PH 221-1D Spring 2013

ROTATION

Lectures 24-26

Chapter 10 (Halliday/Resnick/Walker, Fundamentals of Physics 9th edition)

1

Chapter 10Rotation

In this chapter we will study the rotational motion of rigid bodies about a fixed axis. To describe this type of motion we will introduce the following new concepts:

-Angular displacement -Average and instantaneous angular velocity (symbol: ω )-Average and instantaneous angular acceleration (symbol: α )-Rotational inertia also known as moment of inertia (symbol I ) -Torque (symbol τ )

We will also calculate the kinetic energy associated with rotation, write Newton’s second law for rotational motion, and introduce the work-kinetic energy for rotational motion

2

The Rotational Variables

In this chapter we will study the rotational motion of rigid bodies about fixed axes. A rigid body is defined as one that can rotate with all its parts locked together and without any change of its shape. A fixed axis means that the object rotates about an axis that does not move. We can describe the motion of a rigid body rotating about a fixed axis by specifying just one parameter. Consider the rigid body of the figure.

We take the the z-axis to be the fixed axis of rotation. We define a reference line which is fixed in the rigid body and is perpendicular to the rotational axis. A top view is shown in the lower picture. The angular position of the reference line at any time t is defined by the angle θ(t) that the reference lines makes with the position at t = 0. The angle θ(t) also defines the position of all the points on the rigid body because all the points are locked as they rotate. The angle θ is related to the arc length s traveled by a point at a distance r from the axis via

the equation: Note: The angle θ is measured in radianssr

3

1

2 1 2

2 1

In the picture we show the reference line at a time and at a later time . Between and the body undergoes an angular displacement

An

.

gular Displace

All the points of th

ment

erigid

tt t t

body have the same angular displacement because they

rotate locked together.

1

2 1

2 1

2We define as average angular velocity for the time interval , the ratio:

We define as the instantaneous ang

The SI unit for angular velocity is

An

rad

gular Velocity

ians/secondavg t t t

t t

0

ular velocity the limit of as 0

lim This is the definition of the first deriva

Algerbraic sign of angular f

tive with

If a rigid body rotre atquen es cy: counterclockwise (CC

W

t

tt

tt

) has a positive sign. If on the other hand the rotation is clockwise (CW) has a negative sign

t1

t2

ddt

4

If the angular velocity of a rotating rigid object changes with time we can describe the time rate of change of by defining the

Angula

angula

r Accelerati

r acelera

on

tion

1 2

1 1 2 2

In the figure we show the reference line at a time and at a later time . The angular velocity of the rotating body is equal to at and at . We define as average angular acceleration fo

t tt t

1 2

0

2 1

2

2

1

The SI unit for angular velocity is radia

r the time interval , the ratio:

We define as the instantaneous angular acceleration the limit of

ns/sec

as

lim

ond

0

av

t

g

t t

t

t t

t

t

This is the definition of the first derivative with tt

ω1

ω2

t1

t2

ddt

5

For rotations of rigid bodies about a fixed axis we can describe accurately the angular velocityby asigning an algebraic sigh. Positive for counterclockwi

Angul

se ro

ar Velocity Ve

tation and ne

ctor

gative forclockwise rotation

We can actually use the vector notation to describe rotational motion which is more complicated. The angular velocity vector is defined as follows:The of is along the rotationdirecti axis.

hn

To

e of is defined by the right hand rule (RHL)Curl the right hand so that the fingers point in the direction

of the rotation. The thumb of the right

sense R

hand giight hand rule

ves the sense :

f

o

6

When the angular acceleration is constant we can derive simple e

Rotation with Constant Angular Accelerati

xpressionsthat give us the angular velocity and the angular p

on

osit ion as function of time

We could derive these equations in the same way we did in chapter 2. Instead we will simply write the solutions by exploiting the analogy between translationaland rotational m

Rotatio

otion u

nal Mot

sing the follo

ion

wing correspondance between

Tran

the two motions

slational Motion xv

02

02

2 2 22

(eqs.1)

(eqs.2)

2

(

2

2 2

o o

o oo o

o

av v at

atx x v t

v v x

tt

a x

t

eqs.3) 7

Consider a point P on a rigid body rotating about a fixed axis. At 0 the reference line which connects the origin

Relating the Linear and A

O with point P is on the

ngular Var

x-axis (po

iable

int

s

A)

D

t

uring the time interval point P moves along arc APand covers a distance . At the same time the reference line OP rotates by an angle .

ts

Aθs

O

The arc length s and the angle are connected by the equation:

where is the dista

Relation b

nce OP. T

etween angular velocity a

he speed of point P

nd speed

d rds dr v r

dt dt dts r

v r

circumference 2 2 2The period of revolution is given by: speed

r rT Tv r

2T

1Tf

2 f 8

rO

The acceleration of point P is a vector that has twocomponents. A "radial" componet along the radius and pointing towards point O. We have enountered this component i

The Accele

n chapter

ration

4 where we called it "centripetal" acceleration. Its magnitude is:

22

rva rr

The second component is along the tangent to the circular path of P and is thus known as the "tangential" component. Its magnitude is:

The magnitude of the acceleration vector is

t

d rdv da r rdt dt dt

2 2: t ra a a

ta r

9

10

Example. Motion with constant angular acceleration.

Rotation with constant angular acceleration.

11

12

The equations of rotational kinematics.

13

The equations of kinematics.

14

Angular variables and tangential variables.

15

Angular variables and tangential variables.

16

Centripetal acceleration and tangential acceleration.

Ori

iv

mi

1 2 3

Consider the rotating rigid body shown in the figure. We divide the body into parts of masses , , ,..., ,...The part (or "element") at P

Kinetic Energy of Rotatio

has an index and mass

n

Th

i

i

m m m mi m

2 2 21 1 2 2 3 3

e kinetic energy of rotation is the sum if the kinetic 1 1 1energies of the parts ...2 2 2

K m v m v m v

22

2 2 2 2

1 1 The speed of the -th element 2 2

1 1 The term

rotational i

is known as 2 2

or about the axis nertia moment of inerti of rotation.a The axis

i i i i i ii i

i i i ii i

K m v i v r K m r

K m r I I m r

of

rotation be specified because the value of for a rigid body depends on its mass, its shape as well as on the position of the rotation axis. The rotationalinertia of an object describe

mu

s w

st

ho

I

the mass is distributed about the rotation axis2

i ii

I m r 212

K I2 I r dm 17

In the table below we list the rotational inertias for some rigid bodies2 I r dm

18

An automobile of mass 1400kg has wheels 0.75m in diameter weighing 27kg each. Taking into account the rotational kinetic energy of the wheels about their axes, what is the total kinetic energy of the automobile traveling at 80km/h? What percent of kinetic energy belongs to the rotational motion of the wheels about their axes? Pretend that each wheel has a mass distribution equivalent to that of a uniform disk.

19

2The rotational inertia This expression is useful for a rigid body that

has a discreet disstribution of mass. Fo

Calculating the Rotational In

r a continuous distribution of mass the

a

s

erti

i ii

I m r

2

um

becomes an integral I r dm

We saw earlier that depends on the position of the rotation axisFor a new axis we must recalculate the integral for . A simpler method takes advantage

Parall

of the

el-Axis Theorem

parallel-a he xis t

II

Consider the rigid body of mass M shown in the figure. We assume that we know the rotational inertia about a rotation axis that passes through the center of mass O and is perpendicul

ore

o

m

ar t t

comI

he page. The rotational inertia about an axis parallel to the axis through O that passesthrough point P, a distance h from O is given by the equation:

I

2comI I Mh 20

A We take the origin O to coincide with the center of mass of the rigid body shown in the figure. We assume that we know the ro

Proof of the Parallel-Axis Theo

tational inertia for a

rem

n axis thacomI t is perpendicular to the page and passes through O.

We wish to calculate the rotational ineria about a new axis perpendicular to the page and passes through point P with coodrinates , . Consider

an element of mass at point A with coordinates ,

Ia b

dm x y

2 2

2 22

2 2 2 2

. The distance

between points A and P is:

Rotational Inertia about P:

2 2 The second

and third integrals are zero. The first i

r

r x a y b

I r dm x a y b dm

I x y dm a xdm b ydm a b dm

2 2 2 .

2 2

ntegral is The term

Thus the fourth integral is equal to

comI a b h

h dm Mh

2

comI I Mh 21

According to spectroscopic measurements, the moment of inertia of an oxygen molecule about an axis through the center of mass and perpendicular to the line joining the atoms is 1.95x10-46kg·m2. The mass of an oxygen atom is 2.66x10-26g. What is the distance between atoms?

22

In fig.a we show a body which can rotate about an axis through

point O under the action of a force applied at point P a distance

from O. In fig.b we resolve into two compone

Torq

ts, a

ue

r dial

F

r F

and tangential. The radial component cannot cause any rotationbecause it acts along a line that passes through O. The tangentialcomponent sin on the other hand causes the rotation of the

ob

r

t

F

F F

ject about O. The ability of to rotate the body depends on themagnitude and also on the distance between points P and O.Thus we define as sitorque The distance is known a

nt

trF rF r F

FF r

r

s the and it is the

perpendicular distance between point O and the vector The algebraic sign of the torque is asigned as follows:

If a force tends to rotate an object in

moment arm

the counterc

F

F

lockwise

direction the sign is positive. If a force tends to rotate an object in the clockwise direction the sign is negative.

F

r F 23

For translational motion Newton's second law connects the force acting on a particle with the resulting accelerationThere is a similar r

Newton's Second

elationship betw

Law for Ro

een the to

tation

rque of a forceapplied on a rigid object and the resulting angular acceleration

This equation is known as Newton's second law for rotation. We will explore this law by studying a simple body which consists of a point mass at the end

of a massless rod of length . A force F is

m

r

applied on the particle and rotates

the system about an axis at the origin. As we did earlier, we resolve F into a tangential and a radial component. The tangential component is responsible for the

2

rotation. We first apply Newton's second law for . (eqs.1)The torque acting on the particle is: (eqs.2) We eliminate

between equations 1 and 2:

t t t

t t

t

F F maF r F

ma r m r r mr I

(compare with: )F maI 24

We have derived Newton's second law for rotation for a special case. A rigid body which consists of a pointmass at the end of a massl

Newton's Second Law for Rotation

ess rod of length . We willnow

m r derive the same equation for a general case.

O

12

3

i

ri

Consider the rod-like object shown in the figure which can rotate about an axisthrough point O undet the action of a net torque . We divide the body into parts or "elements" and label them. The e

net

1 2 3

1 2 3

1 1 2 2

3 3

lements have masses , , ,..., and they are located at distances , , ,..., from O. We apply Newton's secondlaw for rotation to each element: (eqs.1), (eqs.2),

(eqs

n

n

m m m mr r r r

I II

21 2 3 1 2 3

1 2 3

.3), etc. If we add all these equations we get:

... ... . Here is the rotational inertiaof the -th element. The sum ... is the net torque applied.

n n i i i

n net

I I I I I m ri

1 2 3The sum ... is the rotational inertia of the body. Thus we end up with the equation:

nI I I I I

net I 25

A block of mass m hangs from a string wrapped around a frictionless pulley of mass M and radius R. If the block descends from rest under the influence of gravity, what is the angular acceleration of the pulley?

26

In chapter 7 we saw that if a force does work on an object, this results in a change of its kinetic energy

. In a similar

Work and Rotational Kin

way, when a torque does

etic Energ

work on ro

y

a

W

K W W tating rigid body, it changes its rotational kinetic

energy by the same amount

Consider the simple rigid body shown in the figure which consists of a mass

at the end of a massless rod of length . The force does work The radial component does zero work becau

t

r

m

r F dW F rd dF

2 2 2 2 2 2

se it is at right angles to the motion.

The work . By virtue of the work-kinetic energy theorem we

1 1 1 1have a change in kinetic energy 2 2 2 2

f

i

t

f i f i

W F rd d

K W mv mv mr

W K

mr

f

i

W d

2 21 12 2f iK I I

W K

27

Conservation of Energy in Rotational MotionA meter stick is initially standing vertically on the floor. If it falls over, with what angular velocity will it hit the floor? Assume that the end in contact with the floor doesn’t slip.

28

Power has been defined as the rate at which work is done by a force and in the case of rotational motion by a torqueWe saw that a torque produces work as it rotates an object

Power

by an angldW d

e .

(Compare with )

ddW d dP d P Fvdt dt dt

Below we summarize the results of the work-rotational kinetic energy theorem

f

i

W d

For constant torque f iW

2 21 12 2f iW K I I Work-Rotational Kinetic Energy Theorem

P 29

Rotational Motion

Analogies betw

een translational and rotational Motio

Translational Motion

n

xv

0

2 22 2

20

2

2

2

22

o o o o

o oo o

av v at

atx x v t

v v a x x

t

tt

2 2

2

2

mvK

mF ma

FP Fv

IK

II

P 30