Tenth Editioneng.sut.ac.th/me/2014/document/EngDynamics/11_Lecture_ppt.pdf · enth Vector Mechanics for Engineers: Dynamics dition Introduction 11 - 4 •Dynamics includes: Kinetics:

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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11 - 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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Concept Quiz

11 - 9

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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11 - 10

• 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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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

11 - 12

a a t 0 0

v t

v

dv a t dt ( )dv

a tdt

v dv 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

ln

v 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

A bowling ball is dropped from a boat so that it

strikes the surface of a lake with a speed of 5 m/s.

Assuming the ball experiences a downward

acceleration of a =10 - 0.01v2 when in the water,

determine the velocity of the ball when it strikes the

bottom of the lake.

Which integral should you choose?

0 0

v t

v

dv a t dt 0 0

v x

v x

v dv a x dx

0 0

v t

v

dvdt

a v

0 0

x v

x v

v dvdx

a v

(a)

(b)

(c)

(d)

+y



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Concept Question

11 - 21

When will the bowling ball start slowing down?

A bowling ball is dropped from a boat so that it

strikes the surface of a lake with a speed of 5 m/s.

Assuming the ball experiences a downward

acceleration of a =10 - 0.01v2 when in the water,

determine the velocity of the ball when it strikes the

bottom of the lake.

The velocity would have to be high

enough for the 0.01 v2 term to be bigger

than 10

+y



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11 - 22

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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11 - 23

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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11 - 24

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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11 - 25

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 - 26

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 - 27

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 - 28

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

constant

v 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 2

v x

v x

dvv a v dv 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

11 - 29

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 - 30

• 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 - 31

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 - 32

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 - 33

• 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 - 34

• 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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Sample Problem 11.5

11 - 35

Pulley D is attached to a collar which

is pulled down at 75 mm/s. At t = 0,

collar A starts moving down from K

with constant acceleration and zero

initial velocity. Knowing that velocity

of collar A is 300 mm/s as it passes L,

determine the change in elevation,

velocity, and acceleration of block B

when block A is at L.

SOLUTION:

• Define origin at upper horizontal surface

with positive displacement downward.

• Collar A has uniformly accelerated

rectilinear motion. Solve for acceleration

and time t to reach L.

• Pulley D has uniform rectilinear motion.

Calculate change of position at time t.

• Block B motion is dependent on motions

of collar A and pulley D. Write motion

relationship and solve for change of block

B position at time t.

• Differentiate motion relation twice to

develop equations for velocity and

acceleration of block B.



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Sample Problem 11.5

11 - 36

SOLUTION:

• Define origin at upper horizontal surface with

positive displacement downward.

• Collar A has uniformly accelerated rectilinear

motion. Solve for acceleration and time t to reach L.

( )0

2

mm mm300 225 1.333 s

s s

= +

= =

A A Av v a t

t t

22

0 0

2

2

2

mm mm300 2 200mm 225

s s

A A A A A

A A

v v a x x

a a



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Sample Problem 11.5

11 - 37

• Pulley D has uniform rectilinear motion. Calculate

change of position at time t.

xD = xD( )0+ vDt

xD - xD( )0= 75

mm

s

æ

èç

ö

ø÷ 1.333s( ) =100 mm

• Block B motion is dependent on motions of collar

A and pulley D. Write motion relationship and

solve for change of block B position at time t.

Total length of cable remains constant,

0 0 0

0 0 0

0

2 2

2 0

200mm 2 100mm 0

A D B A D B

A A D D B B

B B

x x x x x x

x x x x x x

x x

( )0

400mmB Bx x- = -



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Sample Problem 11.5

11 - 38

• Differentiate motion relation twice to develop

equations for velocity and acceleration of block B.

2 constant

2 0

mm mm300 2 75 0

s s

A D B

A D B

B

x x x

v v v

v

mm mm450 450

s sBv = - =

2

2 0

mm225 2(0) 0

s

A D B

B

a a a

a

2 2

mm mm225 225

s sBa = - =



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11 - 39

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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11 - 40

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

11 - 41

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 - 42

• 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 - 43

• 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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11 - 44

• Moment-area method to determine particle position at

time t directly from the a-t curve:

1

0

110

01 curve under area

v

v

dvtttv

tvxx

using dv = a dt ,

1

0

11001

v

v

dtatttvxx

1

0

1

v

v

dtatt first moment of area under a-t curve

with respect to t = t1 line.

Ct

tta-ttvxx

centroid of abscissa

curve under area 11001



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11 - 45

• Method to determine particle acceleration from v-x curve:

BC

AB

dx

dvva

tan

subnormal to v-x curve



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Curvilinear Motion: Position, Velocity & Acceleration

11 - 46

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 - 47

• 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 - 48

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 - 49

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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Derivatives of Vector Functions

11 - 50

uP

• Let be a vector function of scalar variable u,

u

uPuuP

u

P

du

Pd

uu D

D

D

D

DD

00limlim

• Derivative of vector sum,

du

Qd

du

Pd

du

QPd

du

PdfP

du

df

du

Pfd

• Derivative of product of scalar and vector functions,

• Derivative of scalar product and vector product,

du

QdPQ

du

Pd

du

QPd

du

QdPQ

du

Pd

du

QPd

Delete or put in

“bonus” slides



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Rectangular Components of Velocity & Acceleration

11 - 51

• 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 - 52

• 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 - 53

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 - 54

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 - 55

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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Concept Quiz

11 - 56

If you fire a projectile from 150

meters above the ground (see

Ex Problem 11.7), what launch

angle will give you the greatest

horizontal distance x?

a) A launch angle of 45

b) A launch angle less than 45

c) A launch angle greater than 45

d) It depends on the launch velocity



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11 - 57

A baseball pitching machine

“ throws ” baseballs with a

horizontal velocity v0. If you

want the height h to be 1 m,

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

1 m.

• Calculate v0=0



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11 - 58

Analyze the motion in

the y-direction

Given: x= 12 m, yo = 1.5 m,

yf= 1 m.

Find: vo

20

1(0)

2fy y t gt

2 210.5 m (9.81 m/s )

2t

213.5 5

2gt

0.3193 st

Analyze the motion in

the x-direction

0 00 ( )xx v t v t

012 m ( )(0.3193 s)v

0 37.6 m/s 135.3 km/hv



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Motion Relative to a Frame in Translation

11 - 59

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 - 60

• 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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Sample Problem 11.9

11 - 61

Automobile A is traveling east at the

constant speed of 36 km/h. As

automobile A crosses the intersection

shown, automobile B starts from rest

35 m north of the intersection and

moves south with a constant

acceleration of 1.2 m/s2. Determine

the position, velocity, and

acceleration of B relative to A 5 s

after A crosses the intersection.

SOLUTION:

• Define inertial axes for the system

• Determine the position, speed, and

acceleration of car A at t = 5 s

• Using vectors (Eqs 11.31, 11.33, and

11.34) or a graphical approach, determine

the relative position, velocity, and

acceleration

• Determine the position, speed, and

acceleration of car B at t = 5 s



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Sample Problem 11.9

11 - 62

SOLUTION: • Define axes along the road

Given: vA=36 km/h, aA= 0, (xA)0 = 0

(vB)0= 0, aB= - 1.2 m/s2, (yA)0 = 35 m

Determine motion of Automobile A:

We have uniform motion for A so:

At t = 5 s



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Sample Problem 11.9

11 - 63

SOLUTION:

Determine motion of Automobile B:

We have uniform acceleration for B so:

At t = 5 s



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Sample Problem 11.9

11 - 64

SOLUTION:

We can solve the problems geometrically, and apply the arctangent relationship:

Or we can solve the problems using vectors to obtain equivalent results:

B A B/Ar r r B A B/Av v v B A B/Aa a a

20 50

20 50 (m)

B/A

B/A

j i r

r j i

6 10

6 10 (m/s)

B/A

B/A

j i v

v j i 2

1.2 0

1.2 (m/s )

B/ A

B/ A

j i a

a j

Physically, a rider in car A would “see” car B travelling south and west.



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Concept Quiz

11 - 65

If you are sitting in train

B looking out the window,

it which direction does it

appear that train A is

moving?

a)

b)

c)

d)25o

25o



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11 - 66

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 - 67

• 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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11 - 68

• To derive the acceleration vector in tangential

and normal components, define the motion of a

particle as shown in the figure.

• are tangential unit vectors for the

particle path at P and P’. When drawn with

respect to the same origin, and

is the angle between them.

tt ee and

ttt eee

DD

d

ede

eee

e

tn

nnt

t

D

D

D

D

DD

DD 2

2sinlimlim

2sin2

00



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11 - 69

tevv

• With the velocity vector expressed as

the particle acceleration may be written as

dt

ds

ds

d

d

edve

dt

dv

dt

edve

dt

dv

dt

vda tt

but

vdt

dsdsde

d

edn

t r

After substituting,

rr

22 va

dt

dvae

ve

dt

dva ntnt

• The tangential component of acceleration

reflects change of speed and the normal

component reflects change of direction.

• The tangential component may be positive or

negative. Normal component always points

toward center of path curvature.



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11 - 70

rr

22 va

dt

dvae

ve

dt

dva ntnt

• Relations for tangential and normal acceleration

also apply for particle moving along a space curve.

• The plane containing tangential and normal unit

vectors is called the osculating plane.

ntb eee

• The normal to the osculating plane is found from

binormale

normalprincipal e

b

n

• Acceleration has no component along the binormal.



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11 - 71

A motorist is traveling on a curvedsection of highway of radius 750 mat the speed of 90 km/h. Themotorist suddenly applies the brakes,causing the automobile to slowdown at a constant rate. Knowingthat after 8 s the speed has beenreduced to 72 km/h, determine theacceleration of the automobileimmediately after the brakes havebeen 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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21.041m/s=a 53.1a =

2

2

0.833m/stan

0.625m/s

n

t

a

aa = =

2 22(25m/s)

0.833m/s750m

n

va

r= = =

220 m/s 25m/saverage 0.625m/s

8 st t

va a

t

D -= = = = -

D

km 1000 m 1 h90 km/h = 90 25 m/s

h 1 km 3600 s

72 km/h = 20 m/s

Sample Problem 11.8

11 - 72

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 25 m/s

2 2 2 2( 0.625) (0.833)n ta a a



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11 - 73

In 2001, a race scheduled at the Texas Motor Speedway was

cancelled because the normal accelerations were too high and

caused some drivers to experience excessive g-loads (similar to

fighter pilots) and possibly pass out. What are some things that

could be done to solve this problem?

Some possibilities:

Reduce the allowed speed

Increase the turn radius

(difficult and costly)

Have the racers wear g-suits



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11 - 74

SOLUTION:


• 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 - 75

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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11 - 76

a) The accelerations would remain the same

b) The an would increase and the at would decrease

c) The an and at would both increase

d) The an would decrease and the at would increase

Notice that the normal

acceleration is much higher than

the tangential acceleration.

What would happen if, for a

given tangential velocity and

acceleration, the arm radius was

doubled?



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11 - 77

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 - 78

• 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 e is perpendicular to er

v = rer+ rqe

q

• The particle velocity vector is

• The particle acceleration vector is

a = r - rq 2( )er + rq + 2rq( )eq

rerr



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11 - 79

• We can derive the velocity and acceleration

relationships by recognizing that the unit vectors

change direction.

rr e

d

ede

d

ed

dt

de

dt

d

d

ed

dt

ed rr

dt

de

dt

d

d

ed

dt

edr

erer

edt

dre

dt

dr

dt

edre

dt

drer

dt

dv

r

rr

rr

• The particle velocity vector is

• Similarly, the particle acceleration vector is

errerr

dt

ed

dt

dre

dt

dre

dt

d

dt

dr

dt

ed

dt

dre

dt

rd

edt

dre

dt

dr

dt

da

r

rr

r

22

2

2

2

2

rerr



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Concept Quiz

11 - 80

If you are travelling in a perfect

circle, what is always true about

radial/transverse coordinates and

normal/tangential coordinates?

a) The er direction is identical to the en direction.

b) The e direction is perpendicular to the en direction.

c) The e direction is parallel to the er direction.



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11 - 81

• When particle position is given in cylindrical

coordinates, it is convenient to express the

velocity and acceleration vectors using the unit

vectors . and ,, keeR

• Position vector,

kzeRr R

• Velocity vector,

kzeReRdt

rdv R

• Acceleration vector,

kzeRReRRdt

vda R

22



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11 - 82

Rotation of the arm about O is defined

by = 0.15t2 where 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 = 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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11 - 83

SOLUTION:

• Evaluate time t for = 30o.

s 869.1rad524.030

0.15 2

t

t

• 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

t

t



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11 - 84

• Calculate velocity and acceleration.

rr

r

v

vvvv

rv

srv

122 tan

sm270.0srad561.0m481.0

m449.0

0.31sm524.0 v

rr

r

a

aaaa

rra

rra

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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11 - 85

• Evaluate acceleration with respect to arm.

Motion of collar with respect to arm is rectilinear

and defined by coordinate r.

2sm240.0 ra OAB



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11 - 86

SOLUTION:


• Calculate the angular velocity after

three revolutions

• Determine overall acceleration

magnitude

• Calculate the radial and transverse

accelerations

The angular acceleration of the

centrifuge arm varies according to

where is measured in radians. If the

centrifuge starts from rest, determine the

acceleration magnitude after the gondola

has travelled two full rotations.

q = 0.05q (rad/s2)



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11 - 87

In the side view, the transverse

direction points into the “page”

Define your coordinate system

e

er

er

Top View

Determine the angular velocity

Evaluate the integral

q = 0.05q (rad/s2)

qdq =qdqAcceleration is a function

of position, so use:

0.05q 2

20

2(2p )

=q 2

20

q

0.050

(2)(2p )

ò qdq = q dq0

q

ò

q 2 = 0.05 2(2p )éë

ùû

2



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11 - 88

er

Determine the angular velocity

Determine the angular acceleration

q = 2.8099 rad/s

q 2 = 0.05 2(2p )éë

ùû

2

q = 0.05q = 0.05(2)(2p ) = 0.6283 rad/s2

Find the radial and transverse accelerations

a = r - rq 2( )er + rq + 2rq( )eq = 0 - (8)(2.8099)2( ) er + (8)(0.6283) +0( )eq = -63.166 e

r+5.0265e

q (m/s2)

22 2 2( 63.166) + 5.0265 mag ra a a 2 63.365 m/smaga

Magnitude:



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11 - 89

You could now have additional acceleration terms. This might

give you more control over how quickly the acceleration of the

gondola changes (this is known as the G-onset rate).

What would happen if you

designed the centrifuge so

that the arm could extend

from 6 to 10 meters?

r

a = r - rq 2( )er + rq + 2rq( )eq

Documents

Tenth Editioneng.sut.ac.th/me/2014/document/EngDynamics/11_Lecture_ppt.pdf · enth Vector Mechanics for Engineers: Dynamics dition Introduction 11 - 4 •Dynamics includes: Kinetics: