WEBVTT
00:00:00.070 --> 00:00:02.430
The following content is
provided under a Creative
00:00:02.430 --> 00:00:03.810
Commons license.
00:00:03.810 --> 00:00:06.050
Your support will help
MIT OpenCourseWare
00:00:06.050 --> 00:00:10.140
continue to offer high quality
educational resources for free.
00:00:10.140 --> 00:00:12.690
To make a donation or to
view additional materials
00:00:12.690 --> 00:00:16.600
from hundreds of MIT courses,
visit MIT OpenCourseWare
00:00:16.600 --> 00:00:17.258
at ocw.mit.edu.
00:00:27.130 --> 00:00:28.630
PROFESSOR: So today
I'm mostly going
00:00:28.630 --> 00:00:31.780
to talk about two
formulas that we are
00:00:31.780 --> 00:00:34.870
going to make a lot of use of.
00:00:34.870 --> 00:00:37.740
One is the time derivative
of linear momentum,
00:00:37.740 --> 00:00:40.890
which will be used a lot,
has got to be equal to,
00:00:40.890 --> 00:00:41.716
for a particle.
00:00:45.840 --> 00:00:49.160
We'll just do particles today
and then rigid bodies soon.
00:00:49.160 --> 00:00:52.546
So for a particle, this
is just the mass times
00:00:52.546 --> 00:00:53.920
the acceleration
of the particle.
00:00:56.600 --> 00:00:59.830
The rigid body, it's the
total mass times acceleration
00:00:59.830 --> 00:01:00.930
in the center of gravity.
00:01:00.930 --> 00:01:05.650
And the other formula that we've
learned, that we've come up to,
00:01:05.650 --> 00:01:14.770
is the sum of the external
torques with respect to a point
00:01:14.770 --> 00:01:22.770
A on a particle is the time
derivative of the angular
00:01:22.770 --> 00:01:29.670
momentum of the particle
at B with respect to A dt.
00:01:29.670 --> 00:01:34.160
And then it's got this nuisance
term, this velocity of A
00:01:34.160 --> 00:01:40.820
with respect to O cross the
linear momentum at the point B
00:01:40.820 --> 00:01:44.950
with respect to O. So we
talked about this last time.
00:01:44.950 --> 00:01:51.880
So this is a particle
that's located out here
00:01:51.880 --> 00:01:59.410
at B. There's some
intermediate point A. And we're
00:01:59.410 --> 00:02:03.630
computing the angular
momentum of this particle, m,
00:02:03.630 --> 00:02:06.880
with respect to A.
And A can be moving.
00:02:06.880 --> 00:02:09.280
And if it is, when you
go to compute torques,
00:02:09.280 --> 00:02:11.820
you have to use this
kind of messy formula.
00:02:11.820 --> 00:02:15.220
We often try to
simplify the problems
00:02:15.220 --> 00:02:19.450
that we do so that we
can either make A a fixed
00:02:19.450 --> 00:02:21.860
axis of rotation--
make it rotate
00:02:21.860 --> 00:02:25.480
about this point, in which
case this goes to zero--
00:02:25.480 --> 00:02:29.100
or there's another
time when you can make
00:02:29.100 --> 00:02:32.940
the velocity of
this point parallel
00:02:32.940 --> 00:02:34.420
to the velocity of that point.
00:02:34.420 --> 00:02:37.560
And the most common
one of all is
00:02:37.560 --> 00:02:43.020
when you use this formula
about the center of mass.
00:02:43.020 --> 00:02:46.240
Then the velocity of
the center of mass
00:02:46.240 --> 00:02:49.037
is the same direction
as the momentum.
00:02:49.037 --> 00:02:50.620
And therefore, this
term goes to zero.
00:02:50.620 --> 00:02:54.890
So we have these two cases
where this formula simplifies.
00:02:54.890 --> 00:02:57.410
So the way we most
often use it is
00:02:57.410 --> 00:03:00.920
the summation of the
torques with respect to A
00:03:00.920 --> 00:03:07.370
is just the derivative of
the angular momentum dt.
00:03:07.370 --> 00:03:08.250
And that's it.
00:03:08.250 --> 00:03:26.900
This is when vA is zero, or
vA is parallel to P, such
00:03:26.900 --> 00:03:36.560
as A is at the center of mass.
00:03:36.560 --> 00:03:39.500
So these are the two
formulas that we want to use.
00:03:56.580 --> 00:03:58.480
Let's just do a real
simple case first.
00:04:31.130 --> 00:04:38.070
And this is just my mass
on a string spinning
00:04:38.070 --> 00:04:39.640
around, constant speed.
00:04:44.080 --> 00:04:47.199
I'm going to pretend I'm
on a frictionless table
00:04:47.199 --> 00:04:48.740
so I don't have to
deal with gravity.
00:04:48.740 --> 00:04:51.420
This thing is horizontal,
just spinning around.
00:04:51.420 --> 00:04:56.636
And I'll call this point A. This
is my point O. This is x, z.
00:04:59.920 --> 00:05:01.900
No, y.
00:05:01.900 --> 00:05:04.710
It's looking down on it.
00:05:04.710 --> 00:05:09.310
So it's going to have a velocity
in this theta hat direction.
00:05:09.310 --> 00:05:13.977
And you've got your R hat.
00:05:13.977 --> 00:05:15.810
We'll just use polar
coordinates to do this.
00:05:15.810 --> 00:05:19.940
So the velocity of A
with respect to O we
00:05:19.940 --> 00:05:25.310
know-- and we'll give this
just a length magnitude r1.
00:05:25.310 --> 00:05:30.923
So this is going to be
theta dot equals a constant.
00:05:34.880 --> 00:05:36.910
r dot equals 0.
00:05:36.910 --> 00:05:38.690
So it's just fixed length.
00:05:38.690 --> 00:05:40.710
Fixed length is r1.
00:05:40.710 --> 00:05:43.090
So we know the simple
formula for the velocity
00:05:43.090 --> 00:05:50.350
of that is r1 theta dot in
the theta hat direction.
00:05:50.350 --> 00:05:53.910
[INAUDIBLE] hat, excuse me.
00:05:53.910 --> 00:06:00.180
And the momentum then is
just the mass times that.
00:06:04.300 --> 00:06:13.365
And the time derivative
of the linear momentum
00:06:13.365 --> 00:06:16.420
with respect to this
fixed reference frame
00:06:16.420 --> 00:06:18.385
is the time derivative of this.
00:06:18.385 --> 00:06:19.430
This is constant.
00:06:19.430 --> 00:06:20.380
This is constant.
00:06:20.380 --> 00:06:21.790
This is constant.
00:06:21.790 --> 00:06:25.050
The only thing you have to take
is the derivative of theta hat.
00:06:25.050 --> 00:06:34.989
So this is mr1 theta dot
theta hat time derivative.
00:06:34.989 --> 00:06:37.030
But we've worked that one
out two or three times.
00:06:37.030 --> 00:06:45.600
It's theta dot--
minus theta dot r hat.
00:06:56.030 --> 00:06:57.170
We'll put that in.
00:06:57.170 --> 00:07:06.290
You get minus mr1 theta
dot squared r hat.
00:07:06.290 --> 00:07:11.240
And this must be the sum
of the external forces.
00:07:11.240 --> 00:07:13.140
And if we're,
looking down on this,
00:07:13.140 --> 00:07:20.690
draw a free body diagram of
our mass, looking down on it,
00:07:20.690 --> 00:07:25.110
here's r hat.
00:07:25.110 --> 00:07:28.780
The theta direction like this,
z coming out of the board.
00:07:28.780 --> 00:07:33.950
So this is x, y, and o.
00:07:33.950 --> 00:07:37.540
There is a inward
force on it that we
00:07:37.540 --> 00:07:39.830
call the tension in the string.
00:07:39.830 --> 00:07:45.830
And some of external force
is just minus Tr hat.
00:07:45.830 --> 00:07:50.540
So T is mr1 theta
dot squared, which
00:07:50.540 --> 00:07:53.460
is what we talked about
just a second ago.
00:07:53.460 --> 00:07:56.320
So this is just a
demonstration of what
00:07:56.320 --> 00:07:59.050
we were talking about
with the survey questions,
00:07:59.050 --> 00:08:02.030
that when you spin
something, you're
00:08:02.030 --> 00:08:05.760
changing its linear momentum.
00:08:05.760 --> 00:08:07.570
Not in magnitude.
00:08:07.570 --> 00:08:08.900
It stays constant.
00:08:08.900 --> 00:08:12.080
But it's constantly
changing in direction.
00:08:12.080 --> 00:08:15.850
And the direction
change-- you have
00:08:15.850 --> 00:08:20.910
to take this derivative--
gives you the [INAUDIBLE].
00:08:20.910 --> 00:08:26.090
Direction changes causes the
centripetal acceleration.
00:08:26.090 --> 00:08:29.710
To cause that acceleration,
you have to put a force on it.
00:08:29.710 --> 00:08:33.059
And the force is that inward
tension in the string.
00:08:33.059 --> 00:08:37.610
OK, so that's the first
really simple one.
00:08:37.610 --> 00:08:41.479
Now, let's take a [INAUDIBLE].
00:09:01.580 --> 00:09:03.710
Now let's do the same
problem, but let it
00:09:03.710 --> 00:09:05.440
be a little bit more general.
00:09:05.440 --> 00:09:07.900
I've got a tabletop.
00:09:07.900 --> 00:09:10.720
This problem's described
in the book as an example.
00:09:10.720 --> 00:09:12.175
You've got a hole in the table.
00:09:15.176 --> 00:09:17.380
You've got your mass out here.
00:09:17.380 --> 00:09:19.010
So this is looking
down on the table
00:09:19.010 --> 00:09:21.070
and down through
the hole, coming out
00:09:21.070 --> 00:09:23.610
the bottom of the hole,
you have this string.
00:09:23.610 --> 00:09:25.070
And you pull on it.
00:09:31.000 --> 00:09:37.280
So r dot-- so we're
looking down on it.
00:09:37.280 --> 00:09:40.565
This thing, again,
has some theta.
00:09:44.060 --> 00:09:48.755
r hat directions like this,
theta hat directions like that.
00:09:48.755 --> 00:09:55.855
This thing's spinning, going
around and around my table top.
00:09:55.855 --> 00:09:57.820
But there's a hole
in the tabletop,
00:09:57.820 --> 00:09:59.880
and I can pull the string
and shorten the string.
00:10:02.820 --> 00:10:04.040
So r dot's a constant.
00:10:08.700 --> 00:10:11.548
Theta dot will
not be a constant.
00:10:15.030 --> 00:10:16.160
it will change.
00:10:22.743 --> 00:10:24.390
This is my vector r.
00:10:27.610 --> 00:10:30.745
So the velocity, we'll call
this A. This is my origin,
00:10:30.745 --> 00:10:35.790
O. The velocity in this case
of the particle with respect
00:10:35.790 --> 00:10:47.940
to the fixed inertial frame
is r dot r hat plus r theta
00:10:47.940 --> 00:10:50.005
dot theta hat.
00:10:50.005 --> 00:10:53.620
So now it has two
possible velocities,
00:10:53.620 --> 00:10:55.319
and I'm going to
be pulling it in,
00:10:55.319 --> 00:10:57.110
and it's going to be
going round and round.
00:11:10.110 --> 00:11:41.040
So the linear momentum is
just mvA with respect to O.
00:11:41.040 --> 00:11:43.930
So with this problem,
if I take the time
00:11:43.930 --> 00:11:48.920
derivative of the linear
momentum, you recognize this.
00:11:48.920 --> 00:11:52.020
This is just velocity
in polar coordinates.
00:11:52.020 --> 00:11:55.930
We've done this derivative
before to get the acceleration.
00:11:55.930 --> 00:11:58.650
So the acceleration, it's
the time derivative of A
00:11:58.650 --> 00:12:06.160
with respect to
O is acceleration
00:12:06.160 --> 00:12:11.620
of A with respect to O.
And that's a messy formula,
00:12:11.620 --> 00:12:14.270
and we've derived it before.
00:12:14.270 --> 00:12:22.870
So that looks like--
and it's got four terms.
00:12:22.870 --> 00:12:29.520
r double dot minus
r theta dot squared
00:12:29.520 --> 00:12:39.400
and the r hat plus r theta
double dot plus 2r dot theta
00:12:39.400 --> 00:12:43.730
dot in the theta hat direction.
00:12:43.730 --> 00:12:47.740
And if multiplied
by the mass, that's
00:12:47.740 --> 00:12:49.830
the mass times the acceleration.
00:12:49.830 --> 00:12:54.830
And this would be equal to the
sum of all the external forces
00:12:54.830 --> 00:12:56.470
acting on that particle.
00:12:56.470 --> 00:12:57.720
That's Newton's second law.
00:13:02.740 --> 00:13:07.671
But in our problem here,
I've said this is constant.
00:13:07.671 --> 00:13:10.095
So I can throw out this term
that's going to be zero.
00:13:13.360 --> 00:13:14.415
So this term goes away.
00:13:16.935 --> 00:13:20.760
But this term's
certainly not zero.
00:13:20.760 --> 00:13:24.070
This term is not
necessarily zero.
00:13:24.070 --> 00:13:27.820
This term is certainly not zero.
00:13:27.820 --> 00:13:31.300
So what can we do with that?
00:13:31.300 --> 00:13:34.450
So now's the time you draw
some free body diagrams.
00:13:34.450 --> 00:13:37.500
And let's look at the side view.
00:13:41.200 --> 00:13:44.930
The side view,
here's your particle,
00:13:44.930 --> 00:13:52.460
mg down, some normal force
up, and a tension pulling in.
00:13:55.280 --> 00:13:58.490
So the table's supporting
it, gravity down, and just
00:13:58.490 --> 00:14:01.390
the string force pulling it.
00:14:01.390 --> 00:14:08.192
In the top view, you're
looking down on it.
00:14:08.192 --> 00:14:09.650
I'm not allowing
them to-- assuming
00:14:09.650 --> 00:14:12.500
it's a frictionless table,
so there's no friction.
00:14:12.500 --> 00:14:16.090
So the only force seen from
the top is the tension.
00:14:16.090 --> 00:14:24.220
There are no forces in the plane
in the direction theta hat.
00:14:24.220 --> 00:14:27.200
So the sum of the external
forces in the theta hat
00:14:27.200 --> 00:14:30.810
direction are in this direction.
00:14:30.810 --> 00:14:39.770
This is-- so the total forces on
this thing from your free body
00:14:39.770 --> 00:14:41.755
diagram in the theta
hat direction are?
00:14:41.755 --> 00:14:42.570
AUDIENCE: Zero.
00:14:42.570 --> 00:14:43.410
PROFESSOR: Zero.
00:14:43.410 --> 00:14:47.180
And that allows us to take this
term and set it equal to 0.
00:14:57.860 --> 00:14:59.165
This is the x.
00:15:03.040 --> 00:15:07.370
So the theta hat
piece is equal to 0
00:15:07.370 --> 00:15:17.610
is equal to m r theta double
dot plus 2r dot theta dot.
00:15:17.610 --> 00:15:24.180
[INAUDIBLE] solve this
for-- that's equal to 0.
00:15:24.180 --> 00:15:31.410
So r theta double dot equals
minus 2r dot theta dot.
00:15:34.910 --> 00:15:41.470
So in order for this thing
to satisfy Newton's law,
00:15:41.470 --> 00:15:47.150
it happens to be that
as you pull it in,
00:15:47.150 --> 00:15:50.480
the product of r
dot and theta dot
00:15:50.480 --> 00:15:53.230
gives you the
angular acceleration.
00:15:58.340 --> 00:16:00.830
And the other term, we
just went through that.
00:16:00.830 --> 00:16:03.260
So what we did a
couple minutes ago.
00:16:03.260 --> 00:16:08.790
The force in the
r hat direction,
00:16:08.790 --> 00:16:12.020
sum of the forces--
this is a derivative
00:16:12.020 --> 00:16:18.810
of the linear momentum--
is minus mr theta dot
00:16:18.810 --> 00:16:21.450
squared r hat.
00:16:21.450 --> 00:16:24.040
And we know from the
free body diagram
00:16:24.040 --> 00:16:29.680
that that better be equal to
minus T in the r hat direction.
00:16:29.680 --> 00:16:31.140
So just like
before, it tells you
00:16:31.140 --> 00:16:34.900
that T equals mr
theta dot squared.
00:16:34.900 --> 00:16:38.210
So the mass times the
centripetal acceleration,
00:16:38.210 --> 00:16:41.340
in order to make that
centripetal motion happen,
00:16:41.340 --> 00:16:48.190
you have to pull on it with
a force mr theta dot squared.
00:16:48.190 --> 00:16:56.933
So what can you say about
the angular momentum
00:16:56.933 --> 00:17:01.230
of this particle
with respect to O?
00:17:08.960 --> 00:17:11.700
So let's write it out.
00:17:11.700 --> 00:17:17.520
So H of the particle
with respect to O
00:17:17.520 --> 00:17:45.810
is rAO cross P. This
is r r hat cross P.
00:17:45.810 --> 00:17:49.760
So r cross r, you get
nothing from that term.
00:17:49.760 --> 00:17:53.770
r hat cross theta hat gives
you a k, a positive k.
00:17:53.770 --> 00:18:15.096
So you get r r squared theta
dot in the k direction.
00:18:30.950 --> 00:18:32.180
And I'm missing something.
00:18:36.120 --> 00:18:37.510
There we go.
00:18:37.510 --> 00:18:41.000
m r squared theta dot.
00:18:41.000 --> 00:18:44.290
mr theta dot is the
linear momentum times
00:18:44.290 --> 00:18:46.200
another r gives you
the angular momentum,
00:18:46.200 --> 00:18:47.950
m r squared theta dot.
00:18:47.950 --> 00:18:51.830
And because r hat
cross theta dot is k,
00:18:51.830 --> 00:18:56.450
this angular momentum
is directed upward
00:18:56.450 --> 00:18:59.840
about the center of rotation.
00:18:59.840 --> 00:19:02.730
So that's our expression
for our angular momentum.
00:19:02.730 --> 00:19:11.780
But if you take the
time derivative of it,
00:19:11.780 --> 00:19:12.745
what should it tell us?
00:19:12.745 --> 00:19:17.450
Let's go back to our
formulas we started with.
00:19:17.450 --> 00:19:21.340
When the center of
rotation's not moving,
00:19:21.340 --> 00:19:24.580
when the point with respect to
which you're taking the angular
00:19:24.580 --> 00:19:27.740
momentum doesn't
move, then that's
00:19:27.740 --> 00:19:31.520
one of the cases where you can
get rid of those extra terms.
00:19:31.520 --> 00:19:35.120
So in this case, we're
computing the angular momentum
00:19:35.120 --> 00:19:37.330
with respect to O,
which doesn't move.
00:19:37.330 --> 00:19:41.560
The velocity of O here is zero.
00:19:41.560 --> 00:19:46.560
So this allows us just to say
that the sum of the torques
00:19:46.560 --> 00:19:50.090
with respect about O is
just equal to the time
00:19:50.090 --> 00:19:53.820
rate of change of H
with respect to O.
00:19:53.820 --> 00:20:03.960
And to do that, this
term can change,
00:20:03.960 --> 00:20:05.310
and this term can change.
00:20:05.310 --> 00:20:07.790
But how about the
derivative of k?
00:20:07.790 --> 00:20:08.950
It's just constant, right?
00:20:08.950 --> 00:20:11.270
So we're going to get
two terms out of this.
00:20:11.270 --> 00:20:32.170
We're going to get m2r
r dot theta dot k plus r
00:20:32.170 --> 00:20:36.115
squared theta double dot k.
00:20:44.210 --> 00:20:49.134
What are the external
torques in this problem?
00:20:49.134 --> 00:20:50.800
You're going to have
to pretend that I'm
00:20:50.800 --> 00:20:53.710
on a frictionless
table top here.
00:20:53.710 --> 00:20:57.440
When I'm going around
like this, what
00:20:57.440 --> 00:21:02.877
are the torques about
the center point?
00:21:02.877 --> 00:21:04.710
So what are the-- first,
what are the forces
00:21:04.710 --> 00:21:07.590
acting on the mass?
00:21:07.590 --> 00:21:09.040
Just the tension.
00:21:09.040 --> 00:21:12.190
And the tension
cross the moment arm,
00:21:12.190 --> 00:21:13.940
the tension's in
the r hat direction.
00:21:13.940 --> 00:21:16.020
The string is in
the r direction.
00:21:16.020 --> 00:21:17.720
r hat cross r hat is--
00:21:17.720 --> 00:21:18.370
AUDIENCE: Zero.
00:21:18.370 --> 00:21:19.036
PROFESSOR: Zero.
00:21:19.036 --> 00:21:20.870
So there's no torques.
00:21:20.870 --> 00:21:25.770
So for this problem,
this is equal to 0.
00:21:28.500 --> 00:22:06.270
And that then allows us to
write-- the m cancels out,
00:22:06.270 --> 00:22:08.560
obviously, and I
can get rid of-- one
00:22:08.560 --> 00:22:10.350
of these r's goes away.
00:22:10.350 --> 00:22:17.070
And I'm left with
an r dot theta dot.
00:22:17.070 --> 00:22:18.740
And I move this
to the other side
00:22:18.740 --> 00:22:21.300
equal to minus r
theta double dot.
00:22:21.300 --> 00:22:23.700
And that's what we came
up with a minute ago
00:22:23.700 --> 00:22:28.624
when when we did
the time derivative
00:22:28.624 --> 00:22:30.290
of the linear momentum,
we learned this.
00:22:30.290 --> 00:22:31.920
So we haven't learned much more.
00:22:31.920 --> 00:22:34.200
It's just telling us
that, well, this thing's
00:22:34.200 --> 00:22:35.630
going to accelerate.
00:22:35.630 --> 00:22:37.650
It should pull it in.
00:22:37.650 --> 00:22:40.690
And it'll accelerate
at any instant
00:22:40.690 --> 00:22:43.250
in time, whatever r dot
is, theta dot is, this'll
00:22:43.250 --> 00:22:46.465
be the angular acceleration.
00:22:51.030 --> 00:23:02.310
So what will happen then?
00:23:02.310 --> 00:23:05.275
And let's do it at
two points in time.
00:23:21.670 --> 00:23:26.704
And we'll let r2
equals r1 divided by 2.
00:23:26.704 --> 00:23:28.370
So I'm just going to
pull this thing in.
00:23:34.580 --> 00:23:35.705
So let's do the experiment.
00:23:38.800 --> 00:23:41.770
I'm going to pull it in
about half its length.
00:23:41.770 --> 00:23:44.765
It can speed up, slow
down, stay the same speed.
00:23:49.460 --> 00:23:56.402
I'll get it going and then--
and I'll try not to hit you.
00:23:56.402 --> 00:23:59.900
I'll move over here so
I hit him instead, OK?
00:23:59.900 --> 00:24:03.571
Let's try it again.
00:24:03.571 --> 00:24:04.070
All right.
00:24:04.070 --> 00:24:06.500
But there's no torques on it.
00:24:06.500 --> 00:24:09.130
There's no torque being applied.
00:24:09.130 --> 00:24:12.820
The angular momentum
is constant,
00:24:12.820 --> 00:24:14.530
and yet the thing speeds up.
00:24:14.530 --> 00:24:18.290
So I want to ask you a question.
00:24:18.290 --> 00:24:21.760
Do you think the kinetic
energy is staying the same
00:24:21.760 --> 00:24:22.840
or changing?
00:24:22.840 --> 00:24:28.740
So how many think the kinetic
energy as I go from out there
00:24:28.740 --> 00:24:32.500
to in here, the kinetic
energy stays the same?
00:24:32.500 --> 00:24:33.120
OK.
00:24:33.120 --> 00:24:35.820
How many think it's different?
00:24:35.820 --> 00:24:36.320
All right.
00:24:36.320 --> 00:24:38.655
How many are not so sure?
00:24:38.655 --> 00:24:39.280
Let's find out.
00:24:42.950 --> 00:24:47.030
So we've determined
that if this is zero,
00:24:47.030 --> 00:24:52.943
then that means that h with
respect to O is a constant.
00:24:56.950 --> 00:25:06.314
So h-- this is at T1-- is r1.
00:25:10.100 --> 00:25:12.970
Be faster if I look at my notes.
00:25:12.970 --> 00:25:23.240
So the angular momentum
at r1 is mr1 squared
00:25:23.240 --> 00:25:27.420
theta 1 dot in the k direction.
00:25:27.420 --> 00:25:33.580
And that had better
be equal to h at r2.
00:25:36.990 --> 00:25:46.010
And that'll be mr2
squared theta 2 dot.
00:25:46.010 --> 00:25:48.760
And that's also in
the k direction.
00:25:48.760 --> 00:25:54.350
But we know that r2
is r1 divided by 2.
00:25:54.350 --> 00:25:55.735
So we can plug that in.
00:26:07.686 --> 00:26:09.890
I'll bring this over here.
00:26:09.890 --> 00:26:20.720
So mr1 squared theta 1
dot is mr1/4 theta 2 dot.
00:26:20.720 --> 00:26:22.230
So I can solve for theta 2.
00:26:31.780 --> 00:26:35.265
So if I shorten the
length by a factor of 2,
00:26:35.265 --> 00:26:37.770
the angular velocity
goes up by a factor of 4.
00:26:52.590 --> 00:26:54.230
And let's check
the kinetic energy.
00:26:54.230 --> 00:27:00.915
Kinetic energy state
1, 1/2mv1 squared.
00:27:04.170 --> 00:27:12.786
That's 1/2mv1 is r theta
1 dot quantity squared.
00:27:32.204 --> 00:27:34.080
So what do I have?
00:27:34.080 --> 00:27:38.000
I have some
expressions for this.
00:27:38.000 --> 00:27:51.150
r2 is r1/2, and theta 2 is 4
theta 1 dot quantity squared.
00:27:51.150 --> 00:28:27.440
And if you multiply
that out-- yeah.
00:28:27.440 --> 00:28:33.740
So where's the kinetic
energy come from?
00:28:33.740 --> 00:28:35.700
AUDIENCE: You're adding
energy into the system
00:28:35.700 --> 00:28:37.087
by pulling down.
00:28:37.087 --> 00:28:39.170
PROFESSOR: So she says we
add energy to the system
00:28:39.170 --> 00:28:39.850
by pulling down.
00:28:39.850 --> 00:28:41.530
So we're doing some work, right?
00:28:41.530 --> 00:28:46.692
There's tension in that string
equal to mr omega squared.
00:28:46.692 --> 00:28:49.430
If you pull it down a
certain distance-- in fact,
00:28:49.430 --> 00:28:52.560
r1/2-- you're going
to do work that's
00:28:52.560 --> 00:28:54.960
the integral of
the tension times
00:28:54.960 --> 00:28:57.350
dr. You integrate it, right?
00:28:57.350 --> 00:29:01.280
And that work goes into--
there's conservation
00:29:01.280 --> 00:29:02.280
of energy in the system.
00:29:02.280 --> 00:29:04.970
That goes into speeding
up the rotation,
00:29:04.970 --> 00:29:07.730
and yet the angular
momentum has stayed constant
00:29:07.730 --> 00:29:10.740
throughout the action.
00:29:10.740 --> 00:29:12.290
So when I first
saw this years ago,
00:29:12.290 --> 00:29:13.854
I thought, that's really cool.
00:29:13.854 --> 00:29:15.020
That's really quite amazing.
00:29:19.050 --> 00:29:21.190
So a nice application
of conservation
00:29:21.190 --> 00:29:23.580
of angular momentum.
00:29:23.580 --> 00:29:27.080
An application of using
this formula, the time
00:29:27.080 --> 00:29:29.140
rate of change of
angular momentum
00:29:29.140 --> 00:29:32.080
with respect to a
point, it tells you
00:29:32.080 --> 00:29:34.980
about the torques
applied to the system.
00:29:34.980 --> 00:29:37.770
And this is in fact
a pretty simple case.
00:29:37.770 --> 00:29:43.040
So the last-- let's
move on though to doing
00:29:43.040 --> 00:29:45.900
a little more complicated case.
00:29:45.900 --> 00:29:51.700
And this is similar to the
last problem in the homework.
00:30:01.547 --> 00:30:02.755
So this is like the homework.
00:30:14.380 --> 00:30:17.410
The homework, you got this
monkey running up the shaft,
00:30:17.410 --> 00:30:17.910
right?
00:30:17.910 --> 00:30:20.870
So I don't have a monkey, and
it's not running up the shaft.
00:30:20.870 --> 00:30:31.810
But I do have just this
particle on a shaft
00:30:31.810 --> 00:30:35.896
rotating about a central axis.
00:30:35.896 --> 00:30:37.940
Now, this is a
mechanical necessity
00:30:37.940 --> 00:30:38.940
to hold it all together.
00:30:38.940 --> 00:30:40.865
But let's just ignore
the mass of this center
00:30:40.865 --> 00:30:41.740
piece for the moment.
00:30:41.740 --> 00:30:47.340
Just think of this as a massless
arm with a particle on it.
00:30:47.340 --> 00:30:50.680
And I want to calculate
angular momentum.
00:30:50.680 --> 00:30:52.500
I want to calculate forces.
00:30:52.500 --> 00:30:55.245
I want to calculate torques
and see what happens.
00:31:14.590 --> 00:31:25.340
So I'm going to start by
putting my O, x, z frame right
00:31:25.340 --> 00:31:27.910
on the level with this mass.
00:31:31.165 --> 00:31:32.540
What I'm going to
show you now is
00:31:32.540 --> 00:31:38.200
that where you put the reference
point about which you compute
00:31:38.200 --> 00:31:41.510
the angular momentum matters.
00:31:41.510 --> 00:31:44.530
You get different answers
depending on where you put it.
00:31:44.530 --> 00:31:48.810
So I'm going to start
by putting it here.
00:31:48.810 --> 00:31:50.760
And this we'll call
[INAUDIBLE] out here's
00:31:50.760 --> 00:31:59.275
my point A. Here's O. This
is some angle phi here.
00:32:02.390 --> 00:32:10.420
And if this has some length
l, then this up here is my r
00:32:10.420 --> 00:32:16.270
equals l cosine phi.
00:32:16.270 --> 00:32:21.186
And this side would
be l sine phi.
00:32:21.186 --> 00:32:22.810
These are just the
two lengths, but I'm
00:32:22.810 --> 00:32:24.910
going to use polar coordinates.
00:32:24.910 --> 00:32:28.240
So this is going to
be my r hat direction.
00:32:28.240 --> 00:32:29.580
Theta hat's into the board.
00:32:40.070 --> 00:32:43.120
So let's compute--
and it's a particle,
00:32:43.120 --> 00:32:49.050
so I'll continue to
use lowercase h of A
00:32:49.050 --> 00:32:57.840
with respect to O-- it's a
vector-- is r of A with respect
00:32:57.840 --> 00:33:02.450
to O cross P linear
momentum with respect
00:33:02.450 --> 00:33:08.840
to O. This is r r hat.
00:33:08.840 --> 00:33:11.230
We just call this l cosine
theta, just calling it r.
00:33:11.230 --> 00:33:13.920
It's in the r hat direction.
00:33:13.920 --> 00:33:18.930
Cross with P, and P, we'd done
this two or three times now
00:33:18.930 --> 00:33:19.670
today.
00:33:19.670 --> 00:33:26.290
It's the mass times
r times theta dot.
00:33:26.290 --> 00:33:28.040
That's its speed.
00:33:28.040 --> 00:33:31.325
And it's in the
theta hat direction.
00:33:38.190 --> 00:33:42.640
So taking the r hat cross
theta hat gives me k.
00:33:45.220 --> 00:33:53.970
So I get m r
squared theta dot k.
00:34:00.347 --> 00:34:01.305
Very simple expression.
00:34:05.980 --> 00:34:08.560
Now, this is a
fixed axis rotation.
00:34:08.560 --> 00:34:12.440
So I want to
compute the torques.
00:34:12.440 --> 00:34:14.260
Look at my formula.
00:34:14.260 --> 00:34:19.860
The vA in this case
is the velocity of O.
00:34:19.860 --> 00:34:23.870
The point of the axis of
rotation doesn't move.
00:34:23.870 --> 00:34:26.590
So the second terms
go away, and I
00:34:26.590 --> 00:34:34.790
can say that the torque of my
particle at A with respect to O
00:34:34.790 --> 00:34:38.730
is just dhA dt.
00:34:43.210 --> 00:34:45.530
So m's a constant.
00:34:45.530 --> 00:34:46.830
r's a constant.
00:34:46.830 --> 00:34:48.719
k's a constant.
00:34:48.719 --> 00:34:51.534
The only thing that has a
time derivative is theta dot,
00:34:51.534 --> 00:34:54.136
and it becomes theta double dot.
00:34:54.136 --> 00:35:01.690
This is m r squared
theta double dot k.
00:35:01.690 --> 00:35:07.630
And that is equal-- well,
that's equal to the sum
00:35:07.630 --> 00:35:10.790
of the external torques.
00:35:10.790 --> 00:35:14.140
So what physically
does that mean?
00:35:14.140 --> 00:35:17.010
What physically is
that telling us?
00:35:17.010 --> 00:35:20.110
It's telling us theta
double dot is the angular
00:35:20.110 --> 00:35:22.850
acceleration of this
thing speeding up,
00:35:22.850 --> 00:35:25.270
going faster and faster.
00:35:25.270 --> 00:35:29.590
It takes torque to
make that happen.
00:35:29.590 --> 00:35:34.310
If it's going at constant
rate, what's theta double dot?
00:35:34.310 --> 00:35:35.650
Zero.
00:35:35.650 --> 00:35:38.390
So at constant rate,
the torque required
00:35:38.390 --> 00:35:42.070
to make this thing go
constant rate is zero.
00:35:42.070 --> 00:35:43.080
Makes sense.
00:35:43.080 --> 00:35:45.200
But if it were
speeding up, if you're
00:35:45.200 --> 00:35:47.330
making it go faster
and faster and faster,
00:35:47.330 --> 00:35:48.890
it requires torque to drive it.
00:35:48.890 --> 00:35:50.260
And that's the amount of torque.
00:35:50.260 --> 00:35:55.512
And the torque is
around the axis of spin.
00:35:55.512 --> 00:35:56.470
Pretty straightforward.
00:36:02.670 --> 00:36:07.670
So if I did dP [? d, ?]
if I took the time
00:36:07.670 --> 00:36:11.760
derivative of just the linear
momentum for this particle,
00:36:11.760 --> 00:36:13.180
we've done it here today.
00:36:13.180 --> 00:36:17.810
If I took the time derivative
of it, what would I get?
00:36:17.810 --> 00:36:18.440
It's a force.
00:36:18.440 --> 00:36:19.314
And what's the force?
00:36:25.730 --> 00:36:27.650
It's a constant rotation rate.
00:36:27.650 --> 00:36:32.930
Take the time derivative
of P. dP dt gives me
00:36:32.930 --> 00:36:34.340
AUDIENCE: [INAUDIBLE].
00:36:34.340 --> 00:36:35.781
PROFESSOR: Mass times?
00:36:35.781 --> 00:36:37.890
AUDIENCE: [INAUDIBLE].
00:36:37.890 --> 00:36:40.720
PROFESSOR: Mv squared over r
is mass times acceleration.
00:36:40.720 --> 00:36:42.574
The acceleration is which kind?
00:36:42.574 --> 00:36:43.490
AUDIENCE: [INAUDIBLE].
00:36:43.490 --> 00:36:44.060
PROFESSOR: Centripetal.
00:36:44.060 --> 00:36:46.030
So you just get the
same thing back again.
00:36:46.030 --> 00:36:51.660
So there's a force acting
inwards words this thing
00:36:51.660 --> 00:36:56.630
to make it go in a circle that
is the mr theta dot squared
00:36:56.630 --> 00:37:00.270
term that we've
seen so many times.
00:37:00.270 --> 00:37:09.420
But now what I want to do is
move the point about which I
00:37:09.420 --> 00:37:10.940
compute this angular momentum.
00:37:16.740 --> 00:37:19.070
And now I'm going
to put it here,
00:37:19.070 --> 00:37:22.220
the point of
attachment of the arm.
00:37:22.220 --> 00:37:27.280
So here's O, x, z.
00:37:27.280 --> 00:37:28.670
Everything else stays the same.
00:37:28.670 --> 00:37:31.460
All I've done is move the point.
00:37:31.460 --> 00:37:35.960
And I want to compute
the angular momentum
00:37:35.960 --> 00:37:43.570
of this A with respect to O.
00:37:43.570 --> 00:37:44.650
Well, that's r.
00:37:44.650 --> 00:37:54.380
This is now r of A with
respect to O, this vector.
00:37:54.380 --> 00:37:59.140
This distance here, in polar
cylindrical coordinates, is z.
00:38:04.210 --> 00:38:05.430
And this is r.
00:38:05.430 --> 00:38:06.290
Just as before.
00:38:06.290 --> 00:38:08.870
The r hasn't changed.
00:38:08.870 --> 00:38:11.960
And there is an r hat
in this direction,
00:38:11.960 --> 00:38:17.660
theta hat into the board, and
a k hat in the z direction.
00:38:17.660 --> 00:38:19.750
Those are our unit vectors.
00:38:19.750 --> 00:38:34.240
So this is-- rAO is
r r hat plus z k hat.
00:38:34.240 --> 00:38:38.660
That's the position vector.
00:38:38.660 --> 00:38:43.610
And I'm going to cross
that with the momentum of A
00:38:43.610 --> 00:38:55.030
with respect to O.
And the momentum
00:38:55.030 --> 00:39:04.180
is mr. Theta dot
is the velocity.
00:39:04.180 --> 00:39:05.450
And what's its direction?
00:39:08.000 --> 00:39:10.410
Theta hat, right?
00:39:10.410 --> 00:39:12.820
Mass times velocity, momentum.
00:39:12.820 --> 00:39:17.540
And now we need
to carry this out.
00:39:17.540 --> 00:39:22.230
The r hat term times
theta hat gives you a k.
00:39:27.195 --> 00:39:35.760
m r squared theta dot k.
00:39:35.760 --> 00:39:44.000
And this term, k cross theta
hat, gives me a minus r.
00:39:44.000 --> 00:39:57.915
Minus mrz theta dot k.
00:40:03.616 --> 00:40:05.820
AUDIENCE: [INAUDIBLE].
00:40:05.820 --> 00:40:06.820
PROFESSOR: You're right.
00:40:06.820 --> 00:40:08.510
Thank you.
00:40:08.510 --> 00:40:13.240
Because I would have a disaster
if I let that progress.
00:40:13.240 --> 00:40:16.745
This looks like that, right?
k cross theta is minus r hat.
00:40:19.260 --> 00:40:21.346
r cross theta is a k.
00:40:21.346 --> 00:40:23.650
k cross theta is a minus r.
00:40:23.650 --> 00:40:26.850
I get two terms.
00:40:26.850 --> 00:40:30.505
And this is now an expression
for the angular momentum of A
00:40:30.505 --> 00:40:34.960
with respect to O. And let's
see if I have enough room
00:40:34.960 --> 00:40:36.360
to draw it here.
00:40:36.360 --> 00:40:41.120
There is a piece of it here.
00:40:41.120 --> 00:40:49.180
This is h in the z
direction, is this arrow.
00:40:49.180 --> 00:40:55.245
And then there's a piece in
the r hat direction, like this.
00:40:55.245 --> 00:40:59.780
This is h in the
r hat direction.
00:40:59.780 --> 00:41:05.420
And the sum of
those two is that.
00:41:05.420 --> 00:41:10.930
So this is hA with
respect to O, this guy.
00:41:10.930 --> 00:41:15.829
Perpendicular to the shaft.
00:41:15.829 --> 00:41:17.620
It will turn out it
really is perpendicular
00:41:17.620 --> 00:41:20.530
if you work out the numbers.
00:41:20.530 --> 00:41:27.860
Totally different result
than when I did it here.
00:41:27.860 --> 00:41:32.550
So this one, m r squared theta
dot k, m r squared theta dot k.
00:41:32.550 --> 00:41:34.450
Hey, that term's the same.
00:41:34.450 --> 00:41:38.360
So when I did this, I
got just the k term.
00:41:38.360 --> 00:41:40.900
And now I've moved this thing
down, and I get a second term.
00:41:44.550 --> 00:41:47.712
So now what we want to know
is, what about the torques
00:41:47.712 --> 00:41:48.295
in the system?
00:41:58.110 --> 00:42:01.000
So I want to take
the time derivative
00:42:01.000 --> 00:42:04.980
of this guy with
respect to-- the time
00:42:04.980 --> 00:42:08.290
derivative of the
angular momentum, which
00:42:08.290 --> 00:42:11.160
is going to be equal
to the summation
00:42:11.160 --> 00:42:14.870
of the external torques
with respect to O,
00:42:14.870 --> 00:42:18.110
but O is now in a
different place.
00:42:18.110 --> 00:42:21.100
And so I have to carry
out these derivatives.
00:42:21.100 --> 00:42:23.130
And this one I did before.
00:42:23.130 --> 00:42:32.045
This one just gives me my m r
squared theta double dot k hat.
00:42:32.045 --> 00:42:36.530
Now, this term, m is a constant.
00:42:36.530 --> 00:42:38.800
r is a constant.
00:42:38.800 --> 00:42:42.260
z is a constant.
00:42:42.260 --> 00:42:49.036
Theta dot is not
necessarily a constant.
00:42:49.036 --> 00:42:51.600
We're going to let
that be a variable.
00:42:51.600 --> 00:42:55.540
And r hat is certainly
changing direction.
00:42:55.540 --> 00:42:57.320
So when I take the
derivative of this,
00:42:57.320 --> 00:42:59.700
I'm going to get two terms.
00:42:59.700 --> 00:43:10.700
So the first one is minus
mrz theta double dot r hat.
00:43:10.700 --> 00:43:13.505
And that's taking the derivative
of this multiplied by that.
00:43:13.505 --> 00:43:15.520
And the second term
is the derivative
00:43:15.520 --> 00:43:17.810
of this multiplied by that.
00:43:17.810 --> 00:43:22.200
And so the derivative
of r hat is?
00:43:26.480 --> 00:43:27.960
AUDIENCE: [INAUDIBLE].
00:43:27.960 --> 00:43:29.410
PROFESSOR: Theta dot theta hat.
00:43:29.410 --> 00:43:30.080
Right.
00:43:30.080 --> 00:43:30.870
OK.
00:43:30.870 --> 00:43:45.210
So minus mrz theta dot
theta dot theta hat.
00:43:47.720 --> 00:43:50.390
So in other words,
this is squared.
00:43:50.390 --> 00:43:54.210
Now I have three
terms to mess with.
00:43:54.210 --> 00:43:58.100
We know what the
first term means.
00:43:58.100 --> 00:44:00.140
We talked about that.
00:44:00.140 --> 00:44:01.220
This is the torque.
00:44:01.220 --> 00:44:02.910
These are all torques.
00:44:02.910 --> 00:44:06.060
So this is the torque required
to do what, the lead term?
00:44:09.030 --> 00:44:11.792
AUDIENCE: [INAUDIBLE].
00:44:11.792 --> 00:44:13.000
AUDIENCE: [INAUDIBLE] circle.
00:44:13.000 --> 00:44:15.064
PROFESSOR: To make it?
00:44:15.064 --> 00:44:16.030
AUDIENCE: [INAUDIBLE].
00:44:16.030 --> 00:44:16.904
PROFESSOR: Go faster.
00:44:16.904 --> 00:44:18.990
Change its angular speed, right?
00:44:18.990 --> 00:44:23.860
It's just building up the
angular momentum in that spin.
00:44:23.860 --> 00:44:29.190
So this is the angular spin up.
00:44:29.190 --> 00:44:33.405
These other two terms,
these are strange things.
00:44:33.405 --> 00:44:39.100
Well first, let's take a look at
this one. r theta dot squared.
00:44:39.100 --> 00:44:40.528
What's that remind you of?
00:44:43.280 --> 00:44:49.200
What kind of-- torque is usually
some force times a moment arm,
00:44:49.200 --> 00:44:51.920
crossed with a
moment arm, right?
00:44:51.920 --> 00:44:56.040
So we know that there's some
forces acting in this system.
00:44:56.040 --> 00:44:58.080
It's spinning.
00:44:58.080 --> 00:45:03.160
We know that there
is a-- in order
00:45:03.160 --> 00:45:07.030
to make this thing go
round and around-- it
00:45:07.030 --> 00:45:09.060
has centripetal acceleration.
00:45:09.060 --> 00:45:13.720
Therefore, there must be a force
being applied by this shaft
00:45:13.720 --> 00:45:17.840
inward that's equal
to the mass times
00:45:17.840 --> 00:45:22.035
the centripetal acceleration,
mr theta dot squared.
00:45:26.480 --> 00:45:33.786
So here this guy is
mr theta dot squared.
00:45:37.060 --> 00:45:39.650
That's the force.
00:45:39.650 --> 00:45:42.110
And let's do these.
00:45:42.110 --> 00:45:44.770
Let's call this A. Let's Call.
00:45:44.770 --> 00:45:55.140
This term here B, this term C.
So the C term is-- torque C,
00:45:55.140 --> 00:46:01.780
I'll call it-- is
some r cross some F.
00:46:01.780 --> 00:46:10.110
And the F, I'm telling you, is
the centripetal acceleration
00:46:10.110 --> 00:46:11.260
times the mass.
00:46:11.260 --> 00:46:18.110
And that'll probably
be like a minus mr
00:46:18.110 --> 00:46:24.200
theta dot squared r hat, right?
00:46:24.200 --> 00:46:30.650
And what about the moment
arm that that acts about?
00:46:30.650 --> 00:46:32.740
What moment arm is
perpendicular-- so
00:46:32.740 --> 00:46:35.450
that's a force that's acting in.
00:46:35.450 --> 00:46:37.244
What moment arm is
perpendicular to that?
00:46:37.244 --> 00:46:39.160
Because the only thing
that's perpendicular to
00:46:39.160 --> 00:46:42.187
it lead to torques.
00:46:42.187 --> 00:46:42.770
PROFESSOR: Hm?
00:46:42.770 --> 00:46:43.550
AUDIENCE: [INAUDIBLE].
00:46:43.550 --> 00:46:44.091
PROFESSOR: z.
00:46:52.670 --> 00:46:58.525
k cross r should
give me a theta.
00:47:02.100 --> 00:47:07.280
Sure enough, there's
a minus, sure enough.
00:47:07.280 --> 00:47:09.600
And there's the z.
00:47:09.600 --> 00:47:12.690
You multiply this out,
you get this term.
00:47:12.690 --> 00:47:14.820
So this is a strange term.
00:47:14.820 --> 00:47:19.880
It's in the theta hat direction.
00:47:19.880 --> 00:47:21.100
What is that?
00:47:21.100 --> 00:47:25.080
So it's spinning, and
it's lined up like this.
00:47:25.080 --> 00:47:27.900
Theta hat's in that direction.
00:47:27.900 --> 00:47:33.370
Positive k, positive
theta, positive r hat.
00:47:33.370 --> 00:47:34.510
k's in that direction.
00:47:34.510 --> 00:47:36.670
It's telling you
there's a torque
00:47:36.670 --> 00:47:42.684
being applied about this point
in the minus theta direction.
00:47:42.684 --> 00:47:44.670
Does that make sense?
00:47:44.670 --> 00:47:49.850
You have a-- there's a
centripetal acceleration
00:47:49.850 --> 00:47:50.660
times a mass.
00:47:50.660 --> 00:47:53.860
There's a force
times a moment arm.
00:47:53.860 --> 00:47:59.317
This force is trying
to bend this thing out.
00:47:59.317 --> 00:48:01.900
If this thing had a hinge down
here, and I started to spin it,
00:48:01.900 --> 00:48:02.608
what would it do?
00:48:02.608 --> 00:48:04.960
It would just flop out, right?
00:48:04.960 --> 00:48:08.670
There's got to be a torque
keeping that from happening.
00:48:08.670 --> 00:48:10.190
And that's a torque.
00:48:10.190 --> 00:48:11.870
If it wants to go
that way, there's
00:48:11.870 --> 00:48:15.360
got to be a torque going the
other way keeping it in place.
00:48:15.360 --> 00:48:18.330
And that's what that term is.
00:48:18.330 --> 00:48:18.990
So now we know.
00:48:18.990 --> 00:48:20.710
So this is the Euler.
00:48:20.710 --> 00:48:22.630
This is the spin up.
00:48:22.630 --> 00:48:26.410
This is keeping this
thing from flopping out.
00:48:26.410 --> 00:48:27.990
What's this guy?
00:48:27.990 --> 00:48:33.640
This is yet another torque,
and it's in the r, minus r,
00:48:33.640 --> 00:48:34.140
direction.
00:48:36.960 --> 00:48:38.780
So let's see if
we can intuitively
00:48:38.780 --> 00:48:40.600
figure this one out.
00:48:40.600 --> 00:48:44.560
Well, there's an r
theta double dot.
00:48:44.560 --> 00:48:46.260
That should look familiar.
00:48:46.260 --> 00:48:49.350
r theta double dot.
00:48:49.350 --> 00:48:53.060
If this thing is accelerating,
angular acceleration,
00:48:53.060 --> 00:48:59.650
speeding up, out here that
mass says, I'm here right now.
00:48:59.650 --> 00:49:01.765
In order for me to go
a little bit faster,
00:49:01.765 --> 00:49:04.320
I'm accelerating
in that direction.
00:49:04.320 --> 00:49:06.070
There must be a
force being applied
00:49:06.070 --> 00:49:09.260
by this rod in that direction.
00:49:09.260 --> 00:49:15.390
And that force times a moment
arm perpendicular to it, z,
00:49:15.390 --> 00:49:20.870
is a moment in the r direction.
00:49:20.870 --> 00:49:23.770
So as this thing spins
up, if this thing could,
00:49:23.770 --> 00:49:25.740
it would fall back.
00:49:25.740 --> 00:49:27.820
But this rod is stiff
and won't let it do that.
00:49:27.820 --> 00:49:30.980
So this is the
torque down here that
00:49:30.980 --> 00:49:33.660
is required to keep
this thing moving.
00:49:36.200 --> 00:49:38.690
So this is really quite amazing.
00:49:41.200 --> 00:49:46.550
Do either of these torques,
these second ones, this one
00:49:46.550 --> 00:49:50.880
and this one, do they contribute
to-- do they do any work?
00:49:53.460 --> 00:49:56.330
Do they add energy
to the system?
00:49:56.330 --> 00:49:59.720
See, work means force
through a distance.
00:49:59.720 --> 00:50:02.020
They don't do actually any work.
00:50:02.020 --> 00:50:04.260
They are static
torques just required
00:50:04.260 --> 00:50:06.370
to hold the system together.
00:50:06.370 --> 00:50:07.590
This one does some work.
00:50:07.590 --> 00:50:10.715
It actually makes-- this one
leads to energy accumulating,
00:50:10.715 --> 00:50:13.000
just going faster
and faster faster.
00:50:13.000 --> 00:50:15.060
These are just holding
the thing together.
00:50:15.060 --> 00:50:18.720
But the amazing
thing is that you
00:50:18.720 --> 00:50:21.930
can use angular momentum
to calculate things
00:50:21.930 --> 00:50:24.040
like these torques.
00:50:24.040 --> 00:50:30.520
So if you were designing
this, these forces
00:50:30.520 --> 00:50:35.490
acting on this lump out
here are producing torques
00:50:35.490 --> 00:50:39.950
about this point, which are the
same thing as-- in 2.001 you'll
00:50:39.950 --> 00:50:42.200
be doing bending moments.
00:50:42.200 --> 00:50:45.124
It creates a bending
moment in the shaft.
00:50:45.124 --> 00:50:47.040
And if you don't make
the shaft strong enough,
00:50:47.040 --> 00:50:49.270
it'll break it off.
00:50:49.270 --> 00:50:54.250
So the torque about this
point is the bending moment
00:50:54.250 --> 00:50:58.230
in the r direction and
in the theta direction.
00:50:58.230 --> 00:51:00.680
It's trying to be bent
in two different ways.
00:51:00.680 --> 00:51:03.480
And you can calculate the
stresses down here caused
00:51:03.480 --> 00:51:05.020
by those moments.
00:51:05.020 --> 00:51:08.760
And that would help
you design the thing.
00:51:08.760 --> 00:51:11.590
So you not only get
dynamics information
00:51:11.590 --> 00:51:13.770
out of taking things like
the time rate of change
00:51:13.770 --> 00:51:15.250
of angular momentum.
00:51:15.250 --> 00:51:20.050
You get some of the static
information as well.
00:51:20.050 --> 00:51:24.200
The other thing to remember,
the really important point
00:51:24.200 --> 00:51:29.720
of the lecture, is
that angular momentum
00:51:29.720 --> 00:51:34.910
changes depending on where
you pick a reference point.
00:51:34.910 --> 00:51:38.450
So when we picked the reference
point just opposite it,
00:51:38.450 --> 00:51:41.500
we got none of the information
about the torques down here.
00:51:41.500 --> 00:51:43.720
Because with respect
to this point,
00:51:43.720 --> 00:51:47.890
there are no torques
except the one speeded up.
00:51:47.890 --> 00:51:51.940
The centripetal force here
doesn't cause torques.
00:51:51.940 --> 00:51:57.179
The force out here-- there
are no other torques, just
00:51:57.179 --> 00:51:58.470
the one to make it spin faster.
00:51:58.470 --> 00:52:00.220
But as soon as I
move it down here,
00:52:00.220 --> 00:52:04.350
I learn something of
considerable value.
00:52:04.350 --> 00:52:07.404
So the homework problem
has some of the same things
00:52:07.404 --> 00:52:08.820
on it, except the
monkey's moving.
00:52:08.820 --> 00:52:11.780
So you get even
a little bit more
00:52:11.780 --> 00:52:14.760
interesting
information out of it.
00:52:14.760 --> 00:52:15.260
All right.
00:52:28.090 --> 00:52:32.050
So that went faster
than I thought.
00:52:32.050 --> 00:52:34.040
So that gives some time
for some questions.
00:52:34.040 --> 00:52:35.230
I could see several.
00:52:35.230 --> 00:52:38.048
So we'll start there
and then go here.
00:52:38.048 --> 00:52:39.920
AUDIENCE: What
exactly [INAUDIBLE]
00:52:39.920 --> 00:52:42.730
torque B is balancing out?
00:52:42.730 --> 00:52:44.704
PROFESSOR: So say again?
00:52:44.704 --> 00:52:48.600
AUDIENCE: Torque B
[INAUDIBLE], what
00:52:48.600 --> 00:52:51.515
exactly is that balancing out?
00:52:51.515 --> 00:52:52.390
PROFESSOR: This term?
00:52:52.390 --> 00:52:54.017
You're asking about this term?
00:52:54.017 --> 00:52:55.267
AUDIENCE: [INAUDIBLE].
00:52:55.267 --> 00:52:55.850
PROFESSOR: OK.
00:52:55.850 --> 00:52:57.915
You want me to explain
again what this one means?
00:53:02.850 --> 00:53:09.130
This is a term associated with
increasing the angular speed.
00:53:09.130 --> 00:53:21.910
So let's see if we
can't-- so this B term,
00:53:21.910 --> 00:53:27.510
I'll call it the torque
associated with B,
00:53:27.510 --> 00:53:34.120
minus mrz theta
double dot r hat.
00:53:34.120 --> 00:53:42.260
Now, that is going to
be some r cross some F.
00:53:42.260 --> 00:53:44.310
And if we can get
some physical insight,
00:53:44.310 --> 00:53:46.820
if we could figure
out what they are.
00:53:46.820 --> 00:53:55.740
So the mass, the force, is the
mass times an acceleration.
00:53:55.740 --> 00:53:58.450
There's an acceleration
r theta double dot, which
00:53:58.450 --> 00:54:03.690
is the speed of this
thing is increasing speed.
00:54:03.690 --> 00:54:07.490
And the r is going to be z.
00:54:07.490 --> 00:54:11.860
z k hat cross-- and I'm
guessing that it's a force that
00:54:11.860 --> 00:54:18.820
looks like mr theta double dot.
00:54:18.820 --> 00:54:26.415
And this is in the
theta hat direction.
00:54:29.550 --> 00:54:33.440
k cross theta hat
gives me minus r hat.
00:54:33.440 --> 00:54:34.964
r hat and the minus.
00:54:38.130 --> 00:54:40.940
So this looks like a plausible
explanation for where
00:54:40.940 --> 00:54:42.080
this might have come from.
00:54:42.080 --> 00:54:48.360
So this is the force
speeding up, attempting
00:54:48.360 --> 00:54:49.615
to speed up this mass.
00:54:49.615 --> 00:54:53.450
There's a force pushing on it
that's given to it by this rod.
00:54:53.450 --> 00:54:56.950
This rod is pushing on
it to make it go faster.
00:54:56.950 --> 00:55:01.600
Mass times acceleration
would be mr theta double dot.
00:55:01.600 --> 00:55:04.110
And the moment arm
is this distance
00:55:04.110 --> 00:55:06.440
from here down to
the point at which
00:55:06.440 --> 00:55:12.240
I've been computing my reference
point from here to here at z.
00:55:12.240 --> 00:55:16.600
So force times z,
r cross F, puts it
00:55:16.600 --> 00:55:19.985
in the-- ends up in
the minus r direction.
00:55:19.985 --> 00:55:23.636
It's got to be this
direction, k cross theta hat.
00:55:23.636 --> 00:55:25.070
The force is this way.
00:55:31.542 --> 00:55:32.250
Think about this.
00:55:32.250 --> 00:55:34.040
Force is that way.
00:55:34.040 --> 00:55:37.240
The r is this way.
00:55:37.240 --> 00:55:41.840
The r cross F is this way.
00:55:41.840 --> 00:55:43.920
And that's where you
get the minus sign.
00:55:43.920 --> 00:55:47.130
It's in the minus
r hat direction.
00:55:47.130 --> 00:55:51.890
But it comes from trying
to speed up this mass.
00:55:51.890 --> 00:55:58.260
And if this was a floppy, weak
link, as it tries to speed up,
00:55:58.260 --> 00:56:00.300
it would try to bend back.
00:56:00.300 --> 00:56:04.070
It would flop back as this thing
tries to make it go faster.
00:56:04.070 --> 00:56:05.600
It will say, no, I
don't want to go.
00:56:05.600 --> 00:56:06.710
Lay back on me.
00:56:06.710 --> 00:56:11.320
And that would we going
in the-- this way,
00:56:11.320 --> 00:56:12.710
to keep from doing
that, you have
00:56:12.710 --> 00:56:16.420
to put a torque on it this way.
00:56:16.420 --> 00:56:17.460
Just trying to speed up.
00:56:17.460 --> 00:56:19.220
It's trying to lay back,
and you're saying nope,
00:56:19.220 --> 00:56:19.940
can't do that.
00:56:19.940 --> 00:56:21.985
Go like this.
00:56:21.985 --> 00:56:22.860
So that's the B term.
00:56:26.020 --> 00:56:30.194
AUDIENCE: If you do the problem
with a situation like that,
00:56:30.194 --> 00:56:32.150
how do you know where to set it?
00:56:32.150 --> 00:56:33.180
PROFESSOR: How do you
know where to set it?
00:56:33.180 --> 00:56:34.959
Well, that's a good question.
00:56:34.959 --> 00:56:37.250
He's saying, when you're
doing a problem like this, how
00:56:37.250 --> 00:56:39.570
do you know where to
pick the reference frame?
00:56:39.570 --> 00:56:42.856
Well, ask yourself what
it is you want to know.
00:56:42.856 --> 00:56:46.980
And in fact, now that you know
that from angular momentum
00:56:46.980 --> 00:56:50.470
of mechanical things,
you can actually
00:56:50.470 --> 00:56:53.940
get static torques
on the system,
00:56:53.940 --> 00:56:56.640
ask yourself where you
want to know those torques.
00:56:56.640 --> 00:56:58.480
In this case, if
you're designing this,
00:56:58.480 --> 00:57:01.200
you want to know whether you're
going to break this thing off.
00:57:01.200 --> 00:57:04.760
And it's probably going to
break right down at the bottom
00:57:04.760 --> 00:57:07.140
where the moment arms
are the greatest.
00:57:07.140 --> 00:57:08.650
So that's why you
pick that point.
00:57:11.910 --> 00:57:18.460
It comes a lot with experience
will help you choose.
00:57:18.460 --> 00:57:20.990
But the amazing thing
is this information's
00:57:20.990 --> 00:57:23.200
all stored in the
angular momentum
00:57:23.200 --> 00:57:25.120
if you pick it in
the right place.
00:57:25.120 --> 00:57:25.700
Another hand.
00:57:31.100 --> 00:57:32.240
OK.
00:57:32.240 --> 00:57:34.390
Yes, Phillip.
00:57:34.390 --> 00:57:37.785
AUDIENCE: I had a question
about the direction for r hat.
00:57:37.785 --> 00:57:40.210
I thought it had to
come out of the origin.
00:57:40.210 --> 00:57:42.635
But you have it going
in the x direction.
00:57:47.510 --> 00:57:49.350
PROFESSOR: So I've
been using polar-- he's
00:57:49.350 --> 00:57:51.230
asking the direction of r hat.
00:57:51.230 --> 00:57:53.590
So I've just been using
polar coordinates.
00:57:53.590 --> 00:57:58.710
And polar coordinates is
cylindrical, technically.
00:57:58.710 --> 00:58:05.380
This problem has a z
direction upwards, r direction
00:58:05.380 --> 00:58:09.960
radially outwards, r hat,
and theta as drawn here
00:58:09.960 --> 00:58:14.110
would be into the board, given
the position of the mass.
00:58:14.110 --> 00:58:20.530
Looking down on it, here's
my O. And looking down,
00:58:20.530 --> 00:58:23.930
this would be my x and my y.
00:58:23.930 --> 00:58:27.810
And this is some random
arbitrary position here.
00:58:27.810 --> 00:58:31.260
And this is theta.
00:58:31.260 --> 00:58:33.770
Looking down on
it, in this plane
00:58:33.770 --> 00:58:41.690
is r hat theta hat, k hat
coming out of the board.
00:58:41.690 --> 00:58:53.130
Side view, x, z, and
my system is like this.
00:58:53.130 --> 00:59:02.130
Now the theta is into the
board, and the r direction
00:59:02.130 --> 00:59:02.950
is this way.
00:59:02.950 --> 00:59:03.880
That's r hat.
00:59:03.880 --> 00:59:05.850
And this is z.
00:59:05.850 --> 00:59:09.840
This is the
z-coordinate upwards.
00:59:09.840 --> 00:59:15.260
So the position vector, the
thing we call rA with respect
00:59:15.260 --> 00:59:21.410
to O, is indeed the length
of this whole thing.
00:59:21.410 --> 00:59:29.740
But it is made up of a
component in the z direction
00:59:29.740 --> 00:59:39.688
plus a component here that we
call r in the r hat direction.
00:59:39.688 --> 00:59:42.592
AUDIENCE: [INAUDIBLE] example
where we calculated from
00:59:42.592 --> 00:59:45.980
the bottom rather
than the top circle,
00:59:45.980 --> 00:59:50.094
then we got a value for
the angular momentum that
00:59:50.094 --> 00:59:51.788
doesn't have a
theta hat component,
00:59:51.788 --> 00:59:53.040
but as the thing spins--
00:59:53.040 --> 00:59:54.300
PROFESSOR: Which are
you referring to?
00:59:54.300 --> 00:59:54.883
Are you talk--
00:59:54.883 --> 00:59:58.660
AUDIENCE: h about
the point O down--
00:59:58.660 --> 01:00:00.350
PROFESSOR: Which
example, this guy or--
01:00:00.350 --> 01:00:01.516
AUDIENCE: Yeah, [INAUDIBLE].
01:00:01.516 --> 01:00:03.000
PROFESSOR: OK, so point.
01:00:03.000 --> 01:00:04.240
Tell me what you mean here.
01:00:04.240 --> 01:00:05.656
AUDIENCE: Right
here. [INAUDIBLE].
01:00:05.656 --> 01:00:08.326
PROFESSOR: When we computed,
not here, but down here.
01:00:08.326 --> 01:00:11.302
AUDIENCE: We got that there
was no theta component,
01:00:11.302 --> 01:00:14.774
but as this spins around,
theta is changing.
01:00:14.774 --> 01:00:16.975
And if it's always
opposite, shouldn't there
01:00:16.975 --> 01:00:17.850
be a theta component?
01:00:17.850 --> 01:00:18.860
PROFESSOR: A theta
component of what?
01:00:18.860 --> 01:00:20.280
AUDIENCE: Angular momentum.
01:00:20.280 --> 01:00:22.620
PROFESSOR: Angular momentum.
01:00:22.620 --> 01:00:24.710
She's asking, shouldn't
there be a theta component
01:00:24.710 --> 01:00:27.870
of angular momentum?
01:00:27.870 --> 01:00:31.080
So we compute our angular
momentum with this formula
01:00:31.080 --> 01:00:32.360
at the top.
01:00:32.360 --> 01:00:43.350
It's an r cross P. So the
r consists of the z part,
01:00:43.350 --> 01:00:46.680
and the r part is
exactly this right here.
01:00:46.680 --> 01:00:49.080
And the P is only
into the board.
01:00:49.080 --> 01:00:51.350
It's only in the
theta hat direction.
01:00:51.350 --> 01:00:58.000
So you have a term that's r hat
cross theta hat gives you a k,
01:00:58.000 --> 01:01:01.915
and you have a
term k cross theta
01:01:01.915 --> 01:01:03.350
hat, which gives you an r.
01:01:03.350 --> 01:01:05.300
There just are no
cross products that
01:01:05.300 --> 01:01:09.020
come out of this that are
in the theta hat direction.
01:01:09.020 --> 01:01:09.994
Yeah.
01:01:09.994 --> 01:01:12.388
AUDIENCE: So if it's just
at this one position,
01:01:12.388 --> 01:01:13.792
then you don't have it.
01:01:13.792 --> 01:01:15.196
But as it spins--
01:01:15.196 --> 01:01:16.520
PROFESSOR: Ah.
01:01:16.520 --> 01:01:20.210
In this case, that's why
polar coordinates are nice
01:01:20.210 --> 01:01:23.830
because as it spins,
the theta hat's
01:01:23.830 --> 01:01:26.380
just constantly going with it.
01:01:26.380 --> 01:01:29.920
The r hat's constantly
going with it.
01:01:29.920 --> 01:01:32.720
And so the beauty
of this thing is
01:01:32.720 --> 01:01:36.730
this is an axially
symmetric problem.
01:01:36.730 --> 01:01:43.890
It goes round and round, and
the torques are given out
01:01:43.890 --> 01:01:47.550
in this rotating frame.
01:01:47.550 --> 01:01:49.250
I think maybe what's
confusing you,
01:01:49.250 --> 01:01:52.400
is if you wanted to know the
torques in a fixed inertial
01:01:52.400 --> 01:01:56.879
frame, you'd have to break
them down into ijk components,
01:01:56.879 --> 01:01:57.670
which you could do.
01:01:57.670 --> 01:01:59.200
A little tedious.
01:01:59.200 --> 01:02:03.730
But the answers in this one came
out in r hat, theta hat, k hat
01:02:03.730 --> 01:02:04.230
terms.
01:02:12.100 --> 01:02:14.540
Happy to answer.
01:02:14.540 --> 01:02:17.760
This is good stuff, but thick.
01:02:17.760 --> 01:02:20.155
So keep-- other questions?
01:02:31.790 --> 01:02:35.270
So what do you think will happen
in that final homework problem
01:02:35.270 --> 01:02:39.010
with the monkey running--
now he has some velocity.
01:02:39.010 --> 01:02:43.390
How will that problem
differ from what we've done?
01:02:43.390 --> 01:02:44.406
Do you have a question?
01:02:44.406 --> 01:02:45.530
Do you want to answer that?
01:02:45.530 --> 01:02:46.372
Yeah.
01:02:46.372 --> 01:02:48.330
AUDIENCE: Well, [INAUDIBLE]
monkey [INAUDIBLE].
01:02:51.790 --> 01:02:53.040
PROFESSOR: It's going to what?
01:02:53.040 --> 01:02:53.910
AUDIENCE: Look like a circle.
01:02:53.910 --> 01:02:55.730
PROFESSOR: It's going to
look like a circle, OK.
01:02:55.730 --> 01:02:57.360
He'll be going in
a circle, at any--
01:02:57.360 --> 01:02:59.857
AUDIENCE: [INAUDIBLE]
as in a spiral.
01:02:59.857 --> 01:03:00.690
PROFESSOR: A spiral.
01:03:00.690 --> 01:03:02.980
He'll be going
like a helix, huh?
01:03:02.980 --> 01:03:04.000
All right.
01:03:04.000 --> 01:03:07.290
Yeah, the monkey will be
going in a helix, yeah.
01:03:07.290 --> 01:03:08.230
AUDIENCE: [INAUDIBLE].
01:03:08.230 --> 01:03:11.862
PROFESSOR: What forces
will act on that monkey?
01:03:11.862 --> 01:03:13.502
AUDIENCE: [INAUDIBLE].
01:03:13.502 --> 01:03:14.960
PROFESSOR: In what
direction do you
01:03:14.960 --> 01:03:17.890
think there will be forces
acting on that-- he's hanging
01:03:17.890 --> 01:03:19.060
on for dear life, you know.
01:03:19.060 --> 01:03:20.143
This thing's going around.
01:03:20.143 --> 01:03:22.460
They could throw him off, right?
01:03:22.460 --> 01:03:23.930
So what forces act?
01:03:23.930 --> 01:03:25.700
And if you could
figure out what forces
01:03:25.700 --> 01:03:29.030
act-- so you draw a free
body diagram of the monkey.
01:03:29.030 --> 01:03:31.490
There's going to be possibly
forces in the theta hat
01:03:31.490 --> 01:03:36.780
direction, in the z
direction, in the r direction.
01:03:36.780 --> 01:03:40.110
But just think physically
where they come from.
01:03:40.110 --> 01:03:41.930
So we now know
that certainly he's
01:03:41.930 --> 01:03:45.640
hanging on because
he is undergoing
01:03:45.640 --> 01:03:48.080
centripetal acceleration.
01:03:48.080 --> 01:03:50.640
And in order to force
him to go in that circle,
01:03:50.640 --> 01:03:54.870
there has to be an inward force
applied by this shaft to him.
01:03:54.870 --> 01:03:56.400
So there's a force
like that because
01:03:56.400 --> 01:04:00.260
of the centripetal acceleration.
01:04:00.260 --> 01:04:02.970
If it's speeding up,
there's a force pushing
01:04:02.970 --> 01:04:05.370
him to make it go faster.
01:04:05.370 --> 01:04:06.700
But he's running up the shaft.
01:04:06.700 --> 01:04:07.710
What else is there?
01:04:13.056 --> 01:04:14.526
Are there any other
accelerations?
01:04:17.819 --> 01:04:20.110
AUDIENCE: [INAUDIBLE] angular
acceleration [INAUDIBLE]?
01:04:20.110 --> 01:04:21.540
PROFESSOR: Yeah, there's
angular acceleration.
01:04:21.540 --> 01:04:22.890
That's the thing
trying to speed up.
01:04:22.890 --> 01:04:25.306
And so he's hanging on because
this thing is accelerating,
01:04:25.306 --> 01:04:27.970
like pushing you back
in the car seat, right?
01:04:27.970 --> 01:04:29.270
Linear acceleration.
01:04:29.270 --> 01:04:31.690
Well, this is trying to--
at any instant in time,
01:04:31.690 --> 01:04:33.020
it's trying to go faster.
01:04:33.020 --> 01:04:35.420
So he's having to hang
on because of that.
01:04:35.420 --> 01:04:38.400
So that's one of those terms.
01:04:38.400 --> 01:04:40.070
But now he's moving.
01:04:40.070 --> 01:04:42.820
He's also running up the shaft.
01:04:42.820 --> 01:04:45.410
Will that lead to any
other accelerations?
01:04:45.410 --> 01:04:47.450
And force equals mass
times acceleration.
01:04:47.450 --> 01:04:49.230
So every time you
can-- if you can
01:04:49.230 --> 01:04:51.750
account for all the
accelerations in the system,
01:04:51.750 --> 01:04:52.790
multiply it by m.
01:04:52.790 --> 01:04:58.000
You've accounted for
all of the forces, sum
01:04:58.000 --> 01:05:00.580
of the forces of the
mass times acceleration.
01:05:00.580 --> 01:05:03.320
If you add a new acceleration,
you better add a new force.
01:05:03.320 --> 01:05:05.990
If you add a new force, you'll
probably add a new torque.
01:05:05.990 --> 01:05:07.910
AUDIENCE: So we have
to account for gravity?
01:05:07.910 --> 01:05:10.220
PROFESSOR: Well, gravity, yeah.
01:05:10.220 --> 01:05:12.241
What else?
01:05:12.241 --> 01:05:14.730
AUDIENCE: [INAUDIBLE].
01:05:14.730 --> 01:05:17.250
PROFESSOR: So he's suggesting
there might be a Coriolis
01:05:17.250 --> 01:05:19.490
acceleration.
01:05:19.490 --> 01:05:21.720
And the Coriolis
acceleration, in order
01:05:21.720 --> 01:05:25.920
to make the monkey accelerate,
according to that term,
01:05:25.920 --> 01:05:28.750
there will have to
be yet another force.
01:05:28.750 --> 01:05:31.650
And I think-- so we'll
see if that turns up
01:05:31.650 --> 01:05:34.584
in the calculation.
01:05:34.584 --> 01:05:35.750
You've got a couple minutes.
01:05:35.750 --> 01:05:38.680
I want you to do
the money cards.
01:05:38.680 --> 01:05:45.050
Think about-- and
then on your way out,
01:05:45.050 --> 01:05:47.900
I think just pile them up
down here on the table,
01:05:47.900 --> 01:05:50.605
or hand them to me
or one of the TAs.
01:05:53.570 --> 01:05:57.610
And that'll help me
understand what you understood
01:05:57.610 --> 01:05:59.872
or didn't understand today.