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rolling without slipping:
translation + rotation
Smooth Rolling
As seen by an observer at rest:
Smooth rolling,
if a wheel does not slip at the point of contact
ω R = vcm
s =θR
s
+
vcm
2vcm
=
vcm
so, because arc length s is the same as linear
distance moved forward,
θ
0
pure translation
vcm = ω R
s
rolling
pure rotation
acm = α R
Rolling Example
Rolling and Kinetic Energy
Example: Two objects (a solid disk and a solid sphere) are rolling down a ramp.
Both objects start from rest and from the same height. Which object reaches the
bottom of the ramp first?
K tot = K T + K rot
1 2 1 2
mvcm + Iω
2
2
1
⇒ (mR 2 + I )ω 2
2
1
I 2
= ( m + 2 )vcm
2
R
=
ω
If the object rolls without
slipping then vcm = Rω.
h
θ
The object with the largest linear velocity (v) at the bottom of the ramp will win
the race.
4
Example continued:
Example continued:
Apply conservation of mechanical energy:
The moments of inertia are:
Ei = E f
I sphere
U i + Ki = U f + K f
1
1
1
1 v
mgh + 0 = 0 + mv 2 + Iω 2 = mv 2 + I  
2
2
2
2 R
2
For the disk:
1
I 
mgh =  m + 2 v 2
2
R 
Solving for v:
v=
For the sphere:
2mgh
I 

m + 2 
R 

vdisk
4
=
gh
3
vsphere =
1
mR 2
2
2
= mR 2
5
I disk =
Since Vsphere> Vdisk the sphere
wins the race.
10
gh
7
Compare these to a box sliding down the frictionless ramp.
5
vbox = 2 gh
6
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3/12/2014
Acceleration
Add friction:
y
How do objects in the previous example roll?
FBD:
FBD:
N
Fs
y
s
x
N
s
cm
y
θ
w
Also need acm = αR and
x
x
w
∑τ = F r = Iα
∑ F = w sin θ − F = ma
∑ F = N − w cos θ = 0
Both the normal force and the weight act through the center of mass so Στ = 0.
This means that the object cannot rotate when only these two forces are applied.
v 2 = v02 + 2a∆x
The above system of equations can be solved for v at the bottom of the ramp.
The result is the same as when using energy methods. (See text example 8.13.)
It is static friction that makes an object roll.
7
angular momentum
8
Angular Momentum and AMC
REMINDERS
Linear Motion
coordinate
velocity
acceleration
x
v
a
mass
force
1st Newton’s Law
nd
2 Newton’s Law
m
F
F=0: v=const
F = ma
Kinetic Energy
Momentum
K = ½mv2
p = mv
Fnet = lim
∆t → 0
Rotations
θ
ω
α
angle
angular velocity
angular acceleration
Ι
τ
τ=0: ω=const
τ = Iα
rotational inertia
torque
K = ½Iω2
???
Kinetic Energy
Angular Momentum
p = mv
τ net = lim
∆t → 0
L = Iω
When no net external forces act,
the momentum of a system
remains constant (pi = pf)
y-axis
r
Example
∆L
∆t
Units of L are kg m2/s
Units of p are kg m/s
P
p
p⊥
θ
m
When no net external torques act,
the angular momentum of a system
remains constant (Li = Lf).
Angular momentum of a moving particle:
L = rp sin θ = rp⊥ = r⊥ p
x-axis
10
AMC Example
Example (text problem 8.69): A turntable of mass 5.00 kg has a radius of 0.100 m
and spins with a frequency of 0.500 rev/sec. What is the angular momentum?
Assume a uniform disk.
ω = 0.500
∆p
∆t
rev  2π rad 

 = 3.14 rad/sec
sec  1 rev 
Example (text problem 8.75): A skater is initially spinning at a rate of 10.0 rad/sec
with I=2.50 kg m2 when her arms are extended. What is her angular velocity after
she pulls her arms in and reduces I to 1.60 kg m2?
The skater is on ice, so we can ignore external torques.
Li = L f
1

L = Iω =  MR 2 ω = 0.079 kg m 2 /s
2

I i ωi = I f ω f
I 
 2.50 kg m 2 
(10.0 rad/sec) = 15.6 rad/sec
ω f =  i ωi = 
2 
I
 1.60 kg m 
 f 
11
12
2
3/12/2014
Exam 2 Fall 2011: Problem 17
• A cloth tape is wound around the outside of a nonuniform solid cylinder (mass M, radius R) and
fastened to the ceiling as shown in the figure. The
cylinder is held with the tape vertical and then release
from rest. If the acceleration of the center-of-mass of
the cylinder is 3g/5, what is its moment of inertia
about its symmetry axis?
2
2
Answer: 3 MR
% Right: 48%
Mg − FT = Ma y
τ = RFT = Iα =
FT = M ( g − a y )
I=
R 2 FT  g − a y
=
 a
ay
y

• A uniform solid disk with mass M and radius R is
mounted on a vertical shaft with negligible rotational
inertia and is initially rotating with angular speed ω. A
non-rotating uniform solid disk with mass M and
radius R/2 is suddenly dropped onto the same shaft as
shown in the figure. The two disks stick together and
rotate at the same angluar speed. What is the new
angular speed of the two disk system?
R
Answer: 4ω
ω/5
% Right: 47%
Ia y
R
ωf =
y-axis
Mg
PHY 2053
Page 13
University of Florida
M
R
M
Li = I iωi = 12 MR12ω = L f = I f ω f = ( 12 MR12 + 12 MR22 )ω f
FT
R
3



 MR 2 =  g − 5 g  MR 2 = 23 MR 2
 3g 

 5


University of Florida
Exam 2 Fall 2011: Problem 31
R12
R2
ω= 2
ω = 54 ω
R12 + R22
R + ( R / 2) 2
PHY 2053
Page 14
Exam 2 Fall 2010: Problem 16
• A mouse of mass M/6 lies on the rim of a solid uniform
disk of mass M that can rotate freely about its center like
a merry-go-round. Initially the mouse and disk rotate
together with an angular velocity of ω. If the mouse walks
to a new position that is at the center of the disk what is
the new angular velocity
of the mouse-disk system?
Final
Initial
Answer: 4ω
ω/3
% Right: 58%
m
R
L f = I f ω f = I disk ω f
Li = I iωi = ( I disk + mR 2 )ω
I diskω new = ( I disk + mR )ω
2

ω new = 1 +

Note that energy is
not conserved in
this problem!

mR 2 
mR 2 
2m 
ω = 1 +
ω = 1 +
ω = 34 ω
2 
1
I disk 
M 

 2 MR 
University of Florida
R
Li = L f
PM
PHY 2053
m
PM
∆E = E f − Ei
=
L2f
2I f
−
L2i
= 1 MR 2ω 2
2Ii 9
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