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Let's consider examples
of our principle

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that the external
torque about a point, s,

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causes the angular momentum
of a system to change about s.

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We've examined
central force problems

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in which we chose the point s
to be the central point in which

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there was no torque.

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Now, as examples,
let's look at a case

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where we have a pivoted object.

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So I could take an
example of an object.

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Let's see, this will be overhead
view of a ring of radius r.

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And I'm going to have
a mass coming in.

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I'll call this ring
mass m1, this object m2.

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This object is coming in
with an initial velocity,

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and this is my pivot point.

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Now, when the object
m2 is hitting the ring,

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we have a force, F2 on 1.

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This is the collision.

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Here's our pivot point.

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And when this hits
the ring, we'll

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have some type of pivot force.

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I'm just going to
denote the pivot force.

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We're not quite sure
what it will point at,

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but I'll just say F
pivot, for the moment,

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is holding this point in place.

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At the same time, the object
will have a force, F1,2,

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acting on the object.

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Here is our pivot point, p.

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And now, what point should we--

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should we choose to see if
there's no torque [? about? ?]

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Suppose we choose
the pivot point.

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Well-- p.

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Well, clearly, the pivot force
has no torque about the pivot,

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because the vector from
the pivot point to where

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the pivot force is acting is 0.

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Remember, pivot forces have no
torque about the pivot point.

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However, this collision
force will produce

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a torque about the pivot.

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And that's-- the angular
momentum of the ring is not

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constant, because the
ring will start to rotate.

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Similarly, this
object is reversing--

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will reverse
directions, or will do

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something due to the collision.

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And again, this force will
produce a torque that's equal

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and opposite to the
torque on the ring.

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This is the torque
on the particle.

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But if we make our system equal
to the particle and the ring--

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so we'll call this, our system
is now both the particle

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and the ring--

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then these torques are
internal, and because they're

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equal and opposite
forces, the rs vector

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is exactly the same rs vector.

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The internal torques
cancel in pairs.

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The pivot force
produces no torque.

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So the torque on the system
about the pivot is 0.

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And that tells us that
the angular momentum

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of the system about
that pivot point,

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initially, will be equal
to the angular momentum

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about that pivot point,
finally, if we take

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two initial and final states.

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In particular, suppose
our initial state is--

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here's the pivot.

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Our object is coming
in, that was m2 vi.

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The moment arm is r.

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And our vector from here
to the object, rs initial,

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has a moment component that way.

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If we put these
vectors tail to tail

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and figure out that the
angular momentum is pointing

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in this direction, l initial
i, and the moment arm is r,

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then the initial angular
momentum about this pivot point

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is just due to
this moving object.

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So that's m2 vi, and
the moment arm is r,

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and we'll denote its
direction that way.

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Now, the final
angular momentum--

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let's imagine that it sticks.

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So we have m2, pivot,
and now our ring

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is going to be rotating
with some omega final

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because this object hits it.

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And notice that this
object is a distance

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root 2 r from the pivot point.

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So the final angular momentum
can be two different pieces.

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You can think of this
as a system, where

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we have I of the system
about p times omega final,

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because now it's
just a rigid body.

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And the angular
momentum of the system

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is consisting of two pieces.

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It's the angular
momentum of the ring

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about the pivot,
plus the angular

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momentum of the particle about
the pivot, times omega final.

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Now, this is the center of mass.

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This is the pivot point, p.

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The angular momentum of the
ring is the angular momentum

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about the center of-- we'll
use the parallel axis theorem--

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angular momentum about
the center of mass,

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plus the distance from
the center of mass

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to the parallel axis,
which is a distance, r.

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So the first piece is Icm
ring plus m r squared.

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And the angular momentum
about the particle,

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this is going in a
circle of radius r2--

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root 2r, so when
we square that--

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by the way this was,
mass of the ring was 1.

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Mass of the particle
was m2 times r squared,

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which is this distance
squared, which is

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2 r squared times omega final.

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The moment of
inertia of the ring

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is m1 r squared about
the center of mass.

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We have the factor 2.

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We have another factor of 2.

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And so we get m1 plus
m2 times a factor 2 r

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squared omega final,
which we can call omega

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f, pointing out of the board.

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And so now, we have an
angular momentum condition,

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which is that m2 vi r
equals 2 m1 plus m2 r

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squared omega final.

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And that is the statement
that the initial angular

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momentum of our system--

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the ring is at rest here--

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is equal to the final angular
momentum of the system, where

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m2 is stuck to the
ring, and they're

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all rotating about this
pivot point with omega final.

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That's the angular
momentum of the system

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about p, omega final.

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This is the initial
angular momentum.

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And so I can conclude that
omega final is m2 vi over 2

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m1 plus m2 times r.