WEBVTT

00:00:03.190 --> 00:00:06.640
Let's consider examples
of our principle

00:00:06.640 --> 00:00:10.360
that the external
torque about a point, s,

00:00:10.360 --> 00:00:16.270
causes the angular momentum
of a system to change about s.

00:00:16.270 --> 00:00:18.550
We've examined
central force problems

00:00:18.550 --> 00:00:22.300
in which we chose the point s
to be the central point in which

00:00:22.300 --> 00:00:23.700
there was no torque.

00:00:23.700 --> 00:00:26.590
Now, as examples,
let's look at a case

00:00:26.590 --> 00:00:31.450
where we have a pivoted object.

00:00:31.450 --> 00:00:36.020
So I could take an
example of an object.

00:00:36.020 --> 00:00:46.070
Let's see, this will be overhead
view of a ring of radius r.

00:00:46.070 --> 00:00:50.090
And I'm going to have
a mass coming in.

00:00:50.090 --> 00:00:56.430
I'll call this ring
mass m1, this object m2.

00:00:56.430 --> 00:00:59.390
This object is coming in
with an initial velocity,

00:00:59.390 --> 00:01:01.750
and this is my pivot point.

00:01:04.910 --> 00:01:11.380
Now, when the object
m2 is hitting the ring,

00:01:11.380 --> 00:01:15.610
we have a force, F2 on 1.

00:01:15.610 --> 00:01:17.020
This is the collision.

00:01:19.720 --> 00:01:21.420
Here's our pivot point.

00:01:21.420 --> 00:01:23.710
And when this hits
the ring, we'll

00:01:23.710 --> 00:01:26.890
have some type of pivot force.

00:01:26.890 --> 00:01:30.250
I'm just going to
denote the pivot force.

00:01:30.250 --> 00:01:32.560
We're not quite sure
what it will point at,

00:01:32.560 --> 00:01:36.820
but I'll just say F
pivot, for the moment,

00:01:36.820 --> 00:01:39.820
is holding this point in place.

00:01:39.820 --> 00:01:48.250
At the same time, the object
will have a force, F1,2,

00:01:48.250 --> 00:01:50.380
acting on the object.

00:01:50.380 --> 00:01:54.860
Here is our pivot point, p.

00:01:54.860 --> 00:01:58.050
And now, what point should we--

00:01:58.050 --> 00:02:00.960
should we choose to see if
there's no torque [? about? ?]

00:02:00.960 --> 00:02:07.560
Suppose we choose
the pivot point.

00:02:10.360 --> 00:02:13.090
Well-- p.

00:02:13.090 --> 00:02:19.900
Well, clearly, the pivot force
has no torque about the pivot,

00:02:19.900 --> 00:02:21.970
because the vector from
the pivot point to where

00:02:21.970 --> 00:02:25.450
the pivot force is acting is 0.

00:02:25.450 --> 00:02:28.630
Remember, pivot forces have no
torque about the pivot point.

00:02:28.630 --> 00:02:32.140
However, this collision
force will produce

00:02:32.140 --> 00:02:35.494
a torque about the pivot.

00:02:35.494 --> 00:02:37.660
And that's-- the angular
momentum of the ring is not

00:02:37.660 --> 00:02:41.650
constant, because the
ring will start to rotate.

00:02:41.650 --> 00:02:44.590
Similarly, this
object is reversing--

00:02:44.590 --> 00:02:46.180
will reverse
directions, or will do

00:02:46.180 --> 00:02:47.890
something due to the collision.

00:02:47.890 --> 00:02:51.700
And again, this force will
produce a torque that's equal

00:02:51.700 --> 00:02:54.820
and opposite to the
torque on the ring.

00:02:54.820 --> 00:02:57.040
This is the torque
on the particle.

00:02:57.040 --> 00:03:06.430
But if we make our system equal
to the particle and the ring--

00:03:06.430 --> 00:03:10.270
so we'll call this, our system
is now both the particle

00:03:10.270 --> 00:03:11.590
and the ring--

00:03:11.590 --> 00:03:15.760
then these torques are
internal, and because they're

00:03:15.760 --> 00:03:19.210
equal and opposite
forces, the rs vector

00:03:19.210 --> 00:03:21.579
is exactly the same rs vector.

00:03:21.579 --> 00:03:24.040
The internal torques
cancel in pairs.

00:03:24.040 --> 00:03:26.620
The pivot force
produces no torque.

00:03:26.620 --> 00:03:32.200
So the torque on the system
about the pivot is 0.

00:03:32.200 --> 00:03:35.530
And that tells us that
the angular momentum

00:03:35.530 --> 00:03:38.500
of the system about
that pivot point,

00:03:38.500 --> 00:03:42.280
initially, will be equal
to the angular momentum

00:03:42.280 --> 00:03:45.130
about that pivot point,
finally, if we take

00:03:45.130 --> 00:03:48.280
two initial and final states.

00:03:48.280 --> 00:03:54.300
In particular, suppose
our initial state is--

00:03:54.300 --> 00:03:56.730
here's the pivot.

00:03:56.730 --> 00:04:00.970
Our object is coming
in, that was m2 vi.

00:04:03.790 --> 00:04:06.394
The moment arm is r.

00:04:06.394 --> 00:04:13.570
And our vector from here
to the object, rs initial,

00:04:13.570 --> 00:04:16.329
has a moment component that way.

00:04:16.329 --> 00:04:19.420
If we put these
vectors tail to tail

00:04:19.420 --> 00:04:25.000
and figure out that the
angular momentum is pointing

00:04:25.000 --> 00:04:37.350
in this direction, l initial
i, and the moment arm is r,

00:04:37.350 --> 00:04:44.140
then the initial angular
momentum about this pivot point

00:04:44.140 --> 00:04:46.630
is just due to
this moving object.

00:04:46.630 --> 00:04:53.080
So that's m2 vi, and
the moment arm is r,

00:04:53.080 --> 00:04:56.150
and we'll denote its
direction that way.

00:04:56.150 --> 00:04:59.650
Now, the final
angular momentum--

00:04:59.650 --> 00:05:03.010
let's imagine that it sticks.

00:05:03.010 --> 00:05:09.490
So we have m2, pivot,
and now our ring

00:05:09.490 --> 00:05:12.650
is going to be rotating
with some omega final

00:05:12.650 --> 00:05:15.180
because this object hits it.

00:05:15.180 --> 00:05:19.210
And notice that this
object is a distance

00:05:19.210 --> 00:05:25.190
root 2 r from the pivot point.

00:05:25.190 --> 00:05:31.810
So the final angular momentum
can be two different pieces.

00:05:31.810 --> 00:05:34.990
You can think of this
as a system, where

00:05:34.990 --> 00:05:40.960
we have I of the system
about p times omega final,

00:05:40.960 --> 00:05:43.870
because now it's
just a rigid body.

00:05:43.870 --> 00:05:47.080
And the angular
momentum of the system

00:05:47.080 --> 00:05:49.439
is consisting of two pieces.

00:05:49.439 --> 00:05:50.980
It's the angular
momentum of the ring

00:05:50.980 --> 00:05:53.740
about the pivot,
plus the angular

00:05:53.740 --> 00:06:04.700
momentum of the particle about
the pivot, times omega final.

00:06:04.700 --> 00:06:08.360
Now, this is the center of mass.

00:06:08.360 --> 00:06:10.760
This is the pivot point, p.

00:06:10.760 --> 00:06:14.960
The angular momentum of the
ring is the angular momentum

00:06:14.960 --> 00:06:17.540
about the center of-- we'll
use the parallel axis theorem--

00:06:17.540 --> 00:06:19.820
angular momentum about
the center of mass,

00:06:19.820 --> 00:06:22.850
plus the distance from
the center of mass

00:06:22.850 --> 00:06:27.650
to the parallel axis,
which is a distance, r.

00:06:27.650 --> 00:06:34.367
So the first piece is Icm
ring plus m r squared.

00:06:34.367 --> 00:06:36.200
And the angular momentum
about the particle,

00:06:36.200 --> 00:06:39.560
this is going in a
circle of radius r2--

00:06:39.560 --> 00:06:42.380
root 2r, so when
we square that--

00:06:42.380 --> 00:06:45.560
by the way this was,
mass of the ring was 1.

00:06:45.560 --> 00:06:49.560
Mass of the particle
was m2 times r squared,

00:06:49.560 --> 00:06:52.130
which is this distance
squared, which is

00:06:52.130 --> 00:06:57.970
2 r squared times omega final.

00:06:57.970 --> 00:06:59.930
The moment of
inertia of the ring

00:06:59.930 --> 00:07:03.410
is m1 r squared about
the center of mass.

00:07:03.410 --> 00:07:04.700
We have the factor 2.

00:07:04.700 --> 00:07:06.780
We have another factor of 2.

00:07:06.780 --> 00:07:13.770
And so we get m1 plus
m2 times a factor 2 r

00:07:13.770 --> 00:07:17.300
squared omega final,
which we can call omega

00:07:17.300 --> 00:07:21.590
f, pointing out of the board.

00:07:21.590 --> 00:07:25.430
And so now, we have an
angular momentum condition,

00:07:25.430 --> 00:07:34.790
which is that m2 vi r
equals 2 m1 plus m2 r

00:07:34.790 --> 00:07:37.520
squared omega final.

00:07:37.520 --> 00:07:40.430
And that is the statement
that the initial angular

00:07:40.430 --> 00:07:41.659
momentum of our system--

00:07:41.659 --> 00:07:43.250
the ring is at rest here--

00:07:43.250 --> 00:07:46.190
is equal to the final angular
momentum of the system, where

00:07:46.190 --> 00:07:48.020
m2 is stuck to the
ring, and they're

00:07:48.020 --> 00:07:51.800
all rotating about this
pivot point with omega final.

00:07:51.800 --> 00:07:53.570
That's the angular
momentum of the system

00:07:53.570 --> 00:07:55.070
about p, omega final.

00:07:55.070 --> 00:07:56.810
This is the initial
angular momentum.

00:07:56.810 --> 00:08:04.220
And so I can conclude that
omega final is m2 vi over 2

00:08:04.220 --> 00:08:08.090
m1 plus m2 times r.