KinematicsOfParticles.pdf

Introduction to Dynamics (N. Zabaras)

Introduction to Dynamicsa.k.a. Spinning & Shaking

Prof. Nicholas Zabaras

Warwick Centre for Predictive Modelling

University of Warwick

Coventry CV4 7AL

United Kingdom

Email: [email protected]

URL: http://www.zabaras.com/

January 29, 2015

1


Outline

2

Course Web Site

Recommended textbooks

Homework

Course Coverage

This (review) Lecture: Kinematics of a Particle (Chapter 12) Engr. Mechanics,

Dynamics, R. C. Hibbeler, 13th Edition) Kinematics of Particles (Chapters 11 and 12),

Mechanics of Materials, Dynamics, F. Beer, E.R. Johnston, et al. (6th Edt)


HomeHoneowork

The course web site is

The web site will include lecture notes, homework, and course announcements. Visit this link often for useful

information.

Homework problems are assigned on the course web site related to each lecture.

Solution to the homework is not mandatory but highly recommended. We consider these homework activities an

essential part of the course.

Solution to a homework set will be provided on the course web site.

Course Web Site, Lectures & Homework

http://www.zabaras.com/Courses/Dynamics/Dynamics.html

3


Textbook Vector Mechanics for Engineers, F. Beer,

E.R. Johnston and R. Cornwell, 10th

Edition.

Engineering Mechanics: Dynamics, A. Bedford and W. Fowler (course pack)

Engineering Mechanics, Dynamics, R. C. Hibbeler, 13th Edition

Recommended Textbooks

Notes: Our Lectures and Lecture slides will closely follow the textbooks by Beer et al. and

Hibbeler.


Office Hours

Office Hours :

4:30 6:30 pm on Mondays (during the duration of the lectures)

Location

Building: Eng.

Room: 408B (close to the Chemistry bldg.)

Office Hours

5


Attendance

We will work with the hope that no lecture is waste of your valuable time.

We recommend that you download and study the lecture notes from the web before attending so you can

maximize your understanding of the class material.

Even with 300+ students, we encourage your questions, in class or during office hrs.

Out of respect for each other, if you come to class you come to learn. Not allowed in class: Phones on, reading

papers, bla bla with each other, playing computer

games, etc.

Attendance

6


Course coverage

Kinematics of a particle (Hibbeler, Chapter 12) Force and acceleration (Hibbeler, Chapter 13) Work and energy relations (Hibbeler, Chapter 14, B&F Chapter

15)

Impulse and Momentum (Hibbeler, Chapter 15)

Kinematics of a rigid bodies in planar motion (Hibbeler, Chapter 16, B&F Chapter 17)

Force and acceleration of rigid bodies (Hibbeler, Chapter 17) Work and energy relations of rigid bodies (Hibbeler, Chapter 18) Impulse and Momentum of rigid bodies (Hibbeler, Chapter 19)

Vibrations (Hibbeler, Chapter 22, B&F Chapter 21)

Course Coverage

Notes

1. Several lecture notes provide review of things you should know from elementary physics. They are

essential part of the course but will be discussed briefly.

2. The lecture notes follow closely the three recommended textbooks (thus you dont need to purchase them).

7


Introduction

Dynamics includes:

- Kinematics is used to relate displacement, velocity, acceleration,

and time without reference to the cause of motion.

- 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.

8


Areas of mechanics (section12.1) Statics: Study of bodies at rest

Dynamics: Study of bodies in motion

Kinematics, is a study the geometry of the motion s, v, a

Kinetics, is a study of forces F that cause the motion

0

0

0

x

y

F

F

M

x x

y y

F ma

F ma

M I

Introduction to Dynamics

9


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.

10


Frame of ReferenceFrame of reference is a place or object that you

assume is fixed

observations of how objects move are defined in relation to that frame of

reference.

Perception of motion depends on the observers frame of reference.

We usually use the earth as the frame of reference.

11



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

12



Consider particle with velocity v at time tand 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.

13



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

14

Introduction to Dynamics (N. Zabaras) 15


Motion 1Motion 1

16


Motion 2Motion 2

17


Motion 3

slowing

Motion 3

18


Motion 4

bottom

Motion 4

19


Motion 5

speeding up

Motion 5

20


Motion 6

Still rising

Motion 6

21


Motion 7

top

Motion 7

22


1

2

3

4

5

6

7

Velocity Acceleration Velocity Acceleration

bottom 23


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)

24



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

12

21

or or

25



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

26


Relation involving x, v, & a (no time t)

dva

dt

dxdt

v

dxv

dt

dvdt

a

dx dv

v a

adx vdv

Acceleration

Velocity

Position x

27


Example of applying a dx = v dv A car starts from rest and moves along a straight line with an

acceleration of a = ( 3 x -1/3 ) m/s2. where x is in meters. Determine the

cars acceleration when t = 4 s. Rest t = 0 , v = 0

1

33v x

adx vdv

1

3

0 0

3

s v

x dx vdv

2

233 1

(3)2 2

x v

1

33dx

v xdt

1

3 3x dx dt

1

3

0 0

3

x t

x dx dt

2

33

32

x t

3

2(2 )x t

3 9

2 2

19 3

22

(4) (2 4) 2

3(4) 3 2 /

2 2

x

a m s

28


Example Problem

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.

29


Example Problem

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).

30


Example Problem

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

0s

m81.9

s

m10

2

ttv

s019.1t

22

2

2

s019.1s

m905.4s019.1

s

m10m20

s

m905.4

s

m10m20

y

ttty

m1.25y

31


Example Problem

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

32


Example Problem

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 a(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).

33


Example Problem

SOLUTION:

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

ktv

tvdtk

v

dvkv

dt

dva

ttv

v

00

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

34


Example Problem

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

35

0a k v kx


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

36


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

00

0

attvxx

attvxxdtatvdxatvdt

dx tx

x

020

2

020

221

2

constant

00

xxavv

xxavvdxadvvadx

dvv

x

x

v

v

37


Velocity as a Function of Time

c

dva

dt

cdv a dt

0o

v t

c

v

dv a dt

0 cv v a t


Position as a Function of Time

0 c

dxv v a t

dt

0

0

( )

o

x t

c

s

dx v a t dt

2

0 0

1

2cx x v t a t


Velocity as a Function of Position

0 0

v x

c

v s

v dv a dx

cv dv a dx

2 2

0 02 ( )cv v a x x

2 2

0 0

1 1( )

2 2cv v a x x


Example of Constant Acceleration: Free Fall

We throw two balls at the top of a cliff of height H. Ball A is thrown with an initial speed v0 downwards and ball B with the same speed upwards

(as shown).

The speed of the two balls when they hit the ground are vA and vBrespectively. Which of the following is true:

(a) vA < vB (b) vA = vB (c) vA > vB

v0

v0

BA

H

vA vB

41



Intuition should tell you that v = v0

We can prove this:

v0

B

H

v = v0

2 20 2( ) 0v v g H H

This looks just like ball B was thrown down with speed v0, sothe speed at the bottom shouldbe the same as that of ball A!!

y = 042



We can also just use the equation directly:

2 20 2( ) 0v v g H A :

v0

v0

A B

y = 0

2 20 2( ) 0v v g H B :same !!

y = H

43


Summary Time dependent acceleration Constant acceleration

dxv

dt

( )x t

2

2

dv d xa

dt dt

adx vdv

0 cv v a t

2

0 0

1

2cx x v t a t

2 2

0 02 ( )cv v a x x

This applies to a freely falling object:

2 2a g 9.81 / 32.2 /m s ft s

44


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 AABAB vvv

ABAB aaa relative acceleration of

B with respect to AABAB aaa

45


A B

160 km/h 180 km/h

A B A/B

A/B A B

V V V

V V - V 160 180 20 /km h

B A B/A

B/A B A

V V V

V V - V 180 160 20 /km h

Relative Motion

46


The passenger aircraft B is flying east with velocity vB = 800 km/h. A

jet is traveling south with velocity vA = 1200 km/h. What velocity does

A appear to a passenger in B ?

BABA VVV

Solution i800vB

j1200vA

x

y

BAv

BAvi800j1200

jiV BA1200800

Example

47


22/ 1200800 BAAbsolute value :

h/km1442

The direction of BAV

1200

800tan

iVB800

jVA1200

x

y

BAv

7.33 West of

South

Example

48


T

A

60 mi/h

45 mi/h

T A T/AV V V

o o

T/A60 i (45 cos 45 i 45 sin 45 j) V

T/AV { 28.2 i - 31.8 j} mi/h

2 2

/ (28.2) ( 31.8) 42.5 mi/hT A

/

/

( ) 31.8tan

( ) 28.2

T A y

T A x

o5.48

A train travels at a constant speed of 60 mi/

h and crosses over a road. If the automobile

A is traveling at 45 mi /h along the road

determine the magnitude and direction of

the velocity of the train relative to the

automobile.

Example


At the instant shown, the car at A is traveling at 10 m/s around

the curve while increasing its speed at 2 m/s2. The car at B is

traveling at 18.5 m/s along the straightway and increasing its

speed at 5 m/s2. Determine the relative velocity and relative

acceleration of A with respect to B at this instant

A B A/BV V V

A/B10 cos 45i -10 sin 45 j 18.5 i V

2 2o o

A

(10) (10)a {2 cos 45 i 2 sin 45 j- cos 45 i - sin45 j }

100 100

Ba 5 i

A B A/Ba a a

A/BV { -11.42 i 7.07 j} m/s

Example

50


Sample Problem

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.

51


Sample Problem

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

52


Sample Problem

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.

65.325 Ey m3.12Ey

65.381.916

281.918

tv EB

s

m81.19EBv

53


Absolute Dependent Motion Analysis of Two Particles

The motion of one particle will depend on the

corresponding motion of

another particle

Inextensible cords connection


Position

How to establish the position coordinate Reference from a fixed point

(O) or fixed datum.

Measures along each inclined plane in the

direction of motion

Has positive sense from C to A and from D to B.

Establish a total cord length equation

A CD B Ts l s l

55


Velocity

Velocity is the time derivative of the position

From the total cord length equation

Velocity

A CD B Ts l s l

0A B B Ads ds

ordt dt

A B T CDs s l l const

56


Acceleration

Acceleration is the time derivative of the velocity

From the velocity equation

Acceleration

B A

B Aa a

B Ad d

dt dt

57


Example Problem

Position coordinate

Velocity

Acceleration

2 B As h s l

2 B A

2 B Aa a

2 .B As s l h const

1

2B A

58


Example Problem

Position coordinate

Velocity

Acceleration

Positive signs, why?

lshsh AB )(2

2 B A

2 B Aa a

2 3B As s l h Const

59


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

60


Sample Problem

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.

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.61


Sample Problem

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

62


Sample Problem

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

63


Sample Problem

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

s

in.18Bv

0s

in.9

02

2

B

BDA

v

aaa

2s

in.9Ba

64


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 + 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)

65


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.

66


Derivatives of Vector Functions

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

67


Curvilinear MotionWe will investigate particle motion along a curved path

using three coordinate systems

Rectangular Components

Normal and Tangential Components

Polar & Cylindrical Components

68


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

69


Rectangular Components of Velocity & Acceleration

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.

70


Example Problem The position of a particle is described by rA= {2t i +(t

2-1) j} ft. where t is

in seconds.

Determine the position of the point and the speed at 2 seconds.

22(2) (2 1) 4 3Ar i j i j

2 2 2 4AAdr

v i t j i jdt

2 22 4 4.47 ft/sAspeed v

71


Acceleration

Acceleration is the first time derivative of v

Where

Magnitude of acceleration

Direction is not tangent to the path

x y z

dva a i a j a k

dt

x xa x

2 2 2

x y za a a a

y ya y

z za z

72


Position

8(2) 16ftx

2(16) /10 25.6 fty

2 2(16) (25.6) 30.2 ftr

Velocity

(8 ) 8ft/sxd

x tdt

2( /10) 2 /10 2(16)(8) /10 25.6ft/syd

y x xxdt

2 2(8) (25.6) 26.8 ft

1 1 25.6tan tan 72.68

y o

x

Acceleration

(8) 0ft/sxd

a xdt

2(2 /10) 2( ) /10 2 ( ) /10 12.8ft/syd

a y xx x x x xdt

2 2 2(0) (12.8) 12.8 ft/sa

1 1 12.8tan tan 900

y o

a

x

a

a

A x=8t

At any instant the horizontal postition of the weather

balloon is defined by x = (8t) ft, where t is in second . If the equation of the path is y=x2/10 determine the distance

of the balloon from A, the magnitude and direction of the

velocity and the acceleration when t=2sec.

Example Problem

73


Example

74

r(0.75) 0.5sin(1.5)i 0.5cos(1.5) j 0.2(0.75)k

)180( orad

0.5sin(2 )i 0.5cos(2 ) j 0.2 kr t t t

t in seconds, arguments in radians

At t = 0.75 s find location, velocity, and

acceleration

r 0.499i 0.0354j 0.15k

2 2 2(0.499) (0.0354) ( 0.15) 0.522 mr

0.499 0.0354 0.15i j k 0.955i 0.0678j-0.287k

0.522 0.522 0.522ru

1cos (0.955) 17.2o

1cos (0.0678) 86.1o

1cos ( 0.287) 107o

Take derivatives of r(t) w.r.t. time to

define similar eqs for v and a


Motion of a Projectile

Projectile: a body that is given an initial velocity and then follows a path determined by gravitational acceleration

and air resistance.

Trajectory path followed by a projectile

75


Horizontal motion is uniform motion

Notice that the horizontal motion is in no way affected by the vertical motion.

76


Projectile

77


Vertical velocity decreases at a constant rate

due to the influence of gravity.

Horizontal and vertical components of velocity are independent.

Vertical velocity decreases at a constant rate due to the influence

of gravity.

Horizontal and vertical components of velocity

78


Horizontal Motion

Acceleration : ax= 0

Horizontal velocity remains constant

Equal distance covered in equal time intervals

0 0( ) ( )c x xv v a t v v

2

0 0 0 0

1( ) ( )

2c xx x v t a t x x v t

22

0 00( ) 2 ( ) ( )c x xv a s s v vv

0( )x

x xv

t

79


Vertical Motion ac= -g = 9.81 m/s

2 = 32.2 ft/s2

Equal increments of speed gained in equal increments of time

Distance increases in each time interval

0 0( ) ( )c y yv v a t v v gt

2 2

0 0 0 0

1 1( ) ( )

2 2c yy y v t a t y y v t gt

22 22

0 00( ) 2 ( ) 2 ( )

0c y yv a s s g y yvv v

80


Assumptions:

(1) free-fall acceleration

(2) neglect air resistance

Choosing the y direction as positive upward:

ax = 0; ay = - g (a constant)

Take x0= y0 = 0 at t = 0

Initial velocity v0 makes an

angle 0 with the horizontalv0

x

y

0 0 0 0 0 0cos sin

x yv v v v

Projectile Motion

81


At the peak of its trajectory, vy = 0.

From

Time t1 to reach the peak

Substituting into:

0

1

yvt

g

2

0

max2

yvh y

g

0 0y y oyv v gt v gt

2

0

1

2yy v t gt

Maximum Height

82


The optimal angle of projection is dependent on the goal of the activity.

For maximal height the optimal angle is 90o.

For maximal distance the optimal angle is 45o.

Projection Angle

83


Projection angle = 10 degrees

84


10 degrees

30 degrees

40 degrees

45 degrees


85


10 degrees

30 degrees

40 degrees

45 degrees

60 degrees


86


10 degrees

30 degrees

40 degrees

45 degrees

60 degrees

75 degrees

Angle that maximizes range

optimal = 45 degrees


87


Example Problem 1/4 A ball is given an initial velocity of V0 = 37 m/s at an angle of = 53.1.

Find the position of the ball, and the magnitude and direction of its velocity, when t = 2.00 s.

Find the time when the ball reaches the highest point of its flight, and find its height h at this point

The initial velocity of the ball has components:

v0x = v0 cos 0 = (37.0 m/s) cos 53.1 = 22.2 m/s

v0y = v0 sin 0 = (37.0 m/s) sin 53.1= 29.6 m/s

a) position

x = v0xt = (22.2 m/s)(2.00 s) = 44.4 m

y = v0yt - gt2

= (29.6 m/s)(2.00 s) (9.80 m/s2)(2.00 s)2

= 39.6 m

88


Example Problem 2/4

Velocity vx = v0x = 22.2 m/s vy = v0y gt = 29.6 m/s (9.80 m/s

2)(2.00 s) = 10.0 m/s

22 2 222.2 / (10.0 / )

24.3 /

10.0 /arctan arctan 0.450 24.2

22.2 /

x yv v v m s m s

m s

m s

m s

89


Example Problem 3/4

b) Find the time when the ball reaches the highest point of its flight, and find its height H at this point.

0 1

0

1 2

0

29.6 /3.02

9.80 /

y y

y

v v gt

v m st s

g m s

1

2

0 1

2 2

1

2

1(29.6 / )(3.02 ) (9.80 / )(3.02 )

2

44.7

yH v t gt

m s s m s s

m

90


c) Find the horizontal range R, (e.g. the horizontal distance from the starting point to the point at which

the ball hits the ground.)

0 2 (22.2 / )(6.04 ) 134xR v t m s s m

2

0 2 2 2 0 2

1 10 ( )

2 2y yy v t gt t v gt

0

2 2 2

2 2(29.6 / )0 6.04

9.80 /

yv m st and t s

g m s

Example Problem 4/4

91


A ball is traveling at 25 m/s drive off of the edge of a cliff 50 m high. Where does it land?

25 m/s

Vertically

v = v0-gt

y = y0 + v0t + 1/2gt2 .

v2 = v02 - 2g(y-y0).

Horizontally

x = x0 + (v0)x t x = 25 *3.19 = 79.8 m

79.8 m

Initial

Conditions

vx = 25 m/s

vy0 = 0 m/s

a =- 9.8 m/s2

t = 0

y0 = 0 m

y =- 50 m

x0 =0 m

-50 = 0+0+1/2(-9.8)t2 t = 3.19 s

Example Problem

92


Motion Relative to a Frame in Translation

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 Axyz and

ABr

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.

93


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

DD

d

ede

eee

e

tn

nnt

t

D

D

D

D

DD

DD 2

2sinlimlim

2sin2

00

94



tevv

With the velocity vector expressed asthe 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,

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.

95



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.

96


Radius of curvature ()

For the Circular motion : () = radius of the circle

For y = f(x):

2 3/2

2 2

1 ( / )

/

dy dx

d y dx

97


Example Problem

Find the radius of curvature of the parabolic path in the

figure at x = 150 ft.

21 200200

y x

1/

100dy dx x

2 2 1/100

d y dx 3/2

2

3/22

2 2

150

11

1001 ( / )585.9 ft

1/

100

x

xdy dx

d y dx

98


Example Problem A skier travels with a constant

speed of 20 ft/s along the parabolic

path shown. Determine the velocity

at x = 150 ft.

21 200200

y x

1/

100dy dx x

150

1/ 1.5

100 xdy dx x

1tan 1.5 56.3o

15020 ft/s

99


Special case

1- Straight line motion

2- Constant speed curve motion (centripetal acceleration)

0na

ta a

0ta

2

na a

100


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

101


Radial and Transverse Components

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

2

2

102


Sample Problem

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.

103


Sample Problem

ft/s66mph45

ft/s88mph60

SOLUTION:

Calculate tangential and normal components of acceleration.

2

22

2

66 88 ft s ft2.75

8 s s

88ft s ft3.10

2500ft s

t

n

va

t

va

D

D

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

104


Sample Problem

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.

105


Sample Problem

SOLUTION:

Evaluate time t for = 30o.

20.15

30 0.524rad 1.869 s

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

106


Sample Problem

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

107


Sample Problem

Evaluate acceleration with respect to arm.

Motion of collar with respect to arm is

rectilinear and defined by coordinate r.

2sm240.0 ra OAB

108

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