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Rotational Kinetic Energy
Conservation of Angular Momentum
Vector Nature of Angular Quantities
8.7 Rotational Kinetic Energy
All particles making up a rotating rigid body
rotate with the same angular velocity .
The kinetic energy of the rotating body is
KE = (mv2/2) = (m(r)2/2) = (1/2) (mr)22 = I2.
This gives us a starting equation for rotational kinetic
energy:
OSE:
Kp = ½ Ipp2.
The subscript “p” is there to remind you that the moment
of inertia I depends on the pivot point “p,” and therefore
so does the rotational KE.
If an object is rolling (with axis of rotation through its
center of mass) while undergoing translational motion,
then
OSE:
Ktotal = Ktrans + Krot = ½ MVCM2 + ½ ICMCM2.
Example: a solid cylinder and a thin hollow cylinder, both
of mass M and outer radius R, are released from rest at
the start of an inclined plane of height H and length L.
What are the speeds of each cylinder when they reach
the bottom? Which cylinder gets to the bottom first?
“Nearly-official” solution that I scanned for Winter/Spring
2002 is on next page.
Example: Physics 23 Problem on KE of Rotation (from
Test 3 Winter 2001).
A cylinder of mass M, radius R and
length L can freely rotate about a
light metal rod through its center.
The ends of the rod are attached to
a massless yoke that can be used
to pull the cylinder along a surface.
P=2Mg

M r
D
The cylinder is placed at rest on a horizontal surface.
The yoke is then pulled by a worker with a constant
force of magnitude P=2Mg at an angle of  with respect
to the vertical. It rolls without slipping at all times.
Derive an expression for the
linear speed of the cylinder,
in terms of relevant system
parameters, after it has
rolled a distance D.
Vi=0
P=2Mg
Vf ?
 90-
M r
d
D
For an OSE you can use Ef-Ei=[Wother]i→f or [Wnet]i→f=K.
Because the only force that does work is P, in either
equation, the calculation of work is the same. There is
no change of height and there are no springs, so the K
equation is “easier.”
→→
K = WP = P·d
K = PD cos(90-) = PD sin 
The cylinder is solid (if that
is not clear, you should
ask) and has a moment of
inertia ½ Mr2.
Vi=0
P=2Mg
Vf ?
 90-
M r
0
Kf - Ki = PD sin 
d
D
½ M Vcm,f2 + ½ I f2 = PD sin 
½ M Vf2 + ½ (½ Mr2) (Vf /r)2 = (2Mg)D sin 
½ Vf2 + ¼Vf2 = 2gD sin 
¾Vf2 = 2gD sin 
Vf2 = (8gD sin )/3
8.8 Angular Momentum
and its Conservation
Recall that linear momentum is defined by p = mv. The
rotational analog of m is I, and the rotational analog of v
is . We define angular momentum of a rigid, rotating
body by
OSE:
Lz = Iz.
From the definition of torque, if z = 0, then z = 0, z
is constant, and Lz is constant.
Linear momentum p is the fundamental descriptor of
translational motion. It is changed by forces.
Angular momentum L is the fundamental descriptor of
rotational motion. It is changed by torques.
The total angular momentum of a rotating body
remains constant if the net torque acting on it is
zero.
If the net torque is zero, angular momentum is conserved. This is another of the fundamental conservation
laws of physics.
OSE: z,ext = 0

Lz,i = Lz,f.
A particle moving in a straight line with velocity v has an
angular momentum relative to an axis.
R
v
M
z
Lz,particle = RMv.
I will use this in working
example 8.16.
Example 8-16. A mass M attached to the end of a
string revolves in a circle on a frictionless tabletop. The
other end of the string passes through a hole in the
table. Initially the mass revolves with a speed V1 in a
circle or radius R1. The string is then pulled through the
hole, so that the radius is reduced to R2. Calculate the
speed of the mass now.
You can use the equation on the previous slide for the
angular momentum of a particle, or you can use the
moment of inertia of a single particle MR2 from section
8.5, or you can treat the rotating particle like a thin hoop
of mass M and radius R.
8.9 Vector Nature of Angular Quantities
I have already introduced you to the ideas in this
section. You will not be tested on it.
If there is time, I will do demonstrations.
Demonstration: spinning professor makes himself sick.
Demonstration: bicycle wheel gyroscope.
The following demonstration introduces the next chapter.
Demonstration: sacrificial broom.