1r 2r

When the axis of rotation is fixed, all particles move in a circle. Because the object is rigid, they move through the same angular displacement in the same time period.

rx

rx rv rat

dt

d

2

2

dt

d

dt

d Radians only!

Note that as long as there is rotation there will be a radial or centripetal acceleration given by

RR

vacp

22

1-D linear motion

Rotation: fixed axis

∆x ∆

v

a

m

F

F=ma

p=mv

P=Fv

2

2

1mvK

FdxW

2mrI

2

2

1 IK

IIL

dW

P

Linear 1-D vs Fixed Axis Rotation

Constant Angular Acceleration Kinematics

The equations for 1-D motion with constant acceleration are a result the definitions of the quantities; because it immediately follows that if acceleration is constant.

dt

dva

atvv 0

Since the angular variables θ, ω, α are related to each other in exactly the same way as x, v, and a are, it follows that they will obey analogous kinematic equations:

t 0

t02

1

20 2

1tt

220

2

Exercise: A disc initially rotating at 40 rad/s is slowed to 10 rad/s in 5 s. Find the angular acceleration and the angle through which it turns during this time.

t 0

54010

26s

rad

624010 22

rad125

1r 2r

222

2

1

2

1 iiiii rmvmK

222

2

1

2

1 iiiii rmvmKK

2iirmI

2

2

1 IK rot

Moment of inertia about the axis of rotation

Moment of Inertia Examples

2m1m

L C R

R 22

22

21 0 RmRmmIL

21

22

21 0 RmmRmIR

422

2

21

2

2

2

1R

mmR

mR

mIC

2

4

1

2,2 1,2

1,4

2,2origin I and I Find

Continuous Matter Distribution

body

ii dmrrm 22

Example: Uniform rod, length L, mass M. Find ILeft

1) Find contribution of arbitrary little piece dm.

x dx

2xdxL

MdI left

2) Sum up (integrate over correct limits) all the little pieces.

2

0

2

3

1MLdxx

L

MdII

L

body

leftleft

Uniform disc, radius R, mass M. Find IC

1) Find contribution of arbitrary little piece dm.

r

dr

2rdmdI All the mass in ring is same r away.

But what is dm?The area in the ring is an infinitesimal: drrdA 2

Let σ designate mass per area:2R

M

Then drrR

MdAdm

2

2

2) Sum up (integrate over correct limits) all the little pieces.

R

body

rdAdII0

2 drrR

3

0

2 24

2

1

2MR

R

1r 2r

1r 2r

Arel

A rmI 2

CMrel

CM rmI 2

A

CM

Parallel Axis Theorem

2MRII CMA A

CMR

Total mass

Example: Find ICM for a uniform stick of length L, mass M.

2

2

L

MII CMend

L

22

23

1

L

MIML CM2

12

1MLICM

Exercise: Find I about a point P at the rim of a uniform disc of mass M, radius R.

R

P

Torque

From Newton’s 2nd law we know that forces cause accelerations. We might ask what particular quantity, obviously related to force, will cause angular accelerations.Consider a 10 N force applied to a rod pivoted about the left end. We can apply the force in a variety of ways, not all causing the same angular acceleration:

From the examples it should be clear that it is a combination of the force applied, direction of application, and distance from axis that need to be accounted for in determining if a force can cause angular acceleration.

The previous discussion is consistent with focusing on the following quantity:

R

F

RFRFeff sin

Line of action of F

Reff or lever arm effFR

RFRF

sinsin

The greater the torque associated with a force, the more efficient it is in creating angular acceleration.

ir

iF

tiF

tii

ti amF

2ii

tiii

tii rmarmFr Torque on piece

i 2iii rm

Inet Refers to a particular axis

R

I

m

H

Example: Find the acceleration of the mass, the tension in the rope and the speed of the mass having fallen H.

Inet IRT

maTmg

R

aIRT

g

R

Im

ma

2

T T

mg

RT

R

I

m

H

22

2

1

2

1 ImvmgH 2

2

2

1

2

1

R

vImvmgH

2

2

R

Im

mgHv

Rolling

When an object rolls, the point of contact is instantaneous at rest. It can be thought of as an instantaneous axis of rotation, and relative to this point we have pure rotation at that instant.

At this instant, each point shown will have a different speed, but

i

i

r

v

is the same for each point on the object.

v

v2

Rv

rolling condition

CMvv

v

v

v v

v

=

+

Conceptually, we can think of rolling as pure translation at vCM coupled with a simultaneous rotation about CM with

R

vCM

v

v2 2

2

1 cprot IK

22

2

1 MRIK rot

22

2

1

2

1MvIK rot

v

v

v v

v

+

22

2

1

2

1 IMvK rot

We can use energy concepts coupled with kinematics analyze rolling dynamics.

H

v

2MRI

1

2

2

1

2

1 222 gHvIMvMgH

Rv

1

sin

sin202 g

aH

av

sf

N

Mg

1

sinsin

gMMafMg s

1

sinMgf s

Exercise: Find and interpret the work done by fs.

Exercise: At what angle will slipping start?

TorqueA more systematic approach to rotation involves relating the torques associated with forces to the angular accelerations they produce. This can be more complicated if the axis is not fixed, but we shall see that the rolling case is especially simple in this approach.First generalize torque:

rF

Fr

sinrF

Right hand rule, RHR

Note again that torques are calculated relative to a specific origin and will change if we change the origin.Where could I place an origin above so that F would exert zero torque?

Angular Momentum

r

p prl

prrpl sinsin

prlr

1) a vector perpendicular to both and r

p

RHR

2) magnitude equals linear momentum times “closest approach distance”.

3) value clearly depends choice of origin.

What kind of motion might have constant angular momentum?

Why Angular Momentum?

prl

dt

pdrp

dt

rd

dt

ld

Fnetrvvmdt

ld

netdt

ld

For a system of particles we can define the total angular momentum:

ilL

Applying the second and third law we get::

extnetdt

Ld

Rotational analog of 2nd law

Dynamics of rolling and slipping

We will assume that the axis along which the angular momentum points does not change direction. A baseball curve ball is too hard for us to deal with.Just as we could break the kinetic energy into two parts, so too can we break down the angular momentum:

CMaboutCMof LLL

Treat CM as point particlePure rotation around CM axis

v

v

There are two “natural” origins to choose for calculating torques and angular momentum: the center of mass CM and the contact point CP.Let us look at the “L” analysis in each case.

v

ImvRL The second term will always be I

CP

v

CM

IL 0 Note the simplicity

Example:

sf

N

Mg

C`M

IL sRf dt

dLsRfI

mafmg s sinR

a

2

sin

R

Im

mga

2

2sin

R

Im

R

I

mgf s

R

T

mg

Yo-yo

maTmg CM IL RT

dt

dLRTI

R

a

2R

Im

mga

Conservation of Angular Momentum

extnetdt

Ld

Even if there are external forces present, if they exert no torque about a given axis, then L will be conserved.Note that if L is conserved about one axis, it need not be conserved about any other axis.

R

Disc of mass M rotating with ωo . Bog of mass m lands at center and walks out to rim. Find the final ω

02

0 2

1

MRL

ff mRMRL

22

2

1

022

2

0

2

12

1

mRMR

MRLL ff

v

3

v

ML,

m

Frictionless table viewed from above. Stick pivoted at one end. Pivot exerts force but no torque about pivot. Linear momentum not conserved, but angular momentum about pivot will be.

mvLL 0

pf ILv

mL 3

L

v

M

m

ML

mvLLL f 4

3

13

4

20

Was this elastic?

v

3

v

ML,

m

CM

V20L

mvL

ILvmL f

23

L

v

M

m

ML

mvLLL f 8

12

13

2

20

MVv

mmvPP fo 3

vM

mV

3

4

Remove pivot

v

3

v

ML,

m

CM

V00 L

ILMVL f

2

L

V

ML

LMV

LL f 6

12

12

20

MVv

mmvPP fo 3

vM

mV

3

4

Axis at bottom

L

v

M

m8 as before