Chapter 11 : Kinematics of Particlessite.iugaza.edu.ps/mhaiba/files/2012/01/CH-11-Kinematics-of-Particl… · Dynamics: deals with bodies in motion Dr. Mohammad Suliman Abuhaiba,

Chapter 11

Dr. Mohammad Suliman Abuhaiba, P.E. 1

Kinematics of Particles

1/31/2014 8:29 PM

Introduction Mechanics

Mechanics = science which

describes and predicts the

conditions of rest or motion of

bodies under the action of forces

It is divided into three parts:

1. Mechanics of rigid bodies

2. Mechanics of deformable bodies

3. Mechanics of fluids

Dr. Mohammad Suliman Abuhaiba, P.E.

2 1/31/2014 8:29 PM

Introduction

Mechanics of rigid bodies is subdivided into:

1. Statics: deals with bodies at rest

2. Dynamics: deals with bodies in

motion


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Introduction

Dynamics is subdivided into:

1. Kinematics

study of geometry of motion.

relating displacement, velocity,

acceleration, and time without reference to the cause of motion

2. Kinetics

study of the relation existing between the

forces acting on a body, the mass of the body, and the motion of the body


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Introduction

A dynamic study could be done on two levels:

1. Particle

an object whose size and shape can

be ignored when studying its motion.

2. Rigid Body

a collection of particles that remain at

fixed distance from each other at all

times and under all conditions of

loading.


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Motion of Particles

Motion of Particles:

1. Rectilinear Motion

2. Curvilinear Motion


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Rectilinear Motion of Particles

Position


7

Velocity

t

xv

t

x

t

0lim

Average velocity

Instantaneous

velocity

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Acceleration


8

Instantaneous

acceleration t

va

t

0lim

t

v

Average acceleration

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9

• Consider particle with motion given by

326 ttx 2312 tt

dt

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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• Typically, conditions of motion are specified by

type of acceleration experienced by the particle.

• Determination of velocity & position requires two

successive integrations.

Determination of Motion of a Particle


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Three classes of motion may be defined:

1.acceleration is a function of time, a = f(t)

2.acceleration is a function of position, a = f(x)

3.acceleration is a function of velocity, a = f(v)

Determination of Motion of a Particle


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Determination of the Motion of a Particle


12

1. Acceleration is a function of time, a = f(t)

tttx

x

tttv

v

dttvxtxdttvdx

dttvdxtvdt

dx

dttfvtvdttfdv

dttfdvtfadt

dv

0

0

0

0

0

0

0

0

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13

2. Acceleration is a function of position, a = f(x)

x

x

x

x

xv

v

dxxfvxv

dxxfdvvdxxfdvv

xfdx

dvva

dt

dva

v

dxdt

dt

dxv

0

00

2

0212

21

or or

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14

3. Acceleration is 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

dvdt

vf

dv

dtvf

dvvfa

dt

dv

000

00

0

0

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Sample 11.2


15

Determine:

a. velocity & elevation above ground

at time t

b. highest elevation reached by ball

and corresponding time

c. time when ball will hit the ground

and corresponding velocity

Ball tossed with 10 m/s vertical

velocity from window 20 m above

ground.

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Sample 11.2


16

Solution:

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

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

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Solve for t at which velocity equals zero

and evaluate corresponding altitude.

0s

m81.9

s

m10

2

ttv

s019.1t

2

2

2

2

s019.1s

m905.4s019.1

s

m10m20

s

m905.4

s

m10m20

y

ttty

m1.25y


17

Sample 11.2

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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


18

Sample 11.2

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Sample 11.3


19

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; that is a = -kv

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

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Sample 11.3 Solution:

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

kt

v

tv

dtkv

dvkv

dt

dva

ttv

v

0

0

ln

0

ktevtv 0

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

tkt

tkt

tx

kt

ek

vtxdtevdx

evdt

dxtv

00

00

0

0

1

ktek

vtx 10


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

kxvvdxkdv

dxkdvkvdx

dvva

xv

v

0

0

0

kxvv 0

0

0 1v

tv

k

vtx kxvv 0

or 0

ktevtv kte

k

vtx 10


21

Sample 11.3

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Home Work Assignment #11.1

1, 6, 11, 17, 22, 29 Due Sunday 9/2/2014


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Uniform Rectilinear Motion


23

Acceleration is zero and velocity is constant

vtxx

vtxx

dtvdx

vdt

dx

tx

x

0

0

00

constant

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Uniformly Accelerated Rectilinear Motion


24

Acceleration of the particle is constant

atvv

atvvdtadvadt

dv tv

v

0

000

constant

221

00

221

000

00

0

attvxx

attvxxdtatvdxatvdt

dx tx

x

020

2

020

221

2

constant

00

xxavv

xxavvdxadvvadx

dvv

x

x

v

v

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Motion of Several Particles Relative Motion


25

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 wrt A

ABAB xxx

ABAB vvv relative velocity of B wrt A

ABAB vvv

ABAB aaa relative acceleration of B wrt A

ABAB aaa

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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 the

elevator

b. relative velocity of ball and

elevator at contact

Sample 11.4


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27

Solution:

ball moves with uniformly accelerated

rectilinear motion.

2

2

221

00

20

s

m905.4

s

m18m12

s

m81.9

s

m18

ttattvyy

tatvv

B

B

ttvyy

v

EE

E

s

m2m5

s

m2

0

elevator moves with uniform

rectilinear motion.

Sample 11.4

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Write equation for relative position of

ball wrt elevator & solve for zero

relative position (impact)

025905.41812 2 ttty EB

s65.3

smeaningles s39.0

t

t

Substitute impact time into equations for position of

elevator & relative velocity of ball wrt elevator.

65.325 Ey m3.12Ey

65.381.916

281.918

tv EB

s

m81.19EBv


28

Sample 11.4

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Motion of Several Particles: Dependent Motion

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 DOF)


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Motion of Several Particles: Dependent Motion

Positions of three blocks are dependent.

CBA xxx 22 constant (2 DOF)

For linearly related positions,

similar relations hold between

velocities & accelerations.

022or022

022or022

CBACBA

CBACBA

aaadt

dv

dt

dv

dt

dv

vvvdt

dx

dt

dx

dt

dx


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Sample 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.


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Solution:

• Define origin at upper horizontal

surface with +ve 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


32

Sample 11.4

1/31/2014 8:29 PM

cm 4s333.1s

cm30

0

DD

DDD

xx

tvxx

Pulley D has uniform rectilinear motion. Calculate

change of position at time t.


33

Sample 11.4

1/31/2014 8:29 PM

Block B motion is dependent on motions of collar A

& pulley D. Write motion relationship and solve for

change of block B position at time t.

0cm42cm8

02

22

0

000

000

BB

BBDDAA

BDABDA

xx

xxxxxx

xxxxxx

cm

cm

16

16

0

0

BB

BB

xx

xx


34

Sample 11.4

1/31/2014 8:29 PM

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

cm

s

cm

18

18

B

B

v

v

0s

cm9

02

2

B

BDA

a

aaa

2

2

s

cm

s

cm

9

9

B

B

a

a


35

Sample 11.4

1/31/2014 8:29 PM

Home Work Assignment #11.2

33, 38, 42, 47, 52, 57

Due Tuesday 11/2/2014


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• Given x-t curve, v-t curve = x-t curve slope

• Given v-t curve, a-t curve = v-t curve slope

Graphical Solution of

Rectilinear-Motion Problems


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Given a-t curve, change in velocity between t1 &

t2 = area under a-t curve between t1 & t2.

Given v-t curve, change in position between t1 &

t2 = area under v-t curve between t1 & t2.

Graphical Solution of

Rectilinear-Motion Problems


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Sample Problem 11.6 A subway car leaves station A; it gains speed at

the rate of 4 ft/s2 for 6 s and then at the rate of

6 ft/s2 until it has reached the speed of 48 ft/s.

The car maintains the same speed until it

approaches station B; brakes are then applied,

giving the car a constant deceleration and

bringing it to a stop in 6 s. The total running time

from A to B is 40 s. Draw the a−t, v−t, and x−t

curves, and determine the distance between

stations A and B.

(Rewrite)


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Homework Assignment #11.3

61, 67, 73, 79, 87 Due Sunday 16/2/2014


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Curvilinear Motion: Position, Velocity & Acceleration

• Curvilinear motion: Particle moving along a curve

other than a straight line

• Position vector of a particle at time t = a vector

between origin O of a fixed reference frame &

position occupied by particle.


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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) Dr. Mohammad Suliman Abuhaiba, P.E.

42 1/31/2014 8:29 PM


dt

vd

t

va

t

0lim

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 &

velocity vector. Dr. Mohammad Suliman Abuhaiba, P.E.

43 1/31/2014 8:29 PM

• Let be a vector function

of scalar variable u

• Let be a scalar function of

scalar variable u

Derivatives of Vector Functions uQ,uP

uf

u

uPuuP

u

P

du

Pd

uu

00limlim

• Derivative of vector sum,

du

Qd

du

Pd

du

QPd


Derivatives of Vector Functions

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


Rectangular Components of Velocity & Acceleration

Position vector of particle P,

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


Rectangular Components of Velocity & Acceleration

Motion of a projectile

00 zagyaxa zyx

initial conditions:

0000 zyx

Integrating twice:

0

0

221

00

00

zgtyvytvx

vgtvvvv

yx

zyyxx

Motion in horizontal direction is uniform

Motion in vertical direction is uniformly accelerated


Motion Relative to a Frame in Translation

xyz = fixed frame of reference

moving frames of reference: frames not rigidly

attached to the fixed reference frame

Position vectors for particles A and B wrt to the

fixed frame of reference Oxyz are

: position of B wrt

moving frame Ax’y’z’ ABr

ABAB rrr


. and BA rr

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Motion Relative to a Frame in Translation

ABv

velocity of B relative to A ABAB vvv

ABa

acceleration of B relative to A ABAB aaa

Absolute motion of B = combined

motion of A and relative motion of

B wrt moving reference frame

attached to A.


1/31/2014 8:29 PM

Sample Problem 11.7 A projectile is fired from 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. horizontal distance from the gun to the point

where the projectile strikes the ground,

b. greatest elevation above the ground reached

by the projectile.

1/31/2014 8:29 PM


50

Sample Problem 11.9

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.

1/31/2014 8:29 PM


51

Homework Assignment #11.4

89, 95, 101, 107, 113, 120, 126

Due Tuesday 18/2/2014


52 1/31/2014 8:29 PM

Tangential and Normal Components Velocity vector is tangent to path.

Generally, acceleration vector is

not.

= tangential unit

vectors for particle path at P &

P’

tt ee and

ttt eee

d

ede

eee

e

tn

nnt

t

2

2sinlimlim

2sin2

00


Tangential and Normal Components

dt

ds

ds

d

d

edve

dt

dv

dt

edve

dt

dv

dt

vda tt

22 va

dt

dvae

ve

dt

dva ntnt

vdt

dsdsde

d

edn

t


Tangential and Normal Components Tangential component of acceleration reflects

change of speed

Normal component reflects change of direction.

Tangential component may be positive or negative.

Normal component always points toward center of

path curvature.


Tangential and Normal Components

22 va

dt

dvae

ve

dt

dva ntnt

Relations for tangential & normal acceleration

also apply for particle moving along space curve.

Osculating plane: Plane containing

tangential & normal unit vectors

ntb eee

binormale

normalprincipal e

b

n

No Acceleration component along

binormal. Dr. Mohammad Suliman Abuhaiba, P.E.

Radial and Transverse Components

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

rerr


Radial and Transverse Components

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


Radial and Transverse Components – 3D

• Position vector,

kzeRr R

• Velocity vector,

kzeReRdt

rdv R

• Acceleration vector,

kzeRReRRdt

vda R

22


Sample 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 & normal

components of acceleration.

2

22

2

s

m10.3

m2500

sm88

s

m75.2

s 8

sm8866

va

t

va

n

t

Determine acceleration magnitude & direction wrt tangent

to curve.

2222 10.375.2 nt aaa 2s

m14.4a

75.2

10.3tantan 11

t

n

a

a 4.48


Sample 11.10

Sample 11.12 The rotation of the 0.9 m arm OA about

O is defined by the relation 0.15t2

where is expressed in radians and t in

seconds. Collar B slides along the arm

in such a way that its distance from O

is r = 0.9-0.12t2, where r is expressed in

meters and t in seconds. After the arm

OA has rotated through 30o , determine

a. total velocity of the collar

b. total acceleration of the collar

c. relative acceleration of the collar wrt the arm


Solution:

Evaluate time t for = 30o.

s 869.1rad524.030

0.15 2

t

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


Sample 11.12

rr

r

v

vvvv

rv

srv

122 tan

sm270.0srad561.0m481.0

m449.0

0.31sm524.0 v


Sample 11.12

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


Sample 11.12

Evaluate acceleration wrt arm.

Motion of collar wrt arm is rectilinear

and defined by coordinate r.

2sm240.0 ra OAB


Sample 11.12

Howe Work Assignment #11.5

133, 140, 146, 153, 160, 167, 173

Due Sunday 23/2/2014


67 1/31/2014 8:29 PM

Documents

Chapter 11 : Kinematics of Particlessite.iugaza.edu.ps/mhaiba/files/2012/01/CH-11-Kinematics-of-Particl… · Dynamics: deals with bodies in motion Dr. Mohammad Suliman Abuhaiba,