Example: A cylinder of mass M and radius R rolls down an incline of
angle θ with the horizontal. If the cylinder rolls without slipping, what is
its acceleration?
Newton's 2nd law for rotation:
Rolling without slipping:
f s R  I CM 
vCM  R  aCM  R
I CM  12 MR 2
1
f s  I CM  
R

1
2

 aCM  1
MR 

 R R
2
f s  12 MaCM
Mg sin   f s  MaCM
Newton's 2nd law for translation of the CM:
N
Mg sin   32 MaCM
fS
aCM  23 g sin 
compare to g sinθ, the results for
an object sliding without friction
mg
θ
Example: A disk of radius R and mass M that mounted on a massless
shaft of radius r << R and rolling down an incline with a groove.
What is its acceleration?
M
fS
N
R
m~0
r
I CM 
1
MR 2
2
Mg
Mg sin   fS  MaCM
fS r  I CM
aCM  r 
aCM
Mg sin 
g sin 


2
2
M  I CM r
1  12 R r 
Very small if R >> r !
Example: Loop-the-loop. A cart marble of radius r is released from height
h in a roller coaster with a loop of radius R. What is the minimum h to
2 2
keep the cart on the track?
I sphere about CM 
E A  EB
1
1
2
mgh  0  mg 2R  mv CM
 I sphere 2
2
2
5
mr
If no slipping, and r << R:
vCM  r
new term
1
12
2
2 vCM
mgh  mg 2R  mvCM 
mr
2
25
r2
2
At the point B:
A
mg  N  m
2
vCM
R
The minimum velocity is fixed by N = 0:
B
h
2
vCM
 107 g h  2 R  (1)
2
vCM
 gR
(2)
mg+N
R
Combining Eq. 1 & 2:
hmin  2.7 R
(Without rotation the factor is 2.5)
Example: 1 m of cord is wound around a flywheel of radius, 0.25 m and
mass, 2 kg which is concentrated in the rim. The cord is pulled with a
force of 100 N and drives the wheel without slipping. What is the final
angular velocity of the wheel when the cord is exhausted? What is the
period?
d=1 m
F=100 N
m=1 kg
I=mR2
v=Rω
The pulling of the cord implies a fixed amount
of work done on this system. If we equate
this with the rotational energy of the flywheel,
then we can discover how fast it is turning.
W=K
R=0.25m
K  12 mv 2  12 I 2  12 mR 2 2  12 mR 2 2  mR 2 2
W = Fd

1
100 N 1m
10 1

s  40 s 1
0.25m
1kg
0.25
Fd  mR 
2
1

R
2
T
2


Fd
m
2
 0.16 s
1
40 s
Example: Two wheels with fixed hubs, each having a mass of 1 kg, start
from rest, and forces are applied as shown. Assume the hubs and
spokes are massless, so that the moment of inertia is I = mR2.
In order to impart identical angular accelerations, how large must F2 be?
1. 0.25 N
2. 0.5 N
3. 1 N
4. 2 N
5. 4 N
  I
FR  I
I mR 2
F

  mR
R
R
FR
Example: Two identical disks have a string coiled around them. The string
is pulled with a constant force F over a distance d. In one case, the disk
rolls without slipping on a table. In the other case, the disk rotates about
its axis (like a pulley). Compare the final angular speeds of the disks.
Same work (i.e., added energy) in both cases:
W = Fd
In case 2, this energy becomes
only rotational kinetic energy
In case 1, this energy becomes both
translational and rotational kinetic energy
Case 1
A. 1 > 2
B. 1 = 2
C. 1 < 2
F
Case 2
F
Work done by a torque (for pure rotations)
A force F acts on an object as it rotates from θ1 to θ2.
 
dW  F  dl  Ftands  Ftanr d   z d
Work from θ1 to θ2:
W 
2

 z d
1
If torque is constant,
Instantaneous power:
Average power:
W   z 
P 
dW
d
 z
  z z
dt
dt
P 
W
  z z
t
F
Axis
θ2
r
θ1
F
Important! These expressions are only true for pure rotations (with a fixed
axis of rotation, or in the frame of reference where the axis is at rest).
Otherwise, the path by the point where the force is applied is not a circle.
Example: Work by friction
A uniform flywheel moving at 600 rpm comes to a stop after 1000 turns,
mostly due to air resistance. What is the average torque produced by air?
M = 10 kg
R = 1.0 m
0  600
rev 1 min 2 rad
 20 rad/s
min 60 s 1 rev
1
I  MR 2  5 kg m2
2
  1000  2  2000 rad
 z ,average
1
2
I

0
0
W
KE

2




 1.6 Nm