1 7/26/04 midterm 2 – next friday (7/30/04) material from chapters 7-12 i will post a practice...

17/26/04

Midterm 2 – Next Friday (7/30/04)Material from Chapters 7-12

I will post a practice exam on Monday

Announcements

27/26/04

Last Time…

Rotation

Position x

Velocity = d/dtv = dx/dt

Acceleration = d/dta = dv/dt

Translation

Kinetic Energy K = ½I2K=½mv2

Mass I = miri2m

F = ma = INewton’s 2nd Law

37/26/04

Example: (Problem 11.49)

During a launch from the board, a diver’s angular speed about her center of mass changes from zero to 6.20 rad/s in 220 ms. Her rotational inertia about her center of mass is 12.0 kg m2. During the launch, what are the magnitudes of

a) her average angular acceleration and b) the average external torque on her from the board?

47/26/04

Example:

20 /2.28220.0

/0/20.6srad

s

sradsrad

tf

a) Average angular acceleration

b) Average external torque

NmsradmkgI 338)/2.28)(0.12( 22

57/26/04

Example: (Problem 11.53)

The world’s heaviest hinged door has a mass of 44,000 kg, a rotational inertia about a vertical axis through its hinges of 8.7x104 kg m2, and a width of 2.4 m. Neglecting friction, what steady force applied to the door’s outer edge and perpendicular to the plane of the door can move it from rest through an angle of 90° in 30 s?

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Solution:

00

221

0 tt

2

2

t

radf 290

Initial and final positions will be:

Using a standard kinematics equation,

Finally,

IFr 90sin

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Solution:

IFr

2

2

rt

I

r

IF

2

24

)30)(4.2(

)107.8)(2/(2

sm

mkgradF

NF 130

Plugging in our value for :

7/26/04 8

Chapter 12

Rolling, Torque, and Angular Momentum

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Rolling Without Slipping

How do we describe an object that is rolling?

Consider the case of pure translation:

vcm

vcm

vcm

2

2

1cmMvK

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Rolling Without Slipping

Consider the case of pure rotation:

vcm

vcm

2

2

1 cmIK

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Rolling Without Slipping

Combining the two:

vcm

vcm

vcm

22

2

1

2

1 cmcm IMvK

vcm

vcm

+ =vcm

2 vcm

v=0

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Rolling: Another View

Rωdt

dθR

dt

dsvcm

Rα acm

At any instantthe wheel rotates aboutthe point of contact

vc

vc

s = R

P

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2

2

1ωIK P

2

22

2

1

2

1

R

vMRIK cm

cm

R

vω cm

I about point of contact

Parallel Axis Theorem

K = ½IP2

IP = Icm + MR2

P

22

2

1

2

1cmcm MvIK

Rolling: Another View

222

2

1ωMRωIK cm

147/26/04

Rolling Down a Hill

h

vc

Conservation of Energy:

MghvMR

Icm

cm

2

22

1

Ui = Kf2

21

cmMvMgh

For a point mass:

ghvcm 2

For an extended object:

21

2

/MRI

ghv

cmcm

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Rolling Down a Hill

2

2

1MRIcm

2MRIcm

gh/MRMR

ghvcm 3

4

1

222

21

gh/MRMR

ghvcm

221

2

For a disc:

For a ring:

vc

h

Notice there is no mass or radial dependence!

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How to Solve Torque Problems

1) Draw a picture2) Pick an origin (axis of rotation usually a good

choice)3) Sum torques about the origin

net= I4) Sum the forces in every direction (if

necessary)

Fnet = ma5) Solve for unknowns

177/26/04

The Forces of Rolling

Constraint of Rolling without Slipping:

α Racm

Mg

Mg sin

Mg cos

FN

FFBefore when we drew a free body diagram, we drew all the forces from the center of mass. Cannot do that with torques!

Note: Friction causes rolling (otherwise it would just slide)

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The Forces of Rolling

Pick the center of mass asthe origin. Sum the forces and torques.

cmfrxnet MaFθMgF sin,αIRF cmfrnet

Mg

Mg sin

Mg cos

FN

FFFind the linear acceleration of the center of mass.

cmfr MaθMgF sin

cmcm IRMaθMg )sin(

197/26/04

The Forces of Rolling

Mg

Mg sin

Mg cos

FN

FF

cmcm IMRaθMRg sin

α Racm Recall:

cm

cmcm I

aMRθgMRa

22 sin

θgMRMRIa cmcm sin)( 22

2

2 sin

MRI

θgMRa

cmcm

2/1

sin

MRI

θga

cmcm

Linear acceleration doesn’t depend on mass or radius!!

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Another Way to Solve the Problem

What if we pick a different origin? Pick point of contact.

No torque from normal forceor friction!

21

sin

/MRI

θga

cmcm

Same result! (Don’t need to sum forces)

Mg

Mg sin

Mg cos

FN

FF

Pnet IMRg sin2MRII cmP

)(sin 2MRIMRg cm

2

sin

MRI

MRg

cm

217/26/04

Angular Momentum

Recall: linear motion

Using the correspondence with linear motion, angular momentum should have the form:

If no torque, then L is a constant.

We have conservation of angular momentum!

vmp

amdt

pdF

IL dt

LdItot

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Conservation of Angular Momentum

0 fi LL

237/26/04

Conservation of Angular Momentum

IL

If I changes, then must change to keep L constant (no external torques)

Example: Ice skaters pulling arms in

Li=Iii Lf=Iff

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Angular Momentum of a Particle

Can also define angular momentum for a particle with a linear velocity v

r is vector from origin to particle

rv r v L is big

r || v L is 0

(Example: something circling the origin)

)( vrmprL

257/26/04

Angular Momentum of a Particle:

dt

pdrp

dt

rdpr

dt

dL

dt

d

)(

Fr

netLdt

d

Is this new definition of L consistent with the old one?

Same as before!

dt

pdrvmv

267/26/04

Example:Constant velocity particle: Is L really constant?

Direction is constant as well!

)( vrmprL

r

v

θd

const)cos()90sin( mvdrmvmrvL

277/26/04

Torque and Angular Momentum

A) No torque: L is constant

L = Iωif you change I, ω changes to keep L constant

This allows skaters and divers to spin reallyreally fast (they studied their physics!)

L=Iω =dL/dt

B) If I is constant and ω changes, there must have been a torque

287/26/04

Angular Momentum of a Group of Particles:

For a group of particles:

...321 LLLLtot

The net torque on the system will be:

dt

Ld totnet

No need to worry about internal torques

297/26/04

A merry-go-round with radius 3 m and Imgr=90 kg m2 is spinning at 0=5 rad/s. A 100 kg pig is then dropped on it at a radius of 2 meters. His landing takes t=1s.

3 m

2 m

Example: Merry Go Round

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Example: Merry Go Round

What is the final angular velocity of the system?

f

iif I

I 2

22

490

)2)(100()90(

mkg

mkgmkg

III pigmgrf

sradmkg

sradmkg

I

I

f

iif /918.0

490

)/5)(90(2

2

ffii

fi

II

LL

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What is the average torque on the merry-go-round during the landing?

Example: Merry Go Round

t

Lmgravg

t

I fmgravg

)( 0

Nms

sradsradmkgavg 367

1

)/5/918.0)(90( 2

Note: This is an internal torque

327/26/04

What is the average linear tangential acceleration of the rim of the merry-go-round?

Example: Merry Go Round

2

0

/08.4

1

/5/918.0

srad

s

sradsradt

f

22 /2.12)/08.4)(3( smsradmRat

337/26/04

Example:

FOR SALE

m = 5 gM = 2.2 kgv = 300 m/sℓ = 0.2 m

vℓ/2

ℓθ

A bullet hits a sign: how high does it go?

347/26/04

Example:

m = 5 gM = 2.2 kgv = 300 m/sℓ = 0.2 m

Before bullet hits:

After bullet hits:

vℓ/2

ℓ

θsmkgmvlLi /15.0)( 221

ILL if

sradmkg

smkg

I

L f /10.5029.0

/15.02

2

222

029.023

mkgl

mMl

I

357/26/04

Example:

Ki+Ui = Kf+Uf

ℓ/2

ℓ θ

h

221 IK i

ghmMU i )(

ghmMI )(221

)005.2.2)(/8.9(2

)/10.5)(029.0(

)(2 2

222

kgkgsm

sradmkg

mMg

Ih

cmmh 75.10175.0

367/26/04

Precession

FrLdt

d

Right hand rule dL/dt out of the page!

mg

hL

Consider a spinning gyroscope:

377/26/04

Precession

Iω

mgh

L

mgh

Δt

Δφωp

L=I

= mgh

ΔL = Δt

ΔL = mgh Δt

L

tmgh

L

ΔLΔφ

Δ

Lf

LiΔL

mg

hL

387/26/04

The Bicycle

Why is it hard to fall over?

L gets larger as the bike goes faster

A torque is needed to change L

If we lean to one side (i.e. fall over)

= dmg

x

yz

mg

d

in the x direction

Turns the bike!

d