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PROFESSOR: OK.
00:00:22.930 --> 00:00:26.820
I've been giving
out the money cards
00:00:26.820 --> 00:00:30.610
for a few of the lectures,
and two or three questions
00:00:30.610 --> 00:00:32.980
came up in those that I
haven't addressed so far.
00:00:32.980 --> 00:00:35.179
I'm calling them loose ends.
00:00:35.179 --> 00:00:37.220
And I'm going to pick up
a couple of those today,
00:00:37.220 --> 00:00:40.080
I think they'll
help you consolidate
00:00:40.080 --> 00:00:43.010
the knowledge around the quiz.
00:00:43.010 --> 00:00:45.180
So I'm going to tie up a
little loose ends there.
00:00:45.180 --> 00:00:48.710
And then the lecture topic
I started last time, which
00:00:48.710 --> 00:00:51.650
is making this transition
from thinking about angular
00:00:51.650 --> 00:00:56.150
momentum of particles to using
the full angular momentum
00:00:56.150 --> 00:00:58.080
equations for
rigid bodies, where
00:00:58.080 --> 00:01:00.565
we talk about mass
moments of inertia
00:01:00.565 --> 00:01:01.740
and products of inertia.
00:01:01.740 --> 00:01:04.190
And that's where we'll
pick up, there again today.
00:01:04.190 --> 00:01:07.215
Because that's where we're
going for the next few lectures.
00:01:07.215 --> 00:01:07.715
OK.
00:01:18.270 --> 00:01:22.190
Let's pick up with a
first example here.
00:01:22.190 --> 00:01:29.290
This is on the topic
of basically finding
00:01:29.290 --> 00:01:31.310
equations of motion.
00:01:31.310 --> 00:01:32.970
And there's been
a little confusion
00:01:32.970 --> 00:01:38.120
with people, who have asked
me what do you mean find?
00:01:38.120 --> 00:01:39.437
Do you mean solve, et cetera.
00:01:39.437 --> 00:01:42.020
And so I'm going to go through,
and just a real quick example,
00:01:42.020 --> 00:01:47.030
skipping some of the
steps because my purpose
00:01:47.030 --> 00:01:51.810
is emphasizing the steps, not
working out all the details.
00:01:51.810 --> 00:01:55.950
So finding equations of motion.
00:01:55.950 --> 00:01:58.240
Where does it begin?
00:01:58.240 --> 00:01:59.940
One of the really
important steps
00:01:59.940 --> 00:02:07.660
is this, determine the
number of independent
00:02:07.660 --> 00:02:08.770
coordinates you need.
00:02:18.330 --> 00:02:20.980
Because when you've done
that, that tells you,
00:02:20.980 --> 00:02:27.890
basically-- it really
starts finding the number
00:02:27.890 --> 00:02:28.980
of degrees of freedom.
00:02:28.980 --> 00:02:30.730
Should have put this
in a different order.
00:02:30.730 --> 00:02:33.020
Degrees of freedom
tells you the number
00:02:33.020 --> 00:02:35.110
of independent
coordinates you need.
00:02:35.110 --> 00:02:39.590
This is 1, 2, and
then 3, that leads you
00:02:39.590 --> 00:02:43.130
to the number of equations
of motion that you need.
00:02:43.130 --> 00:02:46.520
So this is really
an important step.
00:02:46.520 --> 00:02:52.640
Secondly, draw a
free body diagram.
00:02:55.150 --> 00:03:01.350
And third, apply
summation of forces.
00:03:05.110 --> 00:03:10.320
External vector equations gives
you mass times acceleration,
00:03:10.320 --> 00:03:27.730
and summation of torques gives
you DHDT plus this V cross
00:03:27.730 --> 00:03:30.470
P term.
00:03:30.470 --> 00:03:32.980
So this is just kind
of the step by step.
00:03:32.980 --> 00:03:36.670
So let's apply it briefly.
00:03:46.480 --> 00:03:49.870
We've talked a lot about things
on hills, so here's a cart.
00:03:49.870 --> 00:03:53.680
It's got wheels
attached by a cord
00:03:53.680 --> 00:03:58.630
to a second mass that's sliding.
00:03:58.630 --> 00:04:02.685
m1, m2, doesn't stretch
the cord in between.
00:04:11.240 --> 00:04:14.320
Let's think of these
things as rigid bodies.
00:04:14.320 --> 00:04:15.670
So how many degrees of freedom?
00:04:15.670 --> 00:04:18.890
How many possible
degrees of freedom?
00:04:18.890 --> 00:04:21.529
For the maximum possible, you
have how many rigid bodies?
00:04:21.529 --> 00:04:22.990
2.
00:04:22.990 --> 00:04:26.030
How many degrees of freedom
for rigid bodies possible?
00:04:26.030 --> 00:04:27.860
6 each.
00:04:27.860 --> 00:04:31.750
So we're at 6 times
m plus 3 times
00:04:31.750 --> 00:04:33.750
n minus the number
of constraints
00:04:33.750 --> 00:04:36.300
is the number of independent
degrees of freedom.
00:04:36.300 --> 00:04:39.390
This is the number of rigid
bodies, number of particles,
00:04:39.390 --> 00:04:44.010
so we have 6 times 2, 3
times 0 minus constraints,
00:04:44.010 --> 00:04:47.505
so this one comes out 12
minus the constraints.
00:04:50.032 --> 00:04:51.990
You have to figure out
the constraints quickly.
00:04:51.990 --> 00:04:55.830
We're not going to allow
rotation in any of the three
00:04:55.830 --> 00:04:56.740
directions on either.
00:04:56.740 --> 00:04:58.960
They're on carts, they're
big, they're sliding,
00:04:58.960 --> 00:05:00.680
they're not rolling
or any of that.
00:05:00.680 --> 00:05:04.540
So no rotation for 3,
no rotation for 3 more.
00:05:04.540 --> 00:05:06.390
That's minus 6.
00:05:06.390 --> 00:05:14.000
So c equals minus 6, or
c equals 6 for rotation.
00:05:14.000 --> 00:05:15.650
And then what else can we say?
00:05:19.330 --> 00:05:24.280
There's no-- I'll designate
this the y direction so we
00:05:24.280 --> 00:05:25.670
can talk about directions here.
00:05:25.670 --> 00:05:30.020
And I'll designate
this x in general.
00:05:30.020 --> 00:05:34.620
No acceleration at all in
the y direction, right.
00:05:34.620 --> 00:05:36.010
Can't move in the y.
00:05:36.010 --> 00:05:39.260
So that gives us
1, 2 for each mass,
00:05:39.260 --> 00:05:42.050
plus 2 more, that's
8 constraints
00:05:42.050 --> 00:05:43.630
that we've come up with.
00:05:43.630 --> 00:05:51.340
Now a 9th constraint is
the fact that these two
00:05:51.340 --> 00:05:52.530
are tied together.
00:05:52.530 --> 00:05:54.730
And so if you had
just temporarily
00:05:54.730 --> 00:06:00.510
assigned a coordinate here,
x1, and another one here, x2,
00:06:00.510 --> 00:06:04.284
we know for a fact that x1
has got to be equal to x2,
00:06:04.284 --> 00:06:06.075
and that gives you yet
one more constraint.
00:06:10.800 --> 00:06:13.900
So that's 9.
00:06:13.900 --> 00:06:15.250
So we might just stop there.
00:06:15.250 --> 00:06:18.010
We say OK, that's a
total of 9, so the number
00:06:18.010 --> 00:06:24.980
of degrees of freedom,
12 minus 9 is 3,
00:06:24.980 --> 00:06:29.100
and that implies that you need
three equations of motion.
00:06:29.100 --> 00:06:33.810
Some confusion comes, you
know, if something's not--
00:06:33.810 --> 00:06:37.430
let me rephrase that.
00:06:37.430 --> 00:06:40.810
We haven't talked anything
about the z direction.
00:06:40.810 --> 00:06:44.960
I haven't described any
constraints in the z direction.
00:06:44.960 --> 00:06:50.040
If this is me in a car and I'm
dragging a sled down a hill,
00:06:50.040 --> 00:06:53.460
or I'm in a vehicle and I'm
dragging a sled down a hill.
00:06:53.460 --> 00:06:57.990
I don't know if you've ever
been in a vehicle with a trailer
00:06:57.990 --> 00:07:00.690
on an icy road in the
winter time, that's
00:07:00.690 --> 00:07:05.640
a dicey maneuver, going down a
hill trying to put brakes on.
00:07:05.640 --> 00:07:09.280
So this thing could conceivably
move in this direction.
00:07:09.280 --> 00:07:12.020
And you can either
constrain it to be that way
00:07:12.020 --> 00:07:15.910
to make the problem simple, or
you can just say it's possible,
00:07:15.910 --> 00:07:18.740
so we have three
equations of motion,
00:07:18.740 --> 00:07:23.460
of which two, for now, the
summation of the forces,
00:07:23.460 --> 00:07:27.980
external in the--
since I've got x and y,
00:07:27.980 --> 00:07:32.550
z must be this way--
in the y direction.
00:07:32.550 --> 00:07:35.300
And we'll make it
in z direction,
00:07:35.300 --> 00:07:39.240
this is coordinate system 1,
so this would be z1 or z1.
00:07:39.240 --> 00:07:53.610
The summation of forces,
z1, is m1 z1 double dot.
00:07:53.610 --> 00:07:55.261
But I'm going to
set that equal to 0,
00:07:55.261 --> 00:07:56.510
I just know there's no forces.
00:07:56.510 --> 00:07:59.740
So this becomes a trivial
equation of motion.
00:07:59.740 --> 00:08:03.620
And I add another one, summation
of forces on the second mass.
00:08:03.620 --> 00:08:11.880
This is on m2 in the z2
direction, is m2 z2 double dot,
00:08:11.880 --> 00:08:14.020
and we set that equal to 0 also.
00:08:14.020 --> 00:08:16.410
So what it boils down
to, I had 3 degrees
00:08:16.410 --> 00:08:19.860
of freedom, 2 trivial
equations of motion,
00:08:19.860 --> 00:08:22.350
leaving me with just
1 equation of motion
00:08:22.350 --> 00:08:24.560
that's going to be meaningful.
00:08:24.560 --> 00:08:25.060
Yeah.
00:08:25.060 --> 00:08:31.431
AUDIENCE: If x1 is
equal to x2, would that
00:08:31.431 --> 00:08:36.124
mean that the rod has to be
entirely along the x-axis?
00:08:36.124 --> 00:08:37.606
So that would mean that--
00:08:37.606 --> 00:08:39.971
PROFESSOR: So he
asked if x1 equals x2,
00:08:39.971 --> 00:08:42.179
does that mean they both
have to be along the x-axis?
00:08:42.179 --> 00:08:43.350
I'm assuming that.
00:08:43.350 --> 00:08:46.110
So I really am assuming this
thing's going down the hill.
00:08:46.110 --> 00:08:49.250
I'm making a point about
this z direction thing
00:08:49.250 --> 00:08:51.170
because it's just
a subtlety that you
00:08:51.170 --> 00:08:56.070
have to decide on when you're
figuring out how to actually
00:08:56.070 --> 00:08:57.710
analyze the situation.
00:08:57.710 --> 00:09:00.040
If you really were
thinking about what
00:09:00.040 --> 00:09:04.460
happens when a vehicle is going
down a steep, icy hill towing
00:09:04.460 --> 00:09:08.777
a trailer, maybe put the
brakes on, maybe not.
00:09:08.777 --> 00:09:10.610
You could probably say
well, unless I really
00:09:10.610 --> 00:09:15.690
have a disaster, we're not going
to get rollovers and things
00:09:15.690 --> 00:09:16.200
like that.
00:09:16.200 --> 00:09:18.600
But you could imagine
that it can get out
00:09:18.600 --> 00:09:19.840
of the x direction, right.
00:09:19.840 --> 00:09:22.167
It could start sliding into z.
00:09:22.167 --> 00:09:23.750
It's probably not
going to go anywhere
00:09:23.750 --> 00:09:25.500
in the y, that's still
a good assumption.
00:09:25.500 --> 00:09:28.860
So this is about modeling,
and how complicated
00:09:28.860 --> 00:09:30.790
do you make the modeling.
00:09:30.790 --> 00:09:33.610
You actually have to make quite
a few modeling decisions when
00:09:33.610 --> 00:09:37.510
you go to do this, and we
tend, in class and examples,
00:09:37.510 --> 00:09:39.390
tend to really
oversimplify problems
00:09:39.390 --> 00:09:42.680
so that we can do them.
00:09:42.680 --> 00:09:44.170
So I've boiled
this down to where
00:09:44.170 --> 00:09:46.720
I'm going to end up with one
significant equation of motion.
00:10:04.840 --> 00:10:11.100
And that one's going to say
that the summation of the forces
00:10:11.100 --> 00:10:14.520
in the x direction--
and I'm just
00:10:14.520 --> 00:10:17.050
going to write mx
double dot here,
00:10:17.050 --> 00:10:19.780
not putting down m1s
or m2s because we're
00:10:19.780 --> 00:10:21.750
going to have to go
to free body diagrams
00:10:21.750 --> 00:10:25.310
to figure out how to apply this.
00:10:28.544 --> 00:10:29.335
Free body diagrams.
00:10:32.800 --> 00:10:33.400
First mass.
00:10:39.920 --> 00:10:41.470
So this is m1 here.
00:10:48.420 --> 00:10:53.100
m1g, a normal force, no
doubt a friction force,
00:10:53.100 --> 00:10:58.310
I'll call that f1, and
a tension in the cord.
00:11:01.330 --> 00:11:03.870
And I'm assuming that tension's
going to always be there.
00:11:03.870 --> 00:11:05.530
So if, again, in
a situation where
00:11:05.530 --> 00:11:07.350
the trailer starts
overtaking the car
00:11:07.350 --> 00:11:09.220
and the rope goes
slack, we're not
00:11:09.220 --> 00:11:11.490
going to consider
that one today.
00:11:11.490 --> 00:11:14.020
So there's your free body
diagram for the first one
00:11:14.020 --> 00:11:19.790
and we're going to need to break
mg into a couple of components.
00:11:19.790 --> 00:11:24.690
We're going to need
the slope here,
00:11:24.690 --> 00:11:28.490
and that translates
into an angle here.
00:11:28.490 --> 00:11:31.260
So there's our first
free body diagram.
00:11:31.260 --> 00:11:40.010
And our second free body
diagram, second mass, tension,
00:11:40.010 --> 00:11:46.838
normal force, another m2g.
00:11:49.782 --> 00:11:51.240
And this one's on
wheels, and we're
00:11:51.240 --> 00:11:54.030
going to consider
this one frictionless.
00:11:54.030 --> 00:11:56.280
So we don't have any friction
force holding it back.
00:12:08.950 --> 00:12:10.454
So the reason I've
kind of-- this
00:12:10.454 --> 00:12:12.120
seemed like a ridiculous
simple problem,
00:12:12.120 --> 00:12:13.880
but the point I
want to make is not
00:12:13.880 --> 00:12:17.240
the solution, not the
particular problem,
00:12:17.240 --> 00:12:19.390
but an issue that crops up.
00:12:19.390 --> 00:12:21.280
How many unknowns are
there in this problem.
00:12:24.064 --> 00:12:25.456
AUDIENCE: 2.
00:12:25.456 --> 00:12:29.560
PROFESSOR: Well, you
know n1 immediately,
00:12:29.560 --> 00:12:36.060
or n2, or the friction
force, or the tension,
00:12:36.060 --> 00:12:38.616
or, for that matter, the x
double dot we're looking for.
00:12:38.616 --> 00:12:40.490
There's actually-- you
start off this problem
00:12:40.490 --> 00:12:43.440
with five unknowns.
00:12:43.440 --> 00:12:48.070
But you're only looking to
derive one equation of motion.
00:12:48.070 --> 00:12:50.914
So the fact that we're looking
for one equation of motion
00:12:50.914 --> 00:12:52.330
doesn't say you
don't have to deal
00:12:52.330 --> 00:12:54.610
with several intermediate
equations to get there.
00:12:54.610 --> 00:12:57.860
That's just part of the work.
00:12:57.860 --> 00:13:11.700
So this is five unknowns to
start with, n1, n2, f1, t,
00:13:11.700 --> 00:13:12.670
and x double dot.
00:13:17.340 --> 00:13:21.850
So if you'll find out that the
summation of the forces on y
00:13:21.850 --> 00:13:33.100
and the y1 on the first mass,
this one, this gives you n1.
00:13:33.100 --> 00:13:36.970
Summation of the forces
in the y direction on m2,
00:13:36.970 --> 00:13:41.270
this one gives you m2 directly.
00:13:41.270 --> 00:13:45.400
From this flows directly, you
know that the friction force
00:13:45.400 --> 00:13:47.710
is mu n1.
00:13:47.710 --> 00:13:49.504
See, that's a third equation.
00:13:49.504 --> 00:13:51.420
So this gives you one
equation, this gives you
00:13:51.420 --> 00:13:53.503
another equation, this
gives you a third equation.
00:13:53.503 --> 00:13:55.630
You have 5 unknowns,
you need 5 equations.
00:13:55.630 --> 00:13:58.760
So there's 3 of them
right off the top.
00:13:58.760 --> 00:14:07.790
So this leaves-- you solve
for the those, this leaves t
00:14:07.790 --> 00:14:10.530
and x double dot to solve for.
00:14:18.990 --> 00:14:26.190
So now the sum of the forces
on mass 1 in the x direction
00:14:26.190 --> 00:14:29.280
says m1 x double dot.
00:14:33.040 --> 00:14:37.390
And the sum of the forces on the
second mass in the x direction
00:14:37.390 --> 00:14:42.080
gives you m2 x double dot.
00:14:42.080 --> 00:14:48.060
And I explicitly haven't
said this yet, what I've done
00:14:48.060 --> 00:14:52.780
is, one of my requirements
is x1 equals x2,
00:14:52.780 --> 00:14:56.070
and I'm just going
to call them x.
00:14:56.070 --> 00:14:57.720
Both of these are
exactly the same,
00:14:57.720 --> 00:14:59.870
both masses have to move
with the same motion,
00:14:59.870 --> 00:15:01.030
is the assumption.
00:15:01.030 --> 00:15:05.935
So that means that x1 double
dot equals x2 double dot equals
00:15:05.935 --> 00:15:07.310
x double dot, and
that's what I'm
00:15:07.310 --> 00:15:10.590
assuming when I'm writing
down these two equations.
00:15:10.590 --> 00:15:14.490
I can write those two equations,
one from each free body
00:15:14.490 --> 00:15:15.140
diagram.
00:15:15.140 --> 00:15:18.130
I've already eliminated
three of the unknowns.
00:15:18.130 --> 00:15:23.460
And now because I have two
equations, each have t in them.
00:15:23.460 --> 00:15:35.150
Essentially, you eliminate t
and solve for x double dot.
00:15:35.150 --> 00:15:39.360
And if you do that
in this problem,
00:15:39.360 --> 00:15:40.815
you get your equation of motion.
00:16:14.550 --> 00:16:17.992
You eliminate t, solve
for x1 double dot.
00:16:17.992 --> 00:16:19.700
Look at this, and you
just look at things
00:16:19.700 --> 00:16:21.060
that doesn't make sense.
00:16:21.060 --> 00:16:25.120
This says the total mass times
the acceleration is a system,
00:16:25.120 --> 00:16:27.640
it's one system.
00:16:27.640 --> 00:16:30.040
Mass times the acceleration
of the center of gravity
00:16:30.040 --> 00:16:32.090
of the system, if
you will, has got
00:16:32.090 --> 00:16:33.930
to be equal to the sums
of the forces on it.
00:16:33.930 --> 00:16:39.870
Well, it's got m1 plus m2g sine
theta pulling it down the hill,
00:16:39.870 --> 00:16:47.160
and it has minus m
mu m1g cosine theta
00:16:47.160 --> 00:16:48.780
dragging it back up the hill.
00:16:48.780 --> 00:16:52.180
And that's the entire equation
of motion, it makes sense.
00:16:52.180 --> 00:16:58.650
But the equation of motion,
the thing you're looking for,
00:16:58.650 --> 00:17:01.410
is the one that ends up with
this acceleration term in it.
00:17:01.410 --> 00:17:03.340
If you have multiple
degrees of freedom,
00:17:03.340 --> 00:17:05.240
multiple coordinates--
if you have,
00:17:05.240 --> 00:17:08.470
let's say, three significant
equations of motion
00:17:08.470 --> 00:17:11.000
that result, there
won't necessarily
00:17:11.000 --> 00:17:13.180
be one in terms of
each coordinate.
00:17:13.180 --> 00:17:16.036
They'll have the
coordinates mixed in them.
00:17:16.036 --> 00:17:18.119
Like we did that that two
mass system with springs
00:17:18.119 --> 00:17:22.200
the other day, each equation
of motion had x1 and x2.
00:17:22.200 --> 00:17:24.200
They don't necessarily separate.
00:17:24.200 --> 00:17:26.730
They're coupled through
their coordinates.
00:17:26.730 --> 00:17:29.670
This one, it's one equation
of motion for the system,
00:17:29.670 --> 00:17:31.640
therefore you have
only one coordinate.
00:17:31.640 --> 00:17:34.314
But that's not generally
true of multiple degree
00:17:34.314 --> 00:17:35.105
of freedom systems.
00:17:38.100 --> 00:17:43.020
OK, that's your method,
though, for a simple problem.
00:17:43.020 --> 00:17:47.900
I want to do a little more
difficult problem that
00:17:47.900 --> 00:17:50.450
involves rotation.
00:18:00.964 --> 00:18:02.630
And this is the
problem, I'm sure you've
00:18:02.630 --> 00:18:04.706
done this problem in physics.
00:18:04.706 --> 00:18:07.150
It's a classic problem
that people do.
00:18:09.922 --> 00:18:26.300
A disk, a pulley, really,
supporting two masses
00:18:26.300 --> 00:18:31.390
rotates about this point,
which I'll call a here.
00:18:38.150 --> 00:18:43.760
So at some theta there's
no slip, so theta is going
00:18:43.760 --> 00:18:47.120
to be related to movement, x.
00:18:47.120 --> 00:18:52.820
And I'm going to assign this
one a coordinate, x1 going down.
00:18:52.820 --> 00:18:56.100
This one a coordinate,
x2 going up.
00:18:56.100 --> 00:18:58.674
A little foreknowledge
here, because you've
00:18:58.674 --> 00:18:59.465
worked the problem.
00:19:02.240 --> 00:19:05.020
So I want to solve for
the motion of this system.
00:19:08.120 --> 00:19:13.700
Now again, we need to know the
number of degrees of freedom.
00:19:13.700 --> 00:19:18.890
So it's the maximum possible,
which is our 6m plus
00:19:18.890 --> 00:19:22.040
3n minus the constraints.
00:19:22.040 --> 00:19:27.092
And let's think about how
we want to model this again.
00:19:27.092 --> 00:19:29.300
This time I'm just going to
model these as particles,
00:19:29.300 --> 00:19:31.420
doesn't matter how big they are.
00:19:31.420 --> 00:19:33.780
My problem, really, they
only go up and down.
00:19:33.780 --> 00:19:35.660
So I'm going to model
them as particles.
00:19:35.660 --> 00:19:41.000
This is 6 times 0 plus 3
times 2 minus constraints.
00:19:41.000 --> 00:19:42.620
So 6 minus the constraints.
00:19:42.620 --> 00:19:44.980
So the issue is really
how many constraints.
00:19:47.570 --> 00:19:55.230
Well, we're going to
require x1 equal x2.
00:19:55.230 --> 00:19:57.756
Cord's taught, doesn't stretch.
00:19:57.756 --> 00:19:59.130
If this thing goes
down, that has
00:19:59.130 --> 00:20:01.787
to go up exactly
an equal amount,
00:20:01.787 --> 00:20:02.870
and that's one constraint.
00:20:11.710 --> 00:20:14.690
In that, leaving us 6
minus 1 is 5, leaving us
00:20:14.690 --> 00:20:17.654
with a lot of degrees
of freedom here.
00:20:17.654 --> 00:20:19.570
Kind of back to the issue
I was making before,
00:20:19.570 --> 00:20:24.190
are there any constraints
in the-- I'll call it
00:20:24.190 --> 00:20:29.740
x, y, z directions here.
00:20:29.740 --> 00:20:33.030
Are there any constraints
in the y or z directions
00:20:33.030 --> 00:20:36.280
on either of those masses?
00:20:36.280 --> 00:20:39.520
No, I haven't shown any, no
tracks, no guides, no anything.
00:20:39.520 --> 00:20:42.500
So technically, there are
no additional constraints
00:20:42.500 --> 00:20:43.710
in this problem.
00:20:43.710 --> 00:20:49.560
But if there's no forces in
the y, x, I guess y this way,
00:20:49.560 --> 00:20:51.820
and no forces in
the z, I'm going
00:20:51.820 --> 00:20:54.290
to end up with two trivial
equations of motion
00:20:54.290 --> 00:20:56.980
for this one, 1 in
the y, 1 in the z,
00:20:56.980 --> 00:20:59.460
for this one, 1 in
the y, 1 in the z.
00:20:59.460 --> 00:21:01.880
So back to this
issue of, there's
00:21:01.880 --> 00:21:05.480
a difference between constraints
and trivial equations
00:21:05.480 --> 00:21:05.980
of motion.
00:21:05.980 --> 00:21:08.614
We're going have 4 to
reveal equations of motion.
00:21:08.614 --> 00:21:10.280
So really, again, I'm
going to come down
00:21:10.280 --> 00:21:13.860
to one significant
equation of motion.
00:21:13.860 --> 00:21:23.110
So I have 1 constraint,
4 trivial EOMs,
00:21:23.110 --> 00:21:32.390
and 1 significant
equation of motion.
00:21:32.390 --> 00:21:32.890
OK.
00:21:35.440 --> 00:21:39.700
Now, because I want to
talk about rotation,
00:21:39.700 --> 00:21:41.500
we need to pick a coordinate.
00:21:41.500 --> 00:21:49.020
Now I can pick-- I can
either let x1 equal x2
00:21:49.020 --> 00:21:52.000
and just let it be the sum
x, a single coordinate,
00:21:52.000 --> 00:21:55.880
or theta, the rotation.
00:21:55.880 --> 00:21:57.960
I'm actually going to
use x for a second.
00:22:05.250 --> 00:22:09.990
But there's 2 obvious ways
to approach this problem.
00:22:09.990 --> 00:22:15.140
One is to draw a
free body diagrams
00:22:15.140 --> 00:22:22.160
of each of these masses, sum the
forces on each, and how many--
00:22:22.160 --> 00:22:25.110
if I do that, how many
unknowns do I end up with?
00:22:29.120 --> 00:22:30.990
We can draw the free--
here's the free body
00:22:30.990 --> 00:22:32.280
diagram for mass 1.
00:22:32.280 --> 00:22:33.185
What's it got on it?
00:22:33.185 --> 00:22:34.950
Well, m1g.
00:22:34.950 --> 00:22:37.064
What else is acting on it?
00:22:37.064 --> 00:22:38.972
AUDIENCE: [? Tension. ?]
00:22:38.972 --> 00:22:44.590
PROFESSOR: And here's
the second one, m2g,
00:22:44.590 --> 00:22:47.100
and tension acting on it.
00:22:47.100 --> 00:22:50.090
So now we can sum,
sum of the forces
00:22:50.090 --> 00:22:52.520
equals mass times the
acceleration of each one,
00:22:52.520 --> 00:22:54.840
and the external forces
are going to involve t.
00:22:54.840 --> 00:22:57.810
So you're going to end up
with how many unknowns?
00:23:02.960 --> 00:23:04.290
How many unknowns?
00:23:04.290 --> 00:23:07.380
I can write two equations
for the sum of the forces
00:23:07.380 --> 00:23:08.660
in the x direction.
00:23:08.660 --> 00:23:11.520
x double dot is
certainly an unknown.
00:23:11.520 --> 00:23:12.020
What else?
00:23:12.020 --> 00:23:12.500
AUDIENCE: t.
00:23:12.500 --> 00:23:13.041
PROFESSOR: t.
00:23:13.041 --> 00:23:15.269
So I end up with
this other [? end. ?]
00:23:15.269 --> 00:23:17.560
So that means I'm going to
have to write two equations,
00:23:17.560 --> 00:23:18.900
I'm going to have
to eliminate t,
00:23:18.900 --> 00:23:20.608
going to go through
the same thing there.
00:23:20.608 --> 00:23:22.670
So I don't want to
bother with that.
00:23:22.670 --> 00:23:25.485
Is there another way
to do this problem?
00:23:32.270 --> 00:23:35.980
This is a problem where you
can use angular momentum
00:23:35.980 --> 00:23:38.440
and not have to
deal with t at all.
00:23:38.440 --> 00:23:40.029
So let's set that problem up.
00:23:44.620 --> 00:23:49.120
You know that the sum of
the torques about that point
00:23:49.120 --> 00:23:55.380
a, with respect to point a,
it's going to be derivative,
00:23:55.380 --> 00:23:57.910
and since we're dealing
with particles here,
00:23:57.910 --> 00:24:01.422
of the angular momentum
with respect to-- I'll
00:24:01.422 --> 00:24:04.540
just call it lowercase
h for particles.
00:24:04.540 --> 00:24:09.480
Plus this velocity
of a with respect
00:24:09.480 --> 00:24:16.920
to an inertial frame across a
linear momentum with respect
00:24:16.920 --> 00:24:18.440
to an inertial frame.
00:24:18.440 --> 00:24:21.530
That's the full equation
for sum of torques.
00:24:21.530 --> 00:24:23.830
What's velocity of a with
respect to o in this problem?
00:24:26.930 --> 00:24:27.670
0.
00:24:27.670 --> 00:24:29.420
Fortunately, this is
one of those problems
00:24:29.420 --> 00:24:32.770
where you can get rid of
this difficult second term.
00:24:32.770 --> 00:24:35.990
So it's just torques as the
time derivative of the angular
00:24:35.990 --> 00:24:37.100
momentum.
00:24:37.100 --> 00:24:44.080
So we need an expression, then,
for both the sum of the torques
00:24:44.080 --> 00:24:45.970
with respect to a.
00:24:45.970 --> 00:24:47.390
Let's see, what would that be?
00:24:50.950 --> 00:24:57.290
So now the external torques
with respect to-- I'll
00:24:57.290 --> 00:25:03.220
finish my-- oops, come here
you-- free body diagram.
00:25:03.220 --> 00:25:06.870
So now what I really want
is a free body diagram
00:25:06.870 --> 00:25:07.795
of the whole system.
00:25:10.990 --> 00:25:13.790
So here's the whole system
created as one thing.
00:25:13.790 --> 00:25:20.850
You have a force down, m1g,
another force down, m2g.
00:25:20.850 --> 00:25:25.630
Up here you have
some normal force up,
00:25:25.630 --> 00:25:28.350
that's the support of the pin.
00:25:28.350 --> 00:25:38.820
You have tensions in these,
but now this equation
00:25:38.820 --> 00:25:40.550
applies to the system.
00:25:40.550 --> 00:25:43.910
The ts are internal to the
system, they are irrelevant.
00:25:46.640 --> 00:25:49.360
So I'm talking about this whole
thing treated as a system,
00:25:49.360 --> 00:25:50.860
and I'm going to
compute the moments
00:25:50.860 --> 00:25:53.620
about point a, which is right
there, where that axle is.
00:25:53.620 --> 00:25:56.830
Does n create a
moment at the axle?
00:25:56.830 --> 00:26:03.260
Nope, but the m1 and m2
times g create moments.
00:26:03.260 --> 00:26:04.160
Sure, OK.
00:26:07.467 --> 00:26:09.300
I'm going to have
positive out of the board,
00:26:09.300 --> 00:26:13.370
the positive moment,
positive angular direction.
00:26:13.370 --> 00:26:16.090
So the torques
applied to this system
00:26:16.090 --> 00:26:22.630
are r cross t, so you're
going to end up with-- I
00:26:22.630 --> 00:26:26.350
want to summarize these.
00:26:26.350 --> 00:26:33.734
An m1g, and I didn't write
the radius on this problem,
00:26:33.734 --> 00:26:39.345
but at some radius, capital
R. So the torques that are m1g
00:26:39.345 --> 00:26:49.190
are positive minus
m2gR, k hat direction,
00:26:49.190 --> 00:26:53.460
and that must be equal to the
time derivative of the angular
00:26:53.460 --> 00:26:57.570
momentum about a.
00:26:57.570 --> 00:27:04.470
Now we need an expression
for the angular
00:27:04.470 --> 00:27:06.221
momentum with respect to a.
00:27:09.590 --> 00:27:16.260
Angular momentum
is, in general, this
00:27:16.260 --> 00:27:19.940
is a r cross linear
momentum, right.
00:27:19.940 --> 00:27:28.940
So R for mass 1 with
respect to a cross
00:27:28.940 --> 00:27:34.650
the momentum of that
second mass with respect
00:27:34.650 --> 00:27:35.880
to an inertial frame.
00:27:35.880 --> 00:27:38.340
And a and the inertial
frame are the same thing,
00:27:38.340 --> 00:27:40.340
a sticks in the inertial frame.
00:27:40.340 --> 00:27:42.970
But angular momentum
is always with respect
00:27:42.970 --> 00:27:44.690
to the inertial frame.
00:27:44.690 --> 00:27:50.070
Plus the second piece, which
is R of m2 with respect to a
00:27:50.070 --> 00:27:54.760
crossed with p for mass 2 with
respect to some inertial frame.
00:28:01.230 --> 00:28:03.230
I'm just going to give
you the results for this.
00:28:07.020 --> 00:28:18.516
m1 plus m2 R x dot k.
00:28:18.516 --> 00:28:25.680
So x dot is this velocity,
R cross and mass times
00:28:25.680 --> 00:28:30.360
velocity is momentum, so
the perpendicular radius
00:28:30.360 --> 00:28:33.460
to that is the
radius, R. So it's
00:28:33.460 --> 00:28:35.830
Rx dot times m,
shouldn't surprise you,
00:28:35.830 --> 00:28:36.850
in the k hat direction.
00:28:36.850 --> 00:28:38.940
That's the total
angular momentum
00:28:38.940 --> 00:28:43.970
that comes from these two
particles with respect to a.
00:28:43.970 --> 00:28:45.785
And taking their
time derivative.
00:28:50.000 --> 00:28:53.230
These are constant, that's
a constant, this is not,
00:28:53.230 --> 00:28:55.660
this is a constant, but it
doesn't change direction,
00:28:55.660 --> 00:28:57.420
so this one is pretty simple.
00:29:04.970 --> 00:29:12.050
Now I can set equal the sum
of the external torques, that,
00:29:12.050 --> 00:29:14.990
to the time derivative
of the angular momentum,
00:29:14.990 --> 00:29:16.845
just to fulfill this expression.
00:29:19.450 --> 00:29:26.170
And in so doing, I end up with
a solution for x double dot.
00:29:26.170 --> 00:29:36.520
m1 minus m2, m1
plus m2, times g.
00:29:36.520 --> 00:29:37.935
Turns out the R goes away.
00:29:41.950 --> 00:29:46.130
So one equation, never
had to mess with tension.
00:29:46.130 --> 00:29:52.070
This is a pretty nice, direct
way of solving this problem.
00:29:52.070 --> 00:29:56.674
If you solve for g here, and
you measure x double dot,
00:29:56.674 --> 00:29:59.380
this actually gives
you an experimental way
00:29:59.380 --> 00:30:02.140
of determining
acceleration of gravity.
00:30:02.140 --> 00:30:04.790
It's actually what this thing
was used for a long time
00:30:04.790 --> 00:30:07.930
ago, before they had a lot
of the measurement techniques
00:30:07.930 --> 00:30:09.170
and things that we do today.
00:30:09.170 --> 00:30:12.630
This is a way of determining
the acceleration of gravity.
00:30:12.630 --> 00:30:14.580
So these two masses are
quite close together.
00:30:14.580 --> 00:30:18.514
This number is pretty
small, you can, however
00:30:18.514 --> 00:30:19.805
accurate your timing device is.
00:30:23.970 --> 00:30:27.520
Now, just to mention it, I
neglected something in this.
00:30:27.520 --> 00:30:29.440
I assumed something and
I didn't even say it.
00:30:29.440 --> 00:30:29.970
What was it?
00:30:29.970 --> 00:30:31.511
What would screw up
this measurement?
00:30:31.511 --> 00:30:33.760
I'm trying to measure the
acceleration of gravity,
00:30:33.760 --> 00:30:38.780
if I built this apparatus, would
I get a very good measurement?
00:30:38.780 --> 00:30:41.160
AUDIENCE: The pulley
would have to be massless.
00:30:41.160 --> 00:30:43.510
PROFESSOR: Yeah, the pulley
would have to be massless.
00:30:43.510 --> 00:30:45.260
I've made an assumption
about that, right.
00:30:45.260 --> 00:30:47.040
So how would you
fix up this equation
00:30:47.040 --> 00:30:48.642
to account for the pulley?
00:30:51.946 --> 00:30:54.306
AUDIENCE: You'd have
to take into account
00:30:54.306 --> 00:30:55.260
its moment of inertia.
00:30:55.260 --> 00:30:56.926
PROFESSOR: Yeah, you'd
put in something.
00:30:56.926 --> 00:30:59.750
And where would that
go into the problem?
00:30:59.750 --> 00:31:02.765
How would you account its
inertia, moment of inertia
00:31:02.765 --> 00:31:03.410
in the problem?
00:31:03.410 --> 00:31:04.150
AUDIENCE: ha.
00:31:04.150 --> 00:31:06.250
PROFESSOR: Yeah, you'd
just put it into ha.
00:31:06.250 --> 00:31:11.080
So this expression for h would
end up with one more term,
00:31:11.080 --> 00:31:14.834
it's going to look like--
well, when you take the time
00:31:14.834 --> 00:31:16.250
derivative, you're
going to end up
00:31:16.250 --> 00:31:17.620
with another piece over here.
00:31:17.620 --> 00:31:21.750
Some i about a theta double dot.
00:31:21.750 --> 00:31:24.260
You're going to have to
relate theta double dot to x
00:31:24.260 --> 00:31:28.930
double dot, which you can,
because x equals R theta.
00:31:28.930 --> 00:31:30.660
x double dot is R
theta double dot.
00:31:30.660 --> 00:31:33.890
You could fix that and you'd
have an equation of motion.
00:31:33.890 --> 00:31:37.340
But that means we need
to know about i about a,
00:31:37.340 --> 00:31:42.720
that's where we're going to
at the end of this lecture
00:31:42.720 --> 00:31:45.470
and for the next
several lectures.
00:31:45.470 --> 00:31:46.900
OK.
00:31:46.900 --> 00:31:52.070
That's that example, and
I've got two more brief ones
00:31:52.070 --> 00:31:54.050
that I wanted to talk about.
00:31:54.050 --> 00:31:55.841
Any last questions?
00:31:55.841 --> 00:31:56.340
Yeah.
00:31:56.340 --> 00:31:58.200
AUDIENCE: Can you
explain again why
00:31:58.200 --> 00:31:59.991
you didn't take the
tensions into account
00:31:59.991 --> 00:32:00.990
for your sum of torques?
00:32:00.990 --> 00:32:05.680
PROFESSOR: OK, so why did I not
take the tensions into account?
00:32:12.520 --> 00:32:21.830
So I can write the
equation of motion
00:32:21.830 --> 00:32:26.470
for this thing as
a complete system.
00:32:26.470 --> 00:32:29.170
One, the masses and
the pulley are all
00:32:29.170 --> 00:32:35.300
the same thing, the summation
of the external torques on that,
00:32:35.300 --> 00:32:39.100
they're going to amount up to
taking into account the time
00:32:39.100 --> 00:32:41.870
rate of change of the angular
momentum of the system.
00:32:41.870 --> 00:32:43.550
Now, if I didn't
understand that,
00:32:43.550 --> 00:32:46.614
I could have blindly gone ahead
and put the ts in there, right,
00:32:46.614 --> 00:32:49.030
they would've been exactly
equal and opposite with respect
00:32:49.030 --> 00:32:50.860
to a and it would
have cancelled out.
00:32:50.860 --> 00:32:55.092
So either way, if you're not
sure about that assumption,
00:32:55.092 --> 00:32:56.550
you could just put
them in and they
00:32:56.550 --> 00:33:00.570
would appear in the torque
equation, but as a minus
00:33:00.570 --> 00:33:02.800
tR and a plus tR
and they'd cancel.
00:33:05.420 --> 00:33:11.590
I want to move on
to a third example,
00:33:11.590 --> 00:33:17.850
and this is the third item
that I want to clear up,
00:33:17.850 --> 00:33:19.190
loose ends I'm calling them.
00:33:19.190 --> 00:33:21.370
The muddy cards
are really useful.
00:33:21.370 --> 00:33:25.100
I get questions in those
that spark something.
00:33:25.100 --> 00:33:27.670
And this is a question that
came up two or three times
00:33:27.670 --> 00:33:30.030
in the muddy cards and
I haven't addressed it,
00:33:30.030 --> 00:33:34.540
and that is, we were
working with rotor problems.
00:33:34.540 --> 00:33:37.580
And remember this problem.
00:33:37.580 --> 00:33:39.860
You have the rotor,
it had an arm,
00:33:39.860 --> 00:33:42.770
I did it this way to
make some things obvious.
00:33:42.770 --> 00:33:49.780
But this is the z direction,
it's rotating about that axis.
00:33:49.780 --> 00:33:53.680
I've got a point mass up here.
00:33:53.680 --> 00:33:57.485
r hat, so this is
R-- actually, I'm
00:33:57.485 --> 00:33:59.580
going to make it a
capital R so it's easier
00:33:59.580 --> 00:34:01.200
to distinguish from the r hat.
00:34:01.200 --> 00:34:02.210
And this is z.
00:34:08.620 --> 00:34:18.929
This thing's rotating, it's got
bearings here to keep it going.
00:34:18.929 --> 00:34:24.449
And we talked about
torques, so this
00:34:24.449 --> 00:34:38.030
is my point A. I want to
write the sum of the torques
00:34:38.030 --> 00:34:44.810
about A, time derivative
of the angular momentum.
00:34:44.810 --> 00:34:47.060
We've done this problem
before, so I'm just putting up
00:34:47.060 --> 00:34:51.860
a couple of points
for review to clear up
00:34:51.860 --> 00:34:54.066
some possible misconceptions.
00:34:56.989 --> 00:35:01.140
That's this term, so what about
point A now in this problem?
00:35:01.140 --> 00:35:03.101
What's the velocity at point A?
00:35:03.101 --> 00:35:06.600
0, so again we can
get rid of this guy.
00:35:06.600 --> 00:35:08.600
I'm going to come back
to this and do an example
00:35:08.600 --> 00:35:10.220
one of these days
where this isn't 0,
00:35:10.220 --> 00:35:14.920
where it's really handy
to be able to do a problem
00:35:14.920 --> 00:35:17.710
where that's not 0.
00:35:17.710 --> 00:35:18.210
OK.
00:35:21.100 --> 00:35:22.580
This is true.
00:35:22.580 --> 00:35:25.870
I need a free body diagram
of our little mass,
00:35:25.870 --> 00:35:30.240
so here's my free body diagram.
00:35:30.240 --> 00:35:35.220
And it has possibly a
force in the z direction.
00:35:35.220 --> 00:35:37.410
That comes from the
rod, there's rods that's
00:35:37.410 --> 00:35:39.090
supporting this thing, right.
00:35:39.090 --> 00:35:41.670
There's possibly a force
in the z direction.
00:35:41.670 --> 00:35:45.750
There's a force in the r hat
direction, in the R direction.
00:35:45.750 --> 00:35:54.950
There's a force in theta
direction going into the board.
00:35:54.950 --> 00:35:55.975
And there's mg.
00:36:01.110 --> 00:36:04.660
All sorts of forces
on this thing.
00:36:04.660 --> 00:36:09.420
And the question was asked,
when we did this problem before
00:36:09.420 --> 00:36:13.850
and did the time derivatives
of the angular momentum,
00:36:13.850 --> 00:36:16.660
we found that we got--
there's three terms
00:36:16.660 --> 00:36:19.540
and I'll write them
down here for you.
00:36:19.540 --> 00:36:22.750
I'm just saying in advance
what we're going to do.
00:36:22.750 --> 00:36:24.570
When you solve this
problem, you find out
00:36:24.570 --> 00:36:27.520
that it takes to torque to
accelerate this shaft and spin.
00:36:27.520 --> 00:36:29.840
That the driving one,
that's what makes it happen,
00:36:29.840 --> 00:36:30.910
makes it accelerate.
00:36:30.910 --> 00:36:34.810
We had two more terms that were
torques at this point, that
00:36:34.810 --> 00:36:36.590
is what it takes to
support this system.
00:36:36.590 --> 00:36:39.800
It's trying to bend out,
it's trying to bend back,
00:36:39.800 --> 00:36:41.300
those are torques
that show up here.
00:36:41.300 --> 00:36:43.910
And we actually get them
when we work through this.
00:36:43.910 --> 00:36:48.180
But we don't get
something that tells us
00:36:48.180 --> 00:36:51.062
about the moment
the torque created
00:36:51.062 --> 00:36:52.270
this point caused by gravity.
00:36:55.890 --> 00:36:58.870
The question was, why don't we
get the torque about this point
00:36:58.870 --> 00:36:59.680
caused by gravity.
00:36:59.680 --> 00:37:02.920
There's clearly mg down,
there's clearly a moment arm.
00:37:02.920 --> 00:37:05.490
So mgR is the torque
about this point.
00:37:05.490 --> 00:37:08.770
And if you were doing the
statics problem in 2.001,
00:37:08.770 --> 00:37:10.890
there'd be a torque
around this point caused
00:37:10.890 --> 00:37:13.015
by the weight of this thing
just sitting there, not
00:37:13.015 --> 00:37:14.170
even spinning.
00:37:14.170 --> 00:37:19.430
And what we're doing here
gives you no help with that.
00:37:19.430 --> 00:37:24.730
But just for the quick
review of this problem,
00:37:24.730 --> 00:37:29.030
more in the line of helping
you think about the quiz.
00:37:29.030 --> 00:37:33.660
This then is R, we'll
call this point B
00:37:33.660 --> 00:37:37.270
and this is point A, remember
this is RB with respect
00:37:37.270 --> 00:37:43.850
to A cross p with respect to o.
00:37:43.850 --> 00:37:46.610
And that's where our
angular momentum comes from.
00:37:46.610 --> 00:37:57.460
In this problem that is r hat
plus z k hat cross m times
00:37:57.460 --> 00:38:03.290
the velocity,
which is R omega z.
00:38:03.290 --> 00:38:08.030
And that must be in the
theta hat direction.
00:38:08.030 --> 00:38:19.070
When you multiply these
out, omega z is theta dot.
00:38:19.070 --> 00:38:22.850
They're kind of interchangeable
in this problem.
00:38:22.850 --> 00:38:26.010
So when you multiply this
out, you get two terms.
00:38:26.010 --> 00:38:41.230
mR squared theta dot k hat
minus mRz theta dot r hat.
00:38:41.230 --> 00:38:42.560
Two terms from this.
00:38:45.360 --> 00:38:53.330
And when you do the time
derivative of the dhdt,
00:38:53.330 --> 00:38:54.200
you get three terms.
00:38:56.710 --> 00:39:02.017
mR squared theta
double dot and the k.
00:39:02.017 --> 00:39:03.350
Now, why do you get three terms?
00:39:03.350 --> 00:39:06.970
Because this term has
two variables in it that
00:39:06.970 --> 00:39:10.490
are functions of time,
theta dot has a derivative,
00:39:10.490 --> 00:39:14.550
and r hat has a derivative,
because it rotates.
00:39:14.550 --> 00:39:16.320
So one of the key
bits of mathematics
00:39:16.320 --> 00:39:18.529
you have to learn
in this course,
00:39:18.529 --> 00:39:20.570
I'm kind of giving you a
little quiz review here,
00:39:20.570 --> 00:39:23.030
you need to know how to take
the derivative of a rotating
00:39:23.030 --> 00:39:24.210
vector.
00:39:24.210 --> 00:39:26.250
And that's what we
do here, gives us
00:39:26.250 --> 00:39:34.970
two terms minus mR z
theta double dot r hat
00:39:34.970 --> 00:39:43.500
minus mRz theta dot
squared theta hat.
00:39:43.500 --> 00:39:47.260
So three terms in this time
derivative of the angular
00:39:47.260 --> 00:39:52.400
momentum, and they have to be
equal to the external torques.
00:39:52.400 --> 00:39:56.470
This is equal to the summation
of the torques about A,
00:39:56.470 --> 00:39:58.610
the external torques.
00:39:58.610 --> 00:40:01.980
Well, you'll need a
torque in the k direction.
00:40:01.980 --> 00:40:04.320
That's what it takes to
accelerate the thing,
00:40:04.320 --> 00:40:05.110
make it go faster.
00:40:09.590 --> 00:40:14.890
This mass has a
force on it to make
00:40:14.890 --> 00:40:18.500
it go faster, that's this f
in the theta hat direction.
00:40:18.500 --> 00:40:21.650
And that rods have
a push on that mass,
00:40:21.650 --> 00:40:23.825
the mass pushes back on the rod.
00:40:23.825 --> 00:40:25.200
So if in the theta
direction it's
00:40:25.200 --> 00:40:28.290
like that, the mass pushes back
on the rod, it twists the rod,
00:40:28.290 --> 00:40:29.060
or tries to.
00:40:29.060 --> 00:40:33.936
That's a torque about this
in the r hat direction.
00:40:33.936 --> 00:40:35.435
So there's centripetal
acceleration,
00:40:35.435 --> 00:40:38.590
it takes force to cause
centripetal acceleration.
00:40:38.590 --> 00:40:41.270
It's that force is inward.
00:40:41.270 --> 00:40:44.020
It's about a moment arm
z, and so this gives you
00:40:44.020 --> 00:40:47.000
a torque about the point A
in the theta hat direction.
00:40:47.000 --> 00:40:50.500
So these are three different
terms, each one has a purpose.
00:40:50.500 --> 00:40:52.450
No work is done here,
no work is done here,
00:40:52.450 --> 00:40:53.616
because there's no movement.
00:40:56.990 --> 00:40:59.830
Now, but gravity, we
started this question
00:40:59.830 --> 00:41:01.620
as why doesn't gravity
pop out of this.
00:41:04.220 --> 00:41:07.220
Because this only tells
you about the time rate
00:41:07.220 --> 00:41:09.820
of change of angular momentum.
00:41:09.820 --> 00:41:13.810
Gravity has nothing to
do with angular momentum.
00:41:13.810 --> 00:41:17.870
r cross p is all that
angular momentum is.
00:41:17.870 --> 00:41:19.630
The linear momentum of little
00:41:19.630 --> 00:41:24.060
clumps of mass times the
radius from the point you're
00:41:24.060 --> 00:41:26.000
computing the angular momentum.
00:41:26.000 --> 00:41:28.060
Has nothing to do
with g, never will.
00:41:30.760 --> 00:41:35.970
You'll never get the g
related static moment
00:41:35.970 --> 00:41:38.200
out of this equation.
00:41:38.200 --> 00:41:41.292
It's there, though, and if
you were designing the system,
00:41:41.292 --> 00:41:42.750
you'd have to take
it into account.
00:42:08.530 --> 00:42:12.270
So remember, I didn't bring it
today, but I have my shaker.
00:42:12.270 --> 00:42:13.635
I've bolted it to the floor.
00:42:18.120 --> 00:42:22.720
Inside of that shaker is
a little rotating mass.
00:42:22.720 --> 00:42:25.570
It has a little arm
and eccentricity, it
00:42:25.570 --> 00:42:29.916
has some mass that
I'm going to make m,
00:42:29.916 --> 00:42:35.250
it's rotating theta direction.
00:42:35.250 --> 00:42:39.000
And it rotates a constant speed.
00:42:39.000 --> 00:42:44.940
So it's some constant omega,
theta double dot equals 0.
00:42:44.940 --> 00:42:47.040
So I just got my shaker
bolted to the floors,
00:42:47.040 --> 00:42:49.650
it's putting a lot of
vibration into the floor.
00:42:49.650 --> 00:42:51.920
And the question
that someone came up
00:42:51.920 --> 00:42:55.660
with on a muddy card that was
a really inside insightful
00:42:55.660 --> 00:42:59.960
question, why-- or
they didn't say why--
00:42:59.960 --> 00:43:03.810
they said, shouldn't
the torque required
00:43:03.810 --> 00:43:07.960
to drive this thing somehow
be affected by gravity?
00:43:11.710 --> 00:43:14.320
So does the torque
that it takes to run
00:43:14.320 --> 00:43:17.710
this around and around
depend on gravity,
00:43:17.710 --> 00:43:19.980
was the question that was asked.
00:43:19.980 --> 00:43:21.910
Let's take a quick look at that.
00:43:21.910 --> 00:43:25.000
We just discovered that dhdt
doesn't tell you anything
00:43:25.000 --> 00:43:28.030
about torque from
gravity, right?
00:43:28.030 --> 00:43:30.170
Well, let's see
what happens then.
00:43:30.170 --> 00:43:35.000
So the summation of the external
torques-- I'll call this now
00:43:35.000 --> 00:43:37.080
point A, where it's
rotating about.
00:43:37.080 --> 00:43:39.720
This point now doesn't
move in this problem.
00:43:39.720 --> 00:43:41.430
It's an inertial point.
00:43:41.430 --> 00:43:44.460
Summation of the torques
with respect to A
00:43:44.460 --> 00:43:49.200
is dh with respect to A, dt,
and there's no additional terms
00:43:49.200 --> 00:43:53.370
because that
velocity point is 0.
00:43:53.370 --> 00:44:01.100
And that's d by dt, the
torque is just r cross p,
00:44:01.100 --> 00:44:05.200
so that is me theta dot.
00:44:05.200 --> 00:44:07.435
That's the velocity,
that's the momentum.
00:44:21.100 --> 00:44:22.805
I've left out something.
00:44:31.650 --> 00:44:36.670
So r cross p, I need
an e squared in here.
00:44:36.670 --> 00:44:44.730
me squared theta dot
in the k hat direction.
00:44:44.730 --> 00:44:46.590
I need to take the time
derivative of that.
00:44:46.590 --> 00:44:48.923
That's a constant, that's a
constant, that's a constant,
00:44:48.923 --> 00:44:50.940
it only comes from this term.
00:44:53.470 --> 00:44:58.980
And that gives me me squared
theta double dot k hat
00:44:58.980 --> 00:45:01.960
direction, and that's got to be
equal to the sum of the torques
00:45:01.960 --> 00:45:04.640
in the system, the
external torques.
00:45:04.640 --> 00:45:05.390
And what are they?
00:45:09.790 --> 00:45:11.400
So torques about this point.
00:45:11.400 --> 00:45:16.250
So axial forces in this
thing contribute no torques,
00:45:16.250 --> 00:45:19.290
transverse forces,
external forces only
00:45:19.290 --> 00:45:25.100
come from the mg on this thing.
00:45:25.100 --> 00:45:28.510
So my torques on
this system, there
00:45:28.510 --> 00:45:30.550
is some mechanical
torque being applied.
00:45:30.550 --> 00:45:33.830
That's what I'm looking for.
00:45:33.830 --> 00:45:35.460
I've got a motor
driving this thing,
00:45:35.460 --> 00:45:47.290
so there's some t of t in there,
some torque, minus mge cosine
00:45:47.290 --> 00:45:49.270
theta is the moment arm.
00:45:49.270 --> 00:45:51.700
So there's this force,
there's this moment arm
00:45:51.700 --> 00:45:54.570
is e cosine theta.
00:45:54.570 --> 00:45:57.260
So this is the external
torque caused by gravity,
00:45:57.260 --> 00:46:01.766
but all of this equals what?
00:46:01.766 --> 00:46:03.665
What's theta double dot?
00:46:03.665 --> 00:46:04.165
0.
00:46:08.713 --> 00:46:17.340
The external torque
is mge cosine omega t,
00:46:17.340 --> 00:46:19.950
theta is omega t.
00:46:19.950 --> 00:46:23.560
And so indeed, as this
thing goes around,
00:46:23.560 --> 00:46:27.170
when it's coming up, you've
got to apply enough torque
00:46:27.170 --> 00:46:29.020
to lift it against gravity.
00:46:29.020 --> 00:46:31.269
When it clears the top,
gravity is helping it,
00:46:31.269 --> 00:46:32.560
it's going down the other side.
00:46:32.560 --> 00:46:35.640
So in fact, if you plotted the
torque as a function of time
00:46:35.640 --> 00:46:40.040
for this system, it's like this.
00:46:40.040 --> 00:46:42.220
It's just lifting
that mass up and down.
00:46:42.220 --> 00:46:43.970
Then, of course, if
there's any friction
00:46:43.970 --> 00:46:45.320
in this thing, et
cetera, it's going
00:46:45.320 --> 00:46:47.570
to have to apply a little
bit of torque for that, too.
00:46:47.570 --> 00:46:50.640
But indeed, this is
an insightful question
00:46:50.640 --> 00:46:55.850
that someone asked, is
that the gravity does
00:46:55.850 --> 00:46:57.510
have to enter into this thing.
00:46:57.510 --> 00:47:00.220
So there will be a torque
that the motor has to supply
00:47:00.220 --> 00:47:02.420
to drive this thing in gravity.
00:47:02.420 --> 00:47:02.920
Yeah.
00:47:02.920 --> 00:47:06.798
AUDIENCE: Should that
expression also have--
00:47:06.798 --> 00:47:10.150
the expression for torque also
me squared theta double dot--
00:47:10.150 --> 00:47:12.980
PROFESSOR: Ah, now
theta double dot is?
00:47:12.980 --> 00:47:14.830
Yeah, see, it would.
00:47:14.830 --> 00:47:17.450
If this thing was
spinning up and I
00:47:17.450 --> 00:47:21.180
was trying to account for the
torque required to spin it up,
00:47:21.180 --> 00:47:23.070
then here is.
00:47:23.070 --> 00:47:26.440
Then I would include that, this
would be an equation of motion
00:47:26.440 --> 00:47:28.630
that says all these
things are true,
00:47:28.630 --> 00:47:31.200
and I can solve
for torque again.
00:47:31.200 --> 00:47:34.480
And it will allow me to decide
how fast I could spin it up.
00:47:34.480 --> 00:47:36.230
If I have a dinky
little motor, it
00:47:36.230 --> 00:47:37.770
doesn't spin up
very fast, if I had
00:47:37.770 --> 00:47:40.061
a really powerful motor that
could really put it to it,
00:47:40.061 --> 00:47:41.170
spin it up quickly.
00:47:41.170 --> 00:47:41.670
OK.
00:47:57.410 --> 00:48:00.490
So now I want to move on
to the third topic, which
00:48:00.490 --> 00:48:03.640
is to kind of go back to where
I left off last time, talking
00:48:03.640 --> 00:48:07.270
about we need to move on from
particles to rigid bodies
00:48:07.270 --> 00:48:09.640
so we can do more
interesting problems.
00:48:09.640 --> 00:48:12.530
So I want to pick up with
the subject of angular
00:48:12.530 --> 00:48:13.865
momentum for rigid bodies.
00:48:32.580 --> 00:48:35.990
Now last time I just barely
scratched the surface of this.
00:48:35.990 --> 00:48:38.060
And lots of muddy cards
said I don't get it.
00:48:38.060 --> 00:48:41.010
I didn't expect you to get
it with it being half baked
00:48:41.010 --> 00:48:42.980
and the first time
you've seen it.
00:48:42.980 --> 00:48:46.840
So we're going to continue
and we won't finish today.
00:48:46.840 --> 00:48:53.620
So let's think about
a general rigid body.
00:48:53.620 --> 00:49:05.020
Here's my inertial system,
got a body out here
00:49:05.020 --> 00:49:06.570
that's rotating
about some point,
00:49:06.570 --> 00:49:08.880
A. A could even be
outside the body
00:49:08.880 --> 00:49:10.135
and have it rotate about it.
00:49:13.520 --> 00:49:15.620
And attached to A is
a reference frame.
00:49:21.070 --> 00:49:23.280
Little x, little y, little z.
00:49:23.280 --> 00:49:25.490
So my Axy frame.
00:49:25.490 --> 00:49:32.650
Now, I put up last time, there's
two pages out of Williams
00:49:32.650 --> 00:49:37.190
which gives the
equations for the moment
00:49:37.190 --> 00:49:43.920
and products of inertia in terms
of summations of masses times
00:49:43.920 --> 00:49:45.750
particle locations.
00:49:45.750 --> 00:49:48.090
And in order to do
that, Williams defines
00:49:48.090 --> 00:49:50.250
a coordinate system
on this body,
00:49:50.250 --> 00:49:53.310
and that coordinate system
is fixed to the body,
00:49:53.310 --> 00:49:55.470
rotates to the body,
and Williams calls that
00:49:55.470 --> 00:49:57.050
coordinate system little oxyz.
00:50:00.160 --> 00:50:04.660
In his book, he calls the
inertial frame big Oxyz.
00:50:04.660 --> 00:50:06.670
It's really hard to do
that on a blackboard
00:50:06.670 --> 00:50:09.170
and how you'd be able
to tell it apart, OK.
00:50:09.170 --> 00:50:15.630
So I'm going to depart, and my
frame in here is an Axyz frame.
00:50:15.630 --> 00:50:17.860
But A and o, if you're
reading that handout,
00:50:17.860 --> 00:50:19.220
are the same thing.
00:50:19.220 --> 00:50:20.370
A and little o.
00:50:20.370 --> 00:50:22.960
It's a frame fixed to the
body that's rotating with it.
00:50:34.260 --> 00:50:39.680
We can write angular
momentum for rigid bodies
00:50:39.680 --> 00:50:43.820
as a vector hx having a
component in the i direction, j
00:50:43.820 --> 00:50:54.140
direction, and these
coordinates as the product
00:50:54.140 --> 00:50:58.070
of a matrix of constants.
00:50:58.070 --> 00:51:02.930
And these constants are these
moments of inertia and products
00:51:02.930 --> 00:51:03.765
of inertia terms.
00:51:18.410 --> 00:51:19.580
And so forth.
00:51:19.580 --> 00:51:21.410
I'll write out a couple
more of these, iy.
00:51:31.060 --> 00:51:34.900
It's a symmetric matrix,
and you multiply it
00:51:34.900 --> 00:51:44.000
by the components
of the rotation
00:51:44.000 --> 00:51:48.990
that you are
rotating this object,
00:51:48.990 --> 00:51:52.710
so here's a vector omega.
00:51:52.710 --> 00:51:57.220
This object is rotating about
A, the direction, the axis
00:51:57.220 --> 00:51:59.270
of rotation is like that.
00:51:59.270 --> 00:52:02.060
And you can break
this rotation rate
00:52:02.060 --> 00:52:04.960
into components
in the xyz system.
00:52:04.960 --> 00:52:07.620
And that's what these are,
these are the components of it.
00:52:07.620 --> 00:52:11.290
So you multiply out
this matrix in a vector,
00:52:11.290 --> 00:52:14.030
you will get
individual equations
00:52:14.030 --> 00:52:17.670
for the hx, hy, and hz
components of the angular
00:52:17.670 --> 00:52:19.415
momentum of that object.
00:53:02.550 --> 00:53:04.230
Now let's consider,
let's just do
00:53:04.230 --> 00:53:17.100
a case where the spin is
only about the z-axis.
00:53:17.100 --> 00:53:19.315
We do lots of these
problems, the book
00:53:19.315 --> 00:53:20.856
has a whole chapter
on it and they're
00:53:20.856 --> 00:53:22.150
called planar motion problems.
00:53:22.150 --> 00:53:26.970
We just typically pick the spin
around the z as a convention.
00:53:26.970 --> 00:53:35.350
And if you have a case like
that, then h here is i times
00:53:35.350 --> 00:53:39.170
0, 0, omega z.
00:53:39.170 --> 00:53:53.830
Then you multiply that out, you
get ixz omega z, iyz omega z,
00:53:53.830 --> 00:53:59.090
and izz omega z.
00:53:59.090 --> 00:54:03.560
Vector times the square matrix
gives you back a vector.
00:54:03.560 --> 00:54:06.700
That's what you get back.
00:54:06.700 --> 00:54:13.450
And if you want to write h as
a vector, which we frequently
00:54:13.450 --> 00:54:19.370
do, h, now this is
with respect to A,
00:54:19.370 --> 00:54:23.460
and we'll find that i here
as also with respect to A.
00:54:23.460 --> 00:54:29.140
Have to be very careful in your
construction of this matrix.
00:54:29.140 --> 00:54:33.770
It has to do with the point
about which you are computing
00:54:33.770 --> 00:54:35.410
your angular momentum.
00:54:35.410 --> 00:54:37.590
OK, if you want to
write this as a vector,
00:54:37.590 --> 00:54:47.539
then this becomes hx i hat
plus hy j hat plus hz k hat.
00:54:47.539 --> 00:54:49.330
That's just where the
unit vectors come in.
00:54:49.330 --> 00:54:51.700
When you want to express
this as a vector,
00:54:51.700 --> 00:54:57.740
you take these three components,
and these are hx, hy, hz.
00:55:07.340 --> 00:55:09.650
This little double
subscript, the first one
00:55:09.650 --> 00:55:13.770
tells you the component
of h, this is hx, hy, hz.
00:55:13.770 --> 00:55:17.210
The second one tells
you the axis of rotation
00:55:17.210 --> 00:55:19.720
about which the object
is spinning to give you
00:55:19.720 --> 00:55:22.800
this piece of angular momentum.
00:55:22.800 --> 00:55:30.720
So ixz is hx spinning
at rate omega z.
00:55:33.730 --> 00:55:39.045
Now, the direction of spin was?
00:55:39.045 --> 00:55:40.670
What's the unit vector
in the direction
00:55:40.670 --> 00:55:42.706
of rotation for this problem?
00:55:48.020 --> 00:55:50.520
What's omega?
00:55:50.520 --> 00:55:53.950
We said we're going to start
off with just-- direction,
00:55:53.950 --> 00:55:56.700
it's only spinning
in z direction.
00:55:56.700 --> 00:55:59.420
So it's just spinning
in z direction.
00:55:59.420 --> 00:56:04.000
But I multiply this thing
out, I get three terms.
00:56:04.000 --> 00:56:06.660
And I get a term in
the i, a j, and a k.
00:56:09.230 --> 00:56:19.920
Now these two terms,
so this is i, x is z,
00:56:19.920 --> 00:56:34.210
omega is zi plus iyz omega
zj plus izz omega zk.
00:56:34.210 --> 00:56:38.200
That's these three terms.
00:56:38.200 --> 00:56:44.980
These two terms
exist because I've
00:56:44.980 --> 00:56:50.720
assumed that these off
diagonal terms are not 0.
00:56:50.720 --> 00:56:53.650
The problem we started with,
we started with an example
00:56:53.650 --> 00:56:54.150
last time.
00:56:54.150 --> 00:56:56.850
Our bicycle wheel thing with
the unbalanced masses on it,
00:56:56.850 --> 00:57:02.440
we use the Williams formulas to
compute these different terms.
00:57:02.440 --> 00:57:09.840
If the off diagonal
terms here are not 0,
00:57:09.840 --> 00:57:13.130
then when you write the
angular momentum expression,
00:57:13.130 --> 00:57:15.660
you get parts of
the angular momentum
00:57:15.660 --> 00:57:19.290
that are not in the
direction of spin.
00:57:19.290 --> 00:57:22.000
That's a really
important conclusion.
00:57:22.000 --> 00:57:26.290
So the off diagonal terms
lead to angular momentum
00:57:26.290 --> 00:57:27.997
not in the direction of spin.
00:57:27.997 --> 00:57:29.580
And when you take
the time derivative,
00:57:29.580 --> 00:57:32.280
you end up with torques,
and they're right back
00:57:32.280 --> 00:57:33.520
to this problem up here.
00:57:36.650 --> 00:57:41.280
If you have off diagonal
terms in this matrix,
00:57:41.280 --> 00:57:44.310
when you spin it
around one of its axes,
00:57:44.310 --> 00:57:46.225
it is dynamically unbalanced.
00:57:52.370 --> 00:57:54.890
If these are not 0,
you spin it around one
00:57:54.890 --> 00:57:58.230
of the axes of the system
for which these are defining,
00:57:58.230 --> 00:58:00.810
in which these are
defined, you find out
00:58:00.810 --> 00:58:04.800
that you get unbalanced
torques in the system.
00:58:04.800 --> 00:58:07.290
So those two go together.
00:58:07.290 --> 00:58:30.610
Now, another way of
saying that is any time
00:58:30.610 --> 00:58:34.220
you end up with the angular
momentum vector not pointing
00:58:34.220 --> 00:58:37.790
in the same direction
as the rotation,
00:58:37.790 --> 00:58:40.558
then the system is going to
be dynamically unbalanced.
00:58:57.700 --> 00:59:01.500
Actually, I kind of want to
keep Atwood's machine here.
01:00:15.130 --> 01:00:19.740
So this was our unbalanced
bicycle wheel problem
01:00:19.740 --> 01:00:22.200
we had talked about last time.
01:00:22.200 --> 01:00:25.794
I can simulate that with this.
01:00:31.710 --> 01:00:35.190
I basically have
drawn it like this.
01:00:35.190 --> 01:00:36.230
So this is the problem.
01:00:38.950 --> 01:00:42.140
This thing is
definitely unbalanced,
01:00:42.140 --> 01:00:46.380
it's trying to do this
as it goes around.
01:00:46.380 --> 01:00:48.890
And last time we
actually worked up,
01:00:48.890 --> 01:00:54.830
from the William formulas, what
the moment of inertia matrix
01:00:54.830 --> 01:00:55.350
looked like.
01:00:55.350 --> 01:01:04.190
So now this xyz system
are attached and rotating
01:01:04.190 --> 01:01:06.500
with that frame.
01:01:06.500 --> 01:01:22.400
So my axis of spin is-- this
one's a little exaggerated.
01:01:22.400 --> 01:01:25.260
That drawing is like this.
01:01:25.260 --> 01:01:30.580
The x is like that.
01:01:30.580 --> 01:01:36.240
So x is like this, z is
like that, minus x minus z.
01:01:36.240 --> 01:01:39.100
So when this thing spins, that's
the problem that's drawn there.
01:01:41.810 --> 01:01:46.600
And if I compute with those
with Williams formulas,
01:01:46.600 --> 01:01:52.830
the various quantities--
so i with respect
01:01:52.830 --> 01:01:55.220
to A for this system.
01:02:01.070 --> 01:02:08.280
The first term, the ixx
term, is summation miyi
01:02:08.280 --> 01:02:15.260
squared plus zi
squared, and so forth.
01:02:15.260 --> 01:02:19.990
You get a bunch of terms, and
I will write out one other one
01:02:19.990 --> 01:02:20.490
here.
01:02:20.490 --> 01:02:30.010
This term over in
the corner is ixz,
01:02:30.010 --> 01:02:40.270
and that's minus summation
of the mixizi and so forth.
01:02:40.270 --> 01:02:44.430
And if we went through and
worked up each of these things,
01:02:44.430 --> 01:02:49.290
i with respect to
A for this problem,
01:02:49.290 --> 01:03:04.406
comes out mz1 squared 0
minus mx1z1 0 minus mx1z1.
01:03:07.970 --> 01:03:18.844
The middle term mx1 squared
plus z1 squared, 0, 0,
01:03:18.844 --> 01:03:24.160
and mx1 squared.
01:03:24.160 --> 01:03:27.020
So that's what this mass
moment of inertia matrix
01:03:27.020 --> 01:03:29.090
looks like for
these two particles.
01:04:06.190 --> 01:04:08.230
So now if I want to
write the angular
01:04:08.230 --> 01:04:14.170
momentum of this system
using this new notation,
01:04:14.170 --> 01:04:18.010
I would say that it's
i computed with respect
01:04:18.010 --> 01:04:24.970
to A times my omega, and
our case is 0, 0, omega z.
01:04:28.670 --> 01:04:33.940
And if we write that out, we
do that, multiply that out, we
01:04:33.940 --> 01:04:48.403
end up with a minus mx1z1 omega
z, 0, and mx1 squared omega z.
01:04:51.790 --> 01:04:57.850
These are our three
components, hx, hy, hz.
01:05:08.690 --> 01:05:13.300
And if you wanted to
write it as a vector,
01:05:13.300 --> 01:05:15.420
then you'd add the unit vectors.
01:05:15.420 --> 01:05:22.260
So the hx and the i plus 0
for the hy plus hz in the k.
01:05:30.970 --> 01:05:35.010
So now if you went and took the
time derivative of those terms,
01:05:35.010 --> 01:05:38.010
what do you get?
01:05:38.010 --> 01:05:38.780
AUDIENCE: Torques.
01:05:38.780 --> 01:05:40.560
PROFESSOR: Torques.
01:05:40.560 --> 01:05:43.070
And you'll get 3 terms.
01:05:43.070 --> 01:05:46.000
When we did the example a minute
ago, what we're doing here
01:05:46.000 --> 01:05:48.530
is not very different from that.
01:05:48.530 --> 01:05:51.590
You're going to get the torque
that it takes to accelerate it
01:05:51.590 --> 01:05:53.830
around the spin
axis, but you're also
01:05:53.830 --> 01:05:57.020
going to get the torque two
derivatives of this one, which
01:05:57.020 --> 01:05:58.330
gives you two terms.
01:05:58.330 --> 01:06:00.000
And these are the
moments of torques
01:06:00.000 --> 01:06:03.030
about that center of
the axle, in this case,
01:06:03.030 --> 01:06:05.050
trying to twist
the system around.
01:06:07.970 --> 01:06:12.080
Now, reach some closure here.
01:06:12.080 --> 01:06:13.570
We've got a good stopping point.
01:06:29.090 --> 01:06:34.073
Here's our system one last time.
01:06:34.073 --> 01:06:34.823
Here's the z-axis.
01:06:37.650 --> 01:06:40.550
The angular momentum
that comes out of this,
01:06:40.550 --> 01:06:46.120
you have a component
hz in the z direction,
01:06:46.120 --> 01:06:47.600
and you end up
with a component--
01:06:47.600 --> 01:06:57.150
it's got a minus in it-- in
the x direction like this,
01:06:57.150 --> 01:07:01.910
so that the total h vector with
respect to A looks like that.
01:07:01.910 --> 01:07:04.075
And it's not in the
direction of spin,
01:07:04.075 --> 01:07:10.030
it's actually perpendicular
to our bar here.
01:07:10.030 --> 01:07:13.050
And it's dynamically unbalanced.
01:07:13.050 --> 01:07:19.940
So just to-- how do
we make the transition
01:07:19.940 --> 01:07:27.060
from that to rigid bodies?
01:07:27.060 --> 01:07:31.520
The Williams formulas,
that are these,
01:07:31.520 --> 01:07:34.780
say that if you want the mass
moment of inertia of a body,
01:07:34.780 --> 01:07:36.950
all you have to do is sum
up all the little mass
01:07:36.950 --> 01:07:41.470
bits at the correct
distances off of axes,
01:07:41.470 --> 01:07:42.360
and you will get it.
01:07:42.360 --> 01:07:46.410
So when you have particles,
you can just add them up.
01:07:46.410 --> 01:07:49.130
When you have a rigid
body, those summations
01:07:49.130 --> 01:07:50.980
become integrals.
01:07:50.980 --> 01:08:02.546
And, for example,
izz is the integral
01:08:02.546 --> 01:08:03.670
of-- how should I say this.
01:08:08.340 --> 01:08:13.595
x squared plus y squared
dm, every little mass bit.
01:08:20.250 --> 01:08:22.696
It looks like-- is there
an exam some distance away?
01:08:22.696 --> 01:08:24.029
I see a lot of people vanishing.
01:08:24.029 --> 01:08:26.120
OK, so let me--
I'll tell you what.
01:08:26.120 --> 01:08:29.250
I'll just make it easy
for it and let you go.
01:08:29.250 --> 01:08:33.399
Let me just say one thing
to where we're going.
01:08:33.399 --> 01:08:40.399
For every rigid body,
there is a different set
01:08:40.399 --> 01:08:48.810
of axes for which, when you
go to make up this matrix,
01:08:48.810 --> 01:08:51.790
you can make a diagonal.
01:08:51.790 --> 01:08:53.910
And those are called
the principal axes,
01:08:53.910 --> 01:08:55.520
and that's where
we're going next,
01:08:55.520 --> 01:08:59.189
those play a really important
role in what we want to do.
01:08:59.189 --> 01:09:00.920
OK.