PH 221-3A Fall 2009

ROTATION

Lectures 16-17

Chapter 10 (Halliday/Resnick/Walker, Fundamentals of Physics 8th edition)

1

Chapter 10RotationRotation

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

-Angular displacementAngular 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 )-Rotational inertia also known as moment of inertia (symbol I )

-Torque (symbol τ )

We will also calculate the kinetic energy associated with rotationWe 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 p y gbodies 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 objectany 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 ofspecifying 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 hi h i fi d i th i id b d d i di l t th t ti l i Awhich 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 h i i h l θ( ) l d fi h i i f ll h ithe 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

θ = 3

1In the picture we show the reference line at a time andAngular Displacement

tt2

2 1 2

2 1

at a later time . Between and the body undergoes an angular displacement . All the points of therigid

t t tθ θ θΔ = −

body have the same angular displacement because theyt1

rigid body have the same angular displacement because theyrotate locked together.

( )Angular Velocity

( )1

2 1

2 1

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

The SI unit for angular velocity is radians/secondavg t t t

t tθ θ θω − Δ

= =− Δ2 1

We define as the instantaneous ang

t t tΔ

ular velocity the limit of as 0ttθ

θ

ΔΔ →

ΔΔ dθ

0lim This is the definition of the first deriva

Algerbraic sign of angular f

tive with

If a rigid body rotre atquen es cy: counterclockwise t

ttθω

Δ →

Δ=

Δ d

dtθω =

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

ωω 4

If the angular velocity of a rotating rigid object changesAngular Accelerationω2

t2

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

ωω1

t1

I th fi h th f li t ti d t l t tit t1 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 tω ω

( )1 2r the time interval , the ratio:t tWe define as average angular acceleration fo ( )1 2

2 1

2

2

1

The SI unit for angular velocity is radia

r the time interval , the ratio:

ns/secondavg

t t

t t tω ω ωα − Δ

=Δ−

=

We define as the instantaneous angular acceleration the limit of as

li

0t

t

ω

ωΔΔ →

ΔΔ Thi i th d fi iti f th fi t d i ti ith t dω

0lim

tα

Δ →= This is the definition of the first derivative with t

tΔ ddtωα =

5

For rotations of rigid bodies about a fixed axis Angular Velocity Vector

we can describe accurately the angular velocityby asigning an algebraic sigh. Positive for counterclockwise rotation and negative forcounterclockwise rotation and negative forclockwise rotation

We can actually use the vector notation to describe rotational motion whichWe 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.no ωr

hT 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

ωr

rωof the rotation. The thumb of the right hand gives the sense f o ω

6

Rotation with Constant Angular Acceleration

When the angular acceleration is constant we can derive simple expressionsthat give us the angular velocity and the angular posit

αω ion as function of time

We could derive these equations in the same way we did in chapter 2 Instead weθ

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 motion using the following correspondance between the two motions

Rotational Motion Translational Motion x θ↔

xv

θ↔↔

aωα↔

02

02

(eqs.1)

(eqs.2)

2

2o o o

v v at

atx x v t tt

tω ω α

αθ θ ω

= +

= + +

= +↔

↔ = + +

( ) ( )2 2 22

( q )

2

(2

2 2

o o

o oo o

o

v v xa x ω ω α θ θ− = − − = −↔ eqs.3) 7

Consider a point P on a rigid body rotating about Relating the Linear and Angular Variables

a fixed axis. At 0 the reference line which connects the origin O with point P is on the x-axis (point A)

t =

Aθs

O

During 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

θline OP rotates by an angle . θ

The arc length s and the angle are connected by the equation: Relation between angular velocity and speed

θ

( ) where is the distance OP. The speed of point P d rds dr v r

dt dt dts r

θ θθ = = ==

v rω=

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

r rT Tv rπ π π

ω ω= = = =

2 12T πω

=1Tf

= 2 fω π=8

The acceleration of point P is a vector that has twoThe Acceleration

rO

components. A "radial" componet along the radius and pointing towards point O. We have enountered this component in chapter 4 where we called itthis component in chapter 4 where we called it "centripetal" acceleration. Its magnitude is:

2v 2r

va rr

ω= =

The second component is along the tangent to the circular path of P and is thus

( )

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

d rdv dω ω( )ta r r

dt dt dtα= = = =

ta rα=

The magnitude of the acceleration vector is 2 2: t ra a a= +9

Example. Motion with constant angular acceleration.

10

Rotation with constant angular acceleration.

11

The equations of rotational kinematics.

12

The equations of kinematics.

13

Angular variables and tangential variables.

14

Angular variables and tangential variables.

15

Centripetal acceleration and tangential acceleration.

16

ivr

mi

Consider the rotating rigid body shown in the figure. W di id th b d i t t f

Kinetic Energy of Rotation

Ori

1 2 3We divide the body into parts of masses , , ,..., ,...The part (or "element") at P has an index and mass Th

i

i

m m m mi m

e kinetic energy of rotation is the sum if the kinetic

2 2 21 1 2 2 3 3

gy1 1 1energies of the parts ...2 2 2

K m v m v m v= + + +

( )221 1Th d f h h lK i K∑ ∑ ( )22

2 2 2 2

The speed of the -th element 2 2

1 1 The term is known as

i i i i i ii i

i i i i

K m v i v r K m r

K m r I I m r

ω ω

ω ω

= = → =

⎛ ⎞= = =⎜ ⎟⎝ ⎠

∑ ∑

∑ ∑rotational i

2 2 or about the axis nertia moment of inerti of rotation.a The axis

i i i ii i

⎜ ⎟⎝ ⎠∑ ∑

of rotation be specified because the value of for a rigid body depends onmust Irotation 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 Iω=2 I r dm= ∫ 17

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.

18

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

19

2The rotational inertia This expression is useful for a rigid body that Calculating the Rotational In aerti

i iI m r=∑has a discreet disstribution of mass. For a continuous distribution of mass the s

i

2

um

becomes an integral I r dm= ∫∫

We saw earlier that depends on the position of the rotation axisParallel-Axis Theorem

IFor a new axis we must recalculate the integral for . A simpler method takes advantage of the parallel-a he xis t

I

Consider the rigid body of mass M shown in the figureorem

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

comI

of mass O and is perpendicul oar t the page. The rotational inertia about an axis parallel to the axis through O that passesh h i di h f i i b h i

Ithrough point P, a distance h from O is given by the equation:

2comI I Mh= + 20

A We take the origin O to coincide with the center of mass of the rigid body shownProof of the Parallel-Axis Theorem to coincide with the center of mass of the rigid body shown in the figure. We assume that we know the rotational inertia for an axis thacomI t is perpendicular to the page and passes through O.

We wish to calculate the rotational ineria about a new axis perpendicularI

( )( )

to the page and passes through point P with coodrinates , . Consider

an element of mass at point A with coordinates ,

a b

dm x y . The distance r

( ) ( )

( ) ( )

2 2

2 22

between points A and P is:

Rotational Inertia about P:

r x a y b

I r dm x a y b dm

= − + −

⎡ ⎤= = − + −⎣ ⎦∫ ∫ ( ) ( )

( ) ( )2 2 2 22 2 The second

and third integrals are zero The first integ

I x y dm a xdm b a b dm⎣ ⎦

= + − − + +

∫ ∫∫ ∫ ∫

( )2 2 2ral is The termI a b h+ =and third integrals are zero. The first integ ( ) .

2 2

ral 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

i t O d th ti f f li d t i t P di t

Torque

Fr

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

from O. In fig.b we resolve into two componets, ar dial

F

r Fr

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

r

tF F φ=r

object 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 n

t

rF rF r F

FF r

τ φ ⊥= = =

r

Thus we define as sitorque The distance is known a

ntrF rF r Fr

τ φ ⊥

⊥ s the and it is the

perpendicular distance between point O and the vector

moment arm

Fr

The algebraic sign of the torque is asigned as follows:

If a force tends to rotate an object in the countercFr

lockwise

di ti th i i iti If f t d t t tFr

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

F

r Fτ ⊥= 23

For translational motion Newton's second law connects Newton's Second Law for Rotation

the force acting on a particle with the resulting accelerationThere is a similar relationship between the torque 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 q pthis 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

rr

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 f th

r

t ti W fi t l N t ' d l f ( 1)F Ffor 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

F F maF r F

ma r m r r mr I

τ τ

τ α α α

==

= = = =( ) ( )between equations 1 and 2:

tma r m r r mr Iτ α α α= = = =

(compare with: )F ma=Iτ α= 24

We have derived Newton's second law for rotation Newton's Second Law for Rotation1

23

i for a special case. A rigid body which consists of a pointmass at the end of a massless rod of length . We willnow

m rderive the same equation for a general case

i

ri

now derive the same equation for a general case. O

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 intoτthrough point O undet the action of a net torque . We divide the body into parts or "elements" and label them. The e

netτ

1 2 3

1 2 3

lements have masses , , ,..., and they are located at distances , , ,..., from O. We apply Newton's second

n

n

m m m mr r r r

1 1 2 2

3 3

law for rotation to each element: (eqs.1), (eqs.2), (eqs

I II

τ α τ ατ α

= ==

( ) 2

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

H i h i l i iI I I I I( ) 21 2 3 1 2 3

1 2 3

... ... . 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

τ τ τ τ ατ τ τ τ τ

+ + + + = + + + + =

+ + + +

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

nI I I I I+ + + +

net Iτ α=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 Work and Rotational Kinetic Energy

WW K= Δ

on an object, this results in a change of its kinetic energy . In a similar way, when a torque does work

on roaK W WΔ =

tating rigid body it changes its rotational kineticon roa tating rigid body, it changes its rotational kineticenergy by the same amount

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

θ τ θ= =r

se it is at right angles to the motion.

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

i

tW F rd dθ

θ

θ τ θ= =∫ ∫

2 2 2 2 2 21 1 1 1have a change in kinetic energy 2 2 2 2f i f iK W mv mv mr

W K

mrω ωΔ = = −

Δ

= − →

= fθW KΔ

f

i

W dθ

θ

τ θ= ∫2 21 12 2f iK I Iω ωΔ = −

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 Power

by a force and in the case of rotational motion by a torqueWe saw that a torque produces work as it rotates an object by an angl

dW dτ τ θ=e dθrotates an object by an angl

( )

e .

(Compare with )

ddW d dP d P Fvdt dt dt

θθτ θ τ τω= = = = =

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

f

W dθ

θ∫ ( ) For constant torqueW τ θ θ= i

W dθ

τ θ= ∫ ( ) For constant torque f iW τ θ θ= −

2 21 12 2f iW K I Iω ω= Δ = − Work-Rotational Kinetic Energy Theorem

P τω= 29

Analogies between translational and rotational Motion

Rotational Motion Translational Motion

xv

θω

↔↔

v ω↔

00

a

v v at tαω ω α= + = +

↔↔

( ) ( )2 22 2

22

2

2

2

2

o o o oatx x v t

v v a x x

tt αθ θ ω

ω ω α θ θ

= + = +↔+

−↔−

+

− =− = ( ) ( )

2

2

o oo ov v a x x ω ω α θ θ↔ ==2 2

2 2

mvK IK ω== ↔

mF ma

F

IIτ α

τ=↔

↔=

↔

FP Fv

τ↔=↔

P τω=30