WEBVTT
00:00:03.360 --> 00:00:05.760
Let's consider the
motion of a wheel that's
00:00:05.760 --> 00:00:10.320
rolling along the ground with
some center of mass velocity
00:00:10.320 --> 00:00:11.820
vcm.
00:00:11.820 --> 00:00:13.830
And because the
wheel is rotating
00:00:13.830 --> 00:00:15.970
it has an angular velocity.
00:00:15.970 --> 00:00:18.960
And you can see that
that vector is directed
00:00:18.960 --> 00:00:22.770
into the plane of the board.
00:00:22.770 --> 00:00:25.860
Now, what we'd like to do is
consider the kinetic energy
00:00:25.860 --> 00:00:27.660
of this continuous body.
00:00:27.660 --> 00:00:33.210
A little bit later on, that
body has moved some distance.
00:00:33.210 --> 00:00:35.880
And what we want to
consider is the fact
00:00:35.880 --> 00:00:38.430
that not only is every
point in the body moving
00:00:38.430 --> 00:00:40.650
with the center of
mass speed, but there's
00:00:40.650 --> 00:00:44.820
this additional
rotational energy
00:00:44.820 --> 00:00:47.190
that's associated with
the fact that every point
00:00:47.190 --> 00:00:48.840
in the center of
mass reference frame
00:00:48.840 --> 00:00:50.650
is undergoing circular motion.
00:00:50.650 --> 00:00:52.900
So how do we describe that?
00:00:52.900 --> 00:00:57.190
Well, we'll do that by choosing
some point in the body.
00:00:57.190 --> 00:00:58.900
So let's pick a point.
00:00:58.900 --> 00:01:03.090
We'll call that
mj, with mass mj.
00:01:03.090 --> 00:01:08.910
And the velocity of this point,
remember, has two components.
00:01:08.910 --> 00:01:10.860
To simplify it,
we'll give ourselves
00:01:10.860 --> 00:01:14.820
a little more picture here.
00:01:14.820 --> 00:01:20.250
Every single point in
the object has the vcm.
00:01:20.250 --> 00:01:24.420
But because this object is
undergoing circular motion,
00:01:24.420 --> 00:01:28.020
there is vcmj.
00:01:28.020 --> 00:01:32.320
That's the rotational
circular tangential velocity.
00:01:32.320 --> 00:01:34.950
And so the vector
sum of these two
00:01:34.950 --> 00:01:40.770
is the actual velocity
vj of the j-th object.
00:01:40.770 --> 00:01:45.090
vj is equal to the
center of mass velocity
00:01:45.090 --> 00:01:49.289
plus the tangential rotational
velocity that it has,
00:01:49.289 --> 00:01:53.220
because it's undergoing
circular motion.
00:01:53.220 --> 00:01:56.280
And now what we'd like to do
is calculate the kinetic energy
00:01:56.280 --> 00:01:57.479
of this object.
00:01:57.479 --> 00:02:02.880
Well, the kinetic energy
is the sum j from 1 to n
00:02:02.880 --> 00:02:10.259
of 1/2 mj times the velocity
of this j-th particle squared,
00:02:10.259 --> 00:02:12.910
which we can take
as a dot product.
00:02:12.910 --> 00:02:25.353
So we can write that as vcm
plus vcmj dot vcm plus vcmj.
00:02:27.900 --> 00:02:31.870
And that's just vj squared.
00:02:31.870 --> 00:02:35.430
So when we look at these terms,
it looks complicated at first.
00:02:35.430 --> 00:02:39.480
But there's some nice-- there's
going to be vcm dot vcm.
00:02:39.480 --> 00:02:41.050
There's two cross terms.
00:02:41.050 --> 00:02:42.490
They're identical.
00:02:42.490 --> 00:02:45.030
And vcmj dot vcmj.
00:02:45.030 --> 00:02:48.690
So let's write out
those three terms.
00:02:48.690 --> 00:02:50.970
We have 1/2 mj.
00:02:50.970 --> 00:02:56.070
vcm dot vcm is vcm squared.
00:02:56.070 --> 00:03:00.840
Now, every point in the
object has the same vcm.
00:03:00.840 --> 00:03:05.300
So we can pull that
one out of the sum.
00:03:05.300 --> 00:03:08.520
And now we'll take
these cross terms.
00:03:08.520 --> 00:03:12.780
So we have the sum
over j from 1 to n.
00:03:12.780 --> 00:03:14.940
There's two cross terms.
00:03:14.940 --> 00:03:17.100
So the 2's are going to cancel.
00:03:17.100 --> 00:03:20.430
And inside here,
we have to remember
00:03:20.430 --> 00:03:22.230
to keep our mass element.
00:03:22.230 --> 00:03:23.460
That's important.
00:03:23.460 --> 00:03:29.579
Now, I'm going to write
it as mj vcmj vector.
00:03:29.579 --> 00:03:33.060
Now, remember, when
you dot with vcm,
00:03:33.060 --> 00:03:36.120
every single point
has the same vcm.
00:03:36.120 --> 00:03:38.520
Every j-th element
has the same vm,
00:03:38.520 --> 00:03:42.570
so I can pull that vcm outside.
00:03:42.570 --> 00:03:44.650
And finally, I
have the last term,
00:03:44.650 --> 00:03:54.660
which is the sum over j from
1 to n of 1/2 mj vcmj squared.
00:03:54.660 --> 00:03:57.480
And that's just the dot
product of those two terms.
00:03:57.480 --> 00:04:00.810
And so our kinetic energy
looks rather complicated,
00:04:00.810 --> 00:04:04.380
but let's focus on
this term right here.
00:04:04.380 --> 00:04:07.800
Because recall from our
video on the center of mass
00:04:07.800 --> 00:04:13.000
that the definition of the
center of mass reference frame,
00:04:13.000 --> 00:04:17.040
so if you're moving
in the center of mass,
00:04:17.040 --> 00:04:22.960
that in the center of mass
reference frame, the sum of mj
00:04:22.960 --> 00:04:30.020
vcmj is equal to 0.
00:04:30.020 --> 00:04:34.190
So for instance, if you're
in the center of mass frame,
00:04:34.190 --> 00:04:35.870
you're moving with vcm.
00:04:35.870 --> 00:04:38.090
The only velocity is this.
00:04:38.090 --> 00:04:42.380
And in that frame, the
sum of mj vcmj is 0.
00:04:42.380 --> 00:04:45.180
And we did a video
on that one before.
00:04:45.180 --> 00:04:47.250
And that's exactly
what's in this term.
00:04:47.250 --> 00:04:50.420
So this term is 0.
00:04:50.420 --> 00:04:54.170
So this, remember, was
how we defined the center
00:04:54.170 --> 00:04:55.420
of mass reference frame.
00:05:00.340 --> 00:05:04.360
And therefore, our kinetic
energy consists of two pieces.
00:05:04.360 --> 00:05:11.650
This first piece is just 1/2 the
total mass times vcm squared.
00:05:11.650 --> 00:05:16.430
And our second piece over here,
we'll just write it out now--
00:05:16.430 --> 00:05:24.350
1/2 sum over j mj vcmj squared.
00:05:24.350 --> 00:05:28.070
Now, if you are moving
with the center of mass,
00:05:28.070 --> 00:05:32.690
then this j-th object is just
undergoing circular motion.
00:05:32.690 --> 00:05:36.640
And so we have our
result that we've
00:05:36.640 --> 00:05:40.630
used many times is
that the velocity,
00:05:40.630 --> 00:05:43.300
the tangential
rotational velocity,
00:05:43.300 --> 00:05:46.380
is just equal to
the radius rsj--
00:05:46.380 --> 00:05:51.570
so let's introduce that rsj--
00:05:51.570 --> 00:05:55.485
times the angular speed omega.
00:05:58.470 --> 00:06:00.630
And when we put
that into this term,
00:06:00.630 --> 00:06:04.450
we see our kinetic
energy has two pieces--
00:06:04.450 --> 00:06:12.870
m total v center of mass squared
plus 1/2 j goes from 1 to n--
00:06:12.870 --> 00:06:15.340
I didn't finish that sum there--
00:06:15.340 --> 00:06:19.480
mj rsj squared.
00:06:19.480 --> 00:06:24.900
Now, just remember that every
single point in the object
00:06:24.900 --> 00:06:30.300
has the same angular speed,
and so we can pull out
00:06:30.300 --> 00:06:35.550
the omega squared in there.
00:06:35.550 --> 00:06:41.760
And because this is a continuous
body and we take the limit,
00:06:41.760 --> 00:06:45.480
as we've done before,
as mj goes to 0,
00:06:45.480 --> 00:06:48.390
this quantity of
mass times distance
00:06:48.390 --> 00:06:50.760
squared is just the
moment of inertia
00:06:50.760 --> 00:06:53.250
about the center of
mass of that body.
00:06:53.250 --> 00:06:57.909
And in conclusion, K
is 1/2 m total v center
00:06:57.909 --> 00:07:01.920
of mass squared plus 1/2
the moment of inertia
00:07:01.920 --> 00:07:09.490
about the center of mass times
the angular speed squared.
00:07:09.490 --> 00:07:12.240
Now, this is the same
crucial decomposition
00:07:12.240 --> 00:07:14.190
that we've talked
about many times.
00:07:14.190 --> 00:07:18.450
This first piece is what we
call the translational kinetic
00:07:18.450 --> 00:07:20.940
energy, because it
just represents how
00:07:20.940 --> 00:07:23.310
the center of mass is moving.
00:07:23.310 --> 00:07:25.980
And the second piece
is what we call
00:07:25.980 --> 00:07:28.860
the rotational kinetic
energy, because it's
00:07:28.860 --> 00:07:33.480
a representation of just the
kinetic energy of rotation.
00:07:33.480 --> 00:07:36.090
For example, if you were in
the center of mass frame,
00:07:36.090 --> 00:07:37.830
there would be no
translational energy,
00:07:37.830 --> 00:07:41.132
and this would be the
only kinetic energy.