Transcript Perfect Rolling (print version)

Perfect Rolling (no sliding!)
Consider a disk rolling without slipping with a Uniform Acceleration.
While most points both rotate and move linearly, the center of mass is moving linearly
with a constant acceleration acm

acm
Perfect Rolling (no sliding!)
Both a point on the outside of the
disk and the center of mass must
Rolling Condition – must hold for
move the same linear distance,
an object to roll without
with the same linear velocity and
slipping.
the same linear acceleration for
the disk to roll without slipping!
s  R
vcm   R
Radian measure
acm   R
R
s
s


acm
Perfect Rolling (no sliding!)
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!
Perfect Rolling (no sliding!)
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!
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.
Fg  FN

FN
Both act through the axis of
rotation and therefore both
exert no torque!

Fg
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.
Fg  FN

FN

Fg

Ff
The force of friction does
apply a torque which results
in the rolling object
accelerating both linearly
and rotationally!
Perfect Rolling (no sliding!)
Consider a rolling object smoothly accelerating down a hill.

FN


F fs


Fg
If the hill was frictionless, there would not be a force of static friction and
therefore the object would slide down the hill instead of roll!
Perfect Rolling (no sliding!)
Consider a rolling object smoothly accelerating down a hill.

FN


F fs

Fgy

Fgx
FN  Fgy
Fgx  F fs


Fg
Perfect Rolling (no sliding!)
Consider a rolling object smoothly accelerating down a hill.

FN

F fs

Fgy

 Fy  0
FN  Fgy

Fgx


 Fx  max
Fgx  F fs   ma
FN  Fgy
Fgx  F fs


Fg
  I
FN and Fg exert no torque
since they act through the
axis of rotation (cm)
 F fs R   I
Perfect Rolling (no sliding!)
A hoop and a disk with identical masses and radii roll down an incline. How do their
motions compare?
I hoop  MR
2
I disk
1
 MR 2
2
 F fs R   I
35
So
 disk   hoop

F fs R
I
Since they have the same distance to move
and both start from rest,
tdisk ? thoop
vdisk 
? vhoop
hoop
Perfect Rolling (no sliding!)
Kinetic Energy (Ek) - The ability to produce change due to an object’s motion.
Translational (Linear) Kinetic Energy
Rotational Kinetic Energy
1 2
Ekt  mv
2
1 2
Ekr  I
2
Perfect Rolling (no sliding!)
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
Perfect Rolling (no sliding!)
A hoop and a disk with identical masses and radii roll down an incline. How do their
motions compare?
System = Object, Incline, Earth
Ekf  Egi
So
Ekf disk  Ekf hoop
But
35
1
1 2
2
Ek of rolling  I cm  mv
2
2
1
2
I hoop  MR
I disk  MR 2
2
More of the disk’s kinetic energy will be
translational so,
vdisk 
? vhoop
hoop