Suppose we want to nd the moment of inertia of the arbitrary blob
shaped rigid mass distribution pictured above in gure 76 about the
axis labelled New (Parallel) Axis. This is, by
260 Week 5: Torque and Rotation in One Dimension
r
dm
CM Axis
New (Parallel

special and worth treating in the context of angular momentum
conservation in their own right.
6.4.1: Radial Forces and Angular Momentum Conservation
One of the most important aspects of torque and angular momentum
arises because of a curious feature of t

Iside pivot = MR2 + Icm = MR2 + MR2 = 2MR2 (560)
and were done!
For your homework, you get to evaluate the moment of inertia of a rod
about an axis through its center of mass and about one end of the rod
and compare the two, both using direct integration

v2 =
r1 r2
v1 (600)
Week 6: Vector Torque and Angular Momentum 289
f) Compute the work done by the force from part c) above and identify
the answer as the workkinetic energy theorem. Use this to to nd the
velocity v2. You should get the same answer! Well,

We also know that L = mvr from the previous question! All that
remains is to do some algebra magic to convert one to the other. If we
had one more factor of m on to, and a factor of r2 on top, the top
would magically turn into L2. However, we are only all

dm
lv
zdL
r
dr
Figure 81: The geometry of a rod of mass M and length L, rotating
around a pivot through the end in the x-y plane.
and length l (note that it is now tricky to call its length L as thats also
the symbol for angular momentum, sigh). From the

Figure 73: Atwoods Machine, but this time with a massive pulley of
mass M and radius R (and moment of inertia I = MR2), this time
solving a standard conservation of mechanical energy problem.
We would like to nd the speed v of m1 and m2 (and the angular s

Principle, both beyond the scope of this course. Pauli is arguably coequal with Coulomb in determining atomic and molecular structure.
Week 6: Vector Torque and Angular Momentum 287
where ke is once again a constant of nature.
Both of these are radial for

L2
+ mL (602)
or
xcm =
M/2 + m M + m
L (603)
where Im taking it as obvious that the center of mass of the rod
itself is at L/2.
To nd vf, we note that momentum is conserved (and also recall that
the answer is going to be vcm: pi = mv0 = (M + m)vf = pf (60

In gure 74 a spool of shing line that has a total mass M and a
radius R and is eectively a disk is tied to a pole and released from
rest to fall a height H. Lets nd everything: the acceleration of the
spool, the tension T in the shing line, the speed with

In analyzing the walking the spool problem in class and in the text
above, students often ask how they can predict which direction that
static friction acts on a rolling spool, and I reply that they cant! I
cant, not always, because in this problem it can

32
23
12 31F F
F
FF
F
Figure 80: The coordinates of a small collection of particles, just
enough to illustrate how internal torques work out.
Let us write ~ = d~ L/dt for each particle and sum the whole thing up,
much as we did for
~ F = dp~/dt in chapter

a) Find the torque about the pivot exerted on the pendulum by gravity
at an arbitrary angle .
b) Integrate the torque from = 0 to = 0 to nd the total work done
by the gravitational torque as the pendulum disk falls to its lowest
point. Note that your answ

This spin angular momentum is not classical and does not arise from
the physical motion of mass in some kind of path around an axis
and hence is largely beyond the scope of this class, but we certainly
need to know how to evaluate and alter (via a torque

(which continues moving a the constant speed of the center of mass
before the collision, of course).
All of the terminology developed to describe the energetics of dierent
collisions still holds when we consider conservation of angular
momentum in additio

c) Find the moment of inertia of the leg about the hip/pivot at the
origin. Again, you may leave it in terms of C if you wish or express it
in terms of M, L and x0. Do your answers all have the right units?
d) How might one improve the estimate of the mom

momentum will be the same, but unless we choose the center of mass
of the system to be our pivot we will have to deal with the fact that our
nal angular momentum will have both a translational and a
rotational piece.
This suggests that our best choice is

Mirror symmetry across the axis of rotation and/or Mirror
symmetry across the plane of rotation,
we can write:
L = La = Iaaa = I (578)
where ~ = a a points in the (right handed) direction of the axis of
rotation and where: I = Iaa =X i mir2 i or Z r2dm (

= Lsin()p
or
p =
mgD L In this expression, p is the angular precession frequency of
the top and ~ D is the vector from the point where the tip of the top
rests on the ground to the center of mass of the top. The direction of
precession is determined by th

I was similarly the distance from the axis of rotation of the particular
mass m or mass chunk dm. We considered this to be one dimensional
rotation because the axis of rotation did not change, all rotation was
about that one xed axis.
This is, alas, not t

fvf
L
cmx
Figure 86: A blob of putty of mass m, travelling at initial velocity v0 to
the right, strikes an unpivoted rod of mass M and length L at the end
and sticks to it. No friction or external forces act on the system.
In gure 86 a blob of putty of m

preload your neocortex with lots of quantum cartoons and glances at
the algebra of angular momentum in quantum theory ahead of time.
282 Week 6: Vector Torque and Angular Momentum
~ L, ~ and I with a. Then La=z is the z-component of ~ L, a=x is
the x-comp

v
Figure 83:
A particle of mass m is tied to a string that passes through a hole in a
frictionless table and held. The mass is given a push so that it moves in
a circle of radius r at speed v. Here are several questions that might be
asked and their answe

Thats three results for a single object, and of course we can apply the
parallel axis theorem to the x-rotation or y-rotation as well! The Li =
Ii (for i = x,y,z) result works, but the direction of ~ L and ~ as well
as the value of the scalar moment of in

as the angular momentum vector of a particle of mass m and
momentum p~ located at a vector position ~r with respect to the origin
of coordinates.
In that case Newtons Second Law for a point mass being rotated by a
vector torque is:
~ =
d~ L dt
(572)
which

vertical axis. The professor then pulls the masses in towards the axis
of rotation, reducing their contribution to the total moment of inertia
as illustrated in gure 82
If the moment of inertia of the professor and platform is I0 and the
masses m (includi

a dimensionless number such as 1 2 or 2 5, that might describe a disk
or a solid ball, respectively).
a) Begin by relating v (the speed of the center of mass) to the angular
velocity (for the rolling object). You will use this (and the related two
equatio

Problem 6.
F
F
F
R
r
I (about cm)
mass m
This problem will help you learn required concepts such as:
Direction of torque Rolling Constraint
so please review them before you begin.
In the gure above, a spool of mass m is wrapped with string around
the inn

MR22
Week 5: Torque and Rotation in One Dimension 257
and now we substitute the rolling constraint:
(m2 m1)gH =
12
(m1 + m2)v2 + 1 2
MR2 v2 R2
(m2 m1)gH =
12
(m1 + m2 + M)v2 (532)
to arrive at
v =s2(m2 m1)gH m1 + m2 + M
(533)
You can do this! It isnt real

As long as the object rotates uniformly that is, the object goes
around its own center one time for every time it goes around the axis
of rotation, keeping the same side pointing in towards the center as it
goes then its kinetic energy is just:
K=
12
I2 =