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PROFESSOR: OK, let's get
on with today's lecture.
00:00:25.760 --> 00:00:31.850
And I want to look at a
variety of different problems,
00:00:31.850 --> 00:00:36.156
different classes of problems.
00:00:36.156 --> 00:00:38.530
We're going to look at four
different classes of problems
00:00:38.530 --> 00:00:43.490
and talk about the way you'd
go about approaching them.
00:00:43.490 --> 00:00:52.320
We need very few
fundamental laws.
00:00:52.320 --> 00:00:55.370
We use the same fundamental
laws again and again.
00:01:00.432 --> 00:01:02.890
And the issue is really, how
do you go about applying them?
00:01:02.890 --> 00:01:05.660
One is the sum-- this
fundamental law's
00:01:05.660 --> 00:01:08.145
for rigid bodies.
00:01:13.990 --> 00:01:17.470
So the fundamental laws,
they're always true.
00:01:17.470 --> 00:01:20.190
The sum of the external
forces, vectors.
00:01:26.290 --> 00:01:28.260
Time-rated change
of the momentum
00:01:28.260 --> 00:01:31.500
of the system with respect
to an inertial frame.
00:01:31.500 --> 00:01:36.090
And we recognize this
mass times acceleration.
00:01:36.090 --> 00:01:37.740
Newton's second law.
00:01:37.740 --> 00:01:43.875
The other one is the
summation of external torques.
00:01:47.340 --> 00:01:51.040
And that one is
the time derivative
00:01:51.040 --> 00:01:55.790
of angular momentum with respect
to a point which you choose--
00:01:55.790 --> 00:02:06.960
this is an A-- plus the
velocity of A with respect
00:02:06.960 --> 00:02:12.410
to the inertial frame crossed
with the momentum with respect
00:02:12.410 --> 00:02:13.730
to the inertial frame.
00:02:13.730 --> 00:02:15.780
So that's the torque equation.
00:02:15.780 --> 00:02:19.650
Now we have two special
cases of this where
00:02:19.650 --> 00:02:22.060
the second term goes away.
00:02:22.060 --> 00:02:26.860
One is when you are at
the center of gravity,
00:02:26.860 --> 00:02:30.040
in which case by definition
the velocity of that point
00:02:30.040 --> 00:02:32.020
and the momentum are
in the same direction,
00:02:32.020 --> 00:02:33.910
the cross product disappears.
00:02:33.910 --> 00:02:37.120
So there's two special
cases we use all the time
00:02:37.120 --> 00:02:39.980
when it simplifies to that.
00:02:39.980 --> 00:02:42.120
One is when you're at
the center of gravity.
00:02:42.120 --> 00:02:44.130
The other is when,
for whatever reason,
00:02:44.130 --> 00:02:46.190
this velocity is
parallel to that,
00:02:46.190 --> 00:02:47.960
and the cross
product goes to zero.
00:02:47.960 --> 00:02:50.510
And sometimes the velocity
is just plain zero.
00:02:50.510 --> 00:02:52.410
So there's some
certain cases that we
00:02:52.410 --> 00:02:54.359
use lots of times when
that term goes away.
00:02:54.359 --> 00:02:55.400
But sometimes it doesn't.
00:02:55.400 --> 00:02:57.090
You just have to put up with it.
00:03:01.080 --> 00:03:02.690
So those are our two
fundamental laws.
00:03:02.690 --> 00:03:05.840
And the question really is
how to go about applying them.
00:03:08.890 --> 00:03:11.806
So I'm going to look at
four classes of problems.
00:03:17.320 --> 00:03:19.090
Let's name them right now.
00:03:19.090 --> 00:03:28.780
Pure-- and we'll do simplest
to hardest-- pure rotation
00:03:28.780 --> 00:03:42.150
about a fixed axis through G,
through the center of mass.
00:03:42.150 --> 00:03:42.800
Pretty trivial.
00:03:42.800 --> 00:03:45.050
We can all do those
kind of problems.
00:03:45.050 --> 00:04:05.630
A second class, pure rotation
about a fixed axis at A,
00:04:05.630 --> 00:04:09.680
which is not equal to G
not at the center of mass.
00:04:09.680 --> 00:04:11.675
Again, pretty simple problems.
00:04:15.880 --> 00:04:34.420
Third class, no
external-- I'll call it
00:04:34.420 --> 00:04:35.516
no external constraints.
00:04:43.660 --> 00:04:46.410
We'll have to give an example
to see what I really mean.
00:04:46.410 --> 00:04:51.790
Well, I'll give you an example
right now, which we'll do.
00:04:51.790 --> 00:04:57.800
You have this hockey puck with
a string attached to it force.
00:04:57.800 --> 00:05:01.780
And this whole thing is
on a frictionless surface.
00:05:01.780 --> 00:05:05.760
So it's constrained so it
can't go through the surface.
00:05:05.760 --> 00:05:08.670
So no external constraints
in the direction
00:05:08.670 --> 00:05:09.840
that the motion can happen.
00:05:09.840 --> 00:05:12.340
So this is a 2D problem.
00:05:12.340 --> 00:05:15.050
Can move in the x,
y, and rotation.
00:05:15.050 --> 00:05:16.050
But it's not touching.
00:05:16.050 --> 00:05:17.920
There's no things
constraining it
00:05:17.920 --> 00:05:20.146
in the directions
of movement, which
00:05:20.146 --> 00:05:21.270
are allowed in the problem.
00:05:21.270 --> 00:05:23.230
That's just too much
words to write up here.
00:05:23.230 --> 00:05:24.690
So this kind of problem.
00:05:24.690 --> 00:05:51.935
And the fourth are problems with
moving points of constraint.
00:06:07.690 --> 00:06:12.610
You won't see a textbook with
sections broken out like this.
00:06:12.610 --> 00:06:14.070
I'm using my own terminology.
00:06:14.070 --> 00:06:15.460
I just made it up
last night as I
00:06:15.460 --> 00:06:17.780
was finishing off the lecture.
00:06:17.780 --> 00:06:20.680
But I'm trying to
give you some insight,
00:06:20.680 --> 00:06:24.040
and this is the way I
think about these things.
00:06:24.040 --> 00:06:27.060
So let's quickly
go through examples
00:06:27.060 --> 00:06:29.720
from the first couple
of types because they're
00:06:29.720 --> 00:06:30.810
especially easy.
00:06:33.840 --> 00:06:35.110
So what are we saying?
00:06:35.110 --> 00:06:37.110
They're pure rotation
about a fixed
00:06:37.110 --> 00:06:46.400
axis through G. Center of mass
is somewhere along this axle.
00:06:46.400 --> 00:06:52.420
The axle is fixed so
object can spin around it.
00:06:52.420 --> 00:06:57.210
And these kind of problems
are particularly simple,
00:06:57.210 --> 00:06:58.850
so I'm not going
to dwell on them.
00:06:58.850 --> 00:07:05.080
But here's class one,
class one problems.
00:07:05.080 --> 00:07:11.500
And it's basically-- these
are rotors, typically rotors,
00:07:11.500 --> 00:07:12.595
of almost all kinds.
00:07:17.120 --> 00:07:25.600
So for these problems, you use
momentum, angular momentum,
00:07:25.600 --> 00:07:27.175
with respect to
the center of mass.
00:07:29.910 --> 00:07:43.290
You can express it as IG times
omega x, omega y, omega z.
00:07:43.290 --> 00:07:45.140
All these problems
can be expressed
00:07:45.140 --> 00:07:49.270
with a moment of inertia matrix
times the rotation vector.
00:07:49.270 --> 00:07:52.740
And it will give you HX, HY, HZ.
00:07:52.740 --> 00:07:56.080
And you can use that
second formula up there.
00:07:56.080 --> 00:07:58.640
The torques are dH dt.
00:07:58.640 --> 00:08:01.890
So it's fixed through G. So
the second term doesn't appear.
00:08:04.460 --> 00:08:07.540
So to do these problems,
the sum of the-- see,
00:08:07.540 --> 00:08:09.520
I'll just give an example here.
00:08:09.520 --> 00:08:16.415
For 2D planar motion, they
get especially simple.
00:08:21.880 --> 00:08:31.170
Then, for 2D planar motion,
that means omega here is 0,
00:08:31.170 --> 00:08:37.590
0, and we'll let z
be the rotation axis.
00:08:37.590 --> 00:08:49.050
And this H then with respect
to G just becomes Izz omega z.
00:08:49.050 --> 00:08:54.090
All those 2D planar motion
problems boil down to this.
00:08:54.090 --> 00:09:09.130
And dHG dt then is Izz with
respect to G omega z do,
00:09:09.130 --> 00:09:12.705
or more familiar notation.
00:09:15.370 --> 00:09:19.240
So all those planar
motion problems, axis
00:09:19.240 --> 00:09:20.820
passing through G like this.
00:09:23.590 --> 00:09:31.300
For those problems,
does this matrix
00:09:31.300 --> 00:09:37.400
have to be with respect
to principal axes?
00:09:37.400 --> 00:09:41.560
Do you have your XYZ
coordinate system?
00:09:41.560 --> 00:09:44.940
To do this style
of problem, does
00:09:44.940 --> 00:09:48.869
that actually have to be
expressed in principle
00:09:48.869 --> 00:09:50.285
coordinates so
that it's diagonal?
00:09:55.510 --> 00:09:57.510
It doesn't have to be.
00:09:57.510 --> 00:10:00.330
You're worried about
rotation about z.
00:10:00.330 --> 00:10:02.950
You'll find that that
statement's still always true.
00:10:02.950 --> 00:10:05.180
Now, there may be-- the
thing could definitely
00:10:05.180 --> 00:10:08.560
have imbalances and have
unusual other torques.
00:10:08.560 --> 00:10:11.180
That will fall out
in the problem.
00:10:11.180 --> 00:10:16.710
But for the motion around the
axis of spin, this [INAUDIBLE].
00:10:20.650 --> 00:10:24.310
Where you only have a 1
omega z component, just
00:10:24.310 --> 00:10:28.210
the one component, you will
get-- it'll work out just fine.
00:10:28.210 --> 00:10:28.892
Yeah.
00:10:28.892 --> 00:10:32.748
AUDIENCE: If you do have some
kind of Ixz and Iyz terms,
00:10:32.748 --> 00:10:38.020
you would end up with more--
00:10:38.020 --> 00:10:39.920
PROFESSOR: You will
end up with-- first
00:10:39.920 --> 00:10:43.345
of all, if you have the
off-diagonal z terms, xz,
00:10:43.345 --> 00:10:48.630
yz terms, when you multiply that
out, you will find components
00:10:48.630 --> 00:10:54.160
of H that are in the z direction
as well as perhaps in the x
00:10:54.160 --> 00:10:55.070
and y.
00:10:55.070 --> 00:10:58.790
And those other two
components tell you
00:10:58.790 --> 00:11:02.610
that H is not pointing in the
same direction as omega, right?
00:11:02.610 --> 00:11:06.900
And that it tells you instantly
that the thing is dynamically
00:11:06.900 --> 00:11:11.530
unbalanced and will
have other torques that
00:11:11.530 --> 00:11:15.060
are trying to bend that axle.
00:11:15.060 --> 00:11:16.750
So they'll always appear.
00:11:20.872 --> 00:11:21.790
All right.
00:11:21.790 --> 00:11:22.450
Let's move on.
00:11:22.450 --> 00:11:25.650
I want to make sure we
get through this today.
00:11:25.650 --> 00:11:26.595
Class two problems.
00:11:32.060 --> 00:11:34.370
These are basically--
these are the pure rotation
00:11:34.370 --> 00:11:37.130
around some point
that's not through G.
00:11:37.130 --> 00:11:49.030
And again now, this is a fixed--
key here is fixed axis at A.
00:11:49.030 --> 00:11:50.290
It's not moving.
00:11:50.290 --> 00:11:52.510
This is what makes
these problems simpler.
00:11:52.510 --> 00:11:55.200
For these kinds of
problems, you do
00:11:55.200 --> 00:11:57.350
the sum of the
torques with respect
00:11:57.350 --> 00:11:58.935
to A, the external torques.
00:12:13.060 --> 00:12:16.510
And because that point's fixed,
the second terms don't appear,
00:12:16.510 --> 00:12:19.740
the v cross p terms.
00:12:19.740 --> 00:12:27.320
And you can write these as I,
a moment of inertia matrix,
00:12:27.320 --> 00:12:37.020
times whatever
the rotations are.
00:12:40.410 --> 00:12:43.750
In order to define the mass
moment of inertia matrix,
00:12:43.750 --> 00:12:48.930
you must have chosen
a set of coordinates
00:12:48.930 --> 00:12:52.350
attached to the body.
00:12:52.350 --> 00:12:54.370
And then with those
coordinates, you've
00:12:54.370 --> 00:12:57.490
computed the moments of
inertia for the body.
00:12:57.490 --> 00:13:01.170
And if you chose wisely, you
get principal coordinates
00:13:01.170 --> 00:13:03.720
and you only get diagonal
entries on the matrix.
00:13:03.720 --> 00:13:07.960
If you chose unwisely,
you will get other stuff.
00:13:07.960 --> 00:13:12.350
But it's still a valid mass
moment of inertia matrix.
00:13:12.350 --> 00:13:14.120
It just gives rise--
you have to deal
00:13:14.120 --> 00:13:15.950
with a bunch of other terms.
00:13:15.950 --> 00:13:22.490
So this will still
yield the same answer.
00:13:22.490 --> 00:13:25.170
How do you get I
with respect to A?
00:13:29.410 --> 00:13:33.700
These are opportunities when
you can use parallel axis.
00:13:33.700 --> 00:13:34.765
Yeah.
00:13:34.765 --> 00:13:40.545
AUDIENCE: Isn't I
omega just H dot dH dt?
00:13:40.545 --> 00:13:41.420
PROFESSOR: Excuse me.
00:13:45.280 --> 00:13:46.730
You need to finish that out.
00:13:46.730 --> 00:13:50.250
This is H. So the
time derivative of H
00:13:50.250 --> 00:13:52.220
is the time derivative
of this expression.
00:13:52.220 --> 00:13:54.450
You have to figure out
the moments of inertia
00:13:54.450 --> 00:13:55.970
and the rotation rates.
00:13:55.970 --> 00:14:01.230
And you may get multiple
terms, only one of which--
00:14:01.230 --> 00:14:08.920
let's say that
this will work out.
00:14:08.920 --> 00:14:10.230
You will get multiple terms.
00:14:10.230 --> 00:14:15.130
You will get torques that are
not in the direction of spin.
00:14:15.130 --> 00:14:18.280
Again, these might
be unbalanced.
00:14:18.280 --> 00:14:23.130
On the other hand, it may
be, for 2D problems, which
00:14:23.130 --> 00:14:31.960
is common-- so for the
2D planar motion, which
00:14:31.960 --> 00:14:37.430
most of the problems we do
are, then what you would do
00:14:37.430 --> 00:14:43.070
is you're saying omega
is, say, 0, 0, omega z,
00:14:43.070 --> 00:14:44.640
which simplifies that.
00:14:44.640 --> 00:14:47.800
We're multiplying this
thing out quite a lot.
00:14:47.800 --> 00:15:03.990
And if I with respect
to G is diagonal,
00:15:03.990 --> 00:15:07.535
then that means you your G
you chose principal axes.
00:15:12.740 --> 00:15:16.640
But then how do you get
to I with respect to A?
00:15:16.640 --> 00:15:21.690
For 2D problems,
really simple ones,
00:15:21.690 --> 00:15:23.791
how do you get to I
with respect to A?
00:15:23.791 --> 00:15:24.790
AUDIENCE: Parallel axis.
00:15:24.790 --> 00:15:27.690
PROFESSOR: That's
the classic case
00:15:27.690 --> 00:15:31.670
for using the
parallel axis theorem.
00:15:31.670 --> 00:15:36.320
So for these 2D planar motion
problems-- and planar motion
00:15:36.320 --> 00:15:47.120
problems, then you
can use parallel axis.
00:15:47.120 --> 00:15:48.040
I'll do an example.
00:15:48.040 --> 00:15:51.070
So this is kind of
the set up for this.
00:15:51.070 --> 00:15:53.330
An example of this
on the homework.
00:15:53.330 --> 00:15:57.845
What problem on the homework
is just perfect for this?
00:15:57.845 --> 00:16:02.054
It's 2D, planar motion,
about a fixed point.
00:16:02.054 --> 00:16:03.720
AUDIENCE: Circle with
the square cutout.
00:16:03.720 --> 00:16:07.380
PROFESSOR: Yeah, the cylinder,
this [INAUDIBLE] disk
00:16:07.380 --> 00:16:11.520
with the square cutout with
the pin at the top turning it
00:16:11.520 --> 00:16:13.440
into a pendulum.
00:16:13.440 --> 00:16:14.470
That's the fixed point.
00:16:17.440 --> 00:16:18.770
You can figure this out.
00:16:18.770 --> 00:16:20.870
You can use parallel axis.
00:16:20.870 --> 00:16:23.600
And we're going to do an
example right now that's almost
00:16:23.600 --> 00:16:26.120
identical to that problem.
00:16:26.120 --> 00:16:27.560
And we started it last time.
00:16:27.560 --> 00:16:32.190
So the example I want
to work is very similar.
00:16:32.190 --> 00:16:35.190
It's basically this
problem is that pendulum.
00:16:38.130 --> 00:16:41.170
So let's just work
it through quickly.
00:16:41.170 --> 00:16:44.830
We had done the setup last time.
00:16:58.080 --> 00:17:01.580
So last time I basically
derived an example
00:17:01.580 --> 00:17:05.589
of the parallel axis theorem
for my little stick here.
00:17:05.589 --> 00:17:12.575
And I'll give you some
geometry, some values.
00:17:17.609 --> 00:17:26.690
So here's G. There's the point
I wanted to rotate about, A.
00:17:26.690 --> 00:17:31.650
The distance between
these two points is d.
00:17:31.650 --> 00:17:33.920
That's going to pop up in
my parallel axis theorem.
00:17:36.924 --> 00:17:39.920
We've got a set of
coordinates here.
00:17:39.920 --> 00:17:42.140
My G is, of course, in
the center of this thing,
00:17:42.140 --> 00:17:43.240
the geometric center.
00:17:43.240 --> 00:17:47.870
So I have a body
fixed set of axes,
00:17:47.870 --> 00:17:51.000
which I'm going to call z.
00:17:51.000 --> 00:17:55.520
And my x is in this direction.
00:17:55.520 --> 00:18:00.860
So that would make my
y going into the board.
00:18:00.860 --> 00:18:02.720
So the y is kind of like that.
00:18:06.520 --> 00:18:09.400
And this has some properties.
00:18:09.400 --> 00:18:12.870
The dimension in
this direction means
00:18:12.870 --> 00:18:17.530
A. The direction in
this dimension is B.
00:18:17.530 --> 00:18:30.630
So it's a width of A, a
thickness B, and a length L.
00:18:30.630 --> 00:18:34.580
So when you compute, these
are-- ah, symmetry now.
00:18:37.830 --> 00:18:40.120
So the axes that I've
chosen for this problem
00:18:40.120 --> 00:18:41.175
are they principal axes.
00:18:47.070 --> 00:18:50.160
There's three planes of
symmetry in this problem,
00:18:50.160 --> 00:18:52.615
and I've got one principal
axis perpendicular
00:18:52.615 --> 00:18:53.840
to every one of them.
00:18:53.840 --> 00:18:56.680
And all three pass through
the center of mass,
00:18:56.680 --> 00:18:59.300
and they're orthogonal
to one another.
00:18:59.300 --> 00:19:02.160
So those conditions
are all satisfied
00:19:02.160 --> 00:19:06.890
for these to be principal
axes for this rectangular body
00:19:06.890 --> 00:19:07.973
and its uniform density.
00:19:12.820 --> 00:19:16.580
So I'll give you the--
for bodies like this,
00:19:16.580 --> 00:19:19.060
in your book or any
book on dynamics,
00:19:19.060 --> 00:19:26.322
you can look up then
the properties, the Izz.
00:19:26.322 --> 00:19:27.780
And we're going to
spin this thing.
00:19:27.780 --> 00:19:32.610
This thing, we're going to
have it rotating about-- which
00:19:32.610 --> 00:19:34.240
one am I going to use?
00:19:34.240 --> 00:19:36.180
Yeah, around the z-axis.
00:19:36.180 --> 00:19:37.320
That's how I'll set it up.
00:19:37.320 --> 00:19:39.190
That's the way I
drilled my hole,
00:19:39.190 --> 00:19:41.760
so it's going back and forth.
00:19:41.760 --> 00:19:44.100
So that the wide part
of it's this way.
00:19:47.500 --> 00:19:59.070
So Izz with respect to G M L
squared plus a squared over 12.
00:20:03.740 --> 00:20:04.240
Iyy.
00:20:29.090 --> 00:20:35.190
OK, now just to make a
point, the dimensions
00:20:35.190 --> 00:20:50.560
L 32.1 centimeters,
a 4.71, b, 1.25.
00:20:50.560 --> 00:20:54.990
And eventually my
d, this offset would
00:20:54.990 --> 00:21:02.470
be, for example, where I've
drilled that hole, is at 10.2.
00:21:02.470 --> 00:21:04.140
The point I want
to make here, lots
00:21:04.140 --> 00:21:06.660
of times it would be
nice if I could just
00:21:06.660 --> 00:21:10.440
make the simplification
to call this a slender rod
00:21:10.440 --> 00:21:14.410
and be able to ignore these
a and b dimensions in this,
00:21:14.410 --> 00:21:15.960
just to get quick answers.
00:21:15.960 --> 00:21:19.600
Do you think that's
slender enough?
00:21:19.600 --> 00:21:22.780
It's not even 10 times--
the length of the width
00:21:22.780 --> 00:21:23.630
here isn't even 10.
00:21:23.630 --> 00:21:27.330
It's probably six or seven.
00:21:27.330 --> 00:21:29.810
So the key issue then,
if you look at this,
00:21:29.810 --> 00:21:33.870
is really what's the ratio
of a squared to L squared?
00:21:33.870 --> 00:21:36.650
That would tell you something
about the relative importance
00:21:36.650 --> 00:21:39.540
of the a squared term
to the L squared.
00:21:39.540 --> 00:21:41.170
So let me tell you about that.
00:21:41.170 --> 00:21:51.350
So a squared over
L squared is 0.022.
00:21:51.350 --> 00:21:59.500
b squared over L
squared is 0.002.
00:21:59.500 --> 00:22:02.600
So even with this
kind of fat stick,
00:22:02.600 --> 00:22:05.460
the approximation of ML
squared over 12 is pretty good.
00:22:05.460 --> 00:22:07.500
It's only 2% off.
00:22:07.500 --> 00:22:09.920
And this approximation,
L squared over 12,
00:22:09.920 --> 00:22:14.440
is less than 2/10
of a percent off.
00:22:14.440 --> 00:22:17.600
So for roughly slender
things, we oftentimes
00:22:17.600 --> 00:22:20.880
just say ML squared over 12
for spin about their center.
00:22:26.410 --> 00:22:29.370
We now need to
apply parallel axis.
00:22:29.370 --> 00:22:32.410
I want to spin this around,
let this rotate around,
00:22:32.410 --> 00:22:36.000
not around G, but now around
a point off to the side.
00:22:36.000 --> 00:22:43.790
So we worked out last time
that Izz with respect to A
00:22:43.790 --> 00:22:50.055
is IzzG plus Md squared.
00:23:10.520 --> 00:23:12.730
So in this particular
problem, this Izz about G
00:23:12.730 --> 00:23:16.100
is approximately
ML squared over 12.
00:23:16.100 --> 00:23:20.500
So just by way of example,
to see what we might find out
00:23:20.500 --> 00:23:25.780
here, is let's let d equal L/2.
00:23:25.780 --> 00:23:28.560
That would be if
I move this-- if I
00:23:28.560 --> 00:23:30.300
put my hole right
at the very top,
00:23:30.300 --> 00:23:31.466
how would this thing behave?
00:23:31.466 --> 00:23:34.810
I don't have a hole right at
the top, but I have one close.
00:23:34.810 --> 00:23:38.310
So this is just because the
numbers are easy to work.
00:23:38.310 --> 00:23:41.850
What happens if you put
in L/2 into this formula?
00:23:41.850 --> 00:23:53.650
Well then IzzA is ML squared
over 12 plus M. L over 2
00:23:53.650 --> 00:23:55.150
squared is L squared over 4.
00:23:59.600 --> 00:24:04.510
And that's ML
squared over 3, which
00:24:04.510 --> 00:24:06.960
is a number you'll run into
again and again and again
00:24:06.960 --> 00:24:09.680
in mechanical engineering
because examples like this
00:24:09.680 --> 00:24:10.350
are used a lot.
00:24:10.350 --> 00:24:13.630
The mass moment of inertia
about a slender rod
00:24:13.630 --> 00:24:15.980
pinned at its end,
ML squared over 3.
00:24:34.170 --> 00:24:38.380
So let's take this problem a
little more towards completion.
00:24:38.380 --> 00:24:47.275
The sum of the external moments
with respect to A dHA dt.
00:24:52.580 --> 00:24:56.580
And that's going to be d by dt.
00:24:56.580 --> 00:25:02.420
In this problem, the
only rotation is omega z.
00:25:02.420 --> 00:25:10.380
So this is going to be Izz
with respect to A omega z.
00:25:13.090 --> 00:25:22.380
And that just gives us IzzA
omega z dot, or more familiar,
00:25:22.380 --> 00:25:28.460
IzzA theta double
bond, if you want.
00:25:28.460 --> 00:25:29.730
Here's our problem.
00:25:29.730 --> 00:25:33.770
It simplifies to
this slender rod.
00:25:33.770 --> 00:25:38.600
And let me do the
more general case.
00:25:38.600 --> 00:25:39.970
Here's my rod.
00:25:39.970 --> 00:25:44.300
The pivot point that it's
going around is here.
00:25:44.300 --> 00:25:50.310
This is d still, and this is G.
And it's swinging with respect
00:25:50.310 --> 00:25:51.100
to this point.
00:25:51.100 --> 00:25:56.230
So here's the angle theta.
00:25:56.230 --> 00:25:59.160
So it swings about this
point that you've fixed,
00:25:59.160 --> 00:26:06.490
and that point is d above
the center of gravity,
00:26:06.490 --> 00:26:07.370
center of mass.
00:26:12.710 --> 00:26:18.670
So what are the external
moments about this point?
00:26:18.670 --> 00:26:20.960
There's no torque
right at the point,
00:26:20.960 --> 00:26:26.180
but our free body diagram
of drawing this as a--
00:26:26.180 --> 00:26:28.840
here's our point of rotation.
00:26:28.840 --> 00:26:33.590
Here it is displaced
through an angle theta.
00:26:33.590 --> 00:26:36.835
The weight of the object
can all be concentrated,
00:26:36.835 --> 00:26:39.210
thought of, for the
purposes of the free body
00:26:39.210 --> 00:26:42.950
diagram as acting through G.
00:26:42.950 --> 00:26:49.970
So here's the mass at G,
gravity acting down on it.
00:26:49.970 --> 00:26:56.980
And the length of this arm here
about which it's swinging is D.
00:26:56.980 --> 00:27:04.160
So the torque about this is--
and it's pulling it back--
00:27:04.160 --> 00:27:12.297
minus Mgd sine theta.
00:27:12.297 --> 00:27:14.130
Probably you've seen
that many times before,
00:27:14.130 --> 00:27:17.960
including the recent homework.
00:27:17.960 --> 00:27:25.650
And that must be equal to
Izz about A theta double dot.
00:27:25.650 --> 00:27:29.190
So we have an
equation of motion.
00:27:29.190 --> 00:27:30.695
Just collecting
the terms together.
00:27:40.900 --> 00:27:43.900
So this is a oscillator
that, for this problem,
00:27:43.900 --> 00:27:47.260
has no external excitation.
00:27:47.260 --> 00:27:50.570
This is its equation of motion.
00:27:50.570 --> 00:27:51.850
And I need theta double dot.
00:27:54.670 --> 00:27:55.610
But is it linear?
00:27:59.714 --> 00:28:00.500
Is it linear?
00:28:00.500 --> 00:28:03.230
No, it's not a linear
equation of motion.
00:28:03.230 --> 00:28:05.260
But for sure, it
is an oscillator.
00:28:08.180 --> 00:28:12.190
And for anything
that vibrates, you
00:28:12.190 --> 00:28:14.630
can have lots of
nonlinear problems
00:28:14.630 --> 00:28:16.940
that exhibit vibration.
00:28:16.940 --> 00:28:20.280
You can think of them--
you can pose problems
00:28:20.280 --> 00:28:25.280
where you say, OK, what's their
static equilibrium position?
00:28:25.280 --> 00:28:28.440
And think of a very small motion
about that static equilibrium
00:28:28.440 --> 00:28:29.270
position.
00:28:29.270 --> 00:28:33.570
You can always linearize about
the static equilibrium position
00:28:33.570 --> 00:28:36.270
and be able to come up with a
linearized equation of motion
00:28:36.270 --> 00:28:40.090
that at least from that you can
calculate the natural frequency
00:28:40.090 --> 00:28:45.710
of the system for small motions
around its static equilibrium
00:28:45.710 --> 00:28:46.210
position.
00:28:46.210 --> 00:28:48.420
So in this case, it's
pretty easy to do.
00:28:48.420 --> 00:28:55.215
And you've seen it
before for small theta.
00:28:58.450 --> 00:29:01.070
Sine theta is approximately
equal to theta.
00:29:01.070 --> 00:29:04.630
So we are going to
linearize the equation.
00:29:04.630 --> 00:29:09.140
This theta equals 0 is a
static equilibrium position.
00:29:09.140 --> 00:29:11.710
So we're linearizing around 0.
00:29:11.710 --> 00:29:15.280
And around 0, that's
the approximation.
00:29:15.280 --> 00:29:21.230
So you just substitute
that in, IzzA theta double
00:29:21.230 --> 00:29:26.960
dot plus Mgd theta equals 0.
00:29:26.960 --> 00:29:29.160
There's your linearized
equation of motion.
00:29:29.160 --> 00:29:31.980
I want an estimate of
the natural frequency.
00:29:31.980 --> 00:29:36.300
So find omega n.
00:29:41.010 --> 00:29:42.870
So this is basically
entering into solving
00:29:42.870 --> 00:29:44.220
differential equations.
00:29:44.220 --> 00:29:47.800
But I let mother nature
tell me what the answer is.
00:29:47.800 --> 00:29:50.230
I do the example, and
I say it oscillates.
00:29:50.230 --> 00:29:54.297
Looks a lot like
sine omega t to me.
00:29:54.297 --> 00:29:55.630
Plug in [INAUDIBLE] to make a t.
00:29:55.630 --> 00:29:56.390
Let's find out.
00:30:00.450 --> 00:30:06.080
Some theta amplitude
sine omega t.
00:30:06.080 --> 00:30:08.110
Plug it in.
00:30:08.110 --> 00:30:13.720
So you plug that in [INAUDIBLE]
and you get minus omega squared
00:30:13.720 --> 00:30:21.860
IzzA plus Mgd.
00:30:21.860 --> 00:30:24.700
And all of this, you
can factor out the theta
00:30:24.700 --> 00:30:30.400
0 sine omega t equals 0.
00:30:30.400 --> 00:30:32.150
That's what you get.
00:30:32.150 --> 00:30:34.240
And in general, theta,
that's not 0, or else
00:30:34.240 --> 00:30:37.590
you'd have a trivial
problem, not moving at all.
00:30:37.590 --> 00:30:41.840
But in order for this equation--
and this can be anything.
00:30:41.840 --> 00:30:45.360
We're doing the 0 minus 1 and
plus 1 depending on the time.
00:30:45.360 --> 00:30:47.650
So in order to
satisfy this equation,
00:30:47.650 --> 00:30:54.150
this part inside of the
parentheses has to be 0.
00:30:54.150 --> 00:30:58.120
And when you just
solve that, you
00:30:58.120 --> 00:31:04.365
find that omega squared
equals Mgd/IzzA.
00:31:12.600 --> 00:31:17.490
And the reason I've gone to
the bother of working this out
00:31:17.490 --> 00:31:22.900
in detail right to the end
is that every one degree
00:31:22.900 --> 00:31:28.400
of freedom rotational oscillator
that you will ever encounter--
00:31:28.400 --> 00:31:40.080
sticks, wheels with
static imbalances.
00:31:40.080 --> 00:31:41.148
Let me show you this one.
00:31:45.730 --> 00:31:47.650
It's an oscillator too.
00:31:47.650 --> 00:31:51.300
Any one degree of
freedom oscillator,
00:31:51.300 --> 00:31:53.840
rotational oscillator,
pendulum-- basically
00:31:53.840 --> 00:31:56.412
all pendula-- this
is the formula
00:31:56.412 --> 00:31:57.495
for the natural frequency.
00:32:02.610 --> 00:32:06.781
So it's going to be of
that form for any pendulum.
00:32:06.781 --> 00:32:11.465
So any 1dof pendulum.
00:32:14.740 --> 00:32:17.000
This is the generic answer.
00:32:19.520 --> 00:32:23.920
So that cutout problem for
today has to come down to this.
00:32:23.920 --> 00:32:27.370
Or this is the distance
between the mass center
00:32:27.370 --> 00:32:30.320
and the point of rotation.
00:32:30.320 --> 00:32:31.960
And that's your mass
moment of inertia
00:32:31.960 --> 00:32:34.096
about the point of rotation.
00:32:34.096 --> 00:32:38.110
So that's worth
knowing that one.
00:32:38.110 --> 00:32:39.770
OK, keep moving.
00:32:49.913 --> 00:32:53.120
Now things will begin
to get interesting.
00:32:53.120 --> 00:32:59.340
These latter two classes
are harder conceptually,
00:32:59.340 --> 00:33:01.590
but once you have a
solution method for them,
00:33:01.590 --> 00:33:03.690
they're not all that hard.
00:33:03.690 --> 00:33:08.080
This one is the problem I
described at the beginning.
00:33:08.080 --> 00:33:10.030
We've got this hockey
puck like thing.
00:33:13.270 --> 00:33:16.640
And the string wrapped
around it pulling on it
00:33:16.640 --> 00:33:17.840
with a known force.
00:33:17.840 --> 00:33:21.680
In this problem, they
call it 150 newtons.
00:33:21.680 --> 00:33:27.086
The mass of this
thing is 75 kilograms.
00:33:30.870 --> 00:33:33.135
It's on a frictionless surface.
00:33:36.550 --> 00:33:46.350
And we want to find
its acceleration
00:33:46.350 --> 00:33:51.800
of the center of
mass and the rotation
00:33:51.800 --> 00:33:54.380
around the center of mass.
00:33:54.380 --> 00:33:58.280
So find theta double dot and
find the linear acceleration.
00:33:58.280 --> 00:34:00.960
That's basically the
name of the problem.
00:34:00.960 --> 00:34:06.630
And they give you that it's
75 kilograms, 150 newtons,
00:34:06.630 --> 00:34:15.449
and kappa, the radius of
gyration, is 0.15 meters.
00:34:15.449 --> 00:34:21.404
This is defined as the
radius of gyration.
00:34:24.870 --> 00:34:28.320
So what's that?
00:34:28.320 --> 00:34:29.420
It's a radius of gyration.
00:34:29.420 --> 00:34:34.320
It's really appropriate,
really only useful,
00:34:34.320 --> 00:34:39.949
for mostly 2D
rotational problems
00:34:39.949 --> 00:34:41.960
around an axis of rotation.
00:34:41.960 --> 00:34:51.480
And what it means is this is the
distance away from the center
00:34:51.480 --> 00:34:56.139
of rotation at which you could
concentrate all of the mass
00:34:56.139 --> 00:35:01.580
and have the same mass
moment of inertia.
00:35:01.580 --> 00:35:06.260
So this has-- here's G here
at the center, uniform disk.
00:35:09.040 --> 00:35:17.330
We know that there is an Izz
about G for this problem.
00:35:17.330 --> 00:35:19.260
And I need to pick a
coordinate system so we
00:35:19.260 --> 00:35:20.870
can talk about things here.
00:35:20.870 --> 00:35:23.570
Let me get M out of the center.
00:35:23.570 --> 00:35:27.510
So I'm going to
let z be upwards.
00:35:27.510 --> 00:35:30.690
And because I'm
looking ahead and want
00:35:30.690 --> 00:35:36.140
to keep the equation simple,
I'm going to make my x-axis
00:35:36.140 --> 00:35:41.590
here parallel to f so I only
have to deal with one vector
00:35:41.590 --> 00:35:43.100
component equation.
00:35:43.100 --> 00:35:45.830
And then that makes
the y-axis this way.
00:35:51.550 --> 00:35:56.280
So I'm interested in Izz because
I'm spinning around the z-axis.
00:35:56.280 --> 00:35:59.100
I know that for a uniform
disk, that's a principal axis,
00:35:59.100 --> 00:36:01.870
is the vertical one.
00:36:01.870 --> 00:36:05.830
And what I'm saying is
that you can then set find.
00:36:05.830 --> 00:36:12.990
There's an IzzG that can be
expressed as M kappa squared.
00:36:12.990 --> 00:36:21.700
So kappa, in effect, is just
IzzG over M square root.
00:36:21.700 --> 00:36:23.450
Now, why do we use
that kind of thing?
00:36:23.450 --> 00:36:28.009
Well, the way this
problem was set up-- I
00:36:28.009 --> 00:36:29.300
actually took it out of a book.
00:36:29.300 --> 00:36:32.170
It wasn't a uniform disk.
00:36:32.170 --> 00:36:34.230
It's a pulley
wheel or something.
00:36:34.230 --> 00:36:40.760
And it's got spokes
in here and a rim.
00:36:40.760 --> 00:36:47.577
It's still axially-- it still
has some axial symmetries.
00:36:47.577 --> 00:36:48.910
But it's getting a little messy.
00:36:48.910 --> 00:36:50.326
It's hard for you--
you can't just
00:36:50.326 --> 00:36:52.620
say that's MR squared over 2.
00:36:52.620 --> 00:36:56.250
It's got some other mass moment
of inertia about the center.
00:36:56.250 --> 00:36:59.670
But it's got holes and stuff
in it because of the spokes.
00:36:59.670 --> 00:37:06.010
So oftentimes, you'll be
given the radius of gyration
00:37:06.010 --> 00:37:08.650
because it's a little
difficult to give you
00:37:08.650 --> 00:37:13.316
a mathematical description
of what the actual Izz is.
00:37:13.316 --> 00:37:14.440
That's often why you do it.
00:37:14.440 --> 00:37:17.346
And the thing is,
you can measure.
00:37:17.346 --> 00:37:18.720
Rather than try
to calculate, you
00:37:18.720 --> 00:37:21.320
can actually just
measure the mass moment
00:37:21.320 --> 00:37:22.960
of inertia of something.
00:37:22.960 --> 00:37:26.060
So how would I measure-- let's
say I didn't know any formulas,
00:37:26.060 --> 00:37:28.510
but how would I measure
the mass moment of inertia
00:37:28.510 --> 00:37:30.052
of this in the z direction?
00:37:36.450 --> 00:37:38.055
What experiment would you do?
00:37:38.055 --> 00:37:41.310
AUDIENCE: [INAUDIBLE]
angular acceleration.
00:37:41.310 --> 00:37:42.740
PROFESSOR: She
says apply torque.
00:37:42.740 --> 00:37:44.830
Measure the angular
acceleration.
00:37:44.830 --> 00:37:46.620
Hang a weight off
of it, known weight.
00:37:46.620 --> 00:37:48.590
Wrap a string around it.
00:37:48.590 --> 00:37:49.400
Known mass.
00:37:49.400 --> 00:37:52.120
Known G. Known torque
around the center.
00:37:52.120 --> 00:37:54.620
Measure the angular
acceleration.
00:37:54.620 --> 00:37:56.955
I theta double dot
equals the torque.
00:37:56.955 --> 00:37:59.505
You know the torque,
you know the measure
00:37:59.505 --> 00:38:00.810
of the theta double dot.
00:38:00.810 --> 00:38:03.910
Calculate I. I equals M kappa
squared, and you could just
00:38:03.910 --> 00:38:06.910
say, well, the kappa
for this system is.
00:38:06.910 --> 00:38:08.900
That's how you use it.
00:38:08.900 --> 00:38:11.380
I'll give you a very
common example, really hard
00:38:11.380 --> 00:38:13.180
to calculate mass
moment of inertia.
00:38:13.180 --> 00:38:15.679
A marine propeller.
00:38:15.679 --> 00:38:17.970
You actually do want to know
the mass moment of inertia
00:38:17.970 --> 00:38:21.400
about its center for purposes
of torsional oscillations
00:38:21.400 --> 00:38:23.290
on the shaft, et cetera.
00:38:23.290 --> 00:38:24.620
Hard to calculate.
00:38:24.620 --> 00:38:25.590
Pretty easy to measure.
00:38:28.140 --> 00:38:32.700
So how many degrees of
freedom does this problem has?
00:38:32.700 --> 00:38:36.980
Well, when we say it's
2D, it's a rigid body.
00:38:36.980 --> 00:38:40.870
But it's 2D, which means
it's lying on a plane.
00:38:40.870 --> 00:38:42.290
It's a planar motion problem.
00:38:42.290 --> 00:38:44.390
Only allowed to rotate in z.
00:38:44.390 --> 00:38:48.280
Not allowed to rotate in around
the x-axis or the y-axis.
00:38:48.280 --> 00:38:52.570
So for the rigid body, six
degrees of freedom possible.
00:38:52.570 --> 00:38:55.340
There's two immediately that
you said it can't rotate.
00:38:55.340 --> 00:38:57.770
So we're down to four.
00:38:57.770 --> 00:39:06.070
It cannot translate
in the z direction.
00:39:06.070 --> 00:39:09.750
We're down to three.
00:39:09.750 --> 00:39:11.300
So that leaves us what?
00:39:14.120 --> 00:39:18.340
So the degrees of freedom
for this problem is
00:39:18.340 --> 00:39:20.822
6 minus 3 constraints is 3.
00:39:20.822 --> 00:39:23.030
That means we have to have
three equations of motion.
00:39:23.030 --> 00:39:25.520
And they would account for
what are the possible--
00:39:25.520 --> 00:39:26.895
in other words,
saying this, what
00:39:26.895 --> 00:39:29.170
are the possible motions
now of this problem?
00:39:29.170 --> 00:39:30.520
AUDIENCE: [INAUDIBLE].
00:39:30.520 --> 00:39:31.935
PROFESSOR: x, y, and z.
00:39:31.935 --> 00:39:33.030
Or rotation in z.
00:39:52.260 --> 00:39:54.140
So notice we've
set the problem up.
00:39:54.140 --> 00:40:01.160
Now to go about solving it,
we need a free body diagram.
00:40:01.160 --> 00:40:03.530
So here's my disk.
00:40:03.530 --> 00:40:04.650
Here's the force.
00:40:08.660 --> 00:40:13.420
Any other-- and there certainly
has weight in the z direction,
00:40:13.420 --> 00:40:17.320
but there's no z acceleration.
00:40:17.320 --> 00:40:20.240
So in the plane of
the board, that's
00:40:20.240 --> 00:40:21.970
the only external
forces acting on this.
00:40:25.540 --> 00:40:28.690
And there's our G.
00:40:28.690 --> 00:40:31.510
Now this problem-- and
I'll say generalize
00:40:31.510 --> 00:40:32.580
on this in a few minutes.
00:40:32.580 --> 00:40:40.030
This problem can always
be restated as-- recast,
00:40:40.030 --> 00:40:42.710
let me put it that way--
as, here's your point
00:40:42.710 --> 00:40:49.060
G with a force acting
on this center of mass.
00:40:49.060 --> 00:40:52.209
See, this force doesn't go
through the center of mass.
00:40:52.209 --> 00:40:54.000
This force goes through
the center of mass.
00:40:54.000 --> 00:40:58.070
I'm going to replace that
problem with this problem.
00:40:58.070 --> 00:41:02.760
A pure moment acting
about the center of mass.
00:41:02.760 --> 00:41:04.600
You can always make
this transition.
00:41:04.600 --> 00:41:07.490
And I'll do the general case
for you in just a minute.
00:41:07.490 --> 00:41:08.690
But you can always do this.
00:41:11.230 --> 00:41:14.330
So that's kind of my
second point here.
00:41:14.330 --> 00:41:19.810
Third point, we need then--
this is our free body diagram.
00:41:19.810 --> 00:41:25.211
We need apply our
laws of motion.
00:41:25.211 --> 00:41:31.420
So sum of the forces in
the y are-- now remember,
00:41:31.420 --> 00:41:33.880
this is z coming
out of the board.
00:41:33.880 --> 00:41:36.930
y this way, x this way.
00:41:36.930 --> 00:41:40.210
Sum of the forces in the y?
00:41:40.210 --> 00:41:45.570
Zero, M acceleration of G
in the y direction, zero.
00:41:45.570 --> 00:41:48.020
So you know there's no
acceleration in the y.
00:41:48.020 --> 00:41:49.910
So it has three
degrees of freedom.
00:41:49.910 --> 00:41:52.340
That's the first
equation of motion.
00:41:52.340 --> 00:41:57.020
It gives you a trivial result.
So y double dot is zero.
00:41:57.020 --> 00:41:59.647
That's your first
equation of motion.
00:41:59.647 --> 00:42:01.480
Then you have the second
equation of motion,
00:42:01.480 --> 00:42:04.620
sum of the forces
in the x direction
00:42:04.620 --> 00:42:09.090
equals just our F i hat.
00:42:09.090 --> 00:42:11.480
It's positive x
direction because we
00:42:11.480 --> 00:42:15.310
were clever in how we set
up the coordinate system.
00:42:15.310 --> 00:42:21.450
And that's got to be Mx
double dot, i hat direction.
00:42:21.450 --> 00:42:26.020
So we know right away that
x double dot is the force
00:42:26.020 --> 00:42:33.410
F divided by M. And that's
150 newtons over 75 kilograms,
00:42:33.410 --> 00:42:37.740
or 2 meters per second squared.
00:42:37.740 --> 00:42:38.920
All right.
00:42:38.920 --> 00:42:48.600
The third one,
then, is the moment
00:42:48.600 --> 00:42:52.580
of inertia with respect
to G in this problem--
00:42:52.580 --> 00:43:02.254
excuse me, the angular momentum
in some IzzG times omega z.
00:43:02.254 --> 00:43:03.670
And that's what
we're looking for.
00:43:03.670 --> 00:43:10.920
So this is another way of--
IzzG theta dot k hat direction.
00:43:16.800 --> 00:43:20.680
And we're going to apply that
the external torques, some
00:43:20.680 --> 00:43:28.000
of the torques, is
dH respect to G dt.
00:43:28.000 --> 00:43:33.620
And that's going to give
us IzzG theta double dot.
00:43:38.720 --> 00:43:40.690
So this third class
of problems you
00:43:40.690 --> 00:43:45.110
are best just working with
respect to the center of mass.
00:43:45.110 --> 00:43:46.360
That's kind of the point here.
00:43:46.360 --> 00:43:51.550
There's no points of contact.
00:43:51.550 --> 00:43:54.110
There's just known
external forces.
00:43:54.110 --> 00:43:55.650
You have to deal with them.
00:43:55.650 --> 00:43:59.360
Do your work with respect
to the center of mass.
00:43:59.360 --> 00:44:02.280
So we have force equations.
00:44:02.280 --> 00:44:04.720
We have moment equations.
00:44:04.720 --> 00:44:10.350
And basically you know
Izz for this problem
00:44:10.350 --> 00:44:12.460
is kappa squared M.
And you're given M,
00:44:12.460 --> 00:44:14.200
and you're given kappa.
00:44:14.200 --> 00:44:18.790
So you can now-- and what
we have to-- actually,
00:44:18.790 --> 00:44:21.090
the last thing left here is
to figure out the torque.
00:44:21.090 --> 00:44:24.400
What's the torque?
00:44:24.400 --> 00:44:29.410
Well, it's R in this
direction crossed
00:44:29.410 --> 00:44:30.990
with F in that direction.
00:44:30.990 --> 00:44:41.670
So it's Rj cross with
Fi, j cross i, minus RF,
00:44:41.670 --> 00:44:48.320
minus RF, k hat.
00:44:48.320 --> 00:44:57.160
So you can now solve for theta
double dot as RF over Izz,
00:44:57.160 --> 00:44:59.480
or RF over M kappa squared.
00:45:22.620 --> 00:45:26.890
And that says-- minus says
it's rotating this way, which
00:45:26.890 --> 00:45:28.430
is what you'd expect.
00:45:28.430 --> 00:45:30.570
Right?
00:45:30.570 --> 00:45:32.990
Now I meant to ask you a
question before we started.
00:45:32.990 --> 00:45:34.222
But think about this.
00:45:34.222 --> 00:45:36.180
If I had, right at the
beginning, had said, OK,
00:45:36.180 --> 00:45:37.020
this is a problem.
00:45:39.850 --> 00:45:43.220
If I grab this string and
pull in this direction,
00:45:43.220 --> 00:45:45.717
will there be any motion
in the y direction?
00:45:45.717 --> 00:45:46.800
I meant to start that way.
00:45:46.800 --> 00:45:48.930
I'm really kicking myself
for not doing that.
00:45:48.930 --> 00:45:50.680
Because a lot of people
think that there's possible,
00:45:50.680 --> 00:45:52.420
that it could move
off in the y direction
00:45:52.420 --> 00:45:56.480
because it's not being
loaded symmetrically.
00:45:56.480 --> 00:45:58.100
You're pulling on a side.
00:45:58.100 --> 00:46:01.127
Some people think it'll kind
of try to move away like that.
00:46:01.127 --> 00:46:01.960
It doesn't, does it?
00:46:21.200 --> 00:46:24.026
So an important generalization.
00:46:24.026 --> 00:46:25.315
We've got a rigid body.
00:46:28.780 --> 00:46:30.310
You have a force acting on it.
00:46:33.390 --> 00:46:34.910
Has a mass center here.
00:46:37.460 --> 00:46:43.150
So perpendicular to that
force is some distance.
00:46:43.150 --> 00:46:43.980
We'll call it d.
00:46:46.880 --> 00:46:53.450
You can always equate this
problem to-- and set it up
00:46:53.450 --> 00:46:59.306
as-- a force.
00:46:59.306 --> 00:47:00.680
Conceptually, you
can think of it
00:47:00.680 --> 00:47:02.544
as equal and opposite forces.
00:47:02.544 --> 00:47:03.710
But it's cancel one another.
00:47:03.710 --> 00:47:06.580
It's just like I've done
nothing to this problem.
00:47:06.580 --> 00:47:12.880
And then a force acting
at this distance F.
00:47:12.880 --> 00:47:14.060
This is our distance d.
00:47:14.060 --> 00:47:16.520
So this problem is
identical to that problem.
00:47:16.520 --> 00:47:19.095
I've just added and taken
away two more forces.
00:47:23.510 --> 00:47:25.450
So the total forces
on the system
00:47:25.450 --> 00:47:31.030
are still F. And there's still
an F operating at a lever arm
00:47:31.030 --> 00:47:32.310
d.
00:47:32.310 --> 00:47:36.980
But now, if I had put
these two together,
00:47:36.980 --> 00:47:38.600
they are equal and opposite.
00:47:38.600 --> 00:47:43.630
And they form a couple, a
moment, acting like that.
00:47:43.630 --> 00:47:53.170
So this is equivalent to--
there's G with an F on it
00:47:53.170 --> 00:47:59.480
and a moment M0.
00:47:59.480 --> 00:48:04.750
And this M0 is my D cross F.
00:48:04.750 --> 00:48:08.240
So that's the generalization
for what I did up here.
00:48:08.240 --> 00:48:10.730
I went from that to that.
00:48:10.730 --> 00:48:12.260
And this is why you can do that.
00:48:15.400 --> 00:48:22.180
And so now if you have an object
and lots of different forces
00:48:22.180 --> 00:48:26.570
acting on it, and this is
Fi down here and here's G,
00:48:26.570 --> 00:48:32.770
you can draw a radius
from G to this point.
00:48:32.770 --> 00:48:40.480
So I'll call that Ri
with respect to G. Then
00:48:40.480 --> 00:48:48.840
the way to generalize this is
that this is equal to some F
00:48:48.840 --> 00:49:00.050
total and a moment acting at G.
00:49:00.050 --> 00:49:05.170
And all that you have
to do there is F total
00:49:05.170 --> 00:49:09.430
is the summation of
the Fi's, vector sum.
00:49:09.430 --> 00:49:13.400
And the MG, the M with
respect to G here,
00:49:13.400 --> 00:49:21.010
is the summation of the Ri
with respect to G cross Fi.
00:49:21.010 --> 00:49:26.070
So that's the generalization
for multiple forces on a body.
00:49:26.070 --> 00:49:30.940
So you are making an equivalent
force acting at G and a moment
00:49:30.940 --> 00:49:33.300
acting at G. I shouldn't
call this little g.
00:49:33.300 --> 00:49:34.300
That's really confusing.
00:49:36.980 --> 00:49:38.660
So that's the
generalization when you
00:49:38.660 --> 00:49:43.080
need to do problems like this.
00:50:03.060 --> 00:50:08.950
Catch your breath while
I scrub a board here.
00:50:08.950 --> 00:50:12.365
Now we've got to move on to
these class four problems.
00:50:16.500 --> 00:50:17.155
Moving points.
00:50:30.070 --> 00:50:32.624
An example is the truck
problem that you had.
00:50:38.700 --> 00:50:39.535
Known acceleration.
00:50:45.990 --> 00:50:50.670
So there's two common ways
of doing this problem.
00:50:50.670 --> 00:50:53.470
You can do this problem
by summing forces
00:50:53.470 --> 00:50:57.800
at the center of
mass of that pipe
00:50:57.800 --> 00:51:01.650
and summing moments around it.
00:51:01.650 --> 00:51:04.900
But the moment around this
comes from a friction force
00:51:04.900 --> 00:51:07.490
here, which you don't know.
00:51:07.490 --> 00:51:09.490
So that introduces
an unknown that you
00:51:09.490 --> 00:51:10.830
have to then solve for.
00:51:10.830 --> 00:51:15.830
So if you work around G for
this pipe, you can do it.
00:51:15.830 --> 00:51:18.760
You can work around G. You
could say sum of the moments
00:51:18.760 --> 00:51:20.700
around G, sum of the
force with respect to G.
00:51:20.700 --> 00:51:23.245
But you have to deal
with unknown forces.
00:51:29.060 --> 00:51:41.260
So you're working with respect
to G implies unknown forces, e,
00:51:41.260 --> 00:51:42.460
for example, friction.
00:51:45.760 --> 00:51:49.760
So you'd really maybe rather
around the point of contact,
00:51:49.760 --> 00:51:54.370
A. Because if you sum
your moments about A,
00:51:54.370 --> 00:51:55.980
the friction force
has no moment arm,
00:51:55.980 --> 00:51:57.570
and it doesn't
appear in the answer.
00:52:02.250 --> 00:52:04.940
But this gets trickier.
00:52:04.940 --> 00:52:10.670
This is a little more
sophisticated, I'll just
00:52:10.670 --> 00:52:12.740
call it, step.
00:52:12.740 --> 00:52:18.680
And you need-- to do this,
you need a little theorem.
00:52:18.680 --> 00:52:27.260
So to work with
respect to A, you
00:52:27.260 --> 00:52:31.770
need to be able to say that the
angular momentum with respect
00:52:31.770 --> 00:52:37.110
to A-- you now are working
around a moving point, maybe
00:52:37.110 --> 00:52:38.880
accelerating.
00:52:38.880 --> 00:52:40.750
It's very handy
to be able to say
00:52:40.750 --> 00:52:44.210
it's the angular momentum
around G, which is easier
00:52:44.210 --> 00:52:52.980
to calculate, plus
RGA, the distance
00:52:52.980 --> 00:52:56.490
from the center of mass to
the point you're working on,
00:52:56.490 --> 00:53:00.280
cross the linear momentum
of the system with respect
00:53:00.280 --> 00:53:02.720
to the inertial frame.
00:53:02.720 --> 00:53:03.630
We need this.
00:53:03.630 --> 00:53:05.670
This is a formula we need.
00:53:05.670 --> 00:53:17.690
And see why this is true?
00:53:17.690 --> 00:53:20.910
This is sort of thing--
this is a formula that's
00:53:20.910 --> 00:53:22.750
come out of the blue here.
00:53:22.750 --> 00:53:24.430
And why is it true?
00:53:24.430 --> 00:53:27.760
So I don't usually like to do
proofs, but the proofs of this
00:53:27.760 --> 00:53:29.090
[INAUDIBLE] on the board.
00:53:29.090 --> 00:53:31.555
But the proof of this
is really quite simple.
00:53:37.320 --> 00:53:40.430
Here's a little mass point Mi.
00:53:40.430 --> 00:53:45.940
And this radius is R
of i with respect to A.
00:53:45.940 --> 00:53:53.160
This is R of G
with respect to A.
00:53:53.160 --> 00:54:00.290
And therefore, this is R of i
with respect to G is this one.
00:54:00.290 --> 00:54:04.960
And we know that this
plus this gives us that.
00:54:04.960 --> 00:54:16.120
We can say Ri particle i with
respect to A is RG with respect
00:54:16.120 --> 00:54:20.775
to A plus Ri with respect
to G, all vectors.
00:54:45.170 --> 00:54:52.730
So the angular momentum of that
body from the basic definition
00:54:52.730 --> 00:54:54.530
of angular momentum
is the summation
00:54:54.530 --> 00:54:58.137
of all the little mass bits
of the Ri with respect to A
00:54:58.137 --> 00:55:01.920
cross the linear momentum
of each little mass bit.
00:55:04.560 --> 00:55:07.880
But we can expand
that with that sum.
00:55:07.880 --> 00:55:13.370
So this is the summation
of my RG with respect
00:55:13.370 --> 00:55:22.440
to A plus Ri with respect
to G cross Pio, summation
00:55:22.440 --> 00:55:25.190
over all the mass bits.
00:55:25.190 --> 00:55:28.570
I'm going to expand this.
00:55:28.570 --> 00:55:33.010
And I can expand this into
this times that, summations
00:55:33.010 --> 00:55:35.230
of this times that,
and this times that.
00:55:35.230 --> 00:55:50.610
So that becomes a RGA cross
summation of-- what's Pio?
00:55:50.610 --> 00:55:55.780
This is a little Mi's,
Vi with respect to o.
00:55:55.780 --> 00:55:56.830
Each one has velocity.
00:55:56.830 --> 00:55:58.570
Each one has a mass.
00:55:58.570 --> 00:56:08.340
So this is Mi Vi with respect
to o plus the summation
00:56:08.340 --> 00:56:13.982
RiG's cross Mi Vio's.
00:56:16.890 --> 00:56:19.580
Now, notice I pulled this
one outside the summation.
00:56:19.580 --> 00:56:21.052
That's because
this is a single--
00:56:21.052 --> 00:56:22.010
this is a fixed number.
00:56:22.010 --> 00:56:23.468
It doesn't change
in the summation.
00:56:23.468 --> 00:56:27.500
It's just the distance from
my starting point to G.
00:56:27.500 --> 00:56:29.280
So I can pull it out
and do the summation
00:56:29.280 --> 00:56:30.530
and then do the cross product.
00:56:33.680 --> 00:56:39.155
What is the summation of
all the MVi's for the body?
00:56:44.590 --> 00:56:46.490
That's the momentum of
each little particle.
00:56:46.490 --> 00:56:49.400
Add them all up,
what do you get?
00:56:49.400 --> 00:56:54.520
This is RGA cross P
with respect to o.
00:56:54.520 --> 00:56:56.620
That's this term.
00:56:56.620 --> 00:57:00.730
And this is all the
little distances
00:57:00.730 --> 00:57:08.120
from the center of mass to
the cross with the momentum
00:57:08.120 --> 00:57:09.100
of each little one.
00:57:13.460 --> 00:57:15.370
What's that?
00:57:15.370 --> 00:57:20.190
Well, this looks like a
definition of angular momentum.
00:57:20.190 --> 00:57:24.580
This is the angular momentum
of every little mass particle
00:57:24.580 --> 00:57:28.530
with respect to G added up.
00:57:28.530 --> 00:57:33.900
This is H with
respect to G, which
00:57:33.900 --> 00:57:36.780
is what we set out to prove.
00:57:36.780 --> 00:57:43.920
So the H with respect to A is
H with respect to G plus RGA
00:57:43.920 --> 00:57:46.570
cross the linear
momentum of the body.
00:57:46.570 --> 00:57:47.928
Yeah.
00:57:47.928 --> 00:57:52.374
AUDIENCE: Can you also do this
by writing Pi with respect
00:57:52.374 --> 00:57:54.350
to o, I mean the
velocity part of that,
00:57:54.350 --> 00:57:57.808
as the sum of the velocities?
00:57:57.808 --> 00:58:00.344
PROFESSOR: You got to
keep the M's in there.
00:58:00.344 --> 00:58:01.760
AUDIENCE: Well,
yeah. [INAUDIBLE].
00:58:01.760 --> 00:58:03.736
But instead of writing
out R as a sum,
00:58:03.736 --> 00:58:08.676
you can write out the velocity
as the sum of the velocities
00:58:08.676 --> 00:58:11.991
with respect to the origin.
00:58:11.991 --> 00:58:13.990
PROFESSOR: Are you talking
about this term here?
00:58:13.990 --> 00:58:19.310
AUDIENCE: No, the definition
of angular momentum.
00:58:19.310 --> 00:58:19.810
Yeah.
00:58:24.175 --> 00:58:25.130
The velocity--
00:58:25.130 --> 00:58:26.020
PROFESSOR: If you
could figure out--
00:58:26.020 --> 00:58:27.880
if you had the angular
momentum [INAUDIBLE]
00:58:27.880 --> 00:58:30.520
and multiplied by Ri, that
is the angular momentum
00:58:30.520 --> 00:58:32.940
with respect to
A. But I broke it
00:58:32.940 --> 00:58:37.110
apart so that I could show you
that this formula I want to use
00:58:37.110 --> 00:58:37.875
has two pieces.
00:58:40.490 --> 00:58:42.430
I want to use that.
00:58:42.430 --> 00:58:49.020
So that if I can use that-- it's
easy to get H with respect to G
00:58:49.020 --> 00:58:49.650
sometimes.
00:58:49.650 --> 00:58:52.160
It's really hard to know
what to do with things that
00:58:52.160 --> 00:58:53.980
happen around this point A.
00:58:53.980 --> 00:58:55.832
So now let's go back.
00:58:55.832 --> 00:58:57.790
I think, to understand
this, we need to go back
00:58:57.790 --> 00:59:00.190
to our truck problem.
00:59:00.190 --> 00:59:03.200
We now have a formula that
you know where it comes from.
00:59:03.200 --> 00:59:08.890
We have our truck that is
accelerating at x1 double dot.
00:59:08.890 --> 00:59:16.820
And we want to find out what's
theta double dot for that pipe.
00:59:16.820 --> 00:59:18.272
What's it doing?
00:59:18.272 --> 00:59:19.730
So first we needed
some kinematics.
00:59:25.780 --> 00:59:34.710
And in particular, what is
the x2-- did I label this
00:59:34.710 --> 00:59:35.210
very well?
00:59:35.210 --> 00:59:36.390
I didn't.
00:59:36.390 --> 00:59:38.770
So here's my pipe.
00:59:38.770 --> 00:59:42.310
Here's my truck bed that
it's in contact with.
00:59:42.310 --> 00:59:46.450
Truck's moving at x1
double dot, we know.
00:59:46.450 --> 00:59:49.310
In an inertial
frame, the movement
00:59:49.310 --> 00:59:54.210
of the center of mass
of my pipe is x2.
00:59:54.210 --> 01:00:02.630
And it has some angular
rotation I'll call theta.
01:00:02.630 --> 01:00:06.750
So the movement
of this guy I need
01:00:06.750 --> 01:00:10.260
to be able to express in
terms of x1 and theta.
01:00:10.260 --> 01:00:14.590
Well, if this is fixed to
the truck and the truck move
01:00:14.590 --> 01:00:18.540
forward, then x2
would be equal to x1.
01:00:18.540 --> 01:00:22.360
But in fact, it rolls
back a little bit.
01:00:22.360 --> 01:00:23.820
And the distance
this point moves
01:00:23.820 --> 01:00:25.790
if it rolls through
an angle theta
01:00:25.790 --> 01:00:28.310
is it rolls backwards
an amount R theta.
01:00:32.140 --> 01:00:35.180
And we're going to need to
take two derivatives of that.
01:00:35.180 --> 01:00:42.200
x2 double dot equals x1 double
dot minus r theta double dot.
01:00:42.200 --> 01:00:43.950
So that's a little
kinematic relationship
01:00:43.950 --> 01:00:45.075
we need to do this problem.
01:00:55.440 --> 01:00:58.830
Next we need to apply
a physical law, which
01:00:58.830 --> 01:01:03.720
is the one I've derived.
01:01:03.720 --> 01:01:06.760
So this is our physics
here now, our physical law.
01:01:11.020 --> 01:01:16.680
And that's the external
torques, dH with respect
01:01:16.680 --> 01:01:24.310
to A dt, plus VAo cross Po.
01:01:27.360 --> 01:01:34.656
And in this case, is VAo 0?
01:01:34.656 --> 01:01:38.140
No, in fact, it's x1 dot, right?
01:01:38.140 --> 01:01:40.480
It's in the i direction.
01:01:40.480 --> 01:01:44.164
How about what direction is
P, the momentum of the pipe?
01:01:49.250 --> 01:01:52.355
Does it move in the y direction?
01:01:52.355 --> 01:01:52.855
Up, down?
01:01:52.855 --> 01:01:53.355
No.
01:01:53.355 --> 01:01:54.490
It only moves in the x.
01:01:54.490 --> 01:01:56.920
So this velocity
is only in the x.
01:01:56.920 --> 01:01:59.740
This is in the x, or
the i hat direction.
01:01:59.740 --> 01:02:01.910
This cross product is zero.
01:02:01.910 --> 01:02:03.730
So it just happens
that they're parallel.
01:02:03.730 --> 01:02:04.890
So this thing goes to zero.
01:02:04.890 --> 01:02:07.025
We don't have to deal with it.
01:02:07.025 --> 01:02:08.650
That's because these
guys are parallel.
01:02:13.350 --> 01:02:15.380
Parallel motion.
01:02:15.380 --> 01:02:17.080
But you did have to consider it.
01:02:17.080 --> 01:02:18.371
You did have to think about it.
01:02:18.371 --> 01:02:20.120
It's not just trivially 0.
01:02:20.120 --> 01:02:24.790
OK, so that means that the
torques about A is just dHA dt.
01:02:24.790 --> 01:02:33.070
And HA for this
problem is i-- well,
01:02:33.070 --> 01:02:41.030
it's HG, which is
IzzG theta double dot.
01:02:41.030 --> 01:02:44.480
But I have to put in this--
theta dot, excuse me.
01:02:44.480 --> 01:02:46.550
This is just the
H. I have to put
01:02:46.550 --> 01:02:51.690
in the second term,
RG with respect
01:02:51.690 --> 01:02:55.747
to A cross P with respect to o.
01:02:59.330 --> 01:03:01.280
So here's my HA.
01:03:04.234 --> 01:03:05.400
I'll write it again up here.
01:03:05.400 --> 01:03:13.150
This is IzzG theta dot in the k.
01:03:13.150 --> 01:03:15.020
And now this second term.
01:03:15.020 --> 01:03:19.990
R is Rj.
01:03:19.990 --> 01:03:21.920
My y-axis is up.
01:03:21.920 --> 01:03:26.990
The radius of this thing
is R. Here's the radius.
01:03:26.990 --> 01:03:37.050
So the moment arm is Rj crossed
with the linear momentum
01:03:37.050 --> 01:03:42.070
of that piece of pipe, which
is the mass of the pipe times
01:03:42.070 --> 01:03:48.140
x2 dot in the i direction.
01:03:48.140 --> 01:03:50.350
j cross i is minus k.
01:03:50.350 --> 01:03:56.330
So Izz with respect to G
theta double dot-- theta dot.
01:03:56.330 --> 01:03:59.580
I keep taking the derivative
a little too soon.
01:03:59.580 --> 01:04:06.050
k minus RM x2 dot k.
01:04:06.050 --> 01:04:12.300
And now some of the torques, d
by dt of H with respect to A.
01:04:12.300 --> 01:04:14.380
Take the derivatives.
01:04:14.380 --> 01:04:27.456
IzzG theta double dot k
minus RM x2 double dot k.
01:04:27.456 --> 01:04:29.830
And fortunately, none of these
unit vectors are rotating.
01:04:29.830 --> 01:04:32.070
So we don't have to
deal with any of that.
01:04:32.070 --> 01:04:34.710
And I'm almost to
the end, but now I
01:04:34.710 --> 01:04:39.410
have to go back to that original
kinematic relationship, which
01:04:39.410 --> 01:04:46.380
allows me to express x2 in
terms of x1 and R theta.
01:04:46.380 --> 01:04:51.100
And if I substitute that
in here and solve for it,
01:04:51.100 --> 01:05:10.620
I get theta double dot equals
MR over IzzG plus MR squared.
01:05:13.870 --> 01:05:20.310
x double dot whole
thing, x1 double dot.
01:05:20.310 --> 01:05:22.860
So you've accelerated
x1 double dot.
01:05:22.860 --> 01:05:25.740
The thing starts rolling.
01:05:25.740 --> 01:05:28.440
It's actually rolling
backwards, which is a plus theta
01:05:28.440 --> 01:05:31.820
direction, at this rate.
01:05:31.820 --> 01:05:37.010
We did this using
this formula for HA.
01:05:37.010 --> 01:05:39.700
Now, the beauty
of this formula is
01:05:39.700 --> 01:05:43.170
that it works for
any points that
01:05:43.170 --> 01:05:46.870
are moving, even
can be accelerating,
01:05:46.870 --> 01:05:48.925
all sorts of nasty
conditions, it's true.
01:05:48.925 --> 01:05:54.380
It's based on the fundamental
definition of angular momentum.
01:05:54.380 --> 01:05:55.965
So the book-- yeah.
01:05:55.965 --> 01:05:58.000
AUDIENCE: [INAUDIBLE]
final point here.
01:05:58.000 --> 01:06:00.160
I just want to make
sure i understood.
01:06:00.160 --> 01:06:04.620
In this first line
here, [INAUDIBLE].
01:06:04.620 --> 01:06:07.390
That stroke doesn't
mean that V sub A was 0.
01:06:07.390 --> 01:06:10.940
It meant that that term is 0.
01:06:10.940 --> 01:06:13.370
PROFESSOR: This term's
zero because these
01:06:13.370 --> 01:06:16.772
happen to be parallel, but the
product is zero in this case.
01:06:16.772 --> 01:06:20.390
If it had not turned out to be--
they were different directions.
01:06:20.390 --> 01:06:21.650
It had to be a non-zero term.
01:06:21.650 --> 01:06:25.715
And you would have to bring it
along and take its derivative
01:06:25.715 --> 01:06:26.840
along with the other stuff.
01:06:29.870 --> 01:06:32.430
But this is a really
powerful method.
01:06:32.430 --> 01:06:36.750
Now, the book is reasonably
good in lots of points.
01:06:36.750 --> 01:06:39.750
But in chapter 17,
when it does problems
01:06:39.750 --> 01:06:43.910
that are kind of like this,
it introduces something
01:06:43.910 --> 01:06:46.910
that I find it hard to
digest what they're doing.
01:06:46.910 --> 01:06:48.525
There's particularly--
where's Matt?
01:06:48.525 --> 01:06:53.830
There's equation 1715.
01:06:53.830 --> 01:06:56.340
When you get to that bit of
the book, just ignore it.
01:06:56.340 --> 01:06:59.510
Use this method.
01:06:59.510 --> 01:07:02.360
The author tries to give you a
little trick that you can use.
01:07:02.360 --> 01:07:05.420
But the problem with tricks
is you have to memorize them.
01:07:05.420 --> 01:07:09.800
So what I've shown you today
is based on basic definition
01:07:09.800 --> 01:07:10.740
of angular momentum.
01:07:10.740 --> 01:07:15.950
That expression at the top
is always usable, not just
01:07:15.950 --> 01:07:20.250
special little conditions, which
is what the formula in the book
01:07:20.250 --> 01:07:21.500
is generated for.
01:07:25.710 --> 01:07:27.010
We've got a couple of minutes.
01:07:29.540 --> 01:07:31.780
Let you ask questions,
and then I'll
01:07:31.780 --> 01:07:34.100
just pose a conceptual
problem or two
01:07:34.100 --> 01:07:35.650
and ask you what
method you'd use.
01:07:35.650 --> 01:07:37.420
Yeah.
01:07:37.420 --> 01:07:40.892
AUDIENCE: So [INAUDIBLE]
when you [INAUDIBLE] dHA dt,
01:07:40.892 --> 01:07:44.690
is that equal to 0 because there
is no torque in the system?
01:07:44.690 --> 01:07:49.605
PROFESSOR: Oh, you know, boy,
I'm glad you caught that.
01:07:49.605 --> 01:07:52.080
Yeah, in order to
be able to do this,
01:07:52.080 --> 01:07:56.550
we've got to know
something here, right?
01:07:56.550 --> 01:07:58.020
Important catch.
01:07:58.020 --> 01:07:59.280
Thank you.
01:07:59.280 --> 01:08:02.580
Why is this true?
01:08:02.580 --> 01:08:05.490
AUDIENCE: [INAUDIBLE].
01:08:05.490 --> 01:08:08.890
PROFESSOR: So we
chose-- that's why
01:08:08.890 --> 01:08:16.590
we chose to work around point
A. With respect to that point,
01:08:16.590 --> 01:08:19.350
there are no
external forces that
01:08:19.350 --> 01:08:23.460
create moments on that pipe
with respect to that point.
01:08:23.460 --> 01:08:26.250
And that's why you
can say that the time
01:08:26.250 --> 01:08:28.310
derivative of this
angular momentum
01:08:28.310 --> 01:08:32.060
is equal to zero because
there are no external torques.
01:08:32.060 --> 01:08:35.660
If you had picked G
to do this problem,
01:08:35.660 --> 01:08:39.437
would the sum of the
torques about G be zero?
01:08:39.437 --> 01:08:41.520
No, you'd have to put that
friction force in there
01:08:41.520 --> 01:08:44.600
and have RF and figure out F is.
01:08:44.600 --> 01:08:47.640
We completely avoided having to
calculate the friction force.
01:08:47.640 --> 01:08:50.880
That's the point of being able
to use techniques like this
01:08:50.880 --> 01:08:54.990
and make your computations
around points of contact.
01:09:05.890 --> 01:09:10.390
So textbooks have lots
of problems like this.
01:09:10.390 --> 01:09:14.774
You've got a box on a cart.
01:09:14.774 --> 01:09:17.149
And your kid's pushing it,
and he gets a little exuberant
01:09:17.149 --> 01:09:19.689
and pushes a little too
hard, accelerates the cart
01:09:19.689 --> 01:09:22.710
a little too fast, and the
box falls over and breaks
01:09:22.710 --> 01:09:25.300
the lamp or whatever's in it.
01:09:25.300 --> 01:09:28.580
And if it's this way,
and I accelerate it,
01:09:28.580 --> 01:09:30.620
it's much more tolerant.
01:09:30.620 --> 01:09:32.859
Falls over easier this way.
01:09:32.859 --> 01:09:35.640
But if I asked you,
gave you a problem
01:09:35.640 --> 01:09:40.899
and said, calculate the
maximum acceleration
01:09:40.899 --> 01:09:44.950
that I can put on this
object such that it
01:09:44.950 --> 01:09:49.149
just barely-- just right at
the edge of tipping over,
01:09:49.149 --> 01:09:50.149
but doesn't tip it over.
01:09:50.149 --> 01:09:52.648
What's that maximum acceleration
that you don't tip it over?
01:09:52.648 --> 01:09:57.050
What method would you use?
01:09:57.050 --> 01:10:01.100
I gave you four classes
of approaches to problems.
01:10:07.412 --> 01:10:09.740
AUDIENCE: [INAUDIBLE].
01:10:09.740 --> 01:10:11.230
PROFESSOR: I hear a four.
01:10:11.230 --> 01:10:12.890
Anybody else want to bid here?
01:10:16.090 --> 01:10:17.050
More fours.
01:10:17.050 --> 01:10:20.168
Would three work?
01:10:20.168 --> 01:10:22.580
AUDIENCE: [INAUDIBLE].
01:10:22.580 --> 01:10:23.270
PROFESSOR: Why?
01:10:23.270 --> 01:10:26.060
He says three would
be complicated.
01:10:26.060 --> 01:10:28.400
Three means taking moments
and forces with respect
01:10:28.400 --> 01:10:31.400
to the center of mass.
01:10:31.400 --> 01:10:34.378
If you do that, what do you have
to deal with in this problem?
01:10:34.378 --> 01:10:35.294
AUDIENCE: [INAUDIBLE].
01:10:36.857 --> 01:10:39.190
PROFESSOR: You have to, then--
around the center of mass
01:10:39.190 --> 01:10:40.910
there's a friction force.
01:10:40.910 --> 01:10:44.210
There's a normal
force pushing up here.
01:10:44.210 --> 01:10:47.240
The way you do this problem
is where it's just barely
01:10:47.240 --> 01:10:51.940
about to go, all of the force
is pushing on this corner.
01:10:51.940 --> 01:10:52.440
Think of it.
01:10:52.440 --> 01:10:55.540
It's just lifting up a fraction.
01:10:55.540 --> 01:10:57.710
All of that contact point
moves to right here.
01:10:57.710 --> 01:11:00.230
You have an upward force
and a friction force,
01:11:00.230 --> 01:11:02.500
and they create
a moment about G.
01:11:02.500 --> 01:11:04.120
But if you do the
forces around G,
01:11:04.120 --> 01:11:06.970
you have to solve for those two.
01:11:06.970 --> 01:11:11.680
You do the forces around A.
The trick here is figuring out
01:11:11.680 --> 01:11:15.540
where's A. But if you
realize A is right here,
01:11:15.540 --> 01:11:19.350
and you do what we just
did, this problem's
01:11:19.350 --> 01:11:21.500
just a piece of cake.