Chapter 7: Rotational Motion and the Law of Gravity

Angular Speed & Acceleration A unit of angular measure: radian

y

x

P

r

s = rwhere s,r in m, and in rad(ian)

length of the arc from the x-axis s:

A complete circle: s = 2r

360o = 2 rad

57.3o= 1 rad

Angular displacement and velocity

Momentum and Impulse

y

x

P at tf

r

fi

P at ti

Angular displacement:

fiin a time interval t = tf – ti

Average angular velocity:

if

ifav

ttt

rad/s

Instantaneous angular velocity:

dt

d

tt

0

lim rad/s

counter) clockwise rotation

Angular acceleration

Momentum and Impulse

y

x

P at tf

r

fi

P at ti

Average angular acceleration:

Instantaneous angular acceleration:

if

ifav

ttt

rad/s2

dt

d

tt

0

lim rad/s2

An example of a rigid body

Momentum and Impulse

• The distance of any two points in a rigid object does not change when the body is even in motion.

• When a rigid object rotates about a fixed axis, every portion of the object has the same angular speed and the same angular acceleration.

One-to-one correspondence between linear and angular quantities

Rotational Motion under Constant Angular Acceleration

• Similarity between av and vav

if

ifav

ttt

if

ifav

tt

xx

t

xv

• Similar derivation used for linear quantities can be used for angular quantities.

xavv

attvx

atvv

i

i

i

2

2

1

22

2

2

2

1

22

2

i

i

i

tt

t

An example

Rotational Motion under Constant Angular Acceleration

• Example 7.2 : A rotating wheel

rev 1.75rad) rev/2 rad)(1.00 (11.0

rad 0.112

1 2

tti

A wheel rotates with a constant angular acceleration of 3.50 rad/s2.If the angular speed of the wheel is 2.00 rad/s at ti=0,(a) through what angle does the wheel rotate between t=0 and t=2.00 s?

(b) What is the angular speed of the wheel at t=2.00 s?

rad/s 00.9 ti

and v

Relations between Angular and Linear Quantities

• Consider an object rotating about the z-axis and a point P on it.

r

s

t

s

rtt

s

rt tt

00lim

1lim

1

r

v

rvt tvv

tangential speed

tangent to circle

The tangential speed of a point on a rotating object equalsthe distance of that point from the axis of rotation multipliedby the angular speed.

and

Relations between Angular and Linear Quantities

• Consider an object rotating about the z-axis and a point P on it.

rvt

tr

t

v

tr

t

vt

t

t

t

00

limlim

rat tangential acceleration

tangent to circle

The tangential acceleration of a point on a rotating object equalsthe distance of that point from the axis of rotation multipliedby the angular acceleration.

Acceleration at a constant speed

Centripetal Acceleration

• Consider a car moving in a circular path with constant linear speed v.

t

v

tt

vva

if

ifav

Even though the magnitude of vi andvf are the same, v can be non-zero iftheir directions are different.

This leads to non-zero acceleration called centripetal acceleration.

Centripetal acceleration

Centripetal Acceleration

• Consider a car moving in a circular path with constant linear speed v.

t

v

tt

vva

if

ifav

2

2

22

222

• Triangle OAB and the triangle in Fig. (b) are similar.

sr

vv

r

s

v

v

2222

0r

r

r

r

va

t

s

r

v

t

va c

tav

22ct aaa Total acceleration

Vector nature of angular quantities

Centripetal Acceleration

• Angular quantities are vector and their directions are defined as:

points into the page points out of the page

Centripetal Acceleration Forces causing centripetal acceleration• An object can have a centripetal acceleration only if some external force acts on it.• An example is a ball whirling in a circle at the end of a string. In this case the tension in the string is the force that creates the centripetal force.

r

vmmaF cc

2

Net centripetal force Fc is thesum of the radial componentsof all forces acting on a givenobject.

T=Fc• A net force causing a centripetal acceleration acts toward the center of the circular path. If it vanishes, the object would immediately leave its circular path and move along a straight line tangent to the circle.

Centripetal Acceleration Examples• Example 7.7 : Buckle up for safety

v=13.4 m/sr=50.0 m

Find the minimum coefficient of staticfriction s between tires and roadwayto keep the car from sliding.

nfr

vm ss max,

2

mgnmgn 0

mgr

vm s

2

366.02

rg

vs

Centripetal Acceleration Examples• Example 7.8 : Daytona International Speedway

(a) Find the necessary centripetal acceleration on the banked curve so that the car will not slip due to the inclination (neglect friction).

q=31.0o

r=316 m

gmnFam

cos0cos

mgnmgn

y-component (vertical) :

x-component (horizontal) :

tan

cos

sinsin mg

mgnFc

2m/s 89.5tan/tan/ gmmgmFaFma cccc

(b) Find the speed of the car.

m/s 1.43/2 ccc ravarv

Centripetal Acceleration Examples• Example 7.9 : Riding the tracks

(a) Find the speed at the top.

R=10.0 m

top.at the 0; ngmnam c

gRvmgR

vm top

top 2

(b) Find the speed at the bottom.

2

2

2

1

5.2)2(2

1

2

1

botbot

toptop

mvE

mgRRmgmgRmghmvE

gRvmgRmvEE botbotbottop 55.22

1 2

Centripetal Acceleration Examples• Example 7.9 : Riding the tracks (cont’d)

(a)Find the normal force on a passenger at the bottom if R=10.0 m

gmnam c

mgnR

vm bot

2

mgR

gRmmg

R

vmmgn bot

65

2

n does not depend on R!

Newtonian Gravitation Law of universal gravitation• Using accumulated data on the motions of the Moon and planets, and his first law, Newton deduced the existence of the gravitational force that is responsible for the movement of the Moon and planets and this force acts between any two objects.

221

r

mmGF

-23-111 smkg10673.6 G

If two particles with mass m1 and m2 are separated by a distance r,then a gravitational force acts along a line joining them with magnitude

constant of universal gravity

Newton’s 2nd law

Newtonian Gravitation Law of universal gravitation (cont’d)

• The gravitational force exerted by a uniform sphere on a particle outside the sphere is the same as the force exerted if the entire mass of the sphere were concentrated at its center.

This is a result from Gauss’s law andstems from the fact that the gravitationalforce is inversely proportional to squareof the distance between two particles.

• The expression F=mg is valid only near the surface of Earth and can be derived from Newton’s law of universal gravitation.

Newtonian Gravitation Gravitational potential energy revisited

• Gravitational potential energy near Earth (approximation)

mghPE

• General form of gravitational potential energy due to Earth

EE Rrr

mMGPE for

radius of Earthmass of Earth

This is a special case where the zero level for potential energyis at an infinite distance from the center of Earth.

The gravitational potential energy associated with an object isnothing more than the negative of the work done by the forceof gravity in moving the object.

Newtonian Gravitation Gravitational potential energy revisited (cont’d)

• Derivation of gravitational potential energy near Earth mghPE

)(

11

)(12

hRR

mhGM

RhRmGM

R

mMG

hR

mMGPEPE

EE

E

EEE

E

E

E

E

EEEE

RhRhRR

for 1

)(

12

2212 whereE

E

E

E

R

GMgmghmh

R

GMPEPE

Newtonian Gravitation Escape speed

• If an object is projected upward from Earth’s surface with a large enough speed, it can soar off into space and never return. This speed is called Earth’s escape speed vesc.

The initial mechanical energy of the object-Earth system is:

E

Eiii R

mGMmvPEKE 2

2

1

If we neglect air resistance and assume that the initial speed islarge enough to allow the object to reach infinity with a speed ofzero, this value of vi is the escape speed vesc.

rvPEKEr

mGMmvPEKE fii

Efff ,0;0

2

1 2

E

Eesc R

GMv

2

4.3 km/s for Mercury11.2 for Earth 2.3 for Moon60.0 for Jupiter

Newtonian Gravitation Examples

• Example 7.10 : Billiards m1,2,3 =0.300 kg

(a)Find the net gravitational force on the cue ball.

N 1075.3 112

21

1221

r

mmGFFy

N 1067.6 112

31

1331

r

mmGFFx

N 1065.7 1122 yx FFF

3.29tan 1

x

y

F

F

Newtonian Gravitation Examples

• Example 7.10 : Billiards (cont’d) m1,2,3 =0.300 kg

(b) Find the components of the force of m2 on m3.

N 1040.2 112

23

3223

r

mmGF

800.0sin,600.0cos

N 1044.1cos 112323

FF x

N 1092.1sin 112323

FF y

Newtonian Gravitation Examples

• Example 7.12 : A near-Earth asteroid m1,2,3 =0.300 kg

An asteroid with mass m=1.00x109 kg comes from infinity, and fallstoward Earth.• Find the change in potential energy when it reaches a point 4.00x108

m from Earth. Find the work done by gravity. ri=0.

ifE

i

E

f

Eif rr

mGMr

mGM

r

mGMPEPEPE

11

gravWPE J 1097.9 14

(b) Find the speed of the asteroid when it reaches rf=4.00x108 m.

m/s 1041.10J 1097.9)02

1(0 3142 bb vmvPEKE

(c) Find the work needed to reduce the speed by half.

J 1048.7J 1097.9]0)2/(2

1[ 14142 WvmWPEKEW b