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J. KIM VANDIVER: If I wanted
you to remember three equations,
00:00:26.650 --> 00:00:29.530
really commit them to memory,
that'd allow you to essentially
00:00:29.530 --> 00:00:35.012
do everything that we've done,
they'd be the following three.
00:00:35.012 --> 00:00:41.450
The first would essentially
be Newton's second law.
00:00:49.640 --> 00:00:53.500
We use that all the
time, and in vector form,
00:00:53.500 --> 00:00:58.850
summation of the external
forces on a rigid body
00:00:58.850 --> 00:01:02.710
must be equal to the
time rate of change
00:01:02.710 --> 00:01:07.630
of the linear momentum
of the rigid body.
00:01:07.630 --> 00:01:13.100
And that we know is equal to
its mass times the acceleration
00:01:13.100 --> 00:01:19.060
of its center of mass with
respect to an inertial frame,
00:01:19.060 --> 00:01:22.770
and this slash o means with
respect to an inertial frame.
00:01:22.770 --> 00:01:25.070
That's Newton's second
law, and of course we
00:01:25.070 --> 00:01:27.160
use it all the time.
00:01:27.160 --> 00:01:35.712
Now Euler added to what
Newton worked out for us.
00:01:35.712 --> 00:01:38.020
And let me actually just
point out over here.
00:01:38.020 --> 00:01:44.780
So P with respect to
o then is M velocity
00:01:44.780 --> 00:01:49.955
of G with respect to the
inertial frame-- vector,
00:01:49.955 --> 00:01:52.260
vector, vector.
00:01:52.260 --> 00:01:57.840
OK, what Euler taught us
was that the summation
00:01:57.840 --> 00:02:07.530
of the torques about a point
A is the time rate of change
00:02:07.530 --> 00:02:16.050
of the angular momentum
with respect to that point
00:02:16.050 --> 00:02:24.900
plus the velocity of the point
cross P with respect to o.
00:02:24.900 --> 00:02:32.150
So that's our general
equation for angular momentum
00:02:32.150 --> 00:02:33.460
and [? R ?] for torques.
00:02:33.460 --> 00:02:46.330
And remember, H with respect to
A is the cross product of R--
00:02:46.330 --> 00:02:47.741
how do I want to write this guy?
00:02:57.595 --> 00:02:58.678
I don't want to save that.
00:03:15.130 --> 00:03:17.030
So we define angular momentum.
00:03:17.030 --> 00:03:22.090
It's just a cross product of
the distance from the point
00:03:22.090 --> 00:03:29.810
that we're at to the-- I
didn't write this right.
00:03:29.810 --> 00:03:40.590
This is A with respect
to G cross P o,
00:03:40.590 --> 00:03:44.010
so that's our general
definition of angular momentum.
00:03:44.010 --> 00:03:46.130
And finally, the third
equation is the one
00:03:46.130 --> 00:03:49.760
that I showed you last time,
which does appear in the book
00:03:49.760 --> 00:03:52.190
but not until you
get into chapter 21,
00:03:52.190 --> 00:03:55.950
and that is that you
can write H with respect
00:03:55.950 --> 00:04:11.365
to A as H with respect to G plus
[? RG/A ?] cross P with respect
00:04:11.365 --> 00:04:12.082
to o.
00:04:15.910 --> 00:04:18.420
There's something wrong here.
00:04:18.420 --> 00:04:19.839
Matt, what have I done wrong?
00:04:19.839 --> 00:04:22.840
MATT: Do you want a
sum that will give you
00:04:22.840 --> 00:04:24.302
the angle [INAUDIBLE].
00:04:24.302 --> 00:04:26.260
J. KIM VANDIVER: Yeah,
it just slipped my brain
00:04:26.260 --> 00:04:27.176
because I don't do it.
00:04:27.176 --> 00:04:28.670
I don't think about this.
00:04:28.670 --> 00:04:31.390
This is the-- yeah,
it's a summation.
00:04:31.390 --> 00:04:34.230
There we go.
00:04:34.230 --> 00:04:37.780
In general, this is
a summation, and it
00:04:37.780 --> 00:04:45.040
is the location of all the
little particles i with respect
00:04:45.040 --> 00:04:55.220
to A cross the linear momentum
of each particle [? mi ?] Vi
00:04:55.220 --> 00:04:57.490
with respect to o.
00:04:57.490 --> 00:04:59.890
And we add all that up,
that's the definition
00:04:59.890 --> 00:05:01.300
of angular momentum.
00:05:01.300 --> 00:05:04.330
And we generally for
rigid bodies then
00:05:04.330 --> 00:05:10.140
reduce that to saying that
it's a mass moment of inertia
00:05:10.140 --> 00:05:17.510
matrix computed with respect
to A times omega x, omega y,
00:05:17.510 --> 00:05:19.710
omega z.
00:05:19.710 --> 00:05:23.470
That's how we
generally express this.
00:05:23.470 --> 00:05:28.400
But the third equation I
want to come to is this one.
00:05:28.400 --> 00:05:30.266
Yeah?
00:05:30.266 --> 00:05:34.129
AUDIENCE: That's
not [INAUDIBLE],
00:05:34.129 --> 00:05:36.170
but it's not really with
respect to A in general.
00:05:39.606 --> 00:05:40.106
[INAUDIBLE]
00:05:45.050 --> 00:05:48.160
J. KIM VANDIVER: This is i
with respect to A. Let's just
00:05:48.160 --> 00:05:52.590
do this because I've just
slipped a bit this morning.
00:05:52.590 --> 00:05:56.230
OK, I'm thinking too much
about where I'm going here.
00:05:56.230 --> 00:05:58.670
This equation I
introduced last time.
00:05:58.670 --> 00:06:01.060
The book doesn't make a
big deal about this at all,
00:06:01.060 --> 00:06:03.930
but it's really a
fundamental equation
00:06:03.930 --> 00:06:08.870
that we proved from scratch
very simply last time.
00:06:08.870 --> 00:06:13.120
And this equation allows you
to solve lots of problems
00:06:13.120 --> 00:06:17.050
without even, for example,
knowing the parallel axis
00:06:17.050 --> 00:06:17.780
theorem.
00:06:17.780 --> 00:06:20.850
It'll just get you through it
even without knowing things
00:06:20.850 --> 00:06:21.670
like that.
00:06:21.670 --> 00:06:24.260
So with these three equations,
you can do all the problems
00:06:24.260 --> 00:06:27.680
that we've done, and I'm going
to do two or three examples
00:06:27.680 --> 00:06:30.160
this morning, and
including coming back
00:06:30.160 --> 00:06:33.050
to a discussion of the
parallel axis theorem.
00:06:33.050 --> 00:06:39.849
OK, so these three are the ones
I recommend that you remember
00:06:39.849 --> 00:06:41.890
because everything else
will flow from-- we know,
00:06:41.890 --> 00:06:43.790
like, we have special cases.
00:06:43.790 --> 00:06:58.380
So when A is at
G, then this just
00:06:58.380 --> 00:07:01.490
reduces to H with
respect to G. When
00:07:01.490 --> 00:07:04.840
there's no vel-- when A doesn't
move, this term goes away.
00:07:04.840 --> 00:07:06.944
Those are all special
cases, which you can just
00:07:06.944 --> 00:07:09.360
substitute in the numbers and
discover for yourself if you
00:07:09.360 --> 00:07:10.796
don't remember those.
00:07:10.796 --> 00:07:12.170
But these are the
three equations
00:07:12.170 --> 00:07:14.600
you want to have
in your kit to use.
00:07:36.130 --> 00:07:38.940
So last time, I kind
of classified problems
00:07:38.940 --> 00:07:40.520
in four different
classes depending
00:07:40.520 --> 00:07:42.060
on the simplifications.
00:07:42.060 --> 00:07:45.720
And the class 4 problem
was the more general case,
00:07:45.720 --> 00:07:51.630
and that's when the point A
is allowed to move in general,
00:07:51.630 --> 00:07:56.160
and that last
equation we had can
00:07:56.160 --> 00:07:58.060
be very handy in those cases.
00:07:58.060 --> 00:08:02.080
And I mentioned a problem last
time, but I didn't work it out.
00:08:02.080 --> 00:08:06.450
And that's simply
this problem, where
00:08:06.450 --> 00:08:11.330
you've got a block,
maybe a box on a cart,
00:08:11.330 --> 00:08:15.300
and you're
accelerating the cart.
00:08:15.300 --> 00:08:17.960
At what point will
the block tip over?
00:08:17.960 --> 00:08:21.690
So this one I've put
a nail right here,
00:08:21.690 --> 00:08:27.560
so the block can't slide, but
it will tip about this point.
00:08:27.560 --> 00:08:30.730
So if I accelerate it
slowly, nothing happens.
00:08:30.730 --> 00:08:35.179
If I accelerate it fast enough,
it tips around that point.
00:08:35.179 --> 00:08:36.299
So that's the key.
00:08:36.299 --> 00:08:39.860
Identifying the point's have
the work, so that's the key bit.
00:08:39.860 --> 00:08:42.929
What's the greatest
acceleration that I can have
00:08:42.929 --> 00:08:47.460
so that this block will
just not quite tip?
00:08:47.460 --> 00:08:50.134
That's the problem.
00:08:50.134 --> 00:08:51.550
If you didn't have
the nail there,
00:08:51.550 --> 00:08:53.174
there's also the
possibility that it'll
00:08:53.174 --> 00:08:54.562
slip without tipping over.
00:08:54.562 --> 00:08:56.380
OK, so that's another
problem that we'll
00:08:56.380 --> 00:08:59.740
address at another moment.
00:08:59.740 --> 00:09:01.620
OK, so let's do this
problem quickly.
00:09:05.920 --> 00:09:13.490
So this is our problem,
and I've got a handle here,
00:09:13.490 --> 00:09:16.426
and I'm pushing on
it with a force.
00:09:16.426 --> 00:09:18.480
I've got a couple of wheels.
00:09:18.480 --> 00:09:20.424
This is M1.
00:09:20.424 --> 00:09:28.770
This is M2, and I'll have
an inertial frame here X, Y.
00:09:28.770 --> 00:09:32.640
So the question is
what's the maximum force
00:09:32.640 --> 00:09:36.220
that I can apply such that
the block won't tip over?
00:09:36.220 --> 00:09:41.910
And I've got a little nub there,
so this thing can't slide.
00:09:41.910 --> 00:09:45.880
OK, and this is my point
A because that's the point
00:09:45.880 --> 00:09:48.720
I expect it to rotate around.
00:09:48.720 --> 00:09:52.560
The X, Y, and Z is
coming out of the board.
00:09:52.560 --> 00:09:58.920
OK, so Newton's second,
so we apply it first.
00:09:58.920 --> 00:10:00.630
From Newton's second
law, we say the sum
00:10:00.630 --> 00:10:05.582
of the forces, in this
case, on the system.
00:10:05.582 --> 00:10:10.270
By the system I mean M1 and M2.
00:10:10.270 --> 00:10:12.900
We want to remember that you
can collect things together
00:10:12.900 --> 00:10:13.550
at times.
00:10:13.550 --> 00:10:15.930
So the sum of forces on
the system, in this case,
00:10:15.930 --> 00:10:26.190
is then M1 plus M2, and it's
the acceleration of that point,
00:10:26.190 --> 00:10:33.590
and in this case we are just
going to do in the X-direction
00:10:33.590 --> 00:10:37.720
because we know that's the only
one where there's any action.
00:10:37.720 --> 00:10:40.190
So the sum of the forces
in the X-direction
00:10:40.190 --> 00:10:43.055
is the total mass of the
system times the acceleration
00:10:43.055 --> 00:10:44.930
in the X-direction.
00:10:44.930 --> 00:10:47.390
So the X double
dot we're looking
00:10:47.390 --> 00:10:53.370
for is F over M1 plus M2.
00:10:53.370 --> 00:10:56.120
So this we're trying to find out
what the maximum value for this
00:10:56.120 --> 00:10:56.620
is.
00:11:05.200 --> 00:11:07.400
Now next we can apply
the torque equation
00:11:07.400 --> 00:11:09.010
because that's the key one.
00:11:09.010 --> 00:11:11.550
And that's the sum of
the torques-- now it's
00:11:11.550 --> 00:11:18.930
just on the box-- at
with respect to point A.
00:11:18.930 --> 00:11:20.900
And we can say, well,
looking at that,
00:11:20.900 --> 00:11:23.010
we need a free body
diagram now of our box.
00:11:26.780 --> 00:11:29.790
You've got an M1g down.
00:11:36.760 --> 00:11:41.350
And you might have a normal
force upwards, generally,
00:11:41.350 --> 00:11:44.810
but now we want to
think about where
00:11:44.810 --> 00:11:50.000
are the forces going
to be acting on the box
00:11:50.000 --> 00:11:53.770
when it's just at this point
that it's just barely about--
00:11:53.770 --> 00:11:55.270
just barely starts to tip.
00:11:58.640 --> 00:12:03.110
So we're hypothesizing
that here at A, there'll
00:12:03.110 --> 00:12:07.300
be some upward force, right?
00:12:07.300 --> 00:12:10.330
And it's going to take care
of the static equilibrium,
00:12:10.330 --> 00:12:15.690
but it also could enter
into our moment calculation
00:12:15.690 --> 00:12:17.490
if we weren't clever
with respect to where
00:12:17.490 --> 00:12:18.720
we calculate this point.
00:12:22.960 --> 00:12:25.780
So the sum of the
torques about this point,
00:12:25.780 --> 00:12:29.370
we don't have to bring
this one into it.
00:12:29.370 --> 00:12:31.320
We do have to consider this one.
00:12:31.320 --> 00:12:39.760
So the [? R ?] here cross that
force, so we get a moment, Mg,
00:12:39.760 --> 00:12:42.836
and I need some
dimensions on my box.
00:12:42.836 --> 00:12:53.260
We'll call it b and a height h.
00:12:53.260 --> 00:12:57.080
So the moment arm,
this axon is b/2,
00:12:57.080 --> 00:12:59.780
and it's in the minus
[? z ?] direction,
00:12:59.780 --> 00:13:08.680
so I get a minus M1g b/2 k hat.
00:13:08.680 --> 00:13:10.980
Those are my external
torques on the box
00:13:10.980 --> 00:13:14.650
because the normal force
supporting it on the floor
00:13:14.650 --> 00:13:15.710
is acting at A now.
00:13:19.060 --> 00:13:29.950
And this must be from
our second formula,
00:13:29.950 --> 00:13:34.050
dH with respect to A/dt
plus the velocity of A
00:13:34.050 --> 00:13:39.000
with respect to o, cross
P with respect to o.
00:13:39.000 --> 00:13:49.990
And P with respect to o
was [? M1VG ?] with respect
00:13:49.990 --> 00:13:55.710
to o, all right?
00:13:55.710 --> 00:13:59.060
So we need to know what
is the velocity of A
00:13:59.060 --> 00:14:00.620
with respect to o.
00:14:03.650 --> 00:14:08.220
But we're only allowing
motions in this direction.
00:14:08.220 --> 00:14:10.950
This has to be evaluated
in the inertial frame,
00:14:10.950 --> 00:14:15.200
so we have a coordinate X here.
00:14:15.200 --> 00:14:17.960
So it's going to be
some X dot, and I'll
00:14:17.960 --> 00:14:20.180
give you a capital I
hat just remind you
00:14:20.180 --> 00:14:23.230
that's in the inertial frame.
00:14:23.230 --> 00:14:26.760
What is the velocity
of G with respect to o?
00:14:34.253 --> 00:14:35.544
AUDIENCE: Would it be the same?
00:14:35.544 --> 00:14:37.008
J. KIM VANDIVER: Yeah.
00:14:37.008 --> 00:14:38.740
You know, in this
condition we're
00:14:38.740 --> 00:14:41.130
saying we want the block
not to quite tip over,
00:14:41.130 --> 00:14:43.810
and that means it's moving
with the cart at exactly
00:14:43.810 --> 00:14:45.800
the same velocity,
so it's the same.
00:14:45.800 --> 00:14:49.260
It's also equal to
A with respect to o,
00:14:49.260 --> 00:14:51.260
but if that's the case,
what's the cross product
00:14:51.260 --> 00:14:52.884
between velocity of
A with respect to o
00:14:52.884 --> 00:14:54.811
and G with respect to o?
00:14:54.811 --> 00:14:56.330
Ah, so that one goes to zero.
00:14:59.680 --> 00:15:00.770
That's because of this.
00:15:09.170 --> 00:15:15.630
OK, so that leaves
us with-- we don't
00:15:15.630 --> 00:15:17.780
have to deal with
these terms, but we
00:15:17.780 --> 00:15:25.210
do need to compute
H with respect to A.
00:15:25.210 --> 00:15:28.350
And now I'm going to use this
formulation because it makes
00:15:28.350 --> 00:15:30.630
this problem easy
to do and not have
00:15:30.630 --> 00:15:33.760
to deal with parallel
axes or any of that.
00:15:33.760 --> 00:15:36.020
So I'm going to use
my third equation,
00:15:36.020 --> 00:15:38.860
and it says that this
is H with respect
00:15:38.860 --> 00:15:47.380
to G plus RG with respect to
A cross P with respect to o.
00:15:52.280 --> 00:15:56.590
So you notice I chose this
problem kind of on purpose
00:15:56.590 --> 00:15:59.250
today because I started
saying, here's three
00:15:59.250 --> 00:16:00.990
really important equations.
00:16:00.990 --> 00:16:04.850
We use all three
to do this problem.
00:16:04.850 --> 00:16:06.440
OK, so now we're
exercising the third.
00:16:19.940 --> 00:16:26.720
So my block here,
since we're dealing
00:16:26.720 --> 00:16:28.510
with angular momentum
of rigid bodies,
00:16:28.510 --> 00:16:32.130
I now have to think in terms
of a coordinate system attached
00:16:32.130 --> 00:16:39.980
to the block, and
I'll formulate it
00:16:39.980 --> 00:16:42.840
so it's lined up the same way.
00:16:42.840 --> 00:16:50.480
So this is a x1 I'll
call it, y1, and z1.
00:16:50.480 --> 00:17:03.550
But x1 and big X, these
are in the same direction,
00:17:03.550 --> 00:17:06.599
and so is z and z1
and capital [? E. ?]
00:17:06.599 --> 00:17:12.020
And that means that my I
hats are the same as the unit
00:17:12.020 --> 00:17:15.890
vectors associated with my
coordinate system attached
00:17:15.890 --> 00:17:16.690
to the body.
00:17:16.690 --> 00:17:19.425
Remember, this is this
body-fixed coordinate system
00:17:19.425 --> 00:17:22.260
that allows us to compute
moments of inertia,
00:17:22.260 --> 00:17:25.470
and I'm going to have
this located at G, OK?
00:17:49.660 --> 00:17:53.300
So the only rotation that
we're allowing in this problem.
00:17:53.300 --> 00:17:57.550
This is basically a 2D
planar motion problem.
00:17:57.550 --> 00:18:00.510
It could conceivably
have x and y translations
00:18:00.510 --> 00:18:05.600
and a z rotation
constrained in the y,
00:18:05.600 --> 00:18:09.650
so it has possible
rotation and translation.
00:18:09.650 --> 00:18:16.680
So the only omega we're
considering is this omega z,
00:18:16.680 --> 00:18:22.415
and so we'll proceed
to use it to compute H.
00:18:22.415 --> 00:18:23.880
So we use this formula.
00:18:29.510 --> 00:18:33.507
So H with respect
to G is-- just I'm
00:18:33.507 --> 00:18:35.715
doing a couple of these
things sort of carefully just
00:18:35.715 --> 00:18:38.970
to remind you of where
these things all come from.
00:18:38.970 --> 00:18:46.580
This is I with respect to G,
and in this case, 0 0 omega z.
00:18:46.580 --> 00:18:51.810
And we come out of
this then the only--
00:18:51.810 --> 00:18:55.540
are these principal axes?
00:18:55.540 --> 00:19:00.410
So if they are, this
is diagonal, right,
00:19:00.410 --> 00:19:04.460
so I've located these through G.
00:19:04.460 --> 00:19:06.490
Taken into consideration
the symmetries,
00:19:06.490 --> 00:19:09.460
I know because the symmetry
of these are principal axes.
00:19:09.460 --> 00:19:14.140
Therefore, this is a diagonal
mass moment of inertia matrix,
00:19:14.140 --> 00:19:17.140
so when I calculate to do
out the multiplication,
00:19:17.140 --> 00:19:27.420
I get Izz with respect
to G omega z k-hat.
00:19:27.420 --> 00:19:31.420
There's only one
vector component of H
00:19:31.420 --> 00:19:35.250
that comes out of
that, and that's Hz
00:19:35.250 --> 00:19:41.680
is that Izz G omega z k-hat.
00:19:41.680 --> 00:19:46.880
And there are two other
possible components of H,
00:19:46.880 --> 00:19:48.390
but they're both 0.
00:19:48.390 --> 00:19:52.720
There's no other rotate--
there is no angular momentum
00:19:52.720 --> 00:19:53.950
in the x- or y-directions.
00:19:56.390 --> 00:19:56.890
OK.
00:20:08.970 --> 00:20:14.050
[? RG/A ?] in this problem
is from here-- this
00:20:14.050 --> 00:20:29.960
is point A-- two there, so
RG respect to A is b/2 i plus
00:20:29.960 --> 00:20:33.190
h/2 k.
00:20:33.190 --> 00:20:37.490
It's the location of just the
position of G with respect
00:20:37.490 --> 00:20:44.000
to A. [? b/2 ?] over
[? h/2 up. ?] So in order
00:20:44.000 --> 00:20:46.755
to complete my
calculation of H of A,
00:20:46.755 --> 00:20:50.085
I need to get the second piece
of the [? RG/A ?] cross P/o.
00:21:07.730 --> 00:21:20.170
So [? RG/A ?] is my
b/2 i plus h/2 k cross.
00:21:20.170 --> 00:21:30.560
And then my linear
momentum is M1 x1 dot
00:21:30.560 --> 00:21:33.064
in the i hat direction.
00:21:33.064 --> 00:21:35.230
And it'll either be a
capitalize I or a lowercase i.
00:21:35.230 --> 00:21:37.110
They're the same, all right?
00:21:37.110 --> 00:21:39.970
So this is my second
term the RG cross
00:21:39.970 --> 00:21:46.340
P. i cross i, these two--
this pair gives you nothing,
00:21:46.340 --> 00:21:52.260
so it's k cross i is j, and
so you get Izz with respect
00:21:52.260 --> 00:21:55.590
to G omega z.
00:21:55.590 --> 00:22:11.790
And then a k term
from here M1 h/2
00:22:11.790 --> 00:22:15.390
x dot, and it's also in the k.
00:22:19.190 --> 00:22:25.248
And that's my total angular
momentum now with respect to A.
00:22:25.248 --> 00:22:26.660
AUDIENCE: [INAUDIBLE]?
00:22:26.660 --> 00:22:34.700
J. KIM VANDIVER: And I think--
just a sec-- k-- you're right.
00:22:34.700 --> 00:22:35.820
Wait a second.
00:22:35.820 --> 00:22:38.856
k cross i should
give me a j, right?
00:22:38.856 --> 00:22:39.772
AUDIENCE: [INAUDIBLE].
00:22:46.506 --> 00:22:48.430
Professor?
00:22:48.430 --> 00:22:52.890
J. KIM VANDIVER: Yeah, k cross
i gives me a positive j, right?
00:22:52.890 --> 00:22:53.870
[INTERPOSING VOICES]
00:22:53.870 --> 00:22:57.496
[? AUDIENCE: RGA ?] should
be a positive j based off
00:22:57.496 --> 00:22:58.620
of your [INAUDIBLE] system.
00:22:58.620 --> 00:23:01.340
J. KIM VANDIVER: Ah, yeah, yeah.
00:23:01.340 --> 00:23:06.510
This is a good catch.
00:23:06.510 --> 00:23:08.120
All right, so what happens now?
00:23:08.120 --> 00:23:13.155
Now we get j cross i,
gives you a minus k.
00:23:19.670 --> 00:23:23.290
Now that should be OK,
and let's see if that
00:23:23.290 --> 00:23:24.601
agrees with what I wrote here.
00:23:24.601 --> 00:23:26.100
Yes, they did it
right in the paper.
00:23:26.100 --> 00:23:27.766
Just couldn't put it
right on the board.
00:23:50.060 --> 00:23:54.370
So now the sum of the
torques with respect to A,
00:23:54.370 --> 00:24:01.030
we've already figured out
that it's minus M1 g b/2 k
00:24:01.030 --> 00:24:09.255
hat must be equal to
the time rate of change
00:24:09.255 --> 00:24:11.810
from our second equation dH dt.
00:24:11.810 --> 00:24:14.910
And we've already figured out
that the second term goes away.
00:24:14.910 --> 00:24:17.775
So it's just then the time rate
of change of that expression.
00:24:28.157 --> 00:24:30.830
Oh, I think we could
probably do it straight away,
00:24:30.830 --> 00:24:33.810
so the time rate of change,
the derivative of this,
00:24:33.810 --> 00:24:35.570
you get omega z dot.
00:24:35.570 --> 00:24:37.080
You get an x double dot.
00:24:37.080 --> 00:24:40.590
The k doesn't rotate, so there's
nothing that comes from that.
00:24:40.590 --> 00:24:41.990
If you take that
time derivative,
00:24:41.990 --> 00:24:56.260
you get Izz with respect to
G omega z dot minus M1 h/2
00:24:56.260 --> 00:25:02.090
x double dot, and these are
still all in the k direction.
00:25:02.090 --> 00:25:05.300
And that's an
equation that only has
00:25:05.300 --> 00:25:09.020
k hat terms on the left
hand side, k on the right,
00:25:09.020 --> 00:25:13.500
so this is just one potential
component of the torque.
00:25:13.500 --> 00:25:16.860
And so we can drop
the k's at this point.
00:25:16.860 --> 00:25:20.470
And we now have an expression
for the external torques
00:25:20.470 --> 00:25:24.850
in terms of the
accelerations of the system.
00:25:24.850 --> 00:25:27.480
This is essentially theta
double dot [? if it ?] tips,
00:25:27.480 --> 00:25:30.120
and that's the
linear acceleration.
00:25:30.120 --> 00:25:33.150
But now then the
key to the problem
00:25:33.150 --> 00:25:36.590
is we're trying to find
out when just at the point
00:25:36.590 --> 00:25:42.242
it might tip but doesn't, so
what's omega dot-- omega z dot?
00:25:42.242 --> 00:25:46.640
OK, so what we're looking
for we let omega z
00:25:46.640 --> 00:25:48.985
dot or require that to be 0.
00:25:48.985 --> 00:25:51.150
I mean this makes
this even simpler,
00:25:51.150 --> 00:25:57.260
and you find out then
that M1 g [? b/2 ?]
00:25:57.260 --> 00:26:01.707
equals M1 h/2 x double dot.
00:26:01.707 --> 00:26:03.540
And I have to add minus
signs on both sides,
00:26:03.540 --> 00:26:04.820
so I got rid of them.
00:26:04.820 --> 00:26:10.415
And the M1s go away, and I can
solve for h for x double dot,
00:26:10.415 --> 00:26:13.870
and that's going to be g b/h.
00:26:19.140 --> 00:26:23.290
OK, and we started off saying
what's the maximum force?
00:26:23.290 --> 00:26:28.700
So this is now x max,
the acceleration maximum
00:26:28.700 --> 00:26:38.480
and the force maximum is
just equal to M1 plus M2
00:26:38.480 --> 00:26:51.350
x dot max for-- and let's
see if that makes sense.
00:26:51.350 --> 00:26:59.950
If b and h were equal--
that's sort of a common sense
00:26:59.950 --> 00:27:05.270
argument here-- b
and h were equal,
00:27:05.270 --> 00:27:10.290
that tells you that you could
accelerate 1g at 1g-- just 1g
00:27:10.290 --> 00:27:11.320
and that makes sense.
00:27:11.320 --> 00:27:16.182
The Mg here is [? putting a ?]
restoring moment down of the MG
00:27:16.182 --> 00:27:21.760
[? b/2. ?] And the
acceleration of this is putting
00:27:21.760 --> 00:27:25.870
an overturning moment on it in
the other direction that would
00:27:25.870 --> 00:27:30.750
be Mg [? h/2, ?] and if
h and b are the same,
00:27:30.750 --> 00:27:33.250
then it would be exactly 1g.
00:27:33.250 --> 00:27:36.240
Now so that's the
answer the problem.
00:27:36.240 --> 00:27:41.630
If you want to do live
dangerously from the beginning,
00:27:41.630 --> 00:27:45.410
you could have done this
with a fictitious force.
00:27:45.410 --> 00:27:47.730
I just mention in passing
I'm not recommending
00:27:47.730 --> 00:27:48.910
that you generally go there.
00:27:48.910 --> 00:27:51.260
This was a very straightforward
way of doing it.
00:27:51.260 --> 00:27:54.060
We just started with
these three laws,
00:27:54.060 --> 00:27:58.890
and we just worked our way
through the problem and all
00:27:58.890 --> 00:28:00.980
the way to the end,
and it fell out.
00:28:00.980 --> 00:28:03.260
If you didn't want to
live dangerously though,
00:28:03.260 --> 00:28:15.930
you could say that when
you accelerate this,
00:28:15.930 --> 00:28:19.450
you can think of there being
a fictitious force that
00:28:19.450 --> 00:28:25.340
is equal to minus the
mass of this object
00:28:25.340 --> 00:28:28.460
times the acceleration on the
center of gravity of force
00:28:28.460 --> 00:28:32.400
that is pulling it in
the other direction.
00:28:32.400 --> 00:28:40.584
So you could think of there
being a force M1X double dot
00:28:40.584 --> 00:28:42.500
opposite to the direction
of the acceleration.
00:28:42.500 --> 00:28:45.640
This is minus the mass
times the acceleration.
00:28:45.640 --> 00:28:47.710
It's the fictitious force.
00:28:47.710 --> 00:28:54.860
Here's your point A.
Here's gravity M1 g.
00:28:54.860 --> 00:28:57.040
This is the distance b/2.
00:28:59.570 --> 00:29:03.960
This is the height h/2,
and you could say just
00:29:03.960 --> 00:29:09.140
at that moment of balance,
the sum of those two moment
00:29:09.140 --> 00:29:11.445
should be 0-- the sub
of the external torques.
00:29:15.030 --> 00:29:21.950
And you would end up
with a M1 x double dot
00:29:21.950 --> 00:29:29.100
h/2 minus M1 g [? b/2, ?]
and you come up
00:29:29.100 --> 00:29:30.700
with the same answer.
00:29:30.700 --> 00:29:35.790
But for most of this that
takes a real leap of faith
00:29:35.790 --> 00:29:39.340
to do that to convince yourself
that you're right, right?
00:29:39.340 --> 00:29:40.735
You got to know
a lot of dynamics
00:29:40.735 --> 00:29:43.170
and remember all
the parts and pieces
00:29:43.170 --> 00:29:47.270
to be able to just go there.
00:29:47.270 --> 00:29:50.746
But if you do it just carefully,
those three equations,
00:29:50.746 --> 00:29:51.620
it'll all worked out.
00:29:51.620 --> 00:29:52.280
Yeah?
00:29:52.280 --> 00:29:54.840
AUDIENCE: Why is
omega z [INAUDIBLE]?
00:29:54.840 --> 00:29:56.440
J. KIM VANDIVER:
Because as long as
00:29:56.440 --> 00:30:00.040
the condition satisfied
that it doesn't tip over,
00:30:00.040 --> 00:30:03.336
what's the rotation
rate of this block?
00:30:03.336 --> 00:30:04.260
AUDIENCE: 0.
00:30:04.260 --> 00:30:05.830
J. KIM VANDIVER: 0.
00:30:05.830 --> 00:30:09.820
So we're not solving for
dynamics in this problem
00:30:09.820 --> 00:30:11.040
of the thing tipping.
00:30:11.040 --> 00:30:12.570
It isn't rolling on us.
00:30:12.570 --> 00:30:15.330
We're coming just
up to the point
00:30:15.330 --> 00:30:17.970
that it tips, and say we're
not going any farther.
00:30:17.970 --> 00:30:22.820
And so we just require the two
moments-- the one essentially
00:30:22.820 --> 00:30:26.990
caused by the acceleration,
this fictitious force, that's
00:30:26.990 --> 00:30:30.400
got to be just in equilibrium
with the restoring moment
00:30:30.400 --> 00:30:32.230
provided by Mg.
00:30:32.230 --> 00:30:34.400
That's essentially the
problem we've worked,
00:30:34.400 --> 00:30:36.880
but we've done it really
carefully just using
00:30:36.880 --> 00:30:40.090
these three laws.
00:30:40.090 --> 00:30:43.700
And the real problem though if
you didn't have the nail here
00:30:43.700 --> 00:30:48.210
is if now if you need
to test your solution,
00:30:48.210 --> 00:30:53.330
could you ever actually reach
that rate of acceleration
00:30:53.330 --> 00:30:55.870
before the thing
starts sliding on you?
00:30:55.870 --> 00:30:57.070
And, you know, maybe not.
00:30:57.070 --> 00:31:01.305
This thing slides
before it tips, OK?
00:31:01.305 --> 00:31:06.247
AUDIENCE: Would the fictitious
force take into account M2?
00:31:06.247 --> 00:31:07.830
J. KIM VANDIVER:
Take into account M2?
00:31:07.830 --> 00:31:09.934
AUDIENCE: Yes.
00:31:09.934 --> 00:31:11.600
J. KIM VANDIVER: So
the angular momentum
00:31:11.600 --> 00:31:15.970
was all calculated with respect
to the rigid body M1, right?
00:31:15.970 --> 00:31:19.050
The only reason
M2 entered into it
00:31:19.050 --> 00:31:20.959
this is the way things
happen in real life.
00:31:20.959 --> 00:31:22.000
It's just a real problem.
00:31:22.000 --> 00:31:24.519
It's a box on a cart, and
is it going to tip over,
00:31:24.519 --> 00:31:25.560
and you're pushing on it.
00:31:25.560 --> 00:31:27.770
How hard can I push?
00:31:27.770 --> 00:31:29.790
To do that problem you
just have to consider
00:31:29.790 --> 00:31:32.270
these to begin with so
that you realize that,
00:31:32.270 --> 00:31:36.970
oh, the acceleration involves
both of these masses,
00:31:36.970 --> 00:31:39.230
but the angular momentum
part of the problem
00:31:39.230 --> 00:31:43.670
involves only the boxes tipping.
00:31:43.670 --> 00:31:47.440
So problems even as simple
as this one looks has
00:31:47.440 --> 00:31:49.370
their little
complications, and you've
00:31:49.370 --> 00:31:52.020
got to work through them.
00:31:52.020 --> 00:31:54.610
So just so we had to treat
the body as a two-body system
00:31:54.610 --> 00:31:58.460
to start with, and then look
at angular momentum for just
00:31:58.460 --> 00:31:59.050
the box.
00:31:59.050 --> 00:32:02.598
AUDIENCE: Does the
pin or screw exert
00:32:02.598 --> 00:32:03.990
a force on the [INAUDIBLE]?
00:32:07.880 --> 00:32:10.060
J. KIM VANDIVER: Yes,
absolutely, mm-hmm.
00:32:10.060 --> 00:32:12.430
Does it create a moment?
00:32:12.430 --> 00:32:16.380
No, and that's why we
don't have to consider it.
00:32:16.380 --> 00:32:19.305
I could have put it in there
and probably should have.
00:32:19.305 --> 00:32:27.260
If I had done a free
body diagram here,
00:32:27.260 --> 00:32:31.490
you could still say that there's
a reaction force upwards here.
00:32:31.490 --> 00:32:35.610
And I'll call it Ry and
another one this direction
00:32:35.610 --> 00:32:37.570
I'll call Rx.
00:32:37.570 --> 00:32:41.600
And when computing moments about
this point, neither of them
00:32:41.600 --> 00:32:42.580
enter into the problem.
00:32:42.580 --> 00:32:46.880
And that's why we like to do
our calculations for angular
00:32:46.880 --> 00:32:53.010
momentum with respect to points
around which the body rotates
00:32:53.010 --> 00:32:54.590
because that allows
us to not have
00:32:54.590 --> 00:32:59.040
to solve for those unknown
forces that pop up there,
00:32:59.040 --> 00:33:01.530
so yeah, indeed there
are forces there.
00:33:01.530 --> 00:33:06.520
OK, got it figured out.
00:33:06.520 --> 00:33:09.660
So we've talked a little
tiny bit about parallel axis
00:33:09.660 --> 00:33:13.900
theorem, and so I'm going to
consider this a slender rod.
00:33:17.640 --> 00:33:22.700
I can spin it about one of its
printable axes and calculate
00:33:22.700 --> 00:33:24.790
its angular momentum
of whatever I need,
00:33:24.790 --> 00:33:29.190
and if I move it over to an axis
that's-- I just move it over
00:33:29.190 --> 00:33:30.230
by a little bit.
00:33:30.230 --> 00:33:37.350
Let's say this is the
x-axis here in my body,
00:33:37.350 --> 00:33:40.990
and this is the z,
so it's at omega z.
00:33:40.990 --> 00:33:45.835
And I move my point and around
which I am rotating over
00:33:45.835 --> 00:33:48.460
by a small amount A, and that's
this distance between these two
00:33:48.460 --> 00:33:49.850
holes.
00:33:49.850 --> 00:33:52.200
And by parallel axis
theorem, we could say, well,
00:33:52.200 --> 00:33:55.520
now there's a mass moment of
inertia with respect to A,
00:33:55.520 --> 00:33:58.700
which is the mass moment of
inertia with respect to g,
00:33:58.700 --> 00:34:03.530
which is [? Ml ?] squared over
12, if this is the length,
00:34:03.530 --> 00:34:06.040
plus the mass times
this distance squared.
00:34:06.040 --> 00:34:08.760
We know how to do that.
00:34:08.760 --> 00:34:17.080
But now what happens if I take
this stick, my slender rod,
00:34:17.080 --> 00:34:24.920
and here's my original x, y, z.
00:34:24.920 --> 00:34:29.270
And I want to move it,
my point A, my stick now
00:34:29.270 --> 00:34:38.520
I'm going to move it over by an
amount a and up by an amount c
00:34:38.520 --> 00:34:43.480
so that it's now up here
with respect to this point.
00:34:43.480 --> 00:34:45.889
So this stick has
been like this,
00:34:45.889 --> 00:34:47.960
and now I'm going to then
rotate it about that.
00:34:47.960 --> 00:34:49.350
That's the simple case.
00:34:49.350 --> 00:34:51.250
Now I want to rotate
it about this axis
00:34:51.250 --> 00:34:54.409
when it's moved
over to like that.
00:34:57.780 --> 00:35:04.280
So not only have I moved it--
so here's the original problem--
00:35:04.280 --> 00:35:15.675
like this, but now I move
it up and over-- maybe
00:35:15.675 --> 00:35:17.990
I'll do it there--
and now I spin it.
00:35:17.990 --> 00:35:21.270
And I'm interested in angular
momentum about this point,
00:35:21.270 --> 00:35:24.990
so now it's dynamically
unbalanced for sure.
00:35:24.990 --> 00:35:27.900
This thing is trying
to wobble, and it's
00:35:27.900 --> 00:35:31.550
been pushed up and
over, so is there
00:35:31.550 --> 00:35:34.160
a way to get-- how do
we solve that problem?
00:35:34.160 --> 00:35:38.140
How do we compute the
torques for this problem?
00:35:38.140 --> 00:35:41.010
Well, we could try to go
at it with parallel access
00:35:41.010 --> 00:35:44.110
to start with, but you don't
know how to do that in general
00:35:44.110 --> 00:35:45.130
for this problem.
00:35:45.130 --> 00:35:48.140
So what do we go to?
00:35:48.140 --> 00:35:49.970
We go back to those
three equations,
00:35:49.970 --> 00:35:52.140
and just start there,
and it'll all fall out.
00:35:52.140 --> 00:35:56.800
I'm going to do this kind
of briefly for this problem,
00:35:56.800 --> 00:36:04.420
so I want to compute H with
respect to A H with respect
00:36:04.420 --> 00:36:12.520
to G plus RG with
respect to A cross P.
00:36:12.520 --> 00:36:19.150
And now this is [? RG/A ?]
is this vector here,
00:36:19.150 --> 00:36:21.232
that position vector, OK?
00:36:24.430 --> 00:36:28.600
And the only
rotation is omega z.
00:36:28.600 --> 00:36:30.710
This is our z-axis.
00:36:30.710 --> 00:36:39.030
So I can write this as Izz
with respect to G, 0, 0,
00:36:39.030 --> 00:36:48.800
omega z plus my [? RG/A ?] cross
P/o, and that in this problem--
00:36:48.800 --> 00:36:50.990
let's get the vector
right this time-- it's
00:36:50.990 --> 00:36:59.780
a in the i direction plus
c in the k direction cross.
00:36:59.780 --> 00:37:06.360
Now we need to know what's
P with respect to o,
00:37:06.360 --> 00:37:12.910
but that's just the velocity
of this so the center of mass,
00:37:12.910 --> 00:37:16.270
which is now here.
00:37:16.270 --> 00:37:20.520
The velocity is omega
cross r, so when
00:37:20.520 --> 00:37:22.810
you do that-- you've done
that problem lots of times--
00:37:22.810 --> 00:37:28.140
it's this distance, so
perpendicular distance
00:37:28.140 --> 00:37:30.110
[? across ?] k.
00:37:30.110 --> 00:37:31.740
It's got to be
going into the board
00:37:31.740 --> 00:37:32.864
if it's spinning like this.
00:37:32.864 --> 00:37:35.090
We know the answer's
got to come out plus j,
00:37:35.090 --> 00:37:47.810
and it's r-- it's A omega z in
the j-direction cross Ma omega
00:37:47.810 --> 00:37:50.610
z, but it is in the
j hat direction.
00:37:50.610 --> 00:37:54.240
That's your linear momentum,
and this is [? RG/A ?]
00:37:54.240 --> 00:37:55.535
cross with a linear momentum.
00:37:59.440 --> 00:38:04.150
This here is M velocity of
G here with respect to o.
00:38:04.150 --> 00:38:11.890
That's this term, so we
carry out this calculation
00:38:11.890 --> 00:38:19.000
and we get-- and I need a j, OK.
00:38:19.000 --> 00:38:25.280
So i cross j gives me
a k, so I'll get 2,
00:38:25.280 --> 00:38:30.030
and this is a diagonal
matrix because we
00:38:30.030 --> 00:38:34.710
began with a set of
principal axes for our stick,
00:38:34.710 --> 00:38:40.180
so this term is Izz
with respect to G.
00:38:40.180 --> 00:38:43.390
I shouldn't have written zz
here, just i with respect to z.
00:38:43.390 --> 00:38:44.560
We multiply this out.
00:38:44.560 --> 00:38:54.190
We pick up just this term omega
z k hat plus i cross j is k,
00:38:54.190 --> 00:39:03.370
so that's Ma squared omega
[? zk, ?] and k cross j
00:39:03.370 --> 00:39:12.645
is minus i minus M ac omega
z in the i hat direction.
00:39:15.330 --> 00:39:20.140
So this is my angular momentum,
and reminding, it is a vector.
00:39:20.140 --> 00:39:22.715
This has got components in
the k and the i directions.
00:39:52.570 --> 00:39:54.090
I've rewritten it here.
00:39:54.090 --> 00:39:58.970
This also could be written--
this H with respect to A
00:39:58.970 --> 00:40:11.192
is I with respect to A
times the rotation vector.
00:40:11.192 --> 00:40:12.900
If we had just set
out with this problem,
00:40:12.900 --> 00:40:15.420
and said we're going-- if we
know the mass moment of inertia
00:40:15.420 --> 00:40:18.790
matrix computed with
respect to this point
00:40:18.790 --> 00:40:23.130
and multiplied it by our
rotation rate vector,
00:40:23.130 --> 00:40:26.570
we should have gotten
the same answer.
00:40:26.570 --> 00:40:27.780
So we didn't do that.
00:40:27.780 --> 00:40:30.820
We just worked it out
using just this formula,
00:40:30.820 --> 00:40:33.430
but we've essentially
discovered the answer.
00:40:33.430 --> 00:40:35.250
This is the rotation around k.
00:40:37.760 --> 00:40:42.470
This is the Izz term
with respect to A,
00:40:42.470 --> 00:40:45.850
and it really looks like the
familiar parallel axis piece.
00:40:45.850 --> 00:40:49.640
It's the amount that you
moved this axis over,
00:40:49.640 --> 00:40:53.509
and the axis you're spinning
around, you moved it over A.
00:40:53.509 --> 00:40:55.300
And you know the parallel
axis theorem just
00:40:55.300 --> 00:40:57.980
says add Ma squared, and
you've got the answer.
00:40:57.980 --> 00:41:01.350
But we also moved it up,
and it gave us another term.
00:41:01.350 --> 00:41:04.250
What do you suppose
in that term is?
00:41:04.250 --> 00:41:11.207
So this is Izz with
respect to A, this term.
00:41:11.207 --> 00:41:13.290
By parallel axis theorem,
you're familiar with it.
00:41:13.290 --> 00:41:22.410
This one is I, and
this is xz with respect
00:41:22.410 --> 00:41:26.040
in the A coordinate system.
00:41:26.040 --> 00:41:32.240
OK, we've created-- we've
made this thing unbalanced.
00:41:32.240 --> 00:41:34.010
The angular momentum
vector no longer
00:41:34.010 --> 00:41:36.640
points in the same
direction as the rotation.
00:41:36.640 --> 00:41:38.755
It has a k component
and an I component,
00:41:38.755 --> 00:41:40.880
and sure enough, when I
spend that thing like that,
00:41:40.880 --> 00:41:47.610
I feel that additional
torque out around my hand
00:41:47.610 --> 00:41:49.230
down here where I'm holding it.
00:41:49.230 --> 00:41:52.620
This thing is going-- trying
to pull it back and forth.
00:41:52.620 --> 00:41:55.090
So we have essentially
just shown--
00:41:55.090 --> 00:41:58.790
we have just arrived at a
more complicated parallel axis
00:41:58.790 --> 00:42:02.860
theorem by just using
this basic formula.
00:42:02.860 --> 00:42:07.830
So Williams in the dynamics--
that second handout
00:42:07.830 --> 00:42:12.660
from Williams has actually
given us a general formulation
00:42:12.660 --> 00:42:14.810
for parallel axis theorem.
00:42:18.720 --> 00:42:20.070
So this is just handy to know.
00:42:20.070 --> 00:42:28.200
I'll just give it to you, and
if you have a problem some day,
00:42:28.200 --> 00:42:30.770
where you just like to go
directly to this statement,
00:42:30.770 --> 00:42:32.650
so now I'm just going
to say, in general, we
00:42:32.650 --> 00:42:37.410
could have moved over by
A, up by c and actually
00:42:37.410 --> 00:42:42.980
into the board by b, so x,
I'm moving A. y, I'm moving b.
00:42:42.980 --> 00:42:44.240
z, I'm moving c.
00:42:44.240 --> 00:42:48.310
OK, so you could have all those
possible moves of the new point
00:42:48.310 --> 00:42:49.540
about which you're rotating.
00:42:49.540 --> 00:42:50.948
Yes?
00:42:50.948 --> 00:42:55.214
AUDIENCE: For the [INAUDIBLE]
matrix i with respect to A,
00:42:55.214 --> 00:42:57.435
does that mean that you're
rotating about point A?
00:42:57.435 --> 00:42:58.060
Or what is it--
00:42:58.060 --> 00:43:00.796
J. KIM VANDIVER: You are
rotating about point A.
00:43:00.796 --> 00:43:02.245
AUDIENCE: OK.
00:43:02.245 --> 00:43:04.660
But isn't point A just
the [? immersion? ?]
00:43:11.520 --> 00:43:12.805
J. KIM VANDIVER: Let me think.
00:43:12.805 --> 00:43:14.180
I'm trying to
think of a good way
00:43:14.180 --> 00:43:18.570
to-- when you
started this problem,
00:43:18.570 --> 00:43:22.680
we're spending about the
middle-- about G, OK?
00:43:22.680 --> 00:43:24.840
And we can calculate
angular momentum
00:43:24.840 --> 00:43:29.210
of this object with respect to
G and learn some things, right?
00:43:29.210 --> 00:43:30.950
But now we've built
a different device.
00:43:30.950 --> 00:43:36.520
This is an object now that
is, in fact, spinning.
00:43:42.380 --> 00:43:44.140
Let's imagine it's a
fan blade, and you've
00:43:44.140 --> 00:43:46.440
broken one blade off.
00:43:46.440 --> 00:43:49.230
The motor is here.
00:43:49.230 --> 00:43:52.590
A proper fan blade would
have blades on both sides,
00:43:52.590 --> 00:43:53.760
so it's nice and balanced.
00:43:53.760 --> 00:43:56.460
The shaft has some
length and it sticks out,
00:43:56.460 --> 00:43:58.430
and now you break off
one of the blades.
00:43:58.430 --> 00:44:01.800
And so you have a system
that's doing this,
00:44:01.800 --> 00:44:04.730
but this is going to put a lot
of unusual loads on this point
00:44:04.730 --> 00:44:05.710
down here.
00:44:05.710 --> 00:44:11.790
So it is rotating at the same
rotation rate about this point,
00:44:11.790 --> 00:44:15.820
but it is now a system whose
center of mass is over here,
00:44:15.820 --> 00:44:16.690
and it's up.
00:44:16.690 --> 00:44:20.140
It's above this
point, and it's out
00:44:20.140 --> 00:44:21.940
from this point,
the center of mass.
00:44:21.940 --> 00:44:24.720
And now you compute the angular
momentum with this point,
00:44:24.720 --> 00:44:26.460
and take its time
derivative, you
00:44:26.460 --> 00:44:31.390
will find all the torques
required to make this happen.
00:44:31.390 --> 00:44:32.306
AUDIENCE: [INAUDIBLE]?
00:44:34.710 --> 00:44:36.085
J. KIM VANDIVER:
A little louder.
00:44:36.085 --> 00:44:39.550
AUDIENCE: The movement A,
[? i-hat ?] Mc [? k-hat ?]
00:44:39.550 --> 00:44:40.050
[INAUDIBLE]
00:44:40.050 --> 00:44:42.770
J. KIM VANDIVER: Movement that
you've moved the center of mass
00:44:42.770 --> 00:44:46.560
away from this axis of rotation.
00:44:46.560 --> 00:44:51.660
And you've moved it up and over.
00:44:51.660 --> 00:44:54.342
So actually, let's think
about that for a minute.
00:44:54.342 --> 00:44:55.800
This is our beginning
point, right?
00:44:55.800 --> 00:45:03.090
We're at G. If I just move it
up, does it cause any problems?
00:45:03.090 --> 00:45:05.100
It's still balanced,
and you'll find out
00:45:05.100 --> 00:45:07.760
that nothing's happened
in this problem.
00:45:07.760 --> 00:45:10.700
H with respect to A is
the same as it was before.
00:45:14.190 --> 00:45:15.875
But now you move
it up and over, it
00:45:15.875 --> 00:45:17.230
starts causing complications.
00:45:17.230 --> 00:45:17.750
Yeah?
00:45:17.750 --> 00:45:24.185
AUDIENCE: [INAUDIBLE] H of A
equals [? A ?] times omega z.
00:45:24.185 --> 00:45:28.145
So i doesn't have
like direction,
00:45:28.145 --> 00:45:31.610
but you have an [? ixz ?]
terms and [? xc ?] term,
00:45:31.610 --> 00:45:34.580
and you're multiplying it by
a vector with an omega z term,
00:45:34.580 --> 00:45:37.550
so how are you getting
an i hat from that?
00:45:37.550 --> 00:45:41.890
J. KIM VANDIVER: OK, so
I haven't gone into this,
00:45:41.890 --> 00:45:44.090
and I'm not going to.
00:45:44.090 --> 00:45:48.530
The [? i ?] matrix is a
thing called a tensor.
00:45:48.530 --> 00:45:53.590
And so this is a
convention that you
00:45:53.590 --> 00:45:57.710
can make it a very mathematical
and assigned unit vectors
00:45:57.710 --> 00:46:01.150
to each of these terms inside,
but I haven't done that.
00:46:01.150 --> 00:46:07.810
But the convention here is that
this, the H, angular momentum
00:46:07.810 --> 00:46:09.650
is a vector.
00:46:09.650 --> 00:46:12.526
It consists of three components.
00:46:17.930 --> 00:46:23.590
It has three potential vector
components, Hx in the I,
00:46:23.590 --> 00:46:30.950
Hy in the j, and Hz
in the k directions.
00:46:30.950 --> 00:46:34.630
And we, in a compact form,
say we can express that
00:46:34.630 --> 00:46:37.490
by multiplying its mass
moment of inertia matrix
00:46:37.490 --> 00:46:43.570
with respect to A times the
vector that is its rotation.
00:46:43.570 --> 00:46:45.850
And I only have one
component of this omega,
00:46:45.850 --> 00:46:49.140
but it is in the k direction.
00:46:49.140 --> 00:46:55.050
And when you take this vector
multiplied by the top row, that
00:46:55.050 --> 00:46:58.900
gives you everything that's
[? I. ?] You multiply
00:46:58.900 --> 00:47:00.710
this vector times
the second row,
00:47:00.710 --> 00:47:02.950
it gives you everything
is in the j direction,
00:47:02.950 --> 00:47:06.160
and the result is part
of Hy in the j direction.
00:47:06.160 --> 00:47:09.400
You take this vector, and you
multiply it by the third row,
00:47:09.400 --> 00:47:13.500
it gives you the
z-directed components, OK?
00:47:13.500 --> 00:47:22.110
So this one, because
when we multiply this out
00:47:22.110 --> 00:47:24.790
what's in this matrix--
is this matrix diagonal?
00:47:27.680 --> 00:47:28.870
No way.
00:47:28.870 --> 00:47:32.201
And we move it, and we've gone
and done something strange
00:47:32.201 --> 00:47:32.700
here.
00:47:32.700 --> 00:47:35.110
It's not diagonal,
and when you multiply
00:47:35.110 --> 00:47:38.730
this times this top row,
there's a term here,
00:47:38.730 --> 00:47:41.010
which is not 0--
shouldn't write 0.
00:47:41.010 --> 00:47:42.060
Make an x here.
00:47:42.060 --> 00:47:46.715
This times this gives you an
omega z times that in the i hat
00:47:46.715 --> 00:47:47.215
direction.
00:47:50.430 --> 00:47:55.020
OK this term comes from
there being something there.
00:47:55.020 --> 00:47:57.970
OK, and I'm just going to give
you a generalization that you
00:47:57.970 --> 00:48:01.120
can go back and read the
Williams handout, which
00:48:01.120 --> 00:48:10.620
he says, if you want to rotate
about some fixed point A.
00:48:10.620 --> 00:48:15.120
Then you can say
that this matrix
00:48:15.120 --> 00:48:19.170
is equal to the one with
respect to G, which in our case
00:48:19.170 --> 00:48:21.910
is diagonal.
00:48:21.910 --> 00:48:24.060
The original one that we
had doesn't have to be,
00:48:24.060 --> 00:48:27.590
but if you pick principal
axes, this will be diagonal.
00:48:27.590 --> 00:48:40.580
Plus M b squared plus c
squared minus ab minus ac.
00:48:40.580 --> 00:48:41.900
It's symmetric.
00:48:41.900 --> 00:48:45.555
This is a squared
plus c squared.
00:49:02.070 --> 00:49:04.220
So if we're rotating
a system if you
00:49:04.220 --> 00:49:06.280
know its mass moment
of inertia matrix
00:49:06.280 --> 00:49:10.270
with respect to G [INAUDIBLE]
perhaps principal axes,
00:49:10.270 --> 00:49:13.650
and you want to rotate
it about any other point
00:49:13.650 --> 00:49:17.510
where you've moved it
by an a, by a b, by a c,
00:49:17.510 --> 00:49:20.560
this is essentially the
general parallel axis theorem.
00:49:20.560 --> 00:49:23.450
And you can just plug
it in and use that.
00:49:23.450 --> 00:49:26.600
And you can see what
would happen in this case
00:49:26.600 --> 00:49:29.120
if you multiplied-- if you
add these two together,
00:49:29.120 --> 00:49:30.330
they add term by term.
00:49:30.330 --> 00:49:33.050
This goes into this point on.
00:49:33.050 --> 00:49:39.820
This one adds to this,
so in this problem,
00:49:39.820 --> 00:49:46.150
in our particular
problem, this becomes
00:49:46.150 --> 00:49:55.490
you have an Ixx, Iyy, Izz
with respect to G terms.
00:49:55.490 --> 00:50:01.945
It's diagonal to start with,
and then in our problem
00:50:01.945 --> 00:50:05.220
that we did there,
we didn't have a b.
00:50:05.220 --> 00:50:08.270
No b terms, so the
b terms are all 0,
00:50:08.270 --> 00:50:12.570
but you do end up with a
minus ac term over here.
00:50:18.160 --> 00:50:24.280
And if you multiply this by 0,
0, omega z, see what happens.
00:50:24.280 --> 00:50:29.590
You get minus ac
in the I direction,
00:50:29.590 --> 00:50:32.695
and you need an M here.
00:50:38.110 --> 00:50:41.270
And you come down here,
you get the Izz G omega
00:50:41.270 --> 00:50:43.870
z in the k direction, and
that's those two terms.
00:50:43.870 --> 00:50:44.510
Here they are.
00:50:49.120 --> 00:50:51.280
Yeah, oops, I need the
plus [? A ?] squared.
00:50:55.950 --> 00:50:56.930
Yeah?
00:50:56.930 --> 00:51:00.360
AUDIENCE: When you move it
over to the end of the stick,
00:51:00.360 --> 00:51:05.260
normally like you're just
shifting it parallely,
00:51:05.260 --> 00:51:06.730
but now there's gravity.
00:51:06.730 --> 00:51:08.531
Is that why you have
those little things?
00:51:08.531 --> 00:51:09.655
J. KIM VANDIVER: Say again?
00:51:09.655 --> 00:51:11.613
AUDIENCE: So normally,
if you just take a stick
00:51:11.613 --> 00:51:15.328
and you have a
principal axis and you
00:51:15.328 --> 00:51:17.496
can't move that
parallel to the end,
00:51:17.496 --> 00:51:18.966
you wouldn't get
all of this but--
00:51:18.966 --> 00:51:21.340
J. KIM VANDIVER: Yeah, and if
you just move it-- so yeah,
00:51:21.340 --> 00:51:22.714
I'm glad you asked
that question.
00:51:22.714 --> 00:51:28.870
So normally the types of
problems in which we usually
00:51:28.870 --> 00:51:31.470
use the parallel axis
theorem is you've
00:51:31.470 --> 00:51:34.150
been spinning it
around one axis.
00:51:34.150 --> 00:51:36.600
It's a principal axis,
everything's fine,
00:51:36.600 --> 00:51:40.060
and you want to
just move it over,
00:51:40.060 --> 00:51:43.570
so we do that when we
do pendulum problems.
00:51:43.570 --> 00:51:45.220
Here's the thing
around the center,
00:51:45.220 --> 00:51:48.170
and we just want to know,
how does this thing behave?
00:51:48.170 --> 00:51:50.870
When we move it up here,
it becomes a pendulum,
00:51:50.870 --> 00:51:53.270
and we would like to
compute the angular
00:51:53.270 --> 00:51:54.780
momentum around this point.
00:51:54.780 --> 00:51:57.260
We say, oh, we've
moved it a distance d.
00:51:57.260 --> 00:52:02.870
The new mass moment of
inertia in the z direction
00:52:02.870 --> 00:52:07.820
is the original plus
Md squared, right?
00:52:07.820 --> 00:52:11.520
So when you only make
one move, you only
00:52:11.520 --> 00:52:14.190
have an a, a b, or a c.
00:52:14.190 --> 00:52:16.290
You never get any of
the off-diagonal terms
00:52:16.290 --> 00:52:18.690
because their products, see?
00:52:18.690 --> 00:52:21.950
But if you start
doing more than one,
00:52:21.950 --> 00:52:24.130
you've introduced
complications, and that's
00:52:24.130 --> 00:52:26.550
part of the reason
I put this up is
00:52:26.550 --> 00:52:32.570
there's limits to the simple
parallel axis theorem,
00:52:32.570 --> 00:52:34.800
but there is a general
way of doing it.
00:52:34.800 --> 00:52:35.300
Yes?
00:52:35.300 --> 00:52:39.478
AUDIENCE: I believe her
actual question involved
00:52:39.478 --> 00:52:40.891
what causes the torques?
00:52:40.891 --> 00:52:41.763
Is it gravity?
00:52:41.763 --> 00:52:42.304
For example--
00:52:42.304 --> 00:52:43.720
AUDIENCE: Yeah, I
was asking why--
00:52:43.720 --> 00:52:44.910
AUDIENCE: [INAUDIBLE].
00:52:44.910 --> 00:52:47.900
J. KIM VANDIVER: Ah, so what
causes these torques when
00:52:47.900 --> 00:52:49.067
you have two?
00:52:49.067 --> 00:52:49.650
AUDIENCE: Yes.
00:52:49.650 --> 00:52:52.920
J. KIM VANDIVER: OK, we've
talked about this a little bit
00:52:52.920 --> 00:52:53.420
before.
00:52:56.340 --> 00:52:59.370
When we calculate
the angular momentum,
00:52:59.370 --> 00:53:01.540
it never involves gravity.
00:53:04.470 --> 00:53:08.410
Never gets in there, not in the
angular momentum expression.
00:53:08.410 --> 00:53:11.810
So when you go to compute d, the
time derivative of the angular
00:53:11.810 --> 00:53:13.643
momentum, you will
not find torques
00:53:13.643 --> 00:53:15.532
that are caused by gravity.
00:53:15.532 --> 00:53:16.490
You just can't do them.
00:53:16.490 --> 00:53:19.830
It's not part-- but are the
torques in the System I mean,
00:53:19.830 --> 00:53:20.830
why just hold this here.
00:53:20.830 --> 00:53:22.246
There's [? Mg ?]
down, and there's
00:53:22.246 --> 00:53:25.870
a static moment caused
by this thing trying
00:53:25.870 --> 00:53:27.420
to twist this down.
00:53:27.420 --> 00:53:29.300
When you sum the
torques-- when you
00:53:29.300 --> 00:53:32.060
do the sum of the
torques in the equation,
00:53:32.060 --> 00:53:36.390
you do your free body diagram,
that term will appear.
00:53:36.390 --> 00:53:40.450
But it is balanced by some
physical torque over here
00:53:40.450 --> 00:53:41.920
that balances it.
00:53:41.920 --> 00:53:46.480
You just won't find it
from doing [? dh ?] dt.
00:53:46.480 --> 00:53:49.501
It's just part of the static
equilibrium of the system.
00:53:49.501 --> 00:53:52.868
AUDIENCE: But the reason
when you move the pen over,
00:53:52.868 --> 00:53:56.235
the reason you even have
the shift is because of
00:53:56.235 --> 00:53:57.170
[? gravity? ?]
00:53:57.170 --> 00:53:58.170
J. KIM VANDIVER: No, no.
00:53:58.170 --> 00:54:04.670
So she's asking if is
gravity the reason you--
00:54:04.670 --> 00:54:07.670
I'm not even quite sure.
00:54:07.670 --> 00:54:12.725
AUDIENCE: How did you shift the
z component of your [? plan? ?]
00:54:12.725 --> 00:54:16.200
J. KIM VANDIVER: OK,
so let's just make
00:54:16.200 --> 00:54:18.100
this a real physical problem.
00:54:18.100 --> 00:54:21.470
I'm making an airplane engine,
and I'm making an airplane
00:54:21.470 --> 00:54:23.750
with a propeller on it.
00:54:23.750 --> 00:54:28.320
And the bearing that
supports the propeller shaft,
00:54:28.320 --> 00:54:31.200
if I put it really
close to the propeller,
00:54:31.200 --> 00:54:35.290
OK, then it's doing this,
and everything's fine.
00:54:35.290 --> 00:54:39.080
What if I extended
the propeller shaft?
00:54:39.080 --> 00:54:42.800
OK, so now the
bearing is back here,
00:54:42.800 --> 00:54:45.130
and now the propeller spins.
00:54:45.130 --> 00:54:50.260
Does that cause any torques, any
loads back here on the bearing
00:54:50.260 --> 00:54:52.350
because I've extended it?
00:54:52.350 --> 00:54:55.170
No, and you could prove that
just by going through this.
00:54:55.170 --> 00:54:57.740
You do [? dh ?] dt, and you
find out the only torques
00:54:57.740 --> 00:55:00.330
when it's nice and
balanced are those required
00:55:00.330 --> 00:55:04.700
to drive this propeller
in the direction
00:55:04.700 --> 00:55:07.460
of the axis of rotation.
00:55:07.460 --> 00:55:14.020
But as soon as you do something
to that propeller, like you
00:55:14.020 --> 00:55:18.510
mount it off-center,
which rarely happens,
00:55:18.510 --> 00:55:21.350
but if you dinged--
if you broke off
00:55:21.350 --> 00:55:23.970
the tip of one of the
blades of this propeller,
00:55:23.970 --> 00:55:27.650
you'd now have a propeller
that looks like that, right?
00:55:27.650 --> 00:55:29.800
And now, even in
the original system
00:55:29.800 --> 00:55:34.530
if your bearing is right
close, this is spinning around,
00:55:34.530 --> 00:55:35.930
but is it putting a load?
00:55:35.930 --> 00:55:38.240
Sure there's a
centrifugal force that
00:55:38.240 --> 00:55:39.760
is going around and around.
00:55:39.760 --> 00:55:41.960
It has nothing to
do with gravity,
00:55:41.960 --> 00:55:45.264
nothing at all to
do with gravity.
00:55:45.264 --> 00:55:48.960
OK, just because you now have
the mass centers out here,
00:55:48.960 --> 00:55:51.310
it has momentum, the
time rate of change
00:55:51.310 --> 00:55:56.502
of that linear momentum
is a force making
00:55:56.502 --> 00:55:57.710
this thing going in a circle.
00:55:57.710 --> 00:56:01.770
OK, now if I extend
the propeller shaft
00:56:01.770 --> 00:56:05.920
so it's like that, and I
foolishly designed my airplane
00:56:05.920 --> 00:56:08.700
so it had a long propeller
shaft sticking out there,
00:56:08.700 --> 00:56:11.860
if there's a little bit of
unbalanced in this blade
00:56:11.860 --> 00:56:14.970
so that you're not spinning
about the mass center.
00:56:14.970 --> 00:56:17.650
You're spinning about
some other point.
00:56:17.650 --> 00:56:20.620
Now that centrifugal force
going around is pulling
00:56:20.620 --> 00:56:24.360
is trying to bend this back
and forth around this bearing.
00:56:24.360 --> 00:56:28.930
And because of the way in which
we formulate angular momentum,
00:56:28.930 --> 00:56:31.340
if you formulate
it about that point
00:56:31.340 --> 00:56:36.330
and take its time derivative,
it will reveal those moments.
00:56:36.330 --> 00:56:38.880
It's really amazing that
it can do that for you,
00:56:38.880 --> 00:56:42.433
but it has [? zero, ?] nothing
to do with gravity, OK?
00:56:45.340 --> 00:56:51.400
So handy, hard to remember
this from just a blackboard
00:56:51.400 --> 00:56:55.090
presentation, but it's in that
second reading by Williams.
00:56:55.090 --> 00:56:57.645
He does it in a
very-- he proves it
00:56:57.645 --> 00:57:01.849
in a very simple way using the
summations of the MI [? RI's ?]
00:57:01.849 --> 00:57:02.390
and so forth.
00:57:02.390 --> 00:57:04.970
Proves it in a very simple
way, but a very general
00:57:04.970 --> 00:57:08.880
handy formula that you can use.
00:57:08.880 --> 00:57:14.600
OK, so let's have another topic,
which I'm not going to do.
00:57:14.600 --> 00:57:15.790
We've got a few minutes.
00:57:15.790 --> 00:57:17.059
You have [? money ?] cards.
00:57:17.059 --> 00:57:18.350
We've been thinking about that.
00:57:18.350 --> 00:57:19.741
And let's ask questions, yeah?
00:57:19.741 --> 00:57:20.616
AUDIENCE: [INAUDIBLE]
00:57:23.637 --> 00:57:25.053
J. KIM VANDIVER:
Where did I the--
00:57:25.053 --> 00:57:27.886
AUDIENCE: The plus [? MA ?]
squared and the plus A squared.
00:57:27.886 --> 00:57:30.010
J. KIM VANDIVER: Ah, I was
doing that kind of fast,
00:57:30.010 --> 00:57:31.410
and I probably
even messed it up,
00:57:31.410 --> 00:57:44.550
so let me check the-- so I
was doing for the example
00:57:44.550 --> 00:57:55.100
that [? RG/A ?] equals-- we
moved it over by ai and up
00:57:55.100 --> 00:58:00.840
by ck, so and plus 0j.
00:58:00.840 --> 00:58:03.830
We didn't move it in
the j direction, OK?
00:58:03.830 --> 00:58:06.450
So this is my amount
that I've moved it.
00:58:06.450 --> 00:58:09.520
I've moved it an amount
a, and an amount c.
00:58:09.520 --> 00:58:11.770
So the Williams
formula would say, ah,
00:58:11.770 --> 00:58:15.760
that new mass moment of inertia
matrix with respect to a
00:58:15.760 --> 00:58:20.950
becomes the original plus--
and now every place there's
00:58:20.950 --> 00:58:23.010
a b here, it goes to 0.
00:58:31.810 --> 00:58:41.660
And the remaining
bits I add with those,
00:58:41.660 --> 00:58:51.940
so I should get a M--
whoops, this is a c squared,
00:58:51.940 --> 00:58:53.420
no second term.
00:58:53.420 --> 00:58:57.190
The third term is minus Mac.
00:58:57.190 --> 00:59:00.640
Over here, that one is 0.
00:59:00.640 --> 00:59:03.520
This one becomes a
squared plus c squared--
00:59:03.520 --> 00:59:05.670
AUDIENCE: --minus [INAUDIBLE]?
00:59:05.670 --> 00:59:08.120
J. KIM VANDIVER: Minus, yep, M.
00:59:08.120 --> 00:59:13.960
And this one is minus Mac 0 Izz
and a squared plus b squared,
00:59:13.960 --> 00:59:15.090
so it's just a squared.
00:59:15.090 --> 00:59:17.658
So this is now correct.
00:59:17.658 --> 00:59:18.610
AUDIENCE: [INAUDIBLE]?
00:59:18.610 --> 00:59:19.610
J. KIM VANDIVER: Pardon?
00:59:19.610 --> 00:59:20.840
AUDIENCE: Times M
in the bottom of it?
00:59:20.840 --> 00:59:22.298
J. KIM VANDIVER:
Yeah, all of these
00:59:22.298 --> 00:59:29.750
needs M. Ma
squared's M, M, M, M,
00:59:29.750 --> 00:59:33.800
so that's our problem
when we did two shifts.
00:59:33.800 --> 00:59:37.350
But a second we shifted
it in two directions,
00:59:37.350 --> 00:59:41.310
we get these
off-diagonal terms, which
00:59:41.310 --> 00:59:44.940
means that if you are actually
making this device rotate
00:59:44.940 --> 00:59:47.954
about point A, you will get
these unbalanced moments--
00:59:47.954 --> 00:59:49.120
this dynamically unbalanced.
00:59:51.970 --> 00:59:53.414
Yeah?
00:59:53.414 --> 00:59:56.039
AUDIENCE: Will we be trying
to find our [? mass ?] moments
00:59:56.039 --> 01:00:01.776
of inertia directly, or should
we just do [INAUDIBLE] equation
01:00:01.776 --> 01:00:05.185
and have them fall out like
the integral with x times y
01:00:05.185 --> 01:00:07.679
[? vM? ?] [INAUDIBLE]
01:00:07.679 --> 01:00:10.220
J. KIM VANDIVER: He's kind of
asking advice in whether or not
01:00:10.220 --> 01:00:15.590
we ought to be trying
to find the inertia
01:00:15.590 --> 01:00:18.280
matrix about other
points, right,
01:00:18.280 --> 01:00:21.290
like [? the I ?]
with respect to A.
01:00:21.290 --> 01:00:24.850
Where I started today,
I said you can basically
01:00:24.850 --> 01:00:27.360
do all the problems with
these three equations,
01:00:27.360 --> 01:00:32.080
and this doesn't mention
parallel axis theorem.
01:00:32.080 --> 01:00:36.937
We use this to find, in
fact, to solve this problem.
01:00:36.937 --> 01:00:38.770
We didn't talk parallel
axis theorem at all,
01:00:38.770 --> 01:00:42.220
it just-- the
answer dropped out.
01:00:42.220 --> 01:00:43.800
And if we looked
into that careful,
01:00:43.800 --> 01:00:45.030
we could generalize that.
01:00:45.030 --> 01:00:47.270
We could work now with
that a bit and say, ah,
01:00:47.270 --> 01:00:49.030
there's a pattern to this.
01:00:49.030 --> 01:00:55.650
I'll bet this can be recast
like this, and it can.
01:00:55.650 --> 01:00:59.200
So if you know this
exists, and you
01:00:59.200 --> 01:01:02.500
don't want to have to
grind through finding
01:01:02.500 --> 01:01:08.730
these terms that
come from here, you
01:01:08.730 --> 01:01:14.850
can do the problems by finding
the mass moment of inertia
01:01:14.850 --> 01:01:17.470
matrix with respect to A.
01:01:17.470 --> 01:01:26.460
You can use this form only when
you have fixed axis rotation.
01:01:26.460 --> 01:01:30.220
The thing is the point
A is [? about ?] which
01:01:30.220 --> 01:01:35.060
this rotation is occurring, then
you can write down the formula
01:01:35.060 --> 01:01:37.700
this directly like that.
01:01:37.700 --> 01:01:41.780
So you that got to be
careful when you apply
01:01:41.780 --> 01:01:43.750
the parallel axis theorem.
01:01:46.990 --> 01:01:51.530
The nice thing about
these three forms
01:01:51.530 --> 01:01:54.120
is they're generally true.
01:01:54.120 --> 01:01:55.670
Point A can move.
01:01:55.670 --> 01:01:57.940
Point A can be accelerating.
01:01:57.940 --> 01:02:02.820
The problem that we did here,
point A is accelerating.
01:02:02.820 --> 01:02:06.890
Not only moving, it's not
even an inertial frame.
01:02:06.890 --> 01:02:11.031
We solved this problem-- this
is a non-inertial frame problem.
01:02:11.031 --> 01:02:15.520
We went right at it and
solved it directly, OK?
01:02:15.520 --> 01:02:18.600
So I would say, if you have any
doubt to answer your question,
01:02:18.600 --> 01:02:21.760
when you're in doubt
just use the formula.
01:02:21.760 --> 01:02:25.020
And then you will need to
find moments of inertia
01:02:25.020 --> 01:02:26.940
with respect to
the center of mass,
01:02:26.940 --> 01:02:29.830
and it's in your interest to use
principal axes because they're
01:02:29.830 --> 01:02:30.330
easier.
01:02:30.330 --> 01:02:30.830
Yeah?
01:02:30.830 --> 01:02:32.780
AUDIENCE: If you were
to say [INAUDIBLE]?
01:02:38.662 --> 01:02:40.120
J. KIM VANDIVER:
Ah, so if you took
01:02:40.120 --> 01:02:43.930
the derivative of
this, [? dh ?] dt,
01:02:43.930 --> 01:02:48.050
this one just gives
you omega z dot.
01:02:48.050 --> 01:02:50.900
That's you're spin in the
direction of rotation.
01:02:50.900 --> 01:02:54.530
This term, that
unit vector rotates.
01:02:54.530 --> 01:03:01.720
This gives you two terms-- gives
you an i term, omega dot i,
01:03:01.720 --> 01:03:07.600
and it gives you an
omega j term, OK?
01:03:07.600 --> 01:03:10.730
And what those are,
those are two torques.
01:03:10.730 --> 01:03:14.040
This problem, there's
a torque caused
01:03:14.040 --> 01:03:16.190
by the centrifugal
force trying to bend
01:03:16.190 --> 01:03:20.410
this in the j direction.
01:03:20.410 --> 01:03:24.650
And if this is accelerating, you
know, a theta double dot term,
01:03:24.650 --> 01:03:28.415
there is a torque trying to
bend this thing backwards.
01:03:28.415 --> 01:03:30.410
OK, get them both.
01:03:30.410 --> 01:03:36.610
All right [? money ?] cards
and see you on Thursday.
01:03:36.610 --> 01:03:42.230
Thursday we're going
to do a nasty problem,
01:03:42.230 --> 01:03:44.530
and it is it a lead
in to a problem.
01:03:44.530 --> 01:03:47.130
You can do the same problem
extraordinarily easily
01:03:47.130 --> 01:03:47.940
using energy.
01:03:47.940 --> 01:03:49.731
And that's kind of the
purpose of doing it,
01:03:49.731 --> 01:03:52.060
so you see both
ways of doing it.