chap 7 kine am tics of particle

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VECTOR MECHANICS FOR ENGINEERS:

DYNAMICSCHAPTER

11Kinematics of Particles

7



Vector Mechanics for Engineers: Dynamics

11 - 2

Contents

Introduction

Rectilinear Motion: Position,

Velocity & Acceleration

Determination of the Motion of a

Particle

Sample Problem 7.1

Sample Problem 7.2

Uniform Rectilinear-Motion

Uniformly Accelerated Rectilinear-

Motion

Motion of Several Particles: RelativeMotion

Sample Problem 7.3

Motion of Several Particles:

Dependent Motion

Sample Problem 7.4

Graphical Solution of Rectilinear-Motion

Problems

Other Graphical Methods

Curvilinear Motion: Position, Velocity &

AccelerationRectangular Components of Velocity and

Acceleration

Motion Relative to a Frame in Translation

Tangential and Normal Components

Radial and Transverse Components

Sample Problem 7.5

Sample Problem 7.6




11 - 3

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 usedto 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.




11 - 4

Rectilinear Motion: Position, Velocity & Acceleration

� Particle moving along a straight line is saidto 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.,32

6 t t x !

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




11 - 5


� 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 (

(!!

(

(!

p( 0

lim

Average velocity

Instantaneous velocity

� From the definition of a derivative,

dt

dx

t

xvt

!(

(!p( 0

lim

e.g.,

2

32

312

6

t t

dt

dxv

t t x

!!

!




11 - 6

t dt

dva

t t v

dt

xd

dt

dv

t

va t

612

312e.g.

lim

2

2

2

0

!!

!

!!(

(! p(

� From the definition of a derivative,




11 - 7

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

v¶ at t +(t ,

Instantaneous accelerationt

va

t ((

!!

p( 0

lim

� Instantaneous acceleration may be:

- positive: increasing positive velocity

or decreasing negative velocity

- negative: decreasing positive velocity

or increasing negative velocity.




11 - 8


� Consider particle with motion given by326 t t x !

2312 t t dt

dxv !!

t dt

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




11 - 9

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)




11 - 10


� Acceleration given as a function of time, a = f (t ):

´´´

´´´

!!!!

!!!!!

t t t x

x

t t t v

v

dt t vxt xdt t vdxdt t vdxt vdt

dx

dt t f vt vdt t f dvdt t f dvt f adt

dv

0

0

0

0

0

0

0

0

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

´´´ !!!

!!!!!

x

x

x

x

xv

v

dxxf vxvdxxf dvvdxxf dvv

xf dx

dvva

dt

dva

v

dxdt

dt

dxv

000

202

12

21

or or




11 - 11


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

´

´´

´

´´

!

!!!!

!

!!!!

t v

v

t v

v

t x

x

t v

v

t t v

v

vf dvvxt x

vf

dvvdx

vf

dvvdxvf a

dx

dvv

t vf

dv

dt vf

dvdt

vf

dvvf a

dt

dv

0

00

0

0

0

0




11 - 12

Sample Problem 7.1

Determine:

� velocity and elevation above ground attime 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.




11 - 13

Sample Problem 7.1

t vt vdt dv

adt

dv

t t v

v

81.981.9

sm81.9

0

0

2

0

!!

!!

´´

t t v ¹º

¸©ª

¨!

2s

m81.9

s

m10

2

21

0

0

81.91081.910

81.910

0

t t yt ydt t dy

t v

dt

dy

t t y

y

!!

!!

´´

2

2

s

m905.4

s

m10m20 t t t y ¹

º

¸©ª

¨¹

º

¸©ª

¨!

SOLUTION:

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




11 - 14

Sample Problem 7.1

� Solve for t at which velocity equals zero and evaluate

corresponding altitude.

0

s

m81.9

s

m10

2!¹

º

¸©ª

¨! t t v

s019.1!t

� Solve for t at which altitude equals zero and evaluate

corresponding velocity.

22

2

2

s019.1s

m905.4s019.1

s

m10m20

s

m905.4

s

m10m20

¹º

¸©ª

¨¹

º

¸©ª

¨!

¹

º

¸©

ª

¨¹

º

¸©

ª

¨!

y

t t t y

m1.25!y




11 - 15

Sample Problem 7.1

� Solve for t at which altitude equals zero and

evaluate corresponding velocity.

0

s

m905.4

s

m10m20

2

2!¹

º

¸©ª

¨¹

º

¸©ª

¨! t t t y

s28.3

smeaningles s243.1

!

!

t

t

s28.3s

m81.9s

m10s28.3

s

m81.9

s

m10

2

2

¹º

¸

©ª

¨!

¹º

¸©ª

¨!

v

t t v

s

m2.22!v




11 - 16

Sample Problem 7.2

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 andcylinder 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).




11 - 17

Sample Problem 7.2

SOLUTION:

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

kt

v

t vdt k

v

dvkv

dt

dva

t t v

v

!!!! ´´00

ln

0

kt

evt v

! 0

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

t

kt t

kt t x

kt

ek

vt xdt evdx

evdt

dxt v

0

0

0

0

0

0

1¼½

»¬«!!

!!

´´

kt ek

vt x

! 10




11 - 18

Sample Problem 7.2

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

kxvv

dxk dvdxk dvkvdx

dvva

xv

v

!

!!!! ´´

0

00

kxvv ! 0

� Alternatively,

¹¹º

¸©©ª

¨!

0

0 1v

t v

k

vt x

kxvv ! 0

0

0 or v

t veevt v kt kt !!

kt ek

vt x ! 10with

and

then




11 - 19

Uniform Rectilinear Motion

For particle in uniform rectilinear motion, the acceleration is zero andthe velocity is constant.

vt xx

vt xx

dt vdx

vdt

dx

t x

x

!

!

!

!!

´´

0

0

00

constant




11 - 20

Uniformly Accelerated Rectilinear Motion

For particle in uniformly accelerated rectilinear motion, the acceleration of

the particle is constant.

at vv

at vvdt advadt

dv t v

v

!

!!!! ´´

0

0

00

constant

2

2

100

221

00

0

00

0

at t vxx

at t vxxdt at vdxat vdt

dx t x

x

!

!!! ´´

0

20

2

0

20

2

21

2

constant

00

xxavv

xxavvdxadvvadx

dvv

x

x

v

v

!

!!!! ´´




11 - 21

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 AABAB xxx !

!! ABAB vvv relative velocity of B

with respect to A

ABAB vvv !

!! ABAB aaa relative acceleration of B

with respect to AABAB aaa !




11 - 22

Sample Problem 7.3

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 constantvelocity 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.




11 - 23

Sample Problem 7.3SOLUTION:

� Substitute initial position and velocity and constantacceleration 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

t t at t vyy

t at vv

B

B

¹º

¸©ª

¨¹

º

¸©ª

¨!!

¹º

¸©ª

¨!!

� Substitute initial position and constant velocity of

elevator into equation for uniform rectilinear motion.

t t vyy

v

E E

E

¹º

¸©ª

¨!!

!

s

m2m5

s

m2

0




11 - 24

Sample Problem 7.3

� Write equation for relative position of ball with respect to

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

025905.41812 2!! t t t y E B

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.

65.325!E y

m3.12!E y

65.381.916

281.918

!

! t v E B

s

m81.19!E Bv




11 - 25

Motion of Several Particles: Dependent Motion

�Position of a particle may depend on position of oneor 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.

! C BA xxx 22 constant (two degrees of freedom)

� For linearly related positions, similar relations hold

between velocities and accelerations.

022or 022

022or 022

!!

!!

C BAC BA

C BAC BA

aaad t

d v

d t

d v

d t

d v

vvvd t

d x

d t

d x

d t

d x




11 - 26

Sample Problem 7.4

Pulley D is attached to a collar which

is pulled down at 3 in./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 in./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.

� PulleyD 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.




11 - 27

Sample Problem 7.4SOLUTION:

� Define origin at upper horizontal surface withpositive displacement downward.

� Collar A has uniformly accelerated rectilinear

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

? A

2

2

020

2

s

in.9in.82

s

in.12

2

!!¹º

¸©ª

¨

!

AA

AAAAA

aa

xxavv

s 333.1

s

in.9

s

in.12

2

0

!!

!

t t

t avv AAA




11 - 28

Sample Problem 7.4� PulleyD has uniform rectilinear motion. Calculate

change of position at time t .

in. 4s333.1s

in.3

0

0

!¹º

¸©ª

¨!

!

DD

DDD

xx

t vxx

� 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,

? A ? A ? A ? A 0in.42in.8

02

22

0

000

000

!

!

!

BB

BBDDAA

BDABDA

xx

xxxxxx

xxxxxx

in.160

! BB xx




11 - 29

Sample Problem 7.4

� Differentiate motion relation twice to develop

equations for velocity and acceleration of block B.

0s

in.32

s

in.12

02

constant2

!¹º

¸©ª

¨¹º

¸©ª

¨

!

!

B

BDA

BDA

v

vvv

xxx

s

in.18!Bv

0

s

in.9

02

2!¹

º

¸©ª

¨

!

B

BDA

v

aaa

2s

in.9!Ba




11 - 30

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 T

r dT

!

!(

(!

!

!(

(!

p(

p(

d t

d s

t

sv

d t

r d

t

r v

t

t

0

0

lim

lim

TTT

instantaneous velocity (vector)

instantaneous speed (scalar)




11 - 31

Curvilinear Motion: Position, Velocity & Acceleration

!

!(

(!

p( d t

vd

t

va

t

TTT

0

lim

instantaneous acceleration (vector)

� Consider velocity of particle at time t and velocity

at t + (t ,

vT

vTd

� In general, acceleration vector is not tangent to

particle path and velocity vector.




11 - 32

Rectangular Components of Velocity & Acceleration

� When position vector of particle P is given by its

rectangular components,

k zjyixr TTTT

!

� Velocity vector,

k vjviv

k zjyixk dt dzj

dt dyi

dt dxv

zyx

TTT

TTTTTTT

!

!!

� Acceleration vector,

k ajaia

k zjyixk dt

zd j

dt

yd i

dt

xd a

zyx

TTT

T

T

T

TTTT

!

!!2

2

2

2

2

2




11 - 33

Rectangular Components of Velocity & Acceleration

� Rectangular components particularly effective

when component accelerations can be integratedindependently, e.g., motion of a projectile,

00 !!!!!! zag yaxa zyx

with initial conditions,

0,,0000

000 !!!!zyx

vvvzyx

Integrating twice yields

0

0

221

00

00

!!!

!!!

zgt yvyt vx

vgt vvvv

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.




11 - 34

Motion Relative to a Frame in Translation

� Designate one frame as the fixed frame of reference.

All other frames not rigidly attached to the fixedreference 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 r r TT

� Vector joining A and B defines the position of

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

T

ABAB r r r TTT

!

� Differentiating twice,

!ABvT

velocity of B relative to A.ABAB vvvTTT

!

!ABaT acceleration of B relative

to A.ABAB aaa TTT !

� 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.




11 - 35

Sample Problem 11.7

� A projectile is fired from the edge of a 150m cliff with an initial velocity of 180

m/s at an angle of 30o with the horizontal. Neglecting air resistance, find (a) thehorizontal distance from the gun to the point where the projectile strikes the

ground, (b) the greatest elevation above the ground reached by the projectile.




11 - 36

Sample Problem 11.8

� A projectile is fired with an initial velocity of 240 m/s at a target B located 600

m above the gun A and at a horizontal distance of 3600 m. Neglecting air resistance, determine the value of the firing angle E. (Refer to page 667,

Dynamics)

f




11 - 37


� 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 withrespect to the same origin, and

is the angle between them.

t t eeTTd and

t t t eeeTTT

d!(U(

U

UU

U

U

UU

d

ed e

eee

e

t n

nnt

t

TT

TTT

!

!

((!

((

(!(

p(p( 22sinlimlim

2sin2

00

V M h i f E i D i




11 - 38


t evvTT

!� With the velocity vector expressed as

the particle acceleration may be written as

dt

ds

ds

d

d

ed ve

dt

dv

dt

ed ve

dt

dv

dt

vd a t t

UU

TT

TT

TT

!!!

but

v

dt

dsdsd e

d

ed n

t !!! UV

U

TT

After substituting,

VV

22 va

dt

dvae

ve

dt

dva nt nt !!!

TTT

� Tangential component of acceleration reflects

change of speed and normal component reflectschange of direction.

� Tangential component may be positive or

negative. Normal component always points

toward center of path curvature.

V t M h i f E i D i




11 - 39


VV

22 va

dt

dvae

ve

dt

dva nt nt !!!

TTT

�

Relations for tangential and normal accelerationalso apply for particle moving along space curve.

�Plane containing tangential and normal unitvectors is called the osculating plane.

nt b eeeTTT

v!

� Normal to the osculating plane is found from

binormal e

normal principal e

b

n

!

!

T

T

� Acceleration has no component along binormal.

V t M h i f E i D i




11 - 40

Sample Problem 7.5

A motorist is traveling on curved

section of highway at 60 mph. The

motorist applies brakes causing a

constant deceleration rate.

Knowing that after 8 s the speed has

been reduced to 45 mph, 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.

V t M h i f E i D i




11 - 41

Sample Problem 7.5

ft/s66mph45

ft/s88mph60

!

!

SOLUTION:

� Calculate tangential and normal components of

acceleration.

2

22

2

sft10.3

ft2500sft

88

s

ft75.2

s 8

sft8866

!!!

!

!

((

!

Vva

t

va

n

t

� Determine acceleration magnitude and direction

with respect to tangent to curve.

2222 10.375.2 !! nt aaa 2s

ft

14.4!a

75.2

10.3tantan 11

!!

t

n

a

aE r! 4.48E

V t M h i f E i D i




11 - 42


� When particle position is given in polar coordinates,

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

r r e

d

ed e

d

ed TT

TT

!!UU

UU

dt

d e

dt

d

d

ed

dt

ed r r UUU U

TTT

!!

dt

d e

dt

d

d

ed

dt

ed r

UUUUU TTT

!!

U

U

U

U

er er

e

d t

d r e

d t

d r

d t

ed r e

d t

d r er

d t

d v

r

r r

r r

TT

TTT

TTT

!

!!!

� The particle velocity vector is

� Similarly, the particle acceleration vector is

U

UUU

U

UUU

UUU

U

er r er r

d t

ed

d t

d r e

d t

d r e

d t

d

d t

d r

d t

ed

d t

d r e

d t

r d

e

d t

d r e

d t

d r

d t

d a

r

r r

r

TT

TTT

TT

TTT

22

2

2

2

2

!

!

¹

º

¸©

ª

¨!

r er r TT

!

V t M h i f E i D i




11 - 43


� When particle position is given in cylindrical

coordinates, it is convenient to express thevelocity and acceleration vectors using the unit

vectors . and ,, k eeR

TTTU

� Position vector,

k zeRr R

TTT!

� Velocity vector,

k zeReRdt

r d v R

T

TTT

T!! UU

� Acceleration vector,

k zeRReRRdt

vd a R

T

TTT

T!! UUUU 22

V t M h i f E i D i




11 - 44

Sample Problem 7.6

Rotation of the arm about O is defined

by U = 0.15t 2 where U is in radians and t

in seconds. Collar B slides along the

arm such that r = 0.9 - 0.12t 2 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 U = 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.

V t M h i f E i D i




11 - 45

Sample Problem 7.6

SOLUTION:

� Evaluate time t for U = 30o.

s 869.1rad524.030

0.15 2

!!r!

!

t

t U

� 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

t r

t r

2

2

srad30.0

srad561.030.0

rad524.015.0

!

!!

!!

U

U

U

t

t





11 - 46

Sample Problem 7.6� Calculate velocity and acceleration.

r r

r

v

vvvv

r vsr v

UU

U

F

U

122tan

sm270.0srad561.0m481.0

m449.0

!!

!!!

!!

r!! 0.31sm524.0 Fv

r r

r

a

aaaa

r r a

r r a

UU

U

K

UU

U

122

2

2

2

22

2

tan

sm359.0

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

2

sm391.0

srad561.0m481.0sm240.0

!!

!

!

!

!

!

!

r!! 6.42sm531.0 K a





11 47

Sample Problem 7.6

� Evaluate acceleration with respect to arm.

Motion of collar with respect to arm is rectilinear

and defined by coordinate r .

2sm240.0!! r a OAB

chap 7 kine am tics of particle

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