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Rolling

vcm
s  R
Rolling
ds  d 
  R
dt  dt 

vcm  R
Rolling Condition – must hold for
an object to roll without
slipping.
R
s
s


vcm
Rolling
One way to view rolling is as a combination of pure rotation and pure translation.
Pure Translation
Pure Rotation

vcm



vcm

vcm  R
Rolling

vcm

2vcm

vcm

vcm


vcm
The point that is in
contact with the ground is
not in motion with
respect to the ground!
Rolling
Rolling

2vcm
Since the bottom point is at rest with respect
to the ground, static friction applies if any
friction exists at all. Static friction does not
dissipate energy.
However, there usually is rolling friction
caused by the deformation of the object and
surface as well as the loss of pieces of the
object. Rolling friction does dissipate energy.

vcm
The point that is in
contact with the ground is
not in motion with
respect to the ground!
Rolling
If the disk is moving at constant speed, there is no tendency to slip at the
contact point and so there is no frictional force.
If, however, a force acts on the disk, like when you push on a bike pedal, then
there is a tendency to slide at the point of contact so a frictional force acts at
that point to oppose that tendency.
Rolling
Just as rolling motion can be viewed as a combination of pure rotation and pure
translation, the kinetic energy of a rolling object can be viewed as a combination of pure
rotational kinetic energy and pure translational kinetic energy.
Ek of rolling  Ek of rotation  Ek of translatio n
1
1 2
2
Ek of rolling  I cm  mv
2
2
Pure
Rotation
Pure
Translation
Rolling
1. A hoop of radius 14 cm and mass 2.0 kg rolls across a horizontal table with a
constant speed of 0.10 m .
s
a. What is the kinetic energy of the hoop?
1 2 1 2
Ek  I  Mv
2
2
But
vcm  R
vcm

R
I hoop  MR 2
Note: the v in the Ek equation is vcm


2
1
1 2
2 v
Ek  MR    Mv
2
 R 2
Rolling
1. A hoop of radius 14 cm and mass 2.0 kg rolls across a horizontal table with a
constant speed of 0.10 m .
s
a. What is the kinetic energy of the hoop?
1 2 1 2
Ek  Mv  Mv
2
2
Ek  Mv 2

Ek  2.0 kg  0.10 m
Ek  2.0 102 J
s
2
Rolling
1. A hoop of radius 14 cm and mass 2.0 kg rolls across a horizontal table with a
constant speed of 0.10 m .
s
b. What percentage of the kinetic energy is associated with rotation and what
percentage with translation?
1 2 1 2
Ek  Mv  Mv
2
2
%rotation  50%
%translation  50%
Rolling
1. A hoop of radius 14 cm and mass 2.0 kg rolls across a horizontal table with a
constant speed of 0.10 m .
s
2


c. If the object is instead a solid sphere  I  MR 2 , what percentage of its
5


kinetic energy is associate with rotation and what percentage with translation?
1 2 1 2
Ek  I  Mv
2
2
But
vcm  R
vcm

R
2
I sphere  MR 2
5
2
12
1 2
2  v 
Ek   MR    Mv
2 5
 R  2
Rolling
1. A hoop of radius 14 cm and mass 2.0 kg rolls across a horizontal table with a
constant speed of 0.10 m .
s
2


c. If the object is instead a solid sphere  I  MR 2 , what percentage of its
5


kinetic energy is associate with rotation and what percentage with translation?
1 2 1 2
Ek  Mv  Mv
5
2
2
% rotation  10  100
7
10
2
5
2
Ek  Mv  Mv 2
10
10
%rotation  29%
7
Ek  Mv 2
10
%translation
5
 10  100
7
10
%translation  71%
Rolling
2. A hoop, a disk, and a solid sphere with identical masses and radii roll down an
incline of length 2.4 m and angle 35º.
a. Using dynamics (Newton’s Laws), what is the acceleration of the center of mass
of each object as it rolls down the incline?

FN
35

F fs

Fgy

Fgx
FN  Fgy
Fgx  F fs
35

Fg
Rolling
2. A hoop, a disk, and a solid sphere with identical masses and radii roll down an
incline of length 2.4 m and angle 35º.
a. Using dynamics (Newton’s Laws), what is the acceleration of the center of mass
of each object as it rolls down the incline?


F

m
a
 y
y
FN  Fgy  0


 Fx  max
Fgx  F fs   ma
FN  mg cos mg sin   F fs   ma
  I
FN and Fg exert no
torque since they act
through the axis of
rotation (cm)
 F fs R   I
But
a  R
a

R
Ia
F fs  2
R
Rolling
2. A hoop, a disk, and a solid sphere with identical masses and radii roll down an
incline of length 2.4 m and angle 35º.
a. Using dynamics (Newton’s Laws), what is the acceleration of the center of mass
of each object as it rolls down the incline?
Ia
mg sin   2  ma
R
Ia
mg sin   ma  2
R
I 

mg sin    m  2 a
R 

mg sin 
a
I
m 2
R
g sin 
a
I
1
mR 2
Rolling
2. A hoop, a disk, and a solid sphere with identical masses and radii roll down an
incline of length 2.4 m and angle 35º.
a. Using dynamics (Newton’s Laws), what is the acceleration of the center of mass
of each object as it rolls down the incline?
g sin 
a
I
1
mR 2
But
I hoop  mR
g sin 
ah 
mR 2
1
mR 2
2
1
 mR 2
2
2
I sphere  mR 2
5
g sin 
ad 
1
mR 2
1 2 2
mR
g sin 
as 
2
mR 2
1 5 2
mR
I disk
Rolling
2. A hoop, a disk, and a solid sphere with identical masses and radii roll down an
incline of length 2.4 m and angle 35º.
a. Using dynamics (Newton’s Laws), what is the acceleration of the center of mass
of each object as it rolls down the incline?
g sin 
ah 
11
g sin 
ad 
1
1
2
g sin 
as 
2
1
5
1
ah  g sin 
2
2
ad  g sin 
3
5
as  g sin 
7
a h  2. 8 m
ad  3.7 m
as  4.0 m
s
2
s
2
s2
Rolling
2. A hoop, a disk, and a solid sphere with identical masses and radii roll down an
incline of length 2.4 m and angle 35º.
b. In what order would the hoop, disk, and sphere reach the bottom of the incline?
t ?

vo  0
 
x  xo  L  2.4 m
1 2
  
x  xo  vot  at
2
2L
t 
a
1
2L
2L
2L
ah  g sin  th 
td 
ts 
2
1
5
2
g sin 
g sin 
g sin 
3
2
7
2
ad  g sin 
3
4L
14 L
3L
th 
ts 
td 
5
g sin 
5 g sin 
g sin 
as  g sin 
7
ts  1.09 s
td  1.13 s
th  1.31 s
#3
#2
#1
Rolling
2. A hoop, a disk, and a solid sphere with identical masses and radii roll down an
incline of length 2.4 m and angle 35º.
a. Using energy principles , what is the velocity of the center of mass of each object
as it reaches the bottom of the incline?
35
yi
L

yi  L sin 
Rolling
2. A hoop, a disk, and a solid sphere with identical masses and radii roll down an
incline of length 2.4 m and angle 35º.
a. Using energy principles, what is the velocity of the center of mass of each object
as it reaches the bottom of the incline?
0  Eint
0  Ek  Eg
0  Ekf  Eki  Egf  Egi
Ekf  Egi
1 2 1 2
I f  mv f  mgyi
2
2
But
yi  L sin 
vcm  R
v

R
Rolling
2. A hoop, a disk, and a solid sphere with identical masses and radii roll down an
incline of length 2.4 m and angle 35º.
a. Using energy principles, what is the velocity of the center of mass of each object
as it reaches the bottom of the incline?
2
v
1  f 1 2
I    mv f  mg L sin 
2 R
2

 I v 2  mv 2  2mgL sin 
 2 f
f
R
 
 I v 2  v 2  2 gL sin 

f
2 f
 mR 
1  I v 2  2 gL sin 

2 f
 mR 

Rolling
2. A hoop, a disk, and a solid sphere with identical masses and radii roll down an
incline of length 2.4 m and angle 35º.
a. Using energy principles, what is the velocity of the center of mass of each object
as it reaches the bottom of the incline?
v 2f
2 gL sin 

I
1
mR 2
2 gL sin 
vf 
I
1
mR 2
Rolling
2. A hoop, a disk, and a solid sphere with identical masses and radii roll down an
incline of length 2.4 m and angle 35º.
a. Using energy principles, what is the velocity of the center of mass of each object
as it reaches the bottom of the incline?
2 gL sin 
vf 
I
1
mR 2
But
I hoop  mR
2
2 gL sin 
v fh 
mR 2
1
mR 2
I disk
v fd
1
 mR 2
2
2 gL sin 

1
mR 2
1 2 2
mR
2
I sphere  mR 2
5
2 gL sin 
v fs 
2
mR 2
1 5 2
mR
Rolling
2. A hoop, a disk, and a solid sphere with identical masses and radii roll down an
incline of length 2.4 m and angle 35º.
a. Using energy principles, what is the velocity of the center of mass of each object
as it reaches the bottom of the incline?
v fd
2 gL sin 

1
1
2
2 gL sin 
v fs 
2
1
5
v fh  gl sin 
v fd
4

gl sin 
3
10
v fs 
gl sin 
7
v fh  3.7 m
v fd  4.2 m
2 gL sin 
v fh 
11
s
s
v fs  4.4 m
s
Rolling
2. A hoop, a disk, and a solid sphere with identical masses and radii roll down an
incline of length 2.4 m and angle 35º.
b. In what order would the hoop, disk, and sphere reach the bottom of the incline?
t ?

vo  0
 
x  xo  L  2.4 m
v fh  gl sin 
  1  
x  xo  v f  vo t
2
 
2 x  xo 
t

vf
2L
2L
2L
th 
td 
ts 
4
10
gL sin 
4
gL sin 
gL sin 
v fd 
gl sin 
3
7
3
td  1.13 s
ts  1.09 s
th  1.31 s
10
v fs 
gl sin 
#3
#2
#1
7