Physics 111: Lecture 19
Today’s Agenda
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


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Review
Many body dynamics
Weight and massive pulley
Rolling and sliding examples
Rotation around a moving axis: Puck on ice
Rolling down an incline
Bowling ball: sliding to rolling
Atwood’s Machine with a massive pulley
Physics 111: Lecture 19, Pg 1
Review: Direction & The Right Hand Rule
y

To figure out in which direction the rotation
vector points, curl the fingers of your right
hand the same way the object turns, and
your thumb will point in the direction of the
rotation vector!
x
z
y


We normally pick the z-axis to be the
rotation axis as shown.
= z
= z
= z
x
z
For simplicity we omit the subscripts
unless explicitly needed.
Physics 111: Lecture 19, Pg 2
Review: Torque and Angular Acceleration
NET = I



This is the rotational analogue
of FNET = ma
Torque is the rotational analogue of force:
The amount of “twist” provided by a force.
Moment of inertia I is the rotational analogue of mass
If I is big, more torque is required to achieve a given
angular acceleration.
Physics 111: Lecture 19, Pg 3
Lecture 19, Act 1
Rotations


Two wheels can rotate freely about fixed axles through their
centers. The wheels have the same mass, but one has twice
the radius of the other.
Forces F1 and F2 are applied as shown. What is F2 / F1 if the
angular acceleration of the wheels is the same?
(a) 1
F2
(b) 2
(c) 4
F1
Physics 111: Lecture 19, Pg 4
Lecture 19, Act 1
Solution
We know
but
 I
  FR
so FR  mR 2
F  mR
Since R2 = 2 R1
I  mR 2
and
F2 mR2 R2


F1 mR1 R1
F2
2
F1
F2
F1
Physics 111: Lecture 19, Pg 5
Review: Work & Energy

The work done by a torque  acting through a displacement
 is given by:
W  

The power provided by a constant torque is therefore given
by:
P
dW
d

 
dt
dt
Physics 111: Lecture 19, Pg 6
Falling weight & pulley

A mass m is hung by a string that is
wrapped around a pulley of radius R
attached to a heavy flywheel. The moment
of inertia of the pulley + flywheel is I. The
string does not slip on the pulley.

I
R
T
Starting at rest, how long does it take
for the mass to fall a distance L.
m
a
mg
L
Physics 111: Lecture 19, Pg 7
Falling weight & pulley...

For the hanging mass use F = ma
mg - T = ma
For the pulley + flywheel use  = I

 = TR = I
Realize that a = R


TR  I
Now solve for a using the above
equations.
 mR

a
g
2
 mR  I 
2

I
R
a
R
T
m
a
mg
L
Physics 111: Lecture 19, Pg 8
Flywheel
Falling weight & pulley...

Using 1-D kinematics (Lecture 1) we
can solve for the time required for the
weight to fall a distance L:
w/ weight

I
R
1 2
L  at
2
where
t
2L
a
 mR 2 
a
g
2
 mR  I 
T
m
a
mg
L
Physics 111: Lecture 19, Pg 9
Rotation around a moving axis.

A string is wound around a puck (disk) of mass M and
radius R. The puck is initially lying at rest on a frictionless
horizontal surface. The string is pulled with a force F and
does not slip as it unwinds.
What length of string L has unwound after the puck has
moved a distance D?
M
R
F
Top view
Physics 111: Lecture 19, Pg 10
Rotation around a moving axis...
A

The CM moves according to F = MA

The distance moved by the CM is thus D 
F
M
1 2
F 2
At 
t
2
2M

RF
2F
=
=
I 1
MR
MR 2
2

The disk will rotate about
its CM according to  = I
=

So the angular displacement is  
1 2
F 2
t 
t
2
MR
1
I  MR 2
2

M
A
R
F
Physics 111: Lecture 19, Pg 11
Rotation around a moving axis...

So we know both the distance moved by the CM and the
angle of rotation about the CM as a function of time:
D
F 2
t
2M
Divide (b) by (a):
(a)
 2

D R

F 2
t
MR
R  2 D
(b)
The length of string
pulled out is L = R:
L  2D

F
F
D
L
Physics 111: Lecture 19, Pg 12
Comments on CM acceleration:

We just used  = I for rotation about an axis through the CM
even though the CM was accelerating!
The CM is not an inertial reference frame! Is this OK??
(After all, we can only use F = ma in an inertial reference
frame).

YES! We can always write  = I for an axis through the CM.
This is true even if the CM is accelerating.
We will prove this when we discuss angular momentum!

M
A
R
F
Physics 111: Lecture 19, Pg 13
Rolling

An object with mass M, radius R, and moment of inertia I
rolls without slipping down a plane inclined at an angle 
with respect to horizontal. What is its acceleration?

Consider CM motion and rotation about
the CM separately when solving this
problem (like we did with the last
problem)...
I
M
R

Physics 111: Lecture 19, Pg 14
Rolling...


Static friction f causes rolling. It is an unknown, so we must
solve for it.
First consider the free body diagram of the object and use
FNET = MACM :
In the x direction Mg sin  - f = MA
M

Now consider rotation about the CM
and use  = I realizing that
 = Rf and A = R
Rf  I
A
R
f I
A
R

f
Mg
R2
Physics 111: Lecture 19, Pg 15
Rolling...
Mg sin  - f = ma

We have two equations:

We can combine these to eliminate f:
f I
A
R2
MR 2 sin 
A=g
MR 2 + I
I
A
For a sphere:
MR 2 sin 
A=g
2
2
MR + MR 2
5
5
= gsin 
7
M
R

Physics 111: Lecture 19, Pg 16
Lecture 19, Act 2
Rotations

Two uniform cylinders are machined out of solid aluminum.
One has twice the radius of the other.
If both are placed at the top of the same ramp and released,
which is moving faster at the bottom?
(a) bigger one
(b) smaller one
(c) same
Physics 111: Lecture 19, Pg 17
Lecture 19, Act 2
Solution

Consider one of them. Say it has radius R, mass M and falls
a height H.
Energy conservation: - DU = DK
but
1
I  MR 2
2
and
MgH 

1
1
I  2  MV 2
2
2
V
R
2
1 1
1
2 V
MgH   MR  2  MV 2
R
2 2
2
MgH 
H
1
1
3
MV 2  MV 2  MV 2
4
2
4
Physics 111: Lecture 19, Pg 18
Lecture 19, Act 2
Solution
So:
MgH 
3
MV 2
4
3
gH  V 2
4
V
4
gH
3
So, (c) does not depend on size,
as long as the shape is the same!!
H
Physics 111: Lecture 19, Pg 19
Sliding to Rolling

Roll bowling ball
A bowling ball of mass M and radius R is thrown with initial
velocity v0. It is initially not rotating. After sliding with
kinetic friction along the lane for a distance D it finally rolls
without slipping and has a new velocity vf. The coefficient of
kinetic friction between the ball and the lane is .
What is the final velocity, vf, of the ball?
vf= R

v0
f = Mg
D
Physics 111: Lecture 19, Pg 20
Sliding to Rolling...



While sliding, the force of friction will accelerate the ball in
the -x direction: F = -Mg = Ma so a = -g
The speed of the ball is therefore v = v0 - gt (a)
Friction also provides a torque about the CM of the ball.
Using  = I and remembering that I = 2/5MR2 for a solid
sphere about an axis through its CM:
2
 = MgR = MR 2 
5
x
v f= R
5g
=
2R
 = 0 + t =
5 g
t (b)
2R

v0
f = Mg
D
Physics 111: Lecture 19, Pg 21
Sliding to Rolling...
x
v  v 0  gt (a)
=
5g
t (b)
2R

We have two equations:

Using (b) we can solve for t as a function of 

Plugging this into (a) and using vf = R (the condition for
rolling without slipping):
5
Doesn’t depend
vf  v0
7
on , M, g!!
vf= R
t
2 R
5 g

v0
f = Mg
D
Physics 111: Lecture 19, Pg 22
Lecture 19, Act 3
Rotations

A bowling ball (uniform solid sphere) rolls along the floor
without slipping.
What is the ratio of its rotational kinetic energy to its
translational kinetic energy?
(a)
1
5
(b)
2
5
(c)
1
2
Recall that I  2 MR 2 for a solid sphere about
5
an axis through its CM:
Physics 111: Lecture 19, Pg 23
Lecture 19, Act 3
Solution

The total kinetic energy is partly due to rotation and partly
due to translation (CM motion).
K=
1
1
I  2  MV 2
2
2
rotational
translational
K
K
Physics 111: Lecture 19, Pg 24
Lecture 19, Act 3
Solution
K=
1
1
I  2  MV 2
2
2
Since it rolls without slipping:  
rotational
Translational
K
K
K ROT
KTRANS
V
R
2
1 2
2
2 V
I
 MR  2
5
R  2
2


2
1
2
5
MV
MV
2
Physics 111: Lecture 19, Pg 25
Atwoods Machine with Massive Pulley:



y
A pair of masses are hung over a
massive disk-shaped pulley as shown.
Find the acceleration of the blocks.
For the hanging masses use F = ma
 -m1g + T1 = -m1a
 -m2g + T2 = m2a
a

I
For the pulley use  = I
R
a 1
 T1R - T2R I  MRa
R 2
(Since I 
1
MR 2 for a disk)
2
x
M

R
T2
T1
a
m2
m1
a
m1g
m2g
Physics 111: Lecture 19, Pg 26
Large and small pulleys
Atwoods Machine with Massive Pulley...

y
We have three equations and three
unknowns (T1, T2, a). Solve for a.
x
-m1g + T1 = -m1a
(1)
-m2g + T2 = m2a
(2)
T1 - T2 
1
Ma
2
(3)


m1  m 2
a
g
 m1  m 2  M 2 
M

R
T2
T1
a
m2
m1
a
m1g
m2g
Physics 111: Lecture 19, Pg 27
Recap of today’s lecture

Review
Many body dynamics
Weight and massive pulley
Rolling and sliding examples
Rotation around a moving axis: Puck on ice
Rolling down an incline
Bowling ball: sliding to rolling
Atwood’s Machine with a massive pulley

Look at textbook problems Chapter 9: # 53, 89, 92, 113, 125







(Text: 9-1 to 9-6)
(Text: 9-4)
(Text: 9-6)
(Text: 9-4)
(Text: 9-6)
(Text: 9-4)
Physics 111: Lecture 19, Pg 28