Download - Physics 2 Summer 2002
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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!
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Course Text
• University Physics (10th edition),Young & Freedman
• We will cover Chapters 10-18
• You should be familar with Chapters 1-9
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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)
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Chapter 10
Dynamics of Rotational Motion
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Pure Rotation
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Rotation Revisited
• See Fig. 9-7, 9-8
• Chapter 9: “Rotation happens.”
• Chapter 10: “Why? What causes ?”
Show Fig. 9-7, 9-8
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Torque ()
• Torque = (lever arm) x (force)
• unit = N·m (not J)
• coordinate-dependent:“torque of F about O”
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Torque is a vector
• direction Fig.10-4: R-hand rule
• magnitude
Frτ
tan)sin()sin(
sin
rFFrlFFr
rF
Show Fig. 10-4
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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
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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
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Demonstration:Newton’s 2nd Law for Rotation
I
body
2 rdmI
tension)(stringradius) (spindle(force)arm)(lever
spindle collars rod body rigid
Perform demonstration
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Solving Rotation Problems:Newton’s 2nd Law
• Linear form
• Rotational form
amF
I
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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)
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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
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Rotation
with Translation
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An Example:Rolling Without Slipping
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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)
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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
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Newton’s 2nd Lawfor Pure Rotation
• We used the linear version (single particle)
• To get the rotational version (rigid body)
amF
bodyext I
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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
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Newton’s 2nd Lawfor Rotation and Translation
• ext = Icm
• Two conditions needed for this to hold:
• Axis through CM must be a symmetry axis• Axis must not change direction
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Newton’s 2nd Lawfor Rotation and Translation
• Translation of CM (total mass M)
• Rotation about axis through CM
cmext aMF
cmI
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Example: The Yo-Yo
• Example 10-8:yo-yo = ‘axle’Icm= Icylinder
• More generally:
yo-yo = ‘spool’Icm= Ispool
• But just draw the axle
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Example: The Yo-Yo
• Let’s draw the free body diagram for the yo-yo
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Example: The Yo-Yo
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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
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Exercise 10-15
• We found results for yo-yo with any Icm (not just Icm= Icylinder)
• Exercise 10-15:
yo-yo = hoopIcm= Ihoop (= MR2)
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Exercise 10-15
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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)
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Rotation and Translation:Energy
• Pure rotation
• Translation and rotation
2
21 IK
2cm
2cm 2
121 MvIK
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Exercise 10-16
• Redo Exercise 10-15, but now use energy
• Icm= Ihoop = MR2
• vcm= R
• Find when hoop has fallen a distance h= 0.75 m from rest
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Exercise 10-16
• Redo Exercise 10-15, but now use energy
• Icm= Ihoop = MR2
• vcm= R
• Find when hoop has fallen a distance h= 0.75 m from rest
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Work and Power Done by Torque
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Work and Power Done by Torque
ddRF
dsFdW
) (
tan
tan
2
1
ddWW
dtd
dtdWP
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Work Done by Net Torque (
dI
ddtdI
ddtdI
dIddW
)()(tot
21
22tot 2
1 21
2
1
IIdIW
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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
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Angular Momentum
• for a single particle
• for rigid bodies
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Angular Momentum of a Particle
• angular momentum: L
• rotational analog of linear momentum, p
• you can guess the definition of L?
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Angular Momentum of a Particle
• unit = kg·m2/s
• coordinate-dependent:“angular momentum L about O”
vmrprL
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Angular Momentum of a Particle
• vector
• magnitude
vmrprL
mvlrmv
mvrL
)sin(sin)(
Do Exercise 10-29 (a)
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Angular Momentum of a Particle
• For circular motion in the xy plane, = 90°
Imr
rrmmvrmvrL
2
)(
sin
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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?
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Angular Momentum of a Rigid Body
• consider special case:
• extended rigid body has symmetry axis (here z)
• and also rotates about the symmetry axis
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Angular Momentum of a Rigid Body
• so our special case is:
• If rigid body rotates about a symmetry axis, then
IL
parallel areand L
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Angular Momentum: Particle
• particle
(moving relative to some origin O)
• ‘orbital’ angular momentum
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Angular Momentum: Rigid Body
• rigid body
(rotating about a symmetry axis)
• ‘spin’ angular momentum
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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
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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
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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)
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Torque and Angular Acceleration
• For special case:
• Find relationship:
IdtdI
dtLd
ext
constant and IIL
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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
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Demonstration:
Conservation of Angular Momentum
Go over some explanatory notes
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Conservation ofAngular Momentum
• Good problems to work:
• Exercise 10-34 • Examples 10-13, 10-14
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Exercise 10-34
Do Exercise 10-34
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Example 10-13
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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
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Gyroscopes and Precession
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Demonstration: Precession
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dtLd
...for expressionan find sLet'
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Irw
Ldtd
speedangular alPrecession
Derive the expression for