Your Thoughts
Suppose a ball rolls down a ramp with a coefficient of friction just big enough to keep the ball
from slipping. An identical ball rolls down an identical ramp with a coefficient of friction of 1.
Do both balls reach the bottom at the same time? (y/n)
Does the moment of inertia play a part in determining the amount of time it takes an object to
get to the bottom of an incline.
For the first checkpoint, I know that when a ball is slipping down, the KE is bigger than the
block, since there is some rotational energy but if they are moving up the ramp, should the
rotational energy be negative or positive?
Mechanics Lecture 16, Slide 1
2nd Exam Monday July 15th
Covers Lectures 7-13 only
90 Minutes. 1:30-3:00pm
In Here: 141 Loomis
Mechanics Lecture 9, Slide 2
Physics 211
Lecture 16
Today’s Concepts:
a) Rolling Kinetic Energy
b) Angular Acceleration
Physics 211 Lecture 16, Slide 3
Rolling
What is meant by without slipping?
v = R
2R2 = v2
If there is no slipping:
Mechanics Lecture 16, Slide 4
Two ways of looking at rolling down ramp
PE = MgH
Energy Conservation
KE = 0
PE = 0
H
KE = Ktrans  Krot
Dynamics
a
f
a
F = ma
τ = Iα
Mg
Mechanics Lecture 16, Slide 5
Energy Conservation
Objects of different I rolling down an inclined
plane:
for the first checkpoint, I know that when a
K=0
U = Mgh
R
M
v=0
=0
K=0
h
ball is slipping down, the KE is bigger than
the block, since there is some rotational
energy but if they are moving up the ramp,
should the rotational energy be negative or
positive?
1
1
I 2  Mv 2
2
2
U=0
K=
v = R
K  U = 0
1 2 1 2
MgH = I   Mv
2
2
Mechanics Lecture 16, Slide 6
Lets work out the general case…
Does the moment of inertia play a part in
determining the amount of time it takes
an object to get to the bottom of an
incline?
K=
1 2 1 2
I  Mv
2
2
K=
v
Use v = R and I = cMR2 .
1
1 2 1
2 2
MR


Mv =  c  1Mv 2
c
2
2
2
Hoop: c = 1
Disk: c = 1/2
Sphere: c = 2/5
etc...
So:
1

2
c  1Mv 2 = Mgh
v=
2 gh
1
c 1
Doesn’t depend on M or R, just on c (the shape)
Ramp demo
Mechanics Lecture 16, Slide 7
Clicker Question
A solid cylinder of mass M rolls along the floor without
slipping with speed V. What is its total kinetic energy?
1
2
A) K = MV
2
Recall that I cm
3
B) K = MV 2 C) K = MV 2
4
1
= MR 2 for a solid cylinder:
2
Mechanics Lecture 16, Slide 8
CheckPoint
A block and a ball have the same mass and move with the same
initial velocity across a floor and then encounter identical ramps.
The block slides without friction and the ball rolls without
slipping. Which one makes it furthest up the ramp?
A) Block
B) Ball
C) Both reach the same height.
v

v
Mechanics Lecture 16, Slide 9
CheckPoint
The block slides without friction and the ball rolls without slipping.
Which one makes it furthest up the ramp?
v
A) Block
B) Ball
C) Same

v
A) The block does not have kinetic energy wasted by
rolling. [It actually has less energy!]
B) The ball has both translational KE (equal to that of
the block) and rotational KE. Because it has more KE
to start with, it must have more PE to finish with,
and therefore it makes it further up the ramp. [Yes]
C) since they have same mass and same initial velocity,
so they should reach the same height of the ramp.
[Energy is not just kinetic- mass & velocity]
Mechanics Lecture 16, Slide 10
CheckPoint
A cylinder and a hoop have the same mass and radius. They
are released at the same time and roll down a ramp without
slipping. Which one reaches the bottom first?
A) Cylinder
B) Hoop
C) Both reach the bottom at the
same time
Mechanics Lecture 16, Slide 11
Which one reaches the bottom first?
A) Cylinder
B) Hoop
C) Both reach the bottom at the
same time
A) At the bottom both objects have a combination of kinetic translational and
rotational motion, which are equal because they started with the same
potential energy. Since the hollow cylinder's moment of inertia is greater, its
rotational kinetic energy is greater, and therefore the amount of energy that
becomes translational is less, meaning it has a lower linear velocity than that
of the solid cylinder.
C) Since they both have the same mass, and start at the same height, they will
have the same potential energy, and will have the same kinetic energy at the
bottom.
Mechanics Lecture 16, Slide 12
CheckPoint
A small light cylinder and a large heavy cylinder are
released at the same time and roll down a ramp without
slipping. Which one reaches the bottom first?
A) Small cylinder
B) Large cylinder
C) Both reach the bottom
at the same time
Mechanics Lecture 16, Slide 13
A) Small cylinder
B) Large cylinder
C) Both reach the bottom
at the same time
A) The small cylinder has less inertia than the
larger cylinder.
B) larger cylinder has more potential energy to
begin
C) The mass and radius of each cancel out when
combining the moment of inertia and kinetic
energy equations, which means velocity only
depends on the moment of inertia.
Mechanics Lecture 16, Slide 14
Acceleration
Mechanics Lecture 16, Slide 15
Acceleration
Mechanics Lecture 16, Slide 16
Acceleration
Suppose a ball rolls down a ramp
with a coefficient of friction just big
enough to keep the ball from
slipping. An identical ball rolls down
an identical ramp with a coefficient
of friction of 1. Do both balls reach
the bottom at the same time? (y/n)
Acceleration depends only on the shape, not on mass or radius.
Mechanics Lecture 16, Slide 17
Example Problem
Suppose a cylinder (radius R, mass M) is used as a pulley. Two
masses (m1 > m2) are attached to either end of a string that hangs
over the pulley, and when the system is released it moves as
shown. The string does not slip on the pulley.
What is the acceleration of the blocks?
M
a
R
T2
T1
a
m2
m1
a
Mechanics Lecture 16, Slide 18
Clicker Question
Suppose a cylinder (radius R, mass M) is used as a pulley. Two
masses (m1 > m2) are attached to either end of a string that hangs
over the pulley, and when the system is released it moves as
shown. The string does not slip on the pulley.
Compare the tension in the string
on either side of the pulley:
M
a
A) T1 > T2
B) T1 = T2
C) T1 < T2
R
T2
T1
a2
m2
m1
a1
Mechanics Lecture 16, Slide 19
Atwood's Machine with Massive Pulley:
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
M
a
R
T2 - m2g = m2a
T2
T1
m2
m1
a
m1g
a
m2g
Mechanics Lecture 16, Slide 20
Atwood's Machine with Massive Pulley:
A pair of masses are hung over a
massive disk-shaped pulley as
shown.

Find the acceleration of the blocks.
a
For the pulley use  = Ia = I
R
a 1
T1R - T2R = I
= MRa
R 2
1
(Since I = MR 2 for a disk)
2
M
a
R
T1
T2
a
m2
m1
a
m1g
m2g
Mechanics Lecture 16, Slide 21
Atwood's Machine with Massive Pulley:
We have three equations and three
unknowns (T1, T2, a).
Solve for a.
m1g - T1 = m1a
(1)
T2 - m2g = m2a
(2)
1
T1 - T2 = Ma (3)
2


m1 - m2
 g
a = 
 m1  m2  M 2 
M
a
R
T2
T1
m2
m1
a
m1g
a
m2g
Mechanics Lecture 16, Slide 22
a
f
v
M R
 = Ia

a=
I
5 g
fR

MgR
=
=
=
I
2
2R
2
5
MR
Mechanics Lecture 16, Slide 23
a
f
v
M R
F = Ma
F  Mg
= g
a=
=
M
M
Mechanics Lecture 16, Slide 24
a = g

5 g
a=
2R
v
M R
a
v = R
v
v0
v = v0 - at
t

 = at
t
Once v=R it rolls
without slipping
v0 v = v0 - at
v = R
R = a Rt
t
v0 - at = a tR
5 g
v0 -  gt =
t
2
7 g
v0 =
t
2
t=
2
7 g
v0
Mechanics Lecture 16, Slide 25

a
M R
v
1 2
x = v0t - at
2
v = v0 - at
Plug in a and t found in parts 2) & 3)
a = g
t=
2
7 g
v0
Mechanics Lecture 16, Slide 26

a
M R
a = g
v
Interesting aside: how v is related to v0 :
v = v0 - at
t=
2
7 g
v0
 2v0 
v = v0 - (  g ) 

 7 g 
2
v = v0 - v0
7
5
v = v0
7
Doesn’t depend on 
v =  0.714  v0
We can try this…
Mechanics Lecture 16, Slide 27