WEBVTT
00:00:03.410 --> 00:00:06.680
We've defined moment of inertia
of a rigid body already.
00:00:06.680 --> 00:00:08.540
And we often are
interested in it,
00:00:08.540 --> 00:00:11.180
because when we rotate
a body about an axis,
00:00:11.180 --> 00:00:14.310
say an axis that's passing
through the center of mass.
00:00:14.310 --> 00:00:16.790
Then our kinetic energy depends
in that moment of inertia
00:00:16.790 --> 00:00:18.230
through that axis.
00:00:18.230 --> 00:00:23.120
But we may also want to consider
rotation about another axis.
00:00:23.120 --> 00:00:25.460
So suppose we consider
an axis passing
00:00:25.460 --> 00:00:27.770
through close to the
end and we rotate
00:00:27.770 --> 00:00:30.740
the body around this axis.
00:00:30.740 --> 00:00:35.090
You can see from
overhead like that.
00:00:35.090 --> 00:00:39.140
Then we want to now consider
how the mass is distributed
00:00:39.140 --> 00:00:42.710
about an axis that was
parallel to the axis passing
00:00:42.710 --> 00:00:44.540
through the center of mass.
00:00:44.540 --> 00:00:47.090
So we'll now make a
little calculation.
00:00:47.090 --> 00:00:49.200
But first, we want
to quote the theorem.
00:00:49.200 --> 00:00:53.120
So let's just draw an
example of our rigid body.
00:00:53.120 --> 00:00:55.250
And let's take the
center of mass,
00:00:55.250 --> 00:00:57.710
and let's consider an axis
that's going perpendicular
00:00:57.710 --> 00:00:59.000
to the center of mass.
00:00:59.000 --> 00:01:03.430
We can think of the object
is rotating about that axis.
00:01:03.430 --> 00:01:05.780
And now let's consider
another axis passing
00:01:05.780 --> 00:01:09.830
through a different
point, but also parallel
00:01:09.830 --> 00:01:13.580
to this axis separated
by a distance d.
00:01:13.580 --> 00:01:18.470
Then the result that we want
is at the moment of inertia
00:01:18.470 --> 00:01:20.720
about an axis
passing perpendicular
00:01:20.720 --> 00:01:24.050
to the plane of the
object through the axis s
00:01:24.050 --> 00:01:27.680
is equal to the moment
of inertia about an axis
00:01:27.680 --> 00:01:29.390
passing through the
center of mass--
00:01:29.390 --> 00:01:32.509
notice these are parallel
axes-- plus the mass
00:01:32.509 --> 00:01:38.509
of the object times the distance
between the two parallel axes.
00:01:38.509 --> 00:01:41.420
And this is a result
that is very useful when
00:01:41.420 --> 00:01:42.979
calculating moments of inertia.
00:01:42.979 --> 00:01:45.020
Of course, we could
calculate the moment
00:01:45.020 --> 00:01:46.950
about any axis we wanted.
00:01:46.950 --> 00:01:48.500
And we'll see that in a moment.
00:01:48.500 --> 00:01:54.310
For example, we know that
for a rod of length L
00:01:54.310 --> 00:01:57.560
that the moment of inertia
through the center of mass
00:01:57.560 --> 00:01:58.759
was 1/12.
00:01:58.759 --> 00:02:04.780
And let's call the length of the
rod L. So we have a length L.
00:02:04.780 --> 00:02:08.660
And let's assume it's
a uniform object.
00:02:08.660 --> 00:02:12.050
And we'll calculate a moment
of inertia through an axis
00:02:12.050 --> 00:02:13.670
through the end.
00:02:13.670 --> 00:02:17.540
And so in this case, d
is equal to L over 2.
00:02:17.540 --> 00:02:23.210
And so I about the
end axis is 1/12 mL
00:02:23.210 --> 00:02:28.970
squared plus the mass
times L over 2 square
00:02:28.970 --> 00:02:34.340
and a 12 plus a quarter
is 1/3 mL squared.
00:02:34.340 --> 00:02:37.280
And that means that
all you need to know
00:02:37.280 --> 00:02:39.530
is the moment through
the center of mass,
00:02:39.530 --> 00:02:42.650
and you can calculate the
moment through any other axis.
00:02:42.650 --> 00:02:46.410
Very useful theorem called
the parallel axis theorem.
00:02:49.040 --> 00:02:54.200
Now as I said, we can
calculate the moment s.
00:02:54.200 --> 00:02:58.640
And just to show you very
quickly, if we pick s here,
00:02:58.640 --> 00:03:03.380
and we pick our little dm
there, and we have a distance x,
00:03:03.380 --> 00:03:06.020
then the moment
about s-- and let's
00:03:06.020 --> 00:03:09.860
say that dm is equal to
the total mass divided
00:03:09.860 --> 00:03:14.060
by the length times some
little distance dx--
00:03:14.060 --> 00:03:20.450
then the moment is dm times
x squared, where x we'll
00:03:20.450 --> 00:03:21.980
give this an
integration variable
00:03:21.980 --> 00:03:26.000
x prime goes from 0
to x prime equals l.
00:03:26.000 --> 00:03:31.200
And so we have for our
dm m over L dx prime.
00:03:31.200 --> 00:03:33.829
Remember, that's our
integration variable.
00:03:33.829 --> 00:03:36.000
And we have x squared.
00:03:36.000 --> 00:03:39.110
So that's x prime squared.
00:03:39.110 --> 00:03:43.100
And this is just the
integral of x cubed over 3.
00:03:43.100 --> 00:03:50.690
So we have 1/3 m over L x prime
cubed evaluated from 0 to L,
00:03:50.690 --> 00:03:55.340
and that comes out
to 1/3 mL squared.
00:03:55.340 --> 00:03:58.910
And that's in agreement with
the parallel axis theorem.
00:03:58.910 --> 00:04:01.010
And that's an example of
parallel axis theorems.
00:04:01.010 --> 00:04:06.477
You can do the same thing with
the disk or other objects.