Download Chapter 11

Fundamentals of Physics
Chapter 12 Rolling, Torque & Angular Momentum
1.
2.
Rolling
The Kinetic Energy of Rolling
Rolling as Pure Rotation
3. The Forces of Rolling
Friction & Rolling
Rolling Down a Ramp
4. The Yoyo
5. Torque Revisited
6. Angular Momentum
7. Newton’s 2nd Law in Angular Momentum Form
8. Angular Momentum of a System of Particles
9. Angular Momentum of a Rigid Body Rotating About a Fixed Axis
10. Conservation of Angular Momentum
Review & Summary
Questions
Exercises & Problems
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Physics 2111 Fundamentals of Physics
Chapter 11
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Rolling without Slipping
R
s R
Units: (m)
(radians) (m)
vcom   R
Units: (m/s)
(rad/s) (m)
acom   R
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Rolling
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Example: Value of x for rolling without slipping?


F  ma


  I
x = 0  no initial spin
I 
2
5
m R2
  xF
Rolling without slipping:
a  R
x 
2
R
5
Independent of m and F!
x > 2/5 R  topspin
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The Forces of Rolling
Rolling with slipping initially
No Initial Rotation:
Topspin:
Frictional force reduces the speed &
increases the angular speed until:
vcom   R
Frictional force accelerates the ball in the
direction of motion:
increasing v
decreasing 
Rolling without slipping
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Translation & Rotation
Add:
To get:
Linear velocity due to rotation
Linear velocity of the center-of-mass
Linear velocity of each point on a rolling wheel.
pure rotation
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+
pure translation
=
rolling motion
Physics 2111 Fundamentals of Physics
See p247 in HRW.
Chapter 11
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The Kinetic Energy of Rolling
A rolling object has two types of kinetic energy: a rotational KE due to its rotation about its
center of mass and a translational KE due to the translation of its center of mass.
The instantaneous axis of rotation is at Point P:
1
IP  2
2
I P  I com  M R 2
v
  com
R
1
1
2
K  I com  2 
M vcom
2
2
K 
KR

KT
K 
vcom
R
K 
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1
IP  2
2
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Chapter 11
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The Forces of Rolling
Friction & Rolling: friction is required!
No slipping (aka “smooth rolling”)
acom   R
fs acts on the wheel at P, opposing its tending to slide.
Wheel is rotating about point P.
No work is done by fs!
The wheel does not slow down.

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Physics 2111 Fundamentals of Physics
Chapter 11
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Rolling down a Ramp


F  ma
f S  M g sin   M acom


  I
R f S  I com  com
a   R
 > 0 counter clockwise
a < 0 down the ramp
acom  
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g sin
I com
1
M R2
Any body rolling down a ramp!
Physics 2111 Fundamentals of Physics
Chapter 11
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Uniform Ball Rolling Down
M = 6.00 kg
Radius R
 = 300
h = 1.20 m
v=?
fs = ?
Conservation of energy:
1
2
I 
2
5
M R2
I com  2 
1
2
vcom 
10
7
acom  
vcom   R
Physics 2111 Fundamentals of Physics
gh
g sin
I
1  com 2
MR
a com  
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2
M vcom
 M gh
5
g sin 
7
Chapter 11
10
Hoop, Disk and Sphere rolling down a ramp:
acom  
g sin
I
1  com 2
MR
I ring  M R 2
I cylinder  12 M R 2
I sphere  52 M R2
asphere  acylinder  aring
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Wheel has constant acceleration & no slipping
M  10 kg
fS = ?
I =?
R  0.3 m
a  0.6 m s
fS
a   R


Fnet  m a
F  fS  m a


 net  I 
 fS R   I 
fS   4 N
I  0.6 kg m 2
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Ball rolls down the hill
H = 6.0 m
h = 2.0 m
x=?
x
Energy:
mgH 
1
2
m v 2  12 I  2  m g h
v R
I  52 m R 2
v  7 .5 m s
Projectile: x  4.8 m
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Loop the Loop
I  52 m r 2
h  ? when N  0 at top
N  ? at Q when h  6 R
answer : N 
Energy:
mgh 
1
2
50
7
mg
m v 2  12 I  2  m g 2 R  r 
 v2 
Newton:  N  m g   m a   m  
R
 
R  r
answer : h  2.7 R
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Chapter 11
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Torque “Revisited”
 
  r xF

 
  r F sin 


  r F
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
  r F
Physics 2111 Fundamentals of Physics
Chapter 11
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Angular Momentum

Consider a particle with momentum p
Its angular momentum
with respect to Point O is :

 
l  rp

 
 
l  r x p  m r x v 

 
l  m r v sin 
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All of the particles have the same momentum.
Largest Magnitude of Angular Momentum about O ???
Negative Angular Momentum???
Zero Angular Momentum???
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Physics 2111 Fundamentals of Physics
1 and 3
2 and 3
5
Chapter 11
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Angular Momentum
a)
m = 2.0 kg
r = 3.0 m
v = 4.0 m/s
F = 2.0 N
Angular momentum = ?

 
 
l  r x p  m r x v 

 
l  m r v sin 

m2
l  12 kg
s
b) Torque = ?


  r x


  r
out of the page

F

F sin 

  3.0 N m
2008
Physics 2111 Fundamentals of Physics
out of the page
Chapter 11
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Newton’s 2nd Law in Angular Momentum Form

 
l  m r x v 

l
Time derivative of the angular momentum:




dl
dr
  dv
 m r 

 v


dt
dt
dt

 


dl
 m r  a  0  r  ma
dt

 
dl
 r  Fnet
dt


dl
  net
dt
The vector sum of all the torques acting on a particle is equal to the time rate of
change of the angular momentum of that particle.
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Checkpoint
Largest Torque about Point O???
Positive Torque about Point O???
Zero Torque about Point O???
2008

F3

F1

F2
Physics 2111 Fundamentals of Physics

F4
Chapter 11
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Angular Momentum of a System of Particles

L 
n
l
i 1

dL

dt

dL

dt
Internal torques sum to zero:
i
n
dli

i 1 dt
n

 net ,i
 
 
 1   2  r1  F2 ,1  r2  F1, 2


F2 ,1   F1, 2



 
 1   2  r1  r2   F2 ,1
i 1


The net external torque acting on a system of particles is equal to the time rate of change
of the system’s total angular momentum.

dL 
  net
dt
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Physics 2111 Fundamentals of Physics
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Angular Momentum About a Fixed Axis
v  r


 
 
2 
L  r x p  mr x v   mr   I 
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Chapter 11
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A Rigid Body Rotating About a Fixed Axis

Angular momentum of a mass element wrt the origin:
  
 
li  ri  pi  mi ri  vi
li  mi ri vi



li  ri and vi
Component of angular momentum in z-direction:
liz  li sin 
liz  mi ri vi sin 
liz  mi ri vi
vi  ri i  ri 
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liz  mi ri 
2
Physics 2111 Fundamentals of Physics
Chapter 11
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A Rigid Body Rotating About a Fixed Axis

liz  mi ri i
2
Total angular momentum in z-direction:
Lz 
n
l
i 1
iz

n
 m
i
i 1
n
ri 
2
Lz    mi ri   I z
2
i 1


L  I
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Checkpoint : same radius same mass same force
Largest Angular Momentum about Center???
Newton’s 2nd Law:

 net

dL

 F R  constant
dt
All the angular momenta are the same.
Largest Angular Speed after a time interval???


L  I
largest   smallest I
I sphere  I cylinder  I ring
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Conservation of Angular Momentum
If the net external torque acting on a system is zero, the angular momentum of the system
remains constant, no matter what changes take place within the system.
The Spinning Volunteer:
I f  Ii


L  If f


L  Ii i
 f  i
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Spins faster!
Physics 2111 Fundamentals of Physics
Chapter 11
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Conservation of Angular Momentum
The Diver:


L  I   constant
If the component of the net external torque on a system along a certain axis is zero, then
the component of angular momentum of the system along that axis remains constant, no
matter what changes take place within the system.
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Conservation of Angular Momentum
The system = stool + person + spinning wheel



L  Lb  Lwh


L  Lwh
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Physics 2111 Fundamentals of Physics
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A train starts running on a wheel;  = ?

m
M
IW  M R
v
2
Li  L f  0



vTG  vTW  vW G
L  0  LT  LW
vTG  v   R
m R vTG  IW   0
m R v   R   M R 2   0
 
2008
mv
m  M  R
M  m    0
wheel does not move.
v
M  m   
R
train does not move.
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Chapter 11
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The sun goes out and then shrinks!
Sun goes out
Shrinks size Earth
Period ?
Li  L f

I i i  I f  f
1
2

M R 2 i 

1
2

M r2 f
2
R
 f    i
r
2
T 

 6.37 106 m 
r

T f  Ti    25 days 
8
R
6
.
96

10
m
 


T f  3 minutes revolution
2
2
The sun will become a white dwarf.
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Chapter 11
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