ch11

VECTOR MECHANICS FOR ENGINEERS: DYNAMICS

Eighth Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

Lecture Notes:

J. Walt Oler

Texas Tech University

CHAPTER

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

11Kinematics of Particles


Vector Mechanics for Engineers: Dynamics

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Chapter ObjectivesChapter Objectives

• Concepts of position, displacement, velocity, and acceleration

• Study particle motion along a straight line • Investigate particle motion along a curved path• Analysis of dependent motion of two particles



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Chapter OutlineChapter Outline

11 - 3

1. Introduction2. Rectilinear Kinematics: Continuous Motion3. General Curvilinear Motion4. Curvilinear Motion: Rectangular Components5. Motion of a Projectile6. Curvilinear Motion: Normal and Tangential Components7. Curvilinear Motion: Cylindrical Components8. Absolute Dependent Motion Analysis of Two Particles



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Introduction

• Dynamics includes:

- Kinematics: study of the geometry of motion. Kinematics is used to relate displacement, velocity, acceleration, and time without reference to the cause of motion.

- 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.

• 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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Kinematics of particles

11 - 5

Kinematics of particles

Rectilinear Curvilinear Relative

- Continuous (acceleration varies)Continuous (acceleration varies)- Continuous (Constant acceleration)Continuous (Constant acceleration)

-Rectangular componentsRectangular components- Normal/tangentialNormal/tangential- Radial/transverseRadial/transverse

- Dependent motion- Dependent motion



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Rectilinear Motion: Position, Velocity & Acceleration

• Particle moving along a straight line is said to be in rectilinear motion.

• Position coordinate of a particle is defined by positive or negative distance of particle 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. Motion of the particle 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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• 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+t,

t

xv

t

x

t

0lim

Average velocity

Instantaneous velocity

• From the definition of a derivative,

dt

dx

t

xv

t

0lim

e.g.,

2

32

312

6

ttdt

dxv

ttx



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Rectilinear Motion: Position, Velocity & Acceleration• Consider particle with velocity v at time t

and v’ at t+t,

Instantaneous accelerationt

va

t

0lim

tdt

dva

ttv

dt

xd

dt

dv

t

va

t

612

312e.g.

lim

2

2

2

0

• 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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• Consider particle with motion given by326 ttx

2312 ttdt

dxv

tdt

xd

dt

dva 612

2

2

• at t = 0, x = 0, v = 0, a = 12 m/s2

• 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

• at t = 6 s, x = 0, v = -36 m/s, a = 24 m/s2



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Determination of the Motion of a Particle• Recall, motion of a particle is known if position is known for all time t.

• Typically, conditions of motion are specified by the type of acceleration experienced by the particle. Determination of velocity and position requires two successive integrations.

• Three classes of motion may be defined for:

- 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)



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Determination of the Motion of a Particle• Acceleration given as a function of time, a = f(t):

tttx

x

tttv

v

dttvxtxdttvdxdttvdxtvdt

dx

dttfvtvdttfdvdttfdvtfadt

dv

00

0

00

0

0

0

• Acceleration given as a function of position, a = f(x):

x

x

x

x

xv

v

dxxfvxvdxxfdvvdxxfdvv

xfdx

dvva

dt

dva

v

dxdt

dt

dxv

000

202

1221

or or



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Determination of the Motion of a Particle

• Acceleration given as a function of velocity, a = f(v):

tv

v

tv

v

tx

x

tv

v

ttv

v

vf

dvvxtx

vf

dvvdx

vf

dvvdxvfa

dx

dvv

tvf

dv

dtvf

dvdt

vf

dvvfa

dt

dv

0

00

0

0

0

0



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Sample Problem 11.2

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 at which velocity equals zero (time for maximum elevation) and evaluate corresponding altitude.

• Solve for t at which altitude equals zero (time for ground impact) and evaluate corresponding velocity.



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Sample Problem 11.2

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

22s

m905.4

s

m10m20 ttty

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



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Sample Problem 11.2• Solve for t at which velocity equals zero and evaluate

corresponding altitude.

0s

m81.9

s

m10

2

ttv

s019.1t

• Solve for t at which altitude equals zero and evaluate corresponding velocity.

22

22

s019.1s

m905.4s019.1

s

m10m20

s

m905.4

s

m10m20

y

ttty

m1.25y



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11 - 16

Sample Problem 11.2• Solve for t at which altitude equals zero and

evaluate corresponding velocity.

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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Uniform Rectilinear Motion

For 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



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Uniformly Accelerated Rectilinear Motion

For particle in uniformly accelerated rectilinear motion, the acceleration of the particle is constant.

atvv

atvvdtadvadt

dv tv

v

0

000

constant

221

00

221

000

000

attvxx

attvxxdtatvdxatvdt

dx tx

x

020

2

020

221

2

constant00

xxavv

xxavvdxadvvadx

dvv

x

x

v

v



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EXAMPLE

The car moves in a straight line such that for a short time its velocity is defined by

v = (0.9t2 + 0.6t) m/s, where t is in seconds. Determine its position and acceleration

when t = 3s.



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SOLUTION

Since v = f(t), the car’s position can be determined from v = ds/dt

m

when

tts

tts

dtttds

ttdt

dsv

ts

ts

8.10)3(3.00.3(3)s

3s, t

3.03.0

3.03.0

6.09.0

)6.09.0(

23

23

023

0

0

2

0

2



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Knowing v = f(t), the acceleration is determined from a = dv/dt.

2

2

/ 66.0)3(8.1

,3

6.08.1

6.09.0

sma

stWhen

t

ttdt

d

dt

dva



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EXAMPLE:EXAMPLE:

During a test of a rocket travels upward at 75 m/s, and when it is 40 m from the ground its engine fails. Determine the maximum height sB reached by the rocket and its speed just before it hits the ground. While in motion the rocket is subjected to a constant downward acceleration of 9.81 m/s2 due to gravity. Neglect the effect of air resistance.



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mhss

h

h

asvv

AB

AB

7.3267.28640

m 7.286

)81.9(2750

2 )(2

22

To find sB

To find vC, Consider path BC and assume +ve upward

m/s 1.80m/s 1.80

)7.326)(81.9(20

2 2

22

C

C

BC

v

v

asvv



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To find vC, Consider path BC and assume +ve downward

m/s 1.80

)7.326)(81.9(20

2 2

22

C

C

BC

v

v

asvv

To find vC, Consider path ABC and assume +ve upward

m/s 1.80

)40)(81.9(275

2 2

22

C

C

AC

v

v

asvv



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Motion of Several Particles: Relative Motion

• 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 A

ABAB xxx

ABAB vvv relative velocity of B with respect to A

ABAB vvv

ABAB aaa relative acceleration of B with respect to A

ABAB aaa



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Sample Problem 11.4

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 11.4SOLUTION:

• Substitute initial position and velocity and constant acceleration of ball into general equations for uniformly accelerated rectilinear motion.

22

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

• 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 MotionMotion of Several Particles: Dependent Motion

lengthtotall

lsls

T

TBCDA

Motion of one particle will depend on the corresponding motion of another particle.

The movement of block A downward along the inclined plane will cause a corresponding movement of block B up the other incline.

0 , 00

constantremain ,

BABABA

TCD

aav , vdt

ds

dt

ds

ll



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ExampleExample

ABAB

ABAB

AB

aavv

vvdt

ds

dt

ds

hl

lshs

2 2

02 ,02

constant are ,

2

When B moves downward (+sB), A moves to the left (-sA)



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Motion of Several Particles: Dependent Motion• 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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EXAMPLE

Determine the speed of block A if block B has an upward speed of 2 m/s.

2 m/s



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2 m/s

)( m/s 6

0)2(3

03 ,03

3

A

A

BABA

BA

v

v

vvdt

ds

dt

ds

lss



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EXAMPLE

2 m/s

Determine the speed of block A if block B has an upward speed of 2 m/s.



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35 - 49

2 m/s

/ 8

0)2(4

(upward) /2 when so

04

04

24

get usly tosimultaneo questions theseSolve

)( 2

12

21

smv

v

smv

vvdt

ds

dt

ds

llss

lsssandlss

A

A

B

BA

BA

BA

CBBCA



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Sample Problem 11.5

Pulley D is attached to a collar which is pulled down at 3 cm/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 12 cm/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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• 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.

2

2

020

2

s

cm9cm82

s

cm12

2

AA

AAAAA

aa

xxavv

s 333.1s

cm9

s

cm12

2

0

tt

tavv AAA



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Sample Problem 11.5• Pulley D has uniform rectilinear motion. Calculate

change of position at time t.

cm 4s333.1s

cm30

0

DD

DDD

xx

tvxx

• 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,

0cm42cm8

02

22

0

000

000

BB

BBDDAA

BDABDA

xx

xxxxxx

xxxxxx

cm160 BB xx



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Sample Problem 11.5• Differentiate motion relation twice to develop

equations for velocity and acceleration of block B.

0s

cm32

s

cm12

02

constant2

B

BDA

BDA

v

vvv

xxx

s

cm18Bv

0s

cm9

02

2

B

BDA

v

aaa

2s

cm9Ba



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Graphical Solution of Rectilinear-Motion Problems

• 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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Graphical Solution of Rectilinear-Motion Problems

• 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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EXAMPLEEXAMPLE

A bicycle moves along a straight road such that its position is described the graph shown below. Construct the v-t and a-t graph for 0 30t s



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SOLUTIONSOLUTION

v-tv-t Graph – Graph – since since v = ds/dtv = ds/dt, the , the v-tv-t graph can be determined by differentiating graph can be determined by differentiating the equations defining the the equations defining the s-ts-t graph graph

20 10 ; 2

10 30 ; 20 -100 20

dst s s t v t

dtds

s t s s t vdt



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a-ta-t Graph – Graph – since since a = dv/dta = dv/dt, the , the a-ta-t graph can be determined by differentiating graph can be determined by differentiating the equations defining the the equations defining the v-tv-t graph graph

0 10 ; 2 2

10 30 ; 20 0

ds dvt s v t a

dt dtds dv

s t s v adt dt

SOLUTIONSOLUTION



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EXAMPLEEXAMPLE

A car travels up a hill with the speed shown below. Determine the total distance the car travels until it stops (t = 60 s). Plot the a-t graph.



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Curvilinear Motion: Position, Velocity & Acceleration

• Particle moving along a curve other than a straight line is in curvilinear motion.

• 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 particle which occupies position P defined by at time t and P’ defined by at t + t,

r

r

dt

ds

t

sv

dt

rd

t

rv

t

t

0

0

lim

lim

instantaneous velocity (vector)

instantaneous speed (scalar)



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Curvilinear Motion: Position, Velocity & Acceleration

dt

vd

t

va

t

0

lim

instantaneous acceleration (vector)

• Consider velocity of particle at time t and velocity at t + t,

v

v

• In general, acceleration vector is not tangent to particle path and velocity vector.



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Rectangular Components of Velocity & Acceleration• 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

• The magnitude of r is always positive and defined as

222 zyxr

222zyx aaaa

222zyx vvvv



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Curvilinear Motion: Rectangular Components

11 - 49

PROCEDURE FOR ANALYSIS

Coordinate System • Rectangular coordinate system can be expressed in

terms of its x, y and z components

Kinematic Quantities• Rectilinear motion is found using

v = ds/dt, a = dv/dt or a ds = v ds



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Curvilinear Motion: Rectangular Components Curvilinear Motion: Rectangular Components

11 - 50

Assumption: 1. Air resistance is neglected2. The only force acting on the projectile is its weight.

ax = 0 ay = g = 9.81 m/s2



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Horizontal Motion: ax = 0

x0x022

x002

00

x0x0

)(v v);s-2a(s v v)(

t)(v x x ;at2

1 t v x x )(

)(v v at; v v)(

0

Vertical Motion: Positive y axis is upward, thus ay = -g

)y2g(y)(v v);y-2a(y v v)(

gt2

1-t)(v y y ;at

2

1 t v y y )(

gt)(v v at; v v)(

02

y02

y022

2y00

200

y0y0

0

Horizontal component of velocity remain constant during the motion



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• 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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EXAMPLEEXAMPLE

A projectile is fired from the edge of a 150 m cliff with an initial velocity of 180 m/s at an angle of 30o 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.



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EXAMPLEEXAMPLE

VERTICAL MOTION

2

o0

-9.81m/s-ga

m/s 90(180)sin30)(

yv

Substituting into uniformly accelerated motion

y62.198100 v)y2g(y)(v v

4.90t-90t y gt2

1-t)(v y y

t81.990 v gt )(v v

2y0

2y0

2y

22y00

yy0y



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EXAMPLEEXAMPLE

11 - 55

Horizontal Motion

- Uniform Motion

155.9t x t;)(v x

m/s 9.155(180)cos30 )(v

x0

ox0

a. Horizontal Distance

When the projectile strikes the ground, y = -150 m.From y = 90t – 4.90t2 we have;

t2 - 18.37t - 30.6 = 0 t = 19.91s

From x = 155.9t we have

x = 155.9(19.91) = 3100 m



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EXAMPLEEXAMPLE

11 - 56

b. Greatest Elevation

When the projectile reaches its greatest elevation, vy = 0

From vy = 8100 – 19.62y, we have

0 = 8100 – 19.62y y = 413 m

Greatest elevation above the ground = 150 + 413 = 563 m



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EXAMPLE:EXAMPLE:

11 - 57

The chipping machine is designed to eject wood chips at vO = 7.5 m/s as shown below. If the tube is oriented at 30o from the horizontal, determine how high, h, the chips strike the pile if they land on the pile 6 m from the tube.

vO = 7.5 m/s

2.1 m

6 m



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EXAMPLE:EXAMPLE:

11 - 58

7.5cos30 6.50 m/s

7.5sin 30 3.75 m/s

O x

O y

v

v

Horizontal MotionHorizontal Motion

6 0 6.5

0.9321 s

A O O OAx

OA

OA

x x v t

t

t

Vertical MotionVertical Motion

When the motion is analyzed between points O and A, the three unknowns are represented as the height h, time of flight tOA and vertical component (vA)y.

The initial velocity of the chip has components of

212

212

2.1 0 (3.75)(0.9231) ( 9.81)(0.9231)

1.38

A O O OA c OAyy y v t a t

h

h



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Tangential and Normal Components

• Velocity vector of particle is tangent to path of particle. In general, acceleration vector is not. Wish to express acceleration vector in terms of tangential and normal components.

• 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

d

ede

eee

e

tn

nnt

t

2

2sinlimlim

2sin2

00



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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

After substituting,

22 va

dt

dvae

ve

dt

dva ntnt

• Tangential component of acceleration reflects change of speed and normal component reflects change of direction.

• Tangential component may be positive or negative. Normal component always points toward center of path curvature.



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22 va

dt

dvae

ve

dt

dva ntnt

• Relations for tangential and normal acceleration also apply for particle moving along space curve.

• Plane containing tangential and normal unit vectors is called the osculating plane.

ntb eee

• Normal to the osculating plane is found from

binormale

normalprincipal e

b

n

• Acceleration has no component along binormal.

• The magnitude of the acceleration vector is a = [(at)2 + (an)2]0.5



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There are some special cases of motion to consider.1) The particle moves along a straight line.

=> an = v2/ a = at = v

The tangential component represents the time rate of change in the magnitude of the velocity.

.

2) The particle moves along a curve at constant speed. at = v = 0 => a = an = v2/.

The normal component represents the time rate of change in the direction of the velocity.

SPECIAL CASES OF MOTION



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11 - 63

Sample Problem 11.10

A motorist is traveling on curved section of highway at 88 m/s. The motorist applies brakes causing a constant deceleration rate.

Knowing that after 8 s the speed has been reduced to 66 m/s, determine the acceleration of the automobile immediately after the brakes are applied.

SOLUTION:

• Calculate tangential and normal components of acceleration.

• Determine acceleration magnitude and direction with respect to tangent to curve.



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11 - 64


• Calculate tangential and normal components of acceleration.

2

22

2

s

m10.3

m2500

sm88

s

m75.2

s 8

sm8866

v

a

t

va

n

t

• Determine acceleration magnitude and direction with respect to tangent to curve.

2222 10.375.2 nt aaa 2s

m14.4a

75.2

10.3tantan 11

t

n

a

a 4.48



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ition

EXAMPLE

11 - 65

A race car C travels around the horizontal circular track that has a radius of 90 m. If the car increases its speed at a constant rate of 2.1 m/s2, starting from rest, determine the time needed for it to reach an acceleration of 2.4 m/s2. What is its speed at this instant.

90 m



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EXAMPLE

11 - 66

m/s 10.22.1(4.87)12

4.87s tat time speed The

87.4

049.01.24.2

m/s 2.4reach on toaccelerati for the needed timeThe

049.090

1.2 Thus

1.20)(

/ 1.2

22

22

2

222

0

2222

t.v

st

t

aaa

ttv

a

ttavv

vasmaaaa

nt

n

t

ntnt



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ition

11 - 67

Radial and Transverse Components• When particle position is given in polar coordinates,

it is convenient to express velocity and acceleration with components parallel and perpendicular to OP.

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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Radial and Transverse Components

11 - 68

Time Derivatives

2 common problems:

When coordinates are specified as r = r(t) and θ = θ(t), time derivative can be formed directly

When time parametric equations are not given, need to specify r = f(θ) and relationship between the time derivatives using the chain rule of calculus

2 2

2 2

dr d r d dr r

dt dt dt dt

dr dr dr

dt d dt



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Radial and Transverse Components

11 - 69

rθra rra

rvrv

θr

r

2

2

If the particle is moving in a circle with constant radius, r

θra ra

rvv

θr

r

02



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ition

11 - 70

Sample Problem 11.12

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.

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 - 71


• 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 - 72

Sample Problem 11.12• 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

ch11

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