28.3

smeaningles s243.1

t

t

s28.3s

m81.9

s

m10s28.3

s

m81.9

s

m10

2

2

v

ttv

s

m2.22v



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Sample Problem 11.3

11 - 17

Brake mechanism used to reduce gun

recoil consists of piston attached to barrel

moving in fixed cylinder filled with oil.

As barrel recoils with initial velocity v0,

piston moves and oil is forced through

orifices in piston, causing piston and

cylinder to decelerate at rate proportional

to their velocity.

Determine v(t), x(t), and v(x).

kva

SOLUTION:

• Integrate a = dv/dt = -kv to find v(t).

• Integrate v(t) = dx/dt to find x(t).

• Integrate a = v dv/dx = -kv to find

v(x).



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Sample Problem 11.3

11 - 18

SOLUTION:

• Integrate a = dv/dt = -kv to find v(t).

0 00

lnv t

v

v tdv dva kv k dt kt

dt v v

ktevtv 0

• Integrate v(t) = dx/dt to find x(t).

0

0 0

00 0

1

kt

tx t

kt kt

dxv t v e

dt

dx v e dt x t v ek

ktek

vtx 10



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Sample Problem 11.3

11 - 19

• Integrate a = v dv/dx = -kv to find v(x).

kxvv

dxkdvdxkdvkvdx

dvva

xv

v

0

00

kxvv 0

• Alternatively,

0

0 1v

tv

k

vtx

kxvv 0

00 or

v

tveevtv ktkt

ktek

vtx 10with

and

then



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Group Problem Solving

11 - 20

SOLUTION:

• Determine the proper kinematic

relationship to apply (is acceleration

a function of time, velocity, or

position?

• Determine the total distance the car

travels in one-half lap

• Integrate to determine the velocity

after one-half lap

The car starts from rest and accelerates

according to the relationship

23 0.001a v

It travels around a circular track that has

a radius of 200 meters. Calculate the

velocity of the car after it has travelled

halfway around the track. What is the

car’s maximum possible speed?



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2 - 21

dv

v a vdx

0 0

x v

x v

v dvdx

a v

Choose the proper kinematic relationship

Given:23 0.001a v

vo = 0, r = 200 m

Find: v after ½ lap

Maximum speed

Acceleration is a function of velocity, and

we also can determine distance. Time is not

involved in the problem, so we choose:

Determine total distance travelled

3.14(200) 628.32 mx r



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2 - 22

0 0

x v

x v

v dvdx

a v

Determine the full integral, including limits

628.32

2

0 03 0.001

vv

dx dvv

Evaluate the interval and solve for v

2

0

1628.32 ln 3 0.001

0.002

v

v

2628.32( 0.002) ln 3 0.001 ln 3 0.001(0)v

2ln 3 0.001 1.2566 1.0986= 0.15802v

2 0.158023 0.001v e Take the exponential of each side



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2 - 23

2 0.158023 0.001v e Solve for v

0.158022 3

2146.20.001

ev

46.3268 m/sv

How do you determine the maximum speed the car can reach?

23 0.001a v Velocity is a maximum when

acceleration is zero

This occurs when20.001 3v

30.001maxv

max 54.772 m/sv



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Uniform Rectilinear Motion

11 - 24

For a particle in uniform

rectilinear motion, the

acceleration is zero and

the velocity is constant.

vtxx

vtxx

dtvdx

vdt

dx

tx

x

0

0

00

constant

During free-fall, a parachutist

reaches terminal velocity when

her weight equals the drag

force. If motion is in a straight

line, this is uniform rectilinear

motion.

Careful – these only apply to

uniform rectilinear motion!



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Uniformly Accelerated Rectilinear Motion

11 - 25

If forces applied to a body

are constant (and in a

constant direction), then

you have uniformly

accelerated rectilinear

motion.

Another example is free-

fall when drag is negligible



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Uniformly Accelerated Rectilinear Motion

11 - 26

For a particle in uniformly accelerated rectilinear motion, the

acceleration of the particle is constant. You may recognize these

constant acceleration equations from your physics courses.

0

0

0

constantv t

v

dva dv a dt v v at

dt

0

210 0 0 0 2

0

x t

x

dxv at dx v at dt x x v t at

dt

0 0

2 2

0 0constant 2v x

v x

dvv a vdv a dx v v a x x

dx

Careful – these only apply to uniformly

accelerated rectilinear motion!



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Motion of Several Particles

2 - 27

We may be interested in the motion of several different particles,

whose motion may be independent or linked together.



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Motion of Several Particles: Relative Motion

11 - 28

• For particles moving along the same line, time

should be recorded from the same starting

instant and displacements should be measured

from the same origin in the same direction.

ABAB xxx relative position of B

with respect to AABAB xxx

ABAB vvv relative velocity of B

with respect to AABAB vvv

ABAB aaa relative acceleration of B

with respect to AABAB aaa



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Sample Problem 11.4

11 - 29

Ball thrown vertically from 12 m level

in elevator shaft with initial velocity of

18 m/s. At same instant, open-platform

elevator passes 5 m level moving

upward at 2 m/s.

Determine (a) when and where ball hits

elevator and (b) relative velocity of ball

and elevator at contact.

SOLUTION:

• Substitute initial position and velocity

and constant acceleration of ball into

general equations for uniformly

accelerated rectilinear motion.

• Substitute initial position and constant

velocity of elevator into equation for

uniform rectilinear motion.

• Write equation for relative position of

ball with respect to elevator and solve

for zero relative position, i.e., impact.

• Substitute impact time into equation

for position of elevator and relative

velocity of ball with respect to

elevator.



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Sample Problem 11.4

11 - 30

SOLUTION:

• Substitute initial position and velocity and constant

acceleration of ball into general equations for

uniformly accelerated rectilinear motion.

2

2

221

00

20

s

m905.4

s

m18m12

s

m81.9

s

m18

ttattvyy

tatvv

B

B

• Substitute initial position and constant velocity of

elevator into equation for uniform rectilinear motion.

ttvyy

v

EE

E

s

m2m5

s

m2

0



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Sample Problem 11.4

11 - 31

• Write equation for relative position of ball with respect to

elevator and solve for zero relative position, i.e., impact.

025905.41812 2 ttty EB

s65.3

smeaningles s39.0

t

t

• Substitute impact time into equations for position of elevator

and relative velocity of ball with respect to elevator.

65.325Ey

m3.12Ey

65.381.916

281.918

tv EB

s

m81.19EBv



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Motion of Several Particles: Dependent Motion

11 - 32

• Position of a particle may depend on position of one

or more other particles.

• Position of block B depends on position of block A.

Since rope is of constant length, it follows that sum of

lengths of segments must be constant.

BA xx 2 constant (one degree of freedom)

• Positions of three blocks are dependent.

CBA xxx 22 constant (two degrees of freedom)

• For linearly related positions, similar relations hold

between velocities and accelerations.

022or022

022or022

CBACBA

CBACBA

aaadt

dv

dt

dv

dt

dv

vvvdt

dx

dt

dx

dt

dx



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2 - 33

Slider block A moves to the left with a

constant velocity of 6 m/s. Determine the

velocity of block B.

Solution steps

• Sketch your system and choose

coordinate system

• Write out constraint equation

• Differentiate the constraint equation to

get velocity



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2 - 34

Given: vA= 6 m/s left Find: vB

xA

yB

This length is constant no

matter how the blocks move

Sketch your system and choose coordinates

Differentiate the constraint equation to

get velocity

const nts3 aA Bx y L

Define your constraint equation(s)

6 m/s + 3 0Bv

2 m/sB v

Note that as xA gets bigger, yB gets smaller.



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Graphical Solution of Rectilinear-Motion Problems

2 - 35

Engineers often collect position, velocity, and acceleration

data. Graphical solutions are often useful in analyzing

these data.Data Fideltity / Highest Recorded Punch

0

20

40

60

80

100

120

140

160

180

47.76 47.77 47.78 47.79 47.8 47.81

Time (s)

Accele

rati

on

(g

)

Acceleration data

from a head impact

during a round of

boxing.



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11 - 36

• Given the x-t curve, the v-t curve is

equal to the x-t curve slope.

• Given the v-t curve, the a-t curve is

equal to the v-t curve slope.



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11 - 37

• Given the a-t curve, the change in velocity between t1 and t2 is

equal to the area under the a-t curve between t1 and t2.

• Given the v-t curve, the change in position between t1 and t2 is

equal to the area under the v-t curve between t1 and t2.



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Curvilinear Motion: Position, Velocity & Acceleration

11 - 38

The softball and the car both undergo

curvilinear motion.

• A particle moving along a curve other than a

straight line is in curvilinear motion.



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11 - 39

• The position vector of a particle at time t is defined by a vector between

origin O of a fixed reference frame and the position occupied by particle.

• Consider a particle which occupies position P defined by at time t

and P’ defined by at t + Dt,

r

r



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11 - 40

0lim

t

s dsv

t dtD

D

D

Instantaneous velocity

(vector)

Instantaneous speed

(scalar)

0lim

t

r drv

t dtD

D

D



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11 - 41

0lim

t

v dva

t dtD

D

Dinstantaneous acceleration (vector)

• Consider velocity of a particle at time t and velocity at t + Dt,v

v

• In general, the acceleration vector is not tangent

to the particle path and velocity vector.



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Rectangular Components of Velocity & Acceleration

11 - 42

• When position vector of particle P is given by its

rectangular components,

kzjyixr

• Velocity vector,

kvjviv

kzjyixkdt

dzj

dt

dyi

dt

dxv

zyx

• Acceleration vector,

kajaia

kzjyixkdt

zdj

dt

ydi

dt

xda

zyx

2

2

2

2

2

2



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Rectangular Components of Velocity & Acceleration

11 - 43

• Rectangular components particularly effective

when component accelerations can be integrated

independently, e.g., motion of a projectile,

00 zagyaxa zyx

with initial conditions,

0,,0 000000 zyx vvvzyx

Integrating twice yields

0

0

221

00

00

zgtyvytvx

vgtvvvv

yx

zyyxx

• Motion in horizontal direction is uniform.

• Motion in vertical direction is uniformly accelerated.

• Motion of projectile could be replaced by two

independent rectilinear motions.



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Sample Problem 11.7

11 - 44

A projectile is fired from the edge

of a 150-m cliff with an initial

velocity of 180 m/s at an angle of

30°with the horizontal. Neglecting

air resistance, find (a) the horizontal

distance from the gun to the point

where the projectile strikes the

ground, (b) the greatest elevation

above the ground reached by the

projectile.

SOLUTION:

• Consider the vertical and horizontal motion

separately (they are independent)

• Apply equations of motion in y-direction

• Apply equations of motion in x-direction

• Determine time t for projectile to hit the

ground, use this to find the horizontal

distance

• Maximum elevation occurs when vy=0



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Sample Problem 11.7

11 - 45

SOLUTION:

Given: (v)o =180 m/s (y)o =150 m

(a)y = - 9.81 m/s2 (a)x = 0 m/s2

Vertical motion – uniformly accelerated:

Horizontal motion – uniformly accelerated:

Choose positive x to the right as shown



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Sample Problem 11.7

11 - 46

SOLUTION:

Horizontal distance

Projectile strikes the ground at:

Solving for t, we take the positive root

Maximum elevation occurs when vy=0

Substitute into equation (1) above

Substitute t into equation (4)

Maximum elevation above the ground =



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11 - 47

A baseball pitching machine

“throws” baseballs with a

horizontal velocity v0. If you

want the height h to be 42 in.,

determine the value of v0.

SOLUTION:

• Consider the vertical and horizontal motion

separately (they are independent)

• Apply equations of motion in y-direction

• Apply equations of motion in x-direction

• Determine time t for projectile to fall to 42

inches

• Calculate v0=0



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2 - 48

Analyze the motion in

the y-direction

Given: x= 40 ft, yo = 5 ft,

yf= 42 in.

Find: vo

20

1(0)

2fy y t gt

2 211.5 ft (32.2 ft/s )

2t

213.5 5

2gt

0.305234 st

Analyze the motion in

the x-direction

0 00 ( )xx v t v t

040 ft ( )(0.305234 s)v

0 131.047 ft/s 89.4 mi/hv



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Motion Relative to a Frame in Translation

2 - 49

A soccer player must consider

the relative motion of the ball

and her teammates when

making a pass.

It is critical for a pilot to

know the relative motion

of his aircraft with respect

to the aircraft carrier to

make a safe landing.



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Motion Relative to a Frame in Translation

11 - 50

• Designate one frame as the fixed frame of reference.

All other frames not rigidly attached to the fixed

reference frame are moving frames of reference.

• Position vectors for particles A and B with respect to

the fixed frame of reference Oxyz are . and BA rr

• Vector joining A and B defines the position of

B with respect to the moving frame Ax’y’z’ andABr

ABAB rrr

• Differentiating twice,

ABv

velocity of B relative to A.ABAB vvv

ABa

acceleration of B relative

to A.ABAB aaa

• Absolute motion of B can be obtained by combining

motion of A with relative motion of B with respect to

moving reference frame attached to A.



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2 - 51

If we have an idea of the path of a vehicle, it is often convenient

to analyze the motion using tangential and normal components

(sometimes called path coordinates).



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11 - 52

• The tangential direction (et) is tangent to the path of the

particle. This velocity vector of a particle is in this direction

x

y

et

en

• The normal direction (en) is perpendicular to et and points

towards the inside of the curve.

v= vt et

r= the instantaneous

radius of curvature

2dv v

dt r t na e e

v tv e

• The acceleration can have components in both the en and et directions



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Sample Problem 11.8

11 - 53

A motorist is traveling on a curved

section of highway of radius 2500 ft

at the speed of 60 mi/h. The motorist

suddenly applies the brakes, causing

the automobile to slow down at a

constant rate. Knowing that after 8 s

the speed has been reduced to 45

mi/h, determine the acceleration of

the automobile immediately after the

brakes have been applied.

SOLUTION:

• Define your coordinate system

• Calculate the tangential velocity and

tangential acceleration

• Determine overall acceleration magnitude

after the brakes have been applied

• Calculate the normal acceleration



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Sample Problem 11.8

11 - 54

SOLUTION: • Define your coordinate system

eten

• Determine velocity and acceleration in

the tangential direction

• The deceleration constant, therefore

• Immediately after the brakes are applied,

the speed is still 88 ft/s

2 2 2 22.75 3.10n ta a a



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11 - 55

SOLUTION:

• Define your coordinate system

• Calculate the tangential velocity and

tangential acceleration

• Determine overall acceleration

magnitude

• Calculate the normal acceleration

The tangential acceleration of the

centrifuge cab is given by

where t is in seconds and at is in

m/s2. If the centrifuge starts from

fest, determine the total acceleration

magnitude of the cab after 10

seconds.

20.5 (m/s )ta t



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11 - 56

In the side view, the tangential

direction points into the “page”

Define your coordinate system

et

en

en

Top View

Determine the tangential velocity

0.5ta t

2 2

000.5 0.25 0.25

t t

tv t dt t t

2

0.25 10 25 m/stv

Determine the normal acceleration

2 2

22578.125 m/s

8

t

n

va

r

Determine the total acceleration magnitude

22 2 278.125 + (0.5)(10) mag n ta a a 278.285 m/smaga



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2 - 57

By knowing the distance to the aircraft and the

angle of the radar, air traffic controllers can

track aircraft.

Fire truck ladders can rotate as well as extend;

the motion of the end of the ladder can be

analyzed using radial and transverse

components.



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11 - 58

• The position of a particle P is

expressed as a distance r from the

origin O to P – this defines the

radial direction er. The transverse

direction eq is perpendicular to er

rv r e r eqq

• The particle velocity vector is

• The particle acceleration vector is

2 2ra r r e r r eqq q q

rerr



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11 - 59

Rotation of the arm about O is defined

by q = 0.15t2 where q is in radians and t

in seconds. Collar B slides along the

arm such that r = 0.9 - 0.12t2 where r is

in meters.

After the arm has rotated through 30o,

determine (a) the total velocity of the

collar, (b) the total acceleration of the

collar, and (c) the relative acceleration

of the collar with respect to the arm.

SOLUTION:

• Evaluate time t for q = 30o.

• Evaluate radial and angular positions,

and first and second derivatives at

time t.

• Calculate velocity and acceleration in

cylindrical coordinates.

• Evaluate acceleration with respect to

arm.



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SOLUTION:

• Evaluate time t for q = 30o.

s 869.1rad524.030

0.15 2

t

tq

• Evaluate radial and angular positions, and first

and second derivatives at time t.

2

2

sm24.0

sm449.024.0

m 481.012.09.0

r

tr

tr

2

2

srad30.0

srad561.030.0

rad524.015.0

q

q

q

t

t



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• Calculate velocity and acceleration.

rr

r

v

vvvv

rv

srv

qq

q

q

122 tan

sm270.0srad561.0m481.0

m449.0

0.31sm524.0 v

rr

r

a

aaaa

rra

rra

qq

q

qq

q

122

2

2

2

22

2

tan

sm359.0

srad561.0sm449.02srad3.0m481.0

2

sm391.0

srad561.0m481.0sm240.0

6.42sm531.0 a



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• Evaluate acceleration with respect to arm.

Motion of collar with respect to arm is rectilinear

and defined by coordinate r.

2sm240.0 ra OAB

Ch11 Kinematics of Particles

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Transcript of Ch11 Kinematics of Particles