Physics 2 Summer 2002

Physics 2Summer 2002

Scott Fraser email: [email protected]

office hours: M-F, 2-2:30 or by appt.

Course Information

• Online at http://class.physics.ucsb.edu

• Login: only needed to check grades

• Physics 2 page: click “view list of classes”

• Check the links regularly for updates!

Course Text

• University Physics (10th edition),Young & Freedman

• We will cover Chapters 10-18

• You should be familar with Chapters 1-9

Tips for Success

• Always do the pre-lecture reading

• Work examples in the text for yourself• Algebra first, numbers (if any!) last

• Chapter 10 homework due this Friday!(4pm in locked PHYSICS 2 box in lobby)

Chapter 10

Dynamics of Rotational Motion

Pure Rotation

Rotation Revisited

• See Fig. 9-7, 9-8

• Chapter 9: “Rotation happens.”

• Chapter 10: “Why? What causes ?”

Show Fig. 9-7, 9-8

Torque ()

• Torque = (lever arm) x (force)

• unit = N·m (not J)

• coordinate-dependent:“torque of F about O”

Torque is a vector

• direction Fig.10-4: R-hand rule

• magnitude

Frτ

tan)sin()sin(

sin

rFFrlFFr

rF

Show Fig. 10-4

Newton’s 2nd Law for Rotation

Fixed rotation axis: only F1,tan causes torque

)( 11

tan1,1tan1,

rm

amF

1

211

tan1,11

)(I

rm

Fr

Newton’s 2nd Law for Rotation

• Single particle (m1)

• Rigid body: sum over mi (each i= )

body

2

2 rdmrmI

I

iii

2111

11

rmI

I

Demonstration:Newton’s 2nd Law for Rotation

I

body

2 rdmI

tension)(stringradius) (spindle(force)arm)(lever

spindle collars rod body rigid

Perform demonstration

Solving Rotation Problems:Newton’s 2nd Law

• Linear form

• Rotational form

amF

I

Exercise 10-13

• m1 and m2 move: d =1.20 m during t = 0.800 s

• Values: m1 = 2.00 kg, m2 = 3.00 kg, R = 0.075 m

• Find: T1, T2, I (of pulley about its rotation axis)

See Example 10-4

:slipping)or stretching (no string ideal (a) 21 Raa

111

11 : (b)amTamFx

2222

22 :

amTgm

amFy

)(

:2

12 MRRTRT

I

Now use these equations to do Exercise 10-13

Rotation

with Translation

An Example:Rolling Without Slipping

Rolling without Slipping

• ground sees: wheel translating with speed vcm • center of wheel sees: rim rotating with speed R• no slip: vcm= Rskid: Rvcm , slide: Rvcm)

Rolling without Slipping

• two motions:translation of CM, rotation about CM• superposition: at each point on wheel, v = vcm+ v /

• ground sees: point 1 of wheel momentarily at rest

Newton’s 2nd Lawfor Pure Rotation

• We used the linear version (single particle)

• To get the rotational version (rigid body)

amF

bodyext I

Newton’s 2nd Lawfor Rotation and Translation

• Recall linear relation for CM (total mass M)

• Claim: an analog result (rigid body)

cmext aMF

cmext I


• ext = Icm

• Two conditions needed for this to hold:

• Axis through CM must be a symmetry axis• Axis must not change direction


• Translation of CM (total mass M)

• Rotation about axis through CM

cmext aMF

cmI

Example: The Yo-Yo

• Example 10-8:yo-yo = ‘axle’Icm= Icylinder

• More generally:

yo-yo = ‘spool’Icm= Ispool

• But just draw the axle

Example: The Yo-Yo

• Let’s draw the free body diagram for the yo-yo

Example: The Yo-Yo

Example: The Yo-Yo

• String: no slipping

• Newton’s 2nd Law

cm

cmext

IaMF

RaRv

cm

cm

Use these equations to find acm and T for arbitrary Icm

Exercise 10-15

• We found results for yo-yo with any Icm (not just Icm= Icylinder)

• Exercise 10-15:

yo-yo = hoopIcm= Ihoop (= MR2)

Exercise 10-15

Exercise 10-15

• Icm= Ihoop = MR2

• M = 0.18 kgR = 0.080 m

• Find tension T• Find t, when hoop has

fallen a distance h= 0.75 m from rest

Now use expressions for acm and T for arbitrary Icm (from the Yo-Yo Example)

Rotation and Translation:Energy

• Pure rotation

• Translation and rotation

2

21 IK

2cm

2cm 2

121 MvIK

Exercise 10-16

• Redo Exercise 10-15, but now use energy


• vcm= R

• Find when hoop has fallen a distance h= 0.75 m from rest

Exercise 10-16

• Redo Exercise 10-15, but now use energy


• vcm= R

• Find when hoop has fallen a distance h= 0.75 m from rest

Work and Power Done by Torque

Work and Power Done by Torque

ddRF

dsFdW

) (

tan

tan

2

1

ddWW

dtd

dtdWP

Work Done by Net Torque (

dI

ddtdI

ddtdI

dIddW

)()(tot

21

22tot 2

1 21

2

1

IIdIW

Exercise 10-21

• R = 2.40 mI = 2100 kg·m2

• Initially at rest (1=0)• The child applies:

Ftan = 18.0 N during t = t2 – t1 = 15.0 s

• Find 2 , W , P

Do the calculation, using conservation of energy

Angular Momentum

• for a single particle

• for rigid bodies

Angular Momentum of a Particle

• angular momentum: L

• rotational analog of linear momentum, p

• you can guess the definition of L?


• unit = kg·m2/s

• coordinate-dependent:“angular momentum L about O”

vmrprL


• vector

• magnitude

vmrprL

mvlrmv

mvrL

)sin(sin)(

Do Exercise 10-29 (a)


• For circular motion in the xy plane, = 90°

Imr

rrmmvrmvrL

2

)(

sin

Angular Momentum of a Rigid Body

• consider a rigid body in xy plane(no extent in z)

• sum over particles: L = I for whole body

• what if the body extends in z direction?


• consider special case:

• extended rigid body has symmetry axis (here z)

• and also rotates about the symmetry axis


• so our special case is:

• If rigid body rotates about a symmetry axis, then

IL

parallel areand L

Angular Momentum: Particle

• particle

(moving relative to some origin O)

• ‘orbital’ angular momentum

Angular Momentum: Rigid Body

• rigid body

(rotating about a symmetry axis)

• ‘spin’ angular momentum

Exercise 10-30

• Earth has two kinds of angular momentum:

• orbital Lorb

• (particle on circular orbit: r = 1.5×1011 m)

• spin Lrot • (Earth: M = 6.0×1024 kg, R = 6.4×106 m)

Find the values of the two angular momenta

Angular Momentum of (Extended) Rigid Body

• special case:• If rigid body’s rotation

axis equals its symmetry axis...

• general case:• If no symmetry axis,

or rotation axis is not the symmetry axis...

ILL

axis rot. obeys ofcomponent ajust

parallelnot areand L

IL

parallel areand L

Torque and Angular Momentum

• True for any system of particles!(rigid or not, symmetric or not)

dtLd

ext

Prove this for case of single particle

Do Exercise 10-29 (b)

Torque and Angular Acceleration

• For special case:

• Find relationship:

IdtdI

dtLd

ext

constant and IIL

Conservation ofAngular Momentum

• If ext = 0:

• then L is constant (‘conserved’)

• A deep conservation ‘principle’: • It holds on all scales, from atoms to galaxies

dtLd

ext

Demonstration:

Conservation of Angular Momentum

Go over some explanatory notes

Conservation ofAngular Momentum

• Good problems to work:

• Exercise 10-34 • Examples 10-13, 10-14

Exercise 10-34

Do Exercise 10-34

Example 10-13

Collisions

• Example 10-14:• Rotating objects A, B• Find after collision

• Exercise 10-37 (HW):• A different collision:• B falls onto rim, B=0

Do Example 10-14

Gyroscopes and Precession

Demonstration: Precession

dtLd

...for expressionan find sLet'

Irw

Ldtd

speedangular alPrecession

Derive the expression for

Documents

Physics 2 Summer 2002