Physics 2
Summer 2002
Scott Fraser
email: scottf@physics.ucsb.edu
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
Show Fig. 9-7, 9-8
Rotation Revisited
• See Fig. 9-7, 9-8
• Chapter 9:
“Rotation happens.”
• Chapter 10:
“Why? What causes a?”
Torque (t)
• Torque
= (lever arm) x (force)
• unit = N·m (not J)
• coordinate-dependent:
“torque of F about O”
Show Fig. 10-4
Torque is a vector
• direction
Fig.10-4: R-hand rule
  
τ  r F
• magnitude
t  rF sin 
 (r sin  ) F  lF
 r ( F sin  )  rFtan
Newton’s 2nd Law for Rotation
Fixed rotation axis:
only F1,tan causes torque
F1,tan  m1a1,tan
 m1 (r1a )
t 1  r1 F1,tan
 (m1r12 )a
 I1a
Newton’s 2nd Law for Rotation
• Single particle (m1)
t 1  I1a
I1  m r
2
1 1
• Rigid body: sum over mi (each ai= a )
t  Ia
I   mi ri   dm r
2
i
body
2
Demonstration:
Newton’s 2nd Law for Rotation
t  Ia
rigid body  rod  collars  spindle
t  (lever arm)  (force)
 (spindle radius)  (string tension)
I
2
dm
r

body
Perform demonstration
Solving Rotation Problems:
Newton’s 2nd Law
• Linear form


F  ma
• Rotational form
t  Ia
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
(a) ideal string (no stretching or slipping) : a1  a2  Ra
(b) Fx  m1a1 :
T1  m1a1
t  Ia :
RT2  RT1  ( MR 2 )a
Now use these equations to do Exercise 10-13
Fy  m2 a2 :
m2 g  T2  m2 a2
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 Rw
• no slip: vcm= Rw (skid: Rw < vcm , slide: Rw > vcm)
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 Law
for Pure Rotation
• We used the linear version (single particle)


F  ma
• To get the rotational version (rigid body)
t ext  I bodya
Newton’s 2nd Law
for Rotation and Translation
• Recall linear relation for CM (total mass M)


Fext  Macm
• Claim: an analog result (rigid body)
t ext  I cma
Newton’s 2nd Law
for Rotation and Translation
•
text = Icma
• Two conditions needed for this to hold:
• Axis through CM must be a symmetry axis
• Axis must not change direction
Newton’s 2nd Law
for Rotation and Translation
• Translation of CM (total mass M)


Fext  Macm
• Rotation about axis through CM
t  I cma
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
vcm  Rw
acm  Ra
• Newton’s 2nd Law


Fext  Macm
t  I cma
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 = hoop
Icm= Ihoop (= MR2)
Exercise 10-15
Exercise 10-15
• Icm= Ihoop = MR2
• M = 0.18 kg
R = 0.080 m
• Find tension T
• Find t, w 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
1 2
K  Iw
2
• Translation and rotation
1
1
2
2
K  I cmw  Mvcm
2
2
Exercise 10-16
• Redo Exercise 10-15,
but now use energy
• Icm= Ihoop = MR2
• vcm= Rw
• Find w when hoop
has fallen a distance
h= 0.75 m from rest
Exercise 10-16
• Redo Exercise 10-15,
but now use energy
• Icm= Ihoop = MR2
• vcm= Rw
• Find w when hoop
has fallen a distance
h= 0.75 m from rest
Work and Power Done by Torque
Work and Power Done by Torque
dW  Ftan ds
 Ftan ( R d )
 t d
dW
d
P
t
 tw
dt
dt
2
W   dW   t d
1
Work Done by Net Torque (t)
dWtot  (t )d
 ( Ia )d
dw
I
d
dt
d
I
dw
dt
 Iw dw
w2
1
1
2
2
Wtot   Iw dw  Iw 2  Iw1
2
2
w1
Exercise 10-21
• R = 2.40 m
I = 2100 kg·m2
• Initially at rest (w1=0)
• The child applies:
Ftan = 18.0 N during
Dt = t2 – t1 = 15.0 s
• Find w2 , 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?
Angular Momentum of a Particle
  
Lrp


 r  mv
• unit = kg·m2/s
• coordinate-dependent:
“angular momentum
L about O”
Angular Momentum of a Particle
• vector
  
Lrp


 r  mv
• magnitude
L  r (mv) sin 
 mv(r sin  )
 mvl
Do Exercise 10-29 (a)
Angular Momentum of a Particle
• For circular motion in
the xy plane,  = 90°
L  mvr sin 
 mvr
 m ( rw ) r
 mr w
2
 Iw
Angular Momentum of a Rigid Body
• consider a rigid body
in xy plane
(no extent in z)
• sum over particles:
L = Iw for whole body
• what if the body
extends in z direction?
Angular Momentum of a Rigid Body
• consider special case:
• extended rigid body
has symmetry axis
(here z)
• and also rotates about
the symmetry axis
Angular Momentum of a Rigid Body
• so our special case is:
• If rigid body rotates
about a symmetry
axis, then


L  Iw


L and w are parallel
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...


L  Iw


L and w are parallel
• general case:
• If no symmetry axis,
or rotation axis is not
the symmetry axis...

just a component of L obeys
Lrot. axis  Iw


L and w are not parallel
Torque and Angular Momentum


dL
t ext 
dt
• True for any system of particles!
(rigid or not, symmetric or not)
Prove this for case of single particle
Do Exercise 10-29 (b)
Torque and Angular Acceleration
• For special case:


L  Iw and I  constant
• Find relationship:




dL
dw
t ext 
I
 Ia
dt
dt
Conservation of
Angular Momentum


dL
t ext 
dt
• If text = 0:
• then L is constant (‘conserved’)
• A deep conservation ‘principle’:
• It holds on all scales, from atoms to galaxies
Demonstration:
Conservation of
Angular Momentum
Go over some explanatory notes
Conservation of
Angular 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 w after collision
Do Example 10-14
• Exercise 10-37 (HW):
• A different collision:
• B falls onto rim, wB=0
Gyroscopes and Precession
Demonstration: Precession
 
dL  t dt
Let' s find an expression for ...
Derive the expression for 
d t rw
Precessional angular speed  
 
dt L Iw