Dynamics Kinematics of Particles - جامعة نزوى€¦ · 1 All figures taken from Vector...

1

All figures taken from Vector Mechanics for Engineers: Dynamics, Beer

and Johnston, 2004

Dynamics

Kinematics of Particles

2

Introduction

• Dynamics includes:

- Kinematics: study of the motion (displacement,

velocity, acceleration, & time) without reference to

the cause of motion (i.e. regardless of forces).

- Kinetics: study of the forces acting on a body,

and the resulting motion caused by the given

forces.

• 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.

3

Rectilinear Motion: Position, Velocity & Acceleration

4


• Particle moving along a straight line is said

to be in rectilinear motion.

• Position coordinate of a particle is defined by

(+ or -) 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.

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+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

6


• Consider particle with velocity v at time t and

v’ at t+Dt,

Instantaneous acceleration t

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,

7


• Consider particle with motion given by

326 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

8

Determining the Motion of a Particle

• Recall, motion is defined if position x is known for all time t.

• If the acceleration is given, we can determine velocity and

position by 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)

dxv

dt

dva

dt

2

2

d xa

dt

dv dv dx dva v

dt dx dt dx

9


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

0

0

0 0

( ) ( ) ( ) ( )

v t t

v

dva f t dv f t dt dv f t dt v v f t dt

dt

0

0

0 0

x t t

x

dxv dx vdt dx vdt x x vdt

dt

0 0 0

2 2

0

1 1( ) ( ) ( ) ( )

2 2

v x x

v x x

dva f x v vdv f x dx vdv f x dx v v f x dx

dx

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

0 0

x t

x

dx dx dxv dt dt

dt v v

10


0 00

( )( ) ( ) ( )

v t v

v v

dv dv dv dva f v dt dt t

dt f v f v f v

0 0 0

0( )( ) ( ) ( )

x v v

x v v

dv vdv vdv vdva f v v dx dx x x

dx f v f v f v

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

11

Summary

Procedure:

1. Establish a coordinate system & specify an origin

2. Remember: x,v,a,t are related by:

3. When integrating, either use limits (if known) or add

a constant of integration

dxv

dt

dva

dt

2

2

d xa

dt

dv dv dx dva v

dt dx dt dx

12

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.

13

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

2

2s

m905.4

s

m10m20 ttty

SOLUTION:

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

14

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

2

2

2

2

s019.1s

m905.4s019.1

s

m10m20

s

m905.4

s

m10m20

y

ttty

m1.25y

15

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

16

What if the ball is tossed downwards with the same

speed? (The audience is thinking …)

vo= - 10 m/s

17

Uniform Rectilinear Motion

Uniform rectilinear motion acceleration = 0 velocity = constant

vtxx

vtxx

dtvdx

vdt

dx

tx

x

0

0

00

constant

18

Uniformly Accelerated Rectilinear Motion

Uniformly accelerated motion acceleration = 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

Also:

Application: free fall

19

Motion of Several Particles: Relative Motion

• For particles moving along the same

line, 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

20

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.

Sample Problem 11.4

21

Sample Problem 11.4 SOLUTION:

• Ball: uniformly accelerated motion

(given initial position and velocity).

• Elevator: constant velocity (given

initial position and velocity)

• 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.

22

Sample Problem 11.4

SOLUTION:

• Ball: uniformly accelerated rectilinear motion.

2

2

221

00

20

s

m905.4

s

m18m12

s

m81.9

s

m18

ttattvyy

tatvv

B

B

• Elevator: uniform rectilinear motion.

ttvyy

v

EE

E

s

m2m5

s

m2

0

23

Sample Problem 11.4

• Relative position of ball with respect to elevator:

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.

65.325 Ey

m3.12Ey

65.381.916

281.918

tv EB

s

m81.19EBv

24

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 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

25

Applications

26

Sample Problem 11.5

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.

27

Sample Problem 11.5

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.

2

2

020

2

s

in.9in.82

s

in.12

2

AA

AAAAA

aa

xxavv

s 333.1s

in.9

s

in.12

2

0

tt

tavv AAA

28

Sample Problem 11.5

• Pulley D has uniform rectilinear motion. Calculate

change of position at time t.

in. 4s333.1s

in.30

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,

0in.42in.8

02

22

0

000

000

BB

BBDDAA

BDABDA

xx

xxxxxx

xxxxxx

in.160 BB xx

29

Sample Problem 11.5

• Differentiate motion relation twice to develop

equations for velocity and acceleration of block B.

0s

in.32

s

in.12

02

constant2

B

BDA

BDA

v

vvv

xxx

in.18

sBv

2

2 0

in.9 0

s

A D B

B

a a a

a

2s

in.9Ba

30

Curvilinear Motion

http://news.yahoo.com/photos/ss/441/im:/070123/ids_photos_wl/r2207709100.jpg

A particle moving along a curve other than a straight line

is said to be in curvilinear motion.

31

Curvilinear Motion: Position, Velocity & Acceleration

• 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 + Dt, r

r

D

D

D

D

D

D

dt

ds

t

sv

dt

rd

t

rv

t

t

0

0

lim

lim

instantaneous velocity (vector)

instantaneous speed (scalar)

Velocity is tangent to path

32

Curvilinear Motion: Position, Velocity & Acceleration

D

D

D dt

vd

t

va

t

0lim

instantaneous acceleration (vector)

• Consider velocity of particle at time t and velocity

at t + Dt,

v

v

• In general, acceleration vector is not tangent to

particle path and velocity vector.

33

Rectangular Components of Velocity & Acceleration

• Position vector of particle P 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

34

Rectangular Components of Velocity &

Acceleration

0 0 0 0 00 x yx y z v v given

• Rectangular components are useful when

acceleration components can be integrated

independently, ex: motion of a projectile.

00 zagyaxa zyx

with initial conditions,

Therefore:

• Motion in horizontal direction is uniform.

• Motion in vertical direction is uniformly

accelerated.

• Motion of projectile could be replaced by two

independent rectilinear motions.

0 0

2

0 0

1

2

x x y y

x y

v v v v gt

x v t y v t gt

35

Example

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. Find

(a) the range, and (b) maximum height.

x

y

0v v at 21

0 0 2x x v t at

2 2

0 02v v a x x

Remember:

36

Example

Car A is traveling at a constant speed of 36 km/h. As A crosses

intersection, B starts from rest 35 m north of intersection and

moves with a constant acceleration of 1.2 m/s2. Determine the

speed, velocity and acceleration of B relative to A 5 seconds

after A crosses intersection.

37

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,

tt ee and

t t tde e e

t t te e de

tde dFrom geometry:

t nde d et

n

dee

d

d

tde

38

2 2

t n t n

dv v dv va e e a a

dt dt

Tangential and Normal Components

tevv

• With the velocity vector expressed as

the particle acceleration may be written as

t tt t

de dedv dv dv d dsa e v e v

dt dt dt dt d ds dt

but

vdt

dsdsde

d

edn

t

After substituting,

• 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.

39

rr re

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

we can 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

rr r

d dr dev re e r

dt dt dt • Particle velocity vector:

• Similarly, particle acceleration:

r

rr

r r

da re r e

dt

de dere r r e r e r

dt dt

d dre re r e r e r e

dt dt

rerr

• Particle position vector:

r r

dr dv e r e re r e

dt dt

2 2ra r r e r r e

40

Sample Problem 11.10

A motorist is traveling on curved section of highway at 60 mph. The

motorist applies brakes causing a constant deceleration.

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.

41


ft/s66mph45

ft/s88mph60

SOLUTION:

• Calculate tangential and normal components of

acceleration.

2

22

2

s

ft10.3

ft2500

sft88

s

ft75.2

s 8

sft8866

D

D

va

t

va

n

t

• Determine acceleration magnitude and direction

with respect to tangent to curve.

2222 10.375.2 nt aaa 2s

ft14.4a

75.2

10.3tantan 11

t

n

a

a 4.48

42


Determine the minimum radius of curvature of the trajectory

described by the projectile.

2

n

va

Recall:

2

n

v

a

Minimum , occurs for small v and large an

2

155.92480

9.81m

v is min and an is max

a

an

43


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.

44


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

45


• 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

46


• Evaluate acceleration with respect to arm.

Motion of collar with respect to arm is rectilinear

and defined by coordinate r.

2sm240.0 ra OAB

Dynamics Kinematics of Particles - جامعة نزوى€¦ · 1 All figures taken from Vector...

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Transcript of Dynamics Kinematics of Particles - جامعة نزوى€¦ · 1 All figures taken from Vector...