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PROFESSOR: The technical
topic for today,
00:00:24.590 --> 00:00:26.210
we start-- I rushed,
right at the end,
00:00:26.210 --> 00:00:28.380
a little bit about tangent
and normal unit vectors.
00:00:28.380 --> 00:00:31.120
I'm going to just
recap that quickly.
00:00:31.120 --> 00:00:35.070
And then we're going to go
on really a review, which
00:00:35.070 --> 00:00:37.790
by a review, this is the sort of
stuff that, for the most part,
00:00:37.790 --> 00:00:41.760
I'm sure you've seen in 801
Physics and other Physics
00:00:41.760 --> 00:00:42.550
you've had before.
00:00:42.550 --> 00:00:45.650
And that's impulse-- linear
momentum, and impulse.
00:00:45.650 --> 00:00:47.580
So that'll be a quick review.
00:00:47.580 --> 00:00:51.280
And then the third subject
is one that's much deeper,
00:00:51.280 --> 00:00:53.160
and this is angular momentum.
00:00:53.160 --> 00:00:55.530
And angular momentum
with respect
00:00:55.530 --> 00:00:57.210
to moving points,
which you probably
00:00:57.210 --> 00:00:59.290
haven't encountered before.
00:00:59.290 --> 00:01:02.130
So those are the three
topics for the day.
00:01:02.130 --> 00:01:03.160
Let's get started.
00:01:06.070 --> 00:01:12.000
So last time, the
piece that I rushed
00:01:12.000 --> 00:01:19.105
a bit is this notion of
tangent and normal coordinates.
00:01:22.400 --> 00:01:24.180
So I gave this--
had this example.
00:01:24.180 --> 00:01:32.856
You're driving down the road,
you're drunk or whatever.
00:01:35.770 --> 00:01:39.100
And when you're at this point,
this point, and this point,
00:01:39.100 --> 00:01:41.580
I'd like to know what
the accelerations are.
00:01:41.580 --> 00:01:43.840
So I'll call this 1, 2, 3.
00:01:43.840 --> 00:01:50.225
And this curve, y, is of the
form some y of some f of x.
00:01:52.730 --> 00:01:59.600
And in this case,
it's A sine kx.
00:01:59.600 --> 00:02:02.700
And k is what's
known as wave number.
00:02:02.700 --> 00:02:07.170
This is 2 pi over the wave
length, 2 pi over lambda.
00:02:07.170 --> 00:02:10.830
And the wave length then is,
for example, from here to here.
00:02:17.520 --> 00:02:20.000
Now, the velocity-- I'm
going to pick a point here.
00:02:20.000 --> 00:02:23.270
The velocity at
any point, we know,
00:02:23.270 --> 00:02:26.267
is just the tangent to the path.
00:02:26.267 --> 00:02:28.100
So this is the path,
and this is horizontal.
00:02:28.100 --> 00:02:32.100
So you're driving-- driving
down the road like this.
00:02:32.100 --> 00:02:34.700
That's what we're
trying to do here.
00:02:34.700 --> 00:02:36.390
The gravity's down
into the board.
00:02:36.390 --> 00:02:38.300
It doesn't really
come into the problem.
00:02:41.260 --> 00:02:45.930
So the velocity, at any
point-- at anytime--
00:02:45.930 --> 00:02:50.860
it's a vector-- we can
describe as a magnitude.
00:02:50.860 --> 00:02:54.370
And in a unit vector,
we'll call ut which
00:02:54.370 --> 00:02:56.610
is the tangent unit vector.
00:02:56.610 --> 00:02:59.750
And at any instant in time, it's
just aligned with the tangent
00:02:59.750 --> 00:03:01.980
to the curve.
00:03:01.980 --> 00:03:08.730
And it's perpendicular partner
is a normal unit vector, which
00:03:08.730 --> 00:03:11.450
it points inward on the curve.
00:03:11.450 --> 00:03:14.480
And this will be un.
00:03:14.480 --> 00:03:16.480
Now, we're interested
in accelerations,
00:03:16.480 --> 00:03:18.890
so we'll need to take
a derivative of this.
00:03:27.630 --> 00:03:29.960
So I was going to
say, taken aside.
00:03:29.960 --> 00:03:30.792
But I won't.
00:03:30.792 --> 00:03:32.250
So the acceleration
vector, we have
00:03:32.250 --> 00:03:33.541
to take the derivative of this.
00:03:33.541 --> 00:03:43.500
Well, it's a v dot
ut plus a v ut dot.
00:03:43.500 --> 00:03:44.530
It's a unit vector.
00:03:44.530 --> 00:03:46.080
And we've encountered
this problem
00:03:46.080 --> 00:03:49.780
before because if this
rotates, the unit vector
00:03:49.780 --> 00:03:52.885
has a non-zero derivative
because of this rotation.
00:03:57.020 --> 00:04:03.920
And this derivative
of the unit vector
00:04:03.920 --> 00:04:13.380
is given by theta dot u m
hat Now, what's theta dot?
00:04:13.380 --> 00:04:17.490
Well, on this curve,
at any instant in time,
00:04:17.490 --> 00:04:21.410
here's u-- can't
draw arrows today.
00:04:21.410 --> 00:04:23.200
Here's your ut.
00:04:23.200 --> 00:04:27.970
At any instant when you're
traveling along on a curve,
00:04:27.970 --> 00:04:32.690
you are going around the circle
of some radius we'll call rho.
00:04:32.690 --> 00:04:34.720
So there's some
radius of curvature
00:04:34.720 --> 00:04:37.070
at any instant in time.
00:04:37.070 --> 00:04:44.780
And that radius, as you drive
along in a little time delta t,
00:04:44.780 --> 00:04:51.990
you go forward an
amount, rho theta dot,
00:04:51.990 --> 00:04:59.820
is the velocity at which you're
traveling tangent to the curve.
00:04:59.820 --> 00:05:03.420
There's an angle in here, so
in here, there's some delta.
00:05:03.420 --> 00:05:09.320
You advance some delta theta, so
our omega is a velocity, right?
00:05:09.320 --> 00:05:12.530
So rho theta dot its
velocity, and that gives you
00:05:12.530 --> 00:05:20.234
this velocity
magnitude, v. And we've
00:05:20.234 --> 00:05:21.650
gone through-- I'm
not going to go
00:05:21.650 --> 00:05:24.191
through that little derivative--
that little argument before,
00:05:24.191 --> 00:05:28.260
but the change in direction
of this unit vector,
00:05:28.260 --> 00:05:35.560
ut, when you go forward a little
bit is actually inward, un,
00:05:35.560 --> 00:05:42.080
by an amount rho theta dot-- by
an amount theta dot one times
00:05:42.080 --> 00:05:42.874
theta dot.
00:05:42.874 --> 00:05:44.290
So we've done- I'm
not going to do
00:05:44.290 --> 00:05:46.081
that piece of the
derivative we did before.
00:05:46.081 --> 00:05:49.340
But here is this derivative
of the unit vector.
00:05:49.340 --> 00:05:51.780
It's theta dot un.
00:05:51.780 --> 00:05:54.070
So if we substitute
those back in here,
00:05:54.070 --> 00:05:57.625
we can get an expression
for our acceleration.
00:06:02.600 --> 00:06:11.760
This is just the acceleration
along the path plus v theta dot
00:06:11.760 --> 00:06:14.420
un.
00:06:14.420 --> 00:06:17.550
And then we need to
take into account
00:06:17.550 --> 00:06:22.280
the fact that we know that v,
the magnitude of velocity--
00:06:22.280 --> 00:06:25.650
the speed in other
words-- is rho theta dot.
00:06:28.300 --> 00:06:34.800
And that means theta
dot is v over rho.
00:06:34.800 --> 00:06:37.670
So if we plug that
in up here, then
00:06:37.670 --> 00:06:42.600
we get a final expression
for our acceleration,
00:06:42.600 --> 00:06:52.610
v dot ut plus v
squared over rho un.
00:06:52.610 --> 00:06:56.030
And that v squared
over rho-- this
00:06:56.030 --> 00:06:59.850
is a centripetal
acceleration term.
00:06:59.850 --> 00:07:01.640
You're going around a curve.
00:07:01.640 --> 00:07:04.050
There is an acceleration inward.
00:07:04.050 --> 00:07:07.280
It's just like as we did
from polar coordinates.
00:07:07.280 --> 00:07:09.422
It comes from rotating things.
00:07:09.422 --> 00:07:11.130
Well, as soon as you
go around the curve,
00:07:11.130 --> 00:07:14.030
you're going to generate
an acceleration that
00:07:14.030 --> 00:07:16.340
is of the same
kind, but now we--
00:07:16.340 --> 00:07:19.720
because you tend
to know speed, it's
00:07:19.720 --> 00:07:21.730
easier to express it this
way when you're doing
00:07:21.730 --> 00:07:23.760
these tangent normal problems.
00:07:23.760 --> 00:07:27.120
And this is just the
acceleration along a path,
00:07:27.120 --> 00:07:30.190
the usual hit the gas
petal and speed up.
00:07:30.190 --> 00:07:31.115
That's this term.
00:07:34.240 --> 00:07:37.190
So the piece that
I didn't have time
00:07:37.190 --> 00:07:47.760
to put up last time,
which is in the book as
00:07:47.760 --> 00:07:50.010
is this little derivative--
the derivation of that
00:07:50.010 --> 00:07:53.310
is one page in one of the
early chapters of the book.
00:07:53.310 --> 00:07:55.700
How do you get rho?
00:07:55.700 --> 00:07:58.900
And rho, when you have
y as a function of x,
00:07:58.900 --> 00:08:03.470
there is just this formula from
mathematics-- from calculus--
00:08:03.470 --> 00:08:12.700
that says rho is dy dx 1 plus dy
dx quantity squared to the 3/2
00:08:12.700 --> 00:08:22.710
power all over the
magnitude of d2y dx squared.
00:08:22.710 --> 00:08:25.200
And you just calculate these
quantities and plug them in.
00:08:25.200 --> 00:08:42.850
So for y equals a sine kx,
then dy dx is Ak cosine kx
00:08:42.850 --> 00:08:56.720
and d2y dx squared is
minus a k squared sine kx.
00:08:56.720 --> 00:08:59.580
So we now have these
two quantities.
00:08:59.580 --> 00:09:01.550
We pick a value
of x which we want
00:09:01.550 --> 00:09:04.720
to know the answer, like 0.1.
00:09:04.720 --> 00:09:10.510
So at 0.1, kx is pi over 2
because sine is a maximum.
00:09:17.330 --> 00:09:21.290
X is lambda over 4.
00:09:21.290 --> 00:09:22.970
kx is pi over 2.
00:09:26.370 --> 00:09:30.030
Sine kx is 1.
00:09:30.030 --> 00:09:34.400
Cosine kx is 0.
00:09:34.400 --> 00:09:36.280
And so we can calculate rho.
00:09:39.450 --> 00:09:47.440
1 plus and dy dx the
derivative, is cosine.
00:09:47.440 --> 00:09:48.170
That's 0.
00:09:48.170 --> 00:09:55.560
So this term in here is 0.
00:09:55.560 --> 00:09:56.780
And this is to the 3/2.
00:09:56.780 --> 00:09:58.470
That's pretty easy to calculate.
00:09:58.470 --> 00:10:03.480
This term down here, d2y dx
squared, well, sine of kx is 1.
00:10:03.480 --> 00:10:08.120
So that's minus Ak squared, but
this is an absolute value sign.
00:10:12.650 --> 00:10:19.700
So this just turns out
to be 1 over Ak squared.
00:10:19.700 --> 00:10:22.970
And you plug-in some numbers.
00:10:22.970 --> 00:10:31.036
I'm going to let my
lambda be 150 meters.
00:10:31.036 --> 00:10:34.340
My amplitude here that I'm
swerving back and forth,
00:10:34.340 --> 00:10:35.650
let's make that 5 meters.
00:10:38.790 --> 00:10:44.340
We need a-- that's all
we need to get rho.
00:10:44.340 --> 00:10:50.110
So rho, in this case,
works out to be 114 meters.
00:10:50.110 --> 00:10:53.360
So the radius of curvature,
this road that I'm driving down,
00:10:53.360 --> 00:10:58.900
at that point right at the peak
in the curve is 114 meters.
00:11:05.710 --> 00:11:08.890
So the acceleration
that I'm looking for,
00:11:08.890 --> 00:11:13.600
a, well, it's v dot ut.
00:11:13.600 --> 00:11:15.330
And if I'm not
accelerating-- I'm not
00:11:15.330 --> 00:11:17.010
hitting the gas petal
at constant speed,
00:11:17.010 --> 00:11:19.500
I'll let that would be 0.
00:11:19.500 --> 00:11:21.870
And the other one,
then, is v squared
00:11:21.870 --> 00:11:26.120
over rho of the un direction.
00:11:26.120 --> 00:11:32.140
And let's let v equal-- whatever
I have here in my example--
00:11:32.140 --> 00:11:38.600
20 meters per second,
which is about 40 knots.
00:11:38.600 --> 00:11:42.060
And a knot is 15% more
than a mile per hour,
00:11:42.060 --> 00:11:44.985
so it's somewhere about
45 miles an hour, so
00:11:44.985 --> 00:11:46.100
a typical road speed.
00:11:46.100 --> 00:11:48.650
Driving down the road,
20 meters per second,
00:11:48.650 --> 00:11:49.810
now you can plug-in here.
00:11:49.810 --> 00:11:51.360
You can plug-in
here, and you'll find
00:11:51.360 --> 00:11:54.525
that the acceleration
that I worked out here
00:11:54.525 --> 00:12:00.120
is 3.51 meters
per second squared
00:12:00.120 --> 00:12:01.670
and in what direction is it?
00:12:07.470 --> 00:12:09.636
What direction is
that acceleration?
00:12:12.384 --> 00:12:13.300
AUDIENCE: The normal.
00:12:13.300 --> 00:12:14.670
PROFESSOR: It's in
the normal direction.
00:12:14.670 --> 00:12:17.200
Is it to the inside of the curve
or the outside of the curve?
00:12:17.200 --> 00:12:17.690
AUDIENCE: Inside.
00:12:17.690 --> 00:12:18.440
PROFESSOR: Inside.
00:12:18.440 --> 00:12:20.266
It's always to the
inside of the curve.
00:12:20.266 --> 00:12:23.800
So these tangent
normal coordinates
00:12:23.800 --> 00:12:26.190
are really simple coordinates.
00:12:26.190 --> 00:12:28.620
They're meant for a particular
kind of simple problem
00:12:28.620 --> 00:12:30.650
when you know the path.
00:12:30.650 --> 00:12:34.350
And it's just defined,
the normal un is positive,
00:12:34.350 --> 00:12:36.620
always inward to the
center of the curve,
00:12:36.620 --> 00:12:38.910
pointing toward the
center of rotation
00:12:38.910 --> 00:12:41.450
where that radius of--
where your radius of curve
00:12:41.450 --> 00:12:44.640
is, always pointing
at the origin
00:12:44.640 --> 00:12:48.920
of that radius of your curve.
00:12:48.920 --> 00:12:52.820
So g is order of 10
meters per second squared,
00:12:52.820 --> 00:12:56.265
so that's about a third of a g.
00:12:56.265 --> 00:13:00.690
Is 1/3 of a g enough to notice?
00:13:00.690 --> 00:13:01.737
Yeah, absolutely!
00:13:01.737 --> 00:13:03.820
You'll get-- when you push
to the side of the car,
00:13:03.820 --> 00:13:04.896
you do that .
00:13:08.030 --> 00:13:10.440
So let's move-- any
questions about that?
00:13:10.440 --> 00:13:11.640
I'm going to move on next.
00:13:11.640 --> 00:13:14.435
If not, I'm going to move on
to linear impulse and momentum.
00:13:23.140 --> 00:13:28.860
And this will be a quick
review because this
00:13:28.860 --> 00:13:35.104
consists of the kind of physics
that you've done lots of times.
00:13:35.104 --> 00:13:37.020
We're going to hit it
quickly and then move on
00:13:37.020 --> 00:13:37.853
to angular momentum.
00:13:41.390 --> 00:13:43.880
So we know for a particle--
this is what Newton told us.
00:13:43.880 --> 00:13:47.830
We've got a particle
here and some mass, m.
00:13:47.830 --> 00:13:51.220
We know for a particle-- and it
has some external forces on it.
00:13:54.480 --> 00:13:59.410
We know for a particle that the
sum of these external forces--
00:13:59.410 --> 00:14:03.710
it's a vector sum--
is equal to the mass
00:14:03.710 --> 00:14:07.280
times the acceleration for the
particle, Newton's Second Law.
00:14:13.010 --> 00:14:17.880
And that's the mass times the
derivative of the velocity
00:14:17.880 --> 00:14:19.000
with respect to time.
00:14:19.000 --> 00:14:21.830
That's where we
get acceleration.
00:14:21.830 --> 00:14:24.810
So this formula we can
just rearrange the dt's
00:14:24.810 --> 00:14:25.400
a little bit.
00:14:25.400 --> 00:14:35.360
So the summation of the
forces, dt, is m dv.
00:14:35.360 --> 00:14:36.650
And I want to integrate these.
00:14:39.410 --> 00:14:44.670
So if I integrate this
over time, from t1 to t2,
00:14:44.670 --> 00:14:49.150
this equality says we're going
to find some change in velocity
00:14:49.150 --> 00:14:53.650
from v1 to v2.
00:14:53.650 --> 00:14:55.820
And this is our
impulse-- this is
00:14:55.820 --> 00:14:59.530
the beginnings of our
impulse-momentum relationship.
00:14:59.530 --> 00:15:03.620
We tend to call the integration
of forces over time, impulse.
00:15:03.620 --> 00:15:07.280
And if you have a
nonzero impulse,
00:15:07.280 --> 00:15:10.190
that leads to a change
in the linear momentum.
00:15:13.660 --> 00:15:19.520
So when you carry
it out so writing
00:15:19.520 --> 00:15:22.370
this a little bit different
way, you can do summation
00:15:22.370 --> 00:15:25.940
because these are vectors, and
we have rules about vectors.
00:15:25.940 --> 00:15:29.930
You can bring the summation
outside of the integral.
00:15:29.930 --> 00:15:33.930
So that this says that the
summation of the integral
00:15:33.930 --> 00:15:41.340
from t1 to t2 of
these external forces
00:15:41.340 --> 00:15:46.853
is just mv2 minus mv1
because the integral
00:15:46.853 --> 00:15:51.390
on the right-hand
side is really simple.
00:15:51.390 --> 00:15:56.280
And we'll finally
state this in a way
00:15:56.280 --> 00:16:00.490
that we most commonly use it,
moving this term to the left.
00:16:00.490 --> 00:16:04.950
So you start off with an
initial momentum, mv1.
00:16:04.950 --> 00:16:11.750
You add to it the summation
of the impulses that
00:16:11.750 --> 00:16:18.820
occur over time, t1, t2
of the external forces,
00:16:18.820 --> 00:16:22.670
and you get the
final momentum mv2.
00:16:22.670 --> 00:16:27.544
And this is the way you
usually use the formula.
00:16:27.544 --> 00:16:29.460
So what happens if there's
no external forces?
00:16:32.680 --> 00:16:36.700
That's where we get the law
of conservation of momentum.
00:16:36.700 --> 00:16:39.940
If there are no external
forces on the particle,
00:16:39.940 --> 00:16:44.220
the momentum doesn't change and
mv1 has got to be equal to mv2.
00:17:02.260 --> 00:17:07.390
So just a really trivial
example drawing for something
00:17:07.390 --> 00:17:08.380
that we've done before.
00:17:08.380 --> 00:17:12.069
You got the block
sliding down the hill.
00:17:12.069 --> 00:17:15.444
Draw your free body diagram.
00:17:15.444 --> 00:17:22.120
You got friction, got
gravity, got a normal force.
00:17:27.010 --> 00:17:29.930
And we'll set ourselves up with
a coordinate system aligned
00:17:29.930 --> 00:17:31.590
with the motion
we're interested in.
00:17:31.590 --> 00:17:34.730
So this is an
inertial system, x,y.
00:17:37.810 --> 00:17:39.380
And we've done this
problem before.
00:17:39.380 --> 00:17:41.600
So we said the
summation-- we found out
00:17:41.600 --> 00:17:46.160
that the summation of the forces
in the x direction-- and this
00:17:46.160 --> 00:17:49.460
is a good moment to take
an aside for a second.
00:17:49.460 --> 00:17:53.160
This was a vector expression.
00:17:53.160 --> 00:17:56.456
But you can break it--
you can implement it--
00:17:56.456 --> 00:18:00.140
you can break it down into its
individual vector components.
00:18:00.140 --> 00:18:02.220
So that equation gives
you-- for particles--
00:18:02.220 --> 00:18:06.220
gives you three sub-equations
one in the x, one in the y,
00:18:06.220 --> 00:18:06.870
one in the z.
00:18:06.870 --> 00:18:09.030
And you can use them
each independently.
00:18:09.030 --> 00:18:13.020
So in this case, we're going to
use-- only need the x component
00:18:13.020 --> 00:18:14.320
in order to do the problem.
00:18:14.320 --> 00:18:17.400
So the sum of the forces
in the x direction,
00:18:17.400 --> 00:18:20.860
in this problem which
we had done before,
00:18:20.860 --> 00:18:28.260
is mg sine theta, that
component of gravity pulling it
00:18:28.260 --> 00:18:35.360
down the hill, minus
mu mg cosine theta.
00:18:35.360 --> 00:18:37.270
And that's the friction.
00:18:37.270 --> 00:18:41.490
And none of these are
functions of time.
00:18:41.490 --> 00:18:44.681
So that whole thing
is just some constant.
00:18:44.681 --> 00:18:47.980
I'll just call it k.
00:18:47.980 --> 00:18:51.149
So the total forces on the
system in the x direction
00:18:51.149 --> 00:18:52.065
is just some constant.
00:18:52.065 --> 00:18:54.360
And that makes this integral
up here pretty trivial
00:18:54.360 --> 00:18:56.140
to implement.
00:18:56.140 --> 00:19:03.530
So now we can say that mv1--
I'll make it v1x to emphasize
00:19:03.530 --> 00:19:07.020
that integral from 0, t1 to t2..
00:19:07.020 --> 00:19:10.470
But I'm going to let
t1 be 0 because it
00:19:10.470 --> 00:19:11.700
makes the problems easier.
00:19:11.700 --> 00:19:21.320
So from 0 to t of
k dt equals mv2--
00:19:21.320 --> 00:19:25.580
I can't write this morning--
in the x direction.
00:19:25.580 --> 00:19:34.390
And at v1, x is 0, starts
off at 0 time and 0 velocity
00:19:34.390 --> 00:19:36.490
when you let it
go to start with.
00:19:36.490 --> 00:19:38.350
Then, this term will go away.
00:19:38.350 --> 00:19:42.310
And we just implement
this integral.
00:19:42.310 --> 00:19:46.960
And the integral of k dt is?
00:19:46.960 --> 00:19:47.940
kt, right?
00:19:47.940 --> 00:19:55.270
So this says then that kt
evaluated from 0 to some time
00:19:55.270 --> 00:20:01.810
that we want to know the answer
is mv2 in the x direction.
00:20:01.810 --> 00:20:05.930
And so let's let
t equal 3 seconds.
00:20:08.470 --> 00:20:18.618
You find out that
v 2x is 3k over m.
00:20:22.610 --> 00:20:30.760
And in this problem, then
that looks like 3g sine theta
00:20:30.760 --> 00:20:35.670
minus mu cosine theta.
00:20:35.670 --> 00:20:42.870
So it's really an
almost trivial example.
00:20:42.870 --> 00:20:44.990
It's not a hard example,
but it emphasizes
00:20:44.990 --> 00:20:47.750
all of the key points
in the problem.
00:20:47.750 --> 00:20:49.600
Start off with a
vector equation.
00:20:49.600 --> 00:20:53.520
You can apply it in any one
of the three vector component
00:20:53.520 --> 00:20:54.830
directions.
00:20:54.830 --> 00:20:56.870
We integrate the
forces-- the sum
00:20:56.870 --> 00:21:00.160
of the forces on the object
in that direction over time.
00:21:00.160 --> 00:21:02.860
And you apply the
impulse-momentum formula,
00:21:02.860 --> 00:21:04.590
and you get the answer.
00:21:04.590 --> 00:21:05.320
You can do that.
00:21:05.320 --> 00:21:07.290
So that basic
step-by-step process
00:21:07.290 --> 00:21:11.670
is how basically most
impulse-momentum problems
00:21:11.670 --> 00:21:12.670
are done.
00:21:12.670 --> 00:21:14.390
And if there's no
forces, then you
00:21:14.390 --> 00:21:15.750
have conservation of momentum.
00:21:19.800 --> 00:21:23.030
So let's do-- this
was for particles.
00:21:37.390 --> 00:21:41.780
Just a quick reminder
of we need to ask
00:21:41.780 --> 00:21:45.260
ourselves does this apply to
groups of particles, systems
00:21:45.260 --> 00:21:46.360
of particles.
00:21:46.360 --> 00:21:48.590
Well, remember, that we
said if you've got a bunch
00:21:48.590 --> 00:21:54.210
of particles, a system--
all of these are
00:21:54.210 --> 00:21:59.750
the mi''s-- we've already
figured out that the total mass
00:21:59.750 --> 00:22:05.070
of the system times the
velocity of the center of mass--
00:22:05.070 --> 00:22:14.990
so g is my center of mass with
respect to my inertial frame.
00:22:14.990 --> 00:22:19.330
So this is the momentum of
the system that was just
00:22:19.330 --> 00:22:23.960
a summation of the individual
mivi's with respect
00:22:23.960 --> 00:22:30.130
to-- And furthermore, we
took the derivative of this,
00:22:30.130 --> 00:22:32.220
so this is the
momentum of the system.
00:22:32.220 --> 00:22:34.240
The time derivative
of the momentum
00:22:34.240 --> 00:22:38.810
should be the external forces
so that we were able to--
00:22:38.810 --> 00:22:43.810
by taking that derivative,
you get mt, the total mass,
00:22:43.810 --> 00:22:47.950
times vg with respect to o dot.
00:22:47.950 --> 00:22:55.160
And that had better give you
v equal to the summations
00:22:55.160 --> 00:23:00.466
of the external forces
on all the particles
00:23:00.466 --> 00:23:02.340
because you can do it
one particle at a time.
00:23:02.340 --> 00:23:05.890
And they all add-- each of
these particles has forces.
00:23:05.890 --> 00:23:07.370
You sum them all up.
00:23:07.370 --> 00:23:12.920
You get these forces,
and this allows
00:23:12.920 --> 00:23:17.290
you to say that the sum of
all of the external forces
00:23:17.290 --> 00:23:19.540
on the system is equal
to the total mass
00:23:19.540 --> 00:23:21.260
of the system times
the acceleration
00:23:21.260 --> 00:23:23.860
of the center of mass.
00:23:23.860 --> 00:23:27.420
So this formula
now can-- it looks
00:23:27.420 --> 00:23:31.380
very similar to where we started
with this little derivation
00:23:31.380 --> 00:23:32.810
for a particle.
00:23:32.810 --> 00:23:36.350
You can now say that
the summation-- this all
00:23:36.350 --> 00:23:41.830
leads to the summation
of the integral
00:23:41.830 --> 00:23:52.570
from t1 to t2 of all of
these courses over time--
00:23:52.570 --> 00:23:58.030
gives you the change in
the total linear momentum
00:23:58.030 --> 00:24:05.910
of the system, vg with
respect to o1, 2 rather,
00:24:05.910 --> 00:24:12.160
minus vg with respect
to o2, the second one.
00:24:12.160 --> 00:24:14.540
So this is exactly the
same kind of formulation.
00:24:14.540 --> 00:24:20.660
The change in the linear
momentum of the system
00:24:20.660 --> 00:24:24.730
is equal to the integral of the
forces on the system over time,
00:24:24.730 --> 00:24:26.570
so the impulse to the system.
00:24:26.570 --> 00:24:30.410
So this allows you to do
problems that you might not
00:24:30.410 --> 00:24:33.700
have thought about before.
00:24:33.700 --> 00:24:36.550
So I think about
something like this.
00:24:36.550 --> 00:24:45.345
I've got an old Revolutionary
War earlier period canon.
00:24:50.280 --> 00:24:51.910
I'm trying to draw it
00:24:51.910 --> 00:24:54.740
So here's my cannon barrel.
00:24:54.740 --> 00:25:03.870
And back in those days-- I'll
do a little better job here.
00:25:03.870 --> 00:25:05.610
So we got this
cannon sitting here.
00:25:05.610 --> 00:25:07.870
And in the barrel,
you've got a bunch
00:25:07.870 --> 00:25:13.800
of what's known as-- in the
old days, known as grapeshot.
00:25:13.800 --> 00:25:17.620
So sometimes, for
anti personnel stuff,
00:25:17.620 --> 00:25:20.000
they would just throw a
whole bunch of metal and junk
00:25:20.000 --> 00:25:20.530
into it.
00:25:20.530 --> 00:25:23.090
But just imagine a
bunch of iron balls
00:25:23.090 --> 00:25:26.274
or lead ball rocks or whatever
you want to put in there.
00:25:26.274 --> 00:25:28.940
And you put a whole mess of them
in here, and you have a charge.
00:25:28.940 --> 00:25:31.807
You set it off, and it shoots
them out the end of the gun.
00:25:31.807 --> 00:25:33.265
I want to know
what's the reaction.
00:25:46.190 --> 00:25:48.210
What's the reaction
force on the gun?
00:25:51.409 --> 00:25:53.200
When that gun-- when
that charge goes off--
00:25:53.200 --> 00:25:55.970
if you go down to the
Constitution, which
00:25:55.970 --> 00:26:00.710
is the oldest commissioned
warship in the US Navy
00:26:00.710 --> 00:26:02.850
still afloat-- I don't
know if any of you
00:26:02.850 --> 00:26:05.130
have seen it down here
in the dock in Chelsea.
00:26:05.130 --> 00:26:07.860
It was built around 1799.
00:26:07.860 --> 00:26:11.160
But when they shot
one of those guns,
00:26:11.160 --> 00:26:15.600
the gun would roll back several
feet and get dragged to a stop
00:26:15.600 --> 00:26:19.220
by a bunch of restraining
lines and then dragged back up
00:26:19.220 --> 00:26:19.940
for another load.
00:26:19.940 --> 00:26:22.920
So the reaction force,
the force pushing back--
00:26:22.920 --> 00:26:25.640
these things were usually
sitting on wheels--
00:26:25.640 --> 00:26:26.642
could be pretty large.
00:26:26.642 --> 00:26:27.850
Let's take an estimate of it.
00:26:27.850 --> 00:26:29.845
So this is going to be
a relatively small gun.
00:26:34.180 --> 00:26:40.270
The total mass of
the shot here times g
00:26:40.270 --> 00:26:44.300
is 10 pounds, so
10 pounds a shot.
00:26:44.300 --> 00:26:45.840
A gallon of milk
weighs 8 pounds,
00:26:45.840 --> 00:26:49.650
so this isn't a very big load.
00:26:49.650 --> 00:26:53.700
And let's say that this barrel
here, from here to here,
00:26:53.700 --> 00:26:58.840
the acceleration
length is 7 feet.
00:26:58.840 --> 00:27:02.220
So that shot is accelerated
out of the barrel
00:27:02.220 --> 00:27:05.060
over a distance of 7 feet.
00:27:05.060 --> 00:27:16.270
And it has a muzzle
velocity, an exit velocity,
00:27:16.270 --> 00:27:19.750
of 700 feet per second.
00:27:22.740 --> 00:27:26.140
Most guns, subsonic,
supersonic, projectiles
00:27:26.140 --> 00:27:28.530
when they come out of guns,
have you thought about that?
00:27:28.530 --> 00:27:32.240
I'm just asking-- seeing if
you have a feeling for speed.
00:27:32.240 --> 00:27:36.362
And how close to the
speed of sound is that?
00:27:36.362 --> 00:27:37.278
AUDIENCE: [INAUDIBLE].
00:27:37.278 --> 00:27:38.736
PROFESSOR: So the
speed of sound is
00:27:38.736 --> 00:27:43.100
about 1,100 feet per second,
340 meters per second.
00:27:43.100 --> 00:27:46.030
So this is order of mach
0.6 or something like that.
00:27:52.170 --> 00:27:56.540
Something like 0.6,
not 0.060, 0.6.
00:27:56.540 --> 00:27:58.610
So about 60% of
the speed of sound.
00:27:58.610 --> 00:28:00.490
That's slow actually as guns go.
00:28:03.800 --> 00:28:12.510
So the initial velocity-- the
shot's just sitting theree--
00:28:12.510 --> 00:28:13.640
is 0.
00:28:13.640 --> 00:28:21.350
The final velocity is
700 feet per second.
00:28:21.350 --> 00:28:23.935
The average velocity, which
I'm going to need for a second
00:28:23.935 --> 00:28:25.810
because I'm just making
some estimates here--
00:28:25.810 --> 00:28:28.940
the average velocity,
average of 0 plus 700
00:28:28.940 --> 00:28:30.690
is 350 feet per second.
00:28:33.650 --> 00:28:36.392
The reason I need this is I
need to estimate the delta t.
00:28:36.392 --> 00:28:38.725
How long does it take to get
the shot out of the barrel?
00:28:42.120 --> 00:28:45.120
So distance equals
rate times time, right?
00:28:45.120 --> 00:28:56.260
So the v average times
delta t equals 7 feet.
00:28:56.260 --> 00:28:59.050
At 350 feet per
second, you find out
00:28:59.050 --> 00:29:09.050
delta t is about 0.02 seconds.
00:29:09.050 --> 00:29:12.950
So it gets out the barrel
pretty quickly, 0.02 seconds,
00:29:12.950 --> 00:29:15.110
20 milliseconds.
00:29:15.110 --> 00:29:17.870
Now, that powder,
when it goes off,
00:29:17.870 --> 00:29:21.420
it's putting a lot of
forces on the shot.
00:29:21.420 --> 00:29:24.900
So if the total force on
the shot is 5,000 pounds,
00:29:24.900 --> 00:29:27.030
what's the reaction
force on the cannon?
00:29:35.040 --> 00:29:38.520
So we're going to
apply a principle here
00:29:38.520 --> 00:29:40.890
that we talked about earlier.
00:29:40.890 --> 00:29:43.990
Newton had three laws.
00:29:43.990 --> 00:29:44.980
This is a group.
00:29:44.980 --> 00:29:49.970
This is just massive
balls in there.
00:29:49.970 --> 00:29:53.190
The hot exploding gases are
pushing them out the barrel.
00:29:53.190 --> 00:29:57.920
What must be the push of the
same gas on the canon that's
00:29:57.920 --> 00:29:58.630
containing it?
00:29:58.630 --> 00:29:59.710
AUDIENCE: Equal and opposite.
00:29:59.710 --> 00:30:01.470
PROFESSOR: Equal and
opposite, Newton's Third Law.
00:30:01.470 --> 00:30:01.970
Right.
00:30:01.970 --> 00:30:08.290
So the reaction force that we're
looking for is minus the force
00:30:08.290 --> 00:30:10.290
that it takes to get the
shot out of the barrel.
00:30:13.260 --> 00:30:19.890
So the force on the
balls, on the shot,
00:30:19.890 --> 00:30:22.440
minus the reaction force.
00:30:22.440 --> 00:30:24.040
And that's Newton's Third Law.
00:30:31.400 --> 00:30:33.840
So we're almost there.
00:30:33.840 --> 00:30:35.760
We're just applying
now this concept
00:30:35.760 --> 00:30:36.910
of impulse and momentum.
00:30:40.990 --> 00:30:44.860
So now we can say the integral--
and this is a summation.
00:30:44.860 --> 00:30:47.380
You got a bunch
of balls in there.
00:30:47.380 --> 00:30:49.170
They all got forces
on them, but we're
00:30:49.170 --> 00:30:51.660
treating the group
of balls as a system,
00:30:51.660 --> 00:30:55.370
so we can use this
notion of the total mass
00:30:55.370 --> 00:31:00.250
times the velocity of the center
of gravity to do this problem.
00:31:00.250 --> 00:31:11.280
So this is the force on all of
these little balls integrated
00:31:11.280 --> 00:31:11.970
over time.
00:31:14.560 --> 00:31:19.970
It's going to be the mass total
of the balls times the change
00:31:19.970 --> 00:31:24.000
in the velocity of the center
of gravity, the final velocity
00:31:24.000 --> 00:31:25.860
minus the initial velocity.
00:31:28.460 --> 00:31:29.225
This is 0.
00:31:32.800 --> 00:31:34.010
We know this.
00:31:34.010 --> 00:31:36.900
We know this.
00:31:36.900 --> 00:31:41.570
So this mt is the
10 pounds over g.
00:31:41.570 --> 00:31:44.470
We know this is 700
feet per second.
00:31:44.470 --> 00:31:46.980
We know that all
of this mounts up
00:31:46.980 --> 00:31:54.700
to the force external
on the group delta t.
00:31:58.700 --> 00:32:01.620
And this is the weight
of the ball, w/g.
00:32:04.330 --> 00:32:10.430
And this is 700 feet per second.
00:32:10.430 --> 00:32:15.680
That's 10 pounds divided
by gravity, 32.12.
00:32:15.680 --> 00:32:18.210
And we need to
divide by delta t.
00:32:18.210 --> 00:32:21.990
So the external
force on the balls
00:32:21.990 --> 00:32:32.710
is w/g delta t times
700 feet per second.
00:32:32.710 --> 00:32:45.170
And if you work out those
numbers, you get 10,870 pounds.
00:32:45.170 --> 00:32:56.180
Now, this is an average
force because we--
00:32:56.180 --> 00:32:59.280
that delta t is
the length of time
00:32:59.280 --> 00:33:04.130
it took, assuming we had some--
we assumed an average velocity
00:33:04.130 --> 00:33:05.250
going down the barrel.
00:33:05.250 --> 00:33:07.820
So the peak force is
probably higher than this,
00:33:07.820 --> 00:33:10.340
and the pressure in the barrel
probably isn't constant.
00:33:10.340 --> 00:33:12.610
This actually isn't
a bad estimate.
00:33:12.610 --> 00:33:15.610
If you know the muzzle velocity,
which you can figure out
00:33:15.610 --> 00:33:18.436
probably from the distance
it goes and things like that,
00:33:18.436 --> 00:33:19.560
you can make this estimate.
00:33:19.560 --> 00:33:24.870
So 10,000 pounds of reaction
force is quite a lot.
00:33:24.870 --> 00:33:27.020
This is a dinky gun.
00:33:27.020 --> 00:33:29.955
10 pounds a shot is not much.
00:33:29.955 --> 00:33:32.850
This is the force to get
the balls out the barrel.
00:33:32.850 --> 00:33:35.100
If it's positive, what's
the reaction force?
00:33:39.630 --> 00:33:42.340
Minus 10,870.
00:33:42.340 --> 00:33:45.080
And all through this, I've been
doing this in a single vector
00:33:45.080 --> 00:33:46.920
direction, so I
would have set up
00:33:46.920 --> 00:33:49.460
my coordinate system
like here's o.
00:33:49.460 --> 00:33:52.780
This is the x
direction, y direction.
00:33:52.780 --> 00:33:54.700
Obviously, in the
positive x direction
00:33:54.700 --> 00:33:57.305
is what I've lined up
my positive velocities
00:33:57.305 --> 00:33:59.670
and positive forces to be.
00:33:59.670 --> 00:34:01.241
So this is a really simple way.
00:34:01.241 --> 00:34:02.740
One of the kind of
ways in which you
00:34:02.740 --> 00:34:05.610
use this notion of
impulse and momentum
00:34:05.610 --> 00:34:07.290
that you can use to
make estimates too.
00:34:12.830 --> 00:34:16.921
Any of you ever
fired a shot gun.
00:34:16.921 --> 00:34:17.420
Not many.
00:34:17.420 --> 00:34:18.909
There's-- yeah.
00:34:18.909 --> 00:34:20.131
Any kick?
00:34:20.131 --> 00:34:20.630
Right.
00:34:20.630 --> 00:34:23.240
So what gauge
shotgun did you fire?
00:34:23.240 --> 00:34:23.909
You don't know.
00:34:23.909 --> 00:34:24.409
OK.
00:34:24.409 --> 00:34:25.786
A 12-gauge shot gun.
00:34:25.786 --> 00:34:29.734
12-gauge shot gun, the
cartridge is just about 3/4
00:34:29.734 --> 00:34:31.900
of an inch diameter and
probably has a little amount
00:34:31.900 --> 00:34:32.750
of powder in there.
00:34:32.750 --> 00:34:34.747
It amounts up to
about that much stuff.
00:34:34.747 --> 00:34:36.580
But it can give you a
bruise in the shoulder
00:34:36.580 --> 00:34:38.739
if you don't hold it right.
00:34:38.739 --> 00:34:40.570
So that's what this is about.
00:34:43.610 --> 00:34:46.860
So this quick
review, you've done
00:34:46.860 --> 00:34:49.710
lots of conservation of
momentum problems in your time.
00:34:49.710 --> 00:34:53.690
The homework set this time has
two or three problems on it
00:34:53.690 --> 00:34:56.850
that are conservation of
momentum, linear momentum.
00:34:56.850 --> 00:35:01.990
But now I want to move on to
talking about angular momentum.
00:35:01.990 --> 00:35:05.480
And angular momentum, in a
way in which you probably
00:35:05.480 --> 00:35:09.060
haven't done angular
momentum problems before.
00:35:09.060 --> 00:35:13.290
So anything about
last thing on this.
00:35:13.290 --> 00:35:16.040
So mostly this, I want
you to read the chapter.
00:35:16.040 --> 00:35:19.926
I think it's chapter 15.
00:35:19.926 --> 00:35:22.540
The first few sections of
it are on linear momentum,
00:35:22.540 --> 00:35:24.880
the last few sections
on angular momentum.
00:35:24.880 --> 00:35:28.160
Go read them and just
work the problems.
00:35:28.160 --> 00:35:30.630
That's where you'll get
most of your refresher
00:35:30.630 --> 00:35:33.620
is doing the practice problems.
00:35:33.620 --> 00:35:34.440
Angular momentum.
00:36:02.462 --> 00:36:03.670
So we'll start with particle.
00:36:03.670 --> 00:36:06.560
We're very rapidly going
to get to rigid bodies.
00:36:11.990 --> 00:36:15.250
Now many times, you've done
angular momentum problems
00:36:15.250 --> 00:36:18.610
before, mostly rotations
about fixed axes.
00:36:18.610 --> 00:36:24.300
So here's our inertial
frame, fixed x,y and you have
00:36:24.300 --> 00:36:27.170
a particle out here.
00:36:27.170 --> 00:36:31.240
And it just has some mass, m.
00:36:31.240 --> 00:36:35.009
And it has some total
force on it, the force.
00:36:35.009 --> 00:36:36.550
And this says the
particle is located
00:36:36.550 --> 00:36:41.180
at B here, so the force
at B, total force at B,
00:36:41.180 --> 00:36:44.370
just some vector.
00:36:44.370 --> 00:36:47.580
It also is traveling with
some velocity at that instant
00:36:47.580 --> 00:36:50.720
in time, so it has momentum.
00:36:50.720 --> 00:36:58.480
Momentum is p of this particle
at B with respect to o.
00:37:02.630 --> 00:37:05.640
And I'd like to-- the
standard expression
00:37:05.640 --> 00:37:13.960
then for the angular momentum
of this particle at B
00:37:13.960 --> 00:37:23.700
with respect to o-- we have
a position vector here,
00:37:23.700 --> 00:37:29.190
r Bo, which we've used
lots of times by now,
00:37:29.190 --> 00:37:35.610
is just the cross product
of the position vector
00:37:35.610 --> 00:37:37.540
with the linear momentum.
00:37:43.520 --> 00:37:46.400
And that's the definition
of angular momentum.
00:37:46.400 --> 00:37:48.630
I'm using a lowercase
h here, so they're
00:37:48.630 --> 00:37:50.970
going to do that to
indicate single particles.
00:37:50.970 --> 00:37:53.200
And I'll use a capital
H later on when we're
00:37:53.200 --> 00:37:55.010
referring to rigid bodies.
00:37:55.010 --> 00:37:58.000
So the angular momentum
of this particle,
00:37:58.000 --> 00:38:01.234
with respect to this
point, is given by that.
00:38:05.170 --> 00:38:07.710
And where you've
used this before,
00:38:07.710 --> 00:38:11.630
then, it's the derivative
of hB with respect
00:38:11.630 --> 00:38:17.830
to o dt is-- what's that?
00:38:17.830 --> 00:38:19.400
Remember, where
does this give you?
00:38:19.400 --> 00:38:22.340
The time rate of change of
the angular momentum is the?
00:38:22.340 --> 00:38:23.007
AUDIENCE: Torque
00:38:23.007 --> 00:38:23.756
PROFESSOR: Torque.
00:38:23.756 --> 00:38:24.330
Right.
00:38:24.330 --> 00:38:25.240
And it's the torque.
00:38:25.240 --> 00:38:29.150
It's a sum of the
torques applied
00:38:29.150 --> 00:38:33.710
on that object with respect
to the coordinate system
00:38:33.710 --> 00:38:36.540
in which you're computing
the angular momentum.
00:38:36.540 --> 00:38:38.990
So that's-- we've used
this formula many times.
00:38:38.990 --> 00:38:41.360
And in planar motion
where you only
00:38:41.360 --> 00:38:46.410
have one axis of rotation
using the z-axis of rotation,
00:38:46.410 --> 00:38:50.830
you usually would write
this as I theta double dot.
00:38:50.830 --> 00:38:52.850
In simplest form, if
it's a rigid body,
00:38:52.850 --> 00:38:55.260
it has a mass moment
of inertia times
00:38:55.260 --> 00:38:58.190
theta double dot is equal to
the sum of the external torque.
00:38:58.190 --> 00:39:00.337
So that's where you've
met this before.
00:39:00.337 --> 00:39:02.170
But for the moment,
this is just a particle.
00:39:02.170 --> 00:39:05.480
Let's stick with a particle.
00:39:05.480 --> 00:39:09.080
So the piece that's
new here-- probably new
00:39:09.080 --> 00:39:18.440
for you-- is that what if I want
to know the angular momentum
00:39:18.440 --> 00:39:21.630
with respect to another point?
00:39:21.630 --> 00:39:32.830
So here's a point A. I'd like
to compute hB with respect to A.
00:39:32.830 --> 00:39:45.120
Well, that's rB with respect to
A cross PB with respect to o.
00:39:45.120 --> 00:39:47.460
And this is really
easy to forget.
00:39:47.460 --> 00:39:50.530
The momentum is always
calculated with respect
00:39:50.530 --> 00:39:53.670
to your inertial frame.
00:39:53.670 --> 00:39:56.470
And that's why I keep
in this with respect to
00:39:56.470 --> 00:39:59.710
and telling you what the
frame is is pretty important.
00:39:59.710 --> 00:40:02.310
But we're out here at
some arbitrary point,
00:40:02.310 --> 00:40:06.660
computing the cross product
of the position vector
00:40:06.660 --> 00:40:10.870
from this arbitrary point
to this moving mass.
00:40:10.870 --> 00:40:13.070
And we're defining--
it's just a definition
00:40:13.070 --> 00:40:16.910
defining the angular momentum
with respect to this point A
00:40:16.910 --> 00:40:21.200
as r B/A cross PB
with respect to o.
00:40:31.570 --> 00:40:42.680
And what I want to get to is now
is the torque on this system,
00:40:42.680 --> 00:40:49.810
around this particle,
B with respect to A
00:40:49.810 --> 00:40:57.260
is the time rate of change
of h B/A with respect to t.
00:40:57.260 --> 00:41:00.250
But now it's a little
more complicated.
00:41:00.250 --> 00:41:02.770
Plus, the velocity
of A with respect
00:41:02.770 --> 00:41:12.900
to o-- these are all vectors--
cross PB with respect to o.
00:41:12.900 --> 00:41:13.700
Messy term.
00:41:17.352 --> 00:41:18.810
I was trying to
think of an example
00:41:18.810 --> 00:41:23.630
where you might want to do this.
00:41:31.810 --> 00:41:38.320
So imagine that you've got
an arm which can rotate.
00:41:38.320 --> 00:41:40.710
It might be on a robot
or something like that.
00:41:40.710 --> 00:41:49.510
And it has attached to it
another arm with a mass on it.
00:41:49.510 --> 00:41:54.850
And you've got a motor here,
which can make this rotate.
00:41:54.850 --> 00:41:57.290
And you're trying to
design-- the motor has
00:41:57.290 --> 00:42:00.240
to be able to put out a
certain amount of torque.
00:42:00.240 --> 00:42:02.150
So this is o.
00:42:02.150 --> 00:42:05.930
This is B. This is
A. And you actually--
00:42:05.930 --> 00:42:09.880
I want to know the torque
required in this motor
00:42:09.880 --> 00:42:12.990
to drive this thing around.
00:42:12.990 --> 00:42:15.380
But the motor is
here, and it only
00:42:15.380 --> 00:42:19.420
cares about what it feels,
so the torque at this point
00:42:19.420 --> 00:42:21.240
to drive this thing.
00:42:21.240 --> 00:42:24.550
But this whole system
is now in motion.
00:42:24.550 --> 00:42:25.980
You'd have to use this formula.
00:42:28.900 --> 00:42:30.380
So there are
practical times when
00:42:30.380 --> 00:42:33.360
you'd like to be able to
calculate something like this.
00:42:42.870 --> 00:42:44.930
So there's a--
we're going to show
00:42:44.930 --> 00:42:52.530
you a very brief
derivation of this
00:42:52.530 --> 00:42:54.620
just so you can get a
feeling for where this comes
00:42:54.620 --> 00:42:56.450
from because there's
a couple of outcomes
00:42:56.450 --> 00:42:57.890
that are very important to it.
00:43:10.020 --> 00:43:18.480
So the sum of the forces
there at-- those forces at B,
00:43:18.480 --> 00:43:24.360
give you the time rate
of change of the momentum
00:43:24.360 --> 00:43:26.080
at B with respect to o.
00:43:26.080 --> 00:43:29.930
This is the momentum
vector of our particle.
00:43:29.930 --> 00:43:33.882
And this fB, this is the
total external forces
00:43:33.882 --> 00:43:34.840
acting on the particle.
00:43:39.460 --> 00:43:42.715
And we know that's m.
00:43:42.715 --> 00:43:55.525
It's a single particle
times the velocity of That
00:43:55.525 --> 00:43:56.830
certainly could be right.
00:44:03.920 --> 00:44:08.280
That's our familiar
formula for F equals ma.
00:44:08.280 --> 00:44:11.970
So the time derivative of a
linear momentum gives you this.
00:44:11.970 --> 00:44:17.690
And the torque of
B with respect to A
00:44:17.690 --> 00:44:25.350
is r B/A cross-- I'm going let
this just be a total vector.
00:44:25.350 --> 00:44:28.460
I'll call it fB.
00:44:28.460 --> 00:44:30.120
It's a vector.
00:44:30.120 --> 00:44:32.160
The torque with
respect to A is just
00:44:32.160 --> 00:44:34.590
r B/A cross this
total external force.
00:44:40.940 --> 00:44:53.283
r B/A cross time
derivative of P B/o.
00:44:56.220 --> 00:44:58.680
Because the forces
give us r if it's
00:44:58.680 --> 00:45:01.730
from the time rate of change
of the linear momentum
00:45:01.730 --> 00:45:04.356
of that particle.
00:45:04.356 --> 00:45:06.630
So I can say it like this.
00:45:06.630 --> 00:45:12.640
But now there is just a
little vector identity
00:45:12.640 --> 00:45:17.340
for products of vectors that
I'm going to take advantage.
00:45:17.340 --> 00:45:31.620
And I'll call this Q.
So Q is of the form--
00:45:31.620 --> 00:45:35.290
and I'm going to say this
is a quantity of vector A,
00:45:35.290 --> 00:45:38.600
and this is a quantity
of vector-- time
00:45:38.600 --> 00:45:42.950
derivative of a vector
B. It's A cross dB dt.
00:45:42.950 --> 00:45:47.080
So there's a little
identity that you can use.
00:45:47.080 --> 00:45:55.330
It says A cross dB dt.
00:45:55.330 --> 00:46:01.510
You can alternatively write
that as the time derivative
00:46:01.510 --> 00:46:14.499
of A cross B minus time
derivative of A cross B.
00:46:14.499 --> 00:46:16.040
We're going to take
advantage of this
00:46:16.040 --> 00:46:23.325
and just re-construct this
formula using this expression.
00:46:30.540 --> 00:46:37.860
So that says torque
of B with respect to A
00:46:37.860 --> 00:46:46.695
is the time derivative
of r B/A cross P
00:46:46.695 --> 00:47:02.240
B/o minus the derivative
of r B/A cross P B/o.
00:47:02.240 --> 00:47:05.360
So we've just made this
substitution down here in terms
00:47:05.360 --> 00:47:09.375
of r B/A and P B/o all vectors.
00:47:12.540 --> 00:47:19.520
We know that r B/A, from all
the previous work we've done,
00:47:19.520 --> 00:47:27.620
is just r B/o minus r
A/o so the derivative
00:47:27.620 --> 00:47:43.900
of r B/A with respect to
time is v B/o minus v A/o
00:47:43.900 --> 00:47:45.210
We're almost there.
00:47:45.210 --> 00:47:46.457
We're almost there.
00:47:46.457 --> 00:47:47.790
I'm going to need another board.
00:48:08.210 --> 00:48:15.700
This quantity here,
this is just h B/A.
00:48:15.700 --> 00:48:19.510
This is the angular momentum of
the particle with respect to A.
00:48:19.510 --> 00:48:22.145
It's just r cross
B, if you recall.
00:48:52.360 --> 00:48:56.840
So then we can rewrite this
expression for the torque
00:48:56.840 --> 00:49:02.800
as the time derivative
of h of B with respect
00:49:02.800 --> 00:49:08.190
to A. That's because of
what I pointed out there.
00:49:08.190 --> 00:49:19.217
Now, this is minus v B/o
minus v A/o cross P B/o.
00:49:28.210 --> 00:49:31.670
So P of B with
respect to o is just
00:49:31.670 --> 00:49:38.740
m velocity of B
with respect to o.
00:49:38.740 --> 00:49:42.980
So v B/o cross P
B/o gives you what?
00:49:42.980 --> 00:49:43.480
Nothing.
00:49:43.480 --> 00:49:45.729
Gives you 0 because they are
in parallel of the angle,
00:49:45.729 --> 00:49:48.290
and they're in the
same direction.
00:49:48.290 --> 00:49:53.230
So you only get a nonzero piece
out of this minus times minus
00:49:53.230 --> 00:50:15.030
gives you a plus, and you
end up with-- and that's
00:50:15.030 --> 00:50:16.850
what we set out to find.
00:50:21.010 --> 00:50:23.055
I said this is where
I was trying to get,
00:50:23.055 --> 00:50:26.090
and now we're there.
00:50:26.090 --> 00:50:27.940
Now importantly,
there's a couple
00:50:27.940 --> 00:50:31.410
of special cases of this.
00:50:31.410 --> 00:50:35.570
This can be a nuisance
term to have to deal with.
00:50:35.570 --> 00:50:38.080
Lots of times you'd like
to be able to get rid of it
00:50:38.080 --> 00:50:40.520
and just be able to go back
to that old reliable formula,
00:50:40.520 --> 00:50:44.510
torque is time rate of
change of angular momentum.
00:50:44.510 --> 00:50:48.530
So there are two obvious--
maybe obvious conditions
00:50:48.530 --> 00:50:52.220
in which this will go away.
00:50:52.220 --> 00:50:53.840
What's the most obvious one?
00:50:57.340 --> 00:50:59.010
AUDIENCE: [INAUDIBLE]
change to 0.
00:50:59.010 --> 00:51:00.745
PROFESSOR: Something,
what you say was 0?
00:51:00.745 --> 00:51:02.080
AUDIENCE: One of the terms
00:51:02.080 --> 00:51:02.996
PROFESSOR: Well, yeah.
00:51:02.996 --> 00:51:06.370
This term presumably not.
00:51:06.370 --> 00:51:24.650
This guy, if this
is 0, it's just back
00:51:24.650 --> 00:51:25.885
to our old familiar formula.
00:51:29.820 --> 00:51:31.610
That's one case, so case one.
00:51:34.850 --> 00:51:39.550
But now-- actually, this is
a really important result
00:51:39.550 --> 00:51:49.370
because A can be anywhere
as long as it's not moving.
00:51:49.370 --> 00:51:53.080
So this allows you
to do things, talk
00:51:53.080 --> 00:51:56.200
about rotations
about fixed axes that
00:51:56.200 --> 00:51:57.560
aren't at the center of mass.
00:52:00.640 --> 00:52:03.260
So if you have a fixed axis
of rotation and something
00:52:03.260 --> 00:52:08.810
going around it, that's
what this allows you to do.
00:52:08.810 --> 00:52:09.710
That's one case.
00:52:09.710 --> 00:52:12.100
So this is--
00:52:12.100 --> 00:52:14.150
We'll soon get to rigid bodies.
00:52:14.150 --> 00:52:17.370
Rigid bodies obey exactly
the same formulas.
00:52:17.370 --> 00:52:19.740
And you can have
a rigid body, now,
00:52:19.740 --> 00:52:23.040
that is not rotating
about its center,
00:52:23.040 --> 00:52:26.130
but rotating maybe
about its end like this.
00:52:26.130 --> 00:52:29.130
That formula applies
if the velocity
00:52:29.130 --> 00:52:31.640
of that axis about which it's
rotating, about which you're
00:52:31.640 --> 00:52:34.650
computing, is not
moving, then you can just
00:52:34.650 --> 00:52:38.290
use-- you don't have to
deal with that messy term.
00:52:38.290 --> 00:52:43.490
So this is one case in
which this term goes away.
00:52:43.490 --> 00:52:47.600
The other case is if
this velocity is parallel
00:52:47.600 --> 00:52:49.760
to the direction
of the momentum.
00:52:52.380 --> 00:52:57.240
And there's a really useful
time that that happens.
00:52:57.240 --> 00:53:01.530
So also this formula is true.
00:53:01.530 --> 00:53:09.150
Case two is when
v A/o is parallel
00:53:09.150 --> 00:53:12.370
to P B/o, the direction.
00:53:12.370 --> 00:53:14.370
You're going in
the same direction.
00:53:14.370 --> 00:53:18.840
Now that happens when
A-- it's guaranteed
00:53:18.840 --> 00:53:22.890
to be true if A is
at the center of mass
00:53:22.890 --> 00:53:26.920
because the momentum is defined
as the mass of the object
00:53:26.920 --> 00:53:32.970
times the velocity of its center
of mass, even for rigid bodies.
00:53:32.970 --> 00:53:45.035
So this is true when A
is-- and I'll call it
00:53:45.035 --> 00:53:48.298
G-- at the center of mass.
00:53:53.570 --> 00:54:03.840
So this gives us another
really important generalization
00:54:03.840 --> 00:54:07.500
that we'll use-- that we make
great use of in dynamics.
00:54:22.330 --> 00:54:27.440
And that says that the
torque, with respect
00:54:27.440 --> 00:54:34.030
to the center of mass,
is time rate of change
00:54:34.030 --> 00:54:42.040
of h with respect
to G. Now, I'm not
00:54:42.040 --> 00:54:44.860
going to go-- I did the proof--
went through this little proof
00:54:44.860 --> 00:54:46.340
just for a particle.
00:54:46.340 --> 00:54:49.540
But by summing a
bunch of particles
00:54:49.540 --> 00:54:51.730
and going through all
the summations, as we
00:54:51.730 --> 00:54:54.870
did, to prove the
center of mass formula,
00:54:54.870 --> 00:54:59.490
you can show-- this allows
you to very quickly show
00:54:59.490 --> 00:55:02.705
that this formulation is
also true for rigid bodies.
00:55:05.730 --> 00:55:21.050
So the way you say it
for rigid bodies is
00:55:21.050 --> 00:55:24.040
that the sum of the
torques, with respect
00:55:24.040 --> 00:55:33.930
to some point for a rigid body,
is capital H dot with respect
00:55:33.930 --> 00:55:42.960
to A plus the velocity of A
with respect to o cross P.
00:55:42.960 --> 00:55:46.890
And now I'm going to say G,
its center of mass, center
00:55:46.890 --> 00:55:50.588
of gravity, with respect to o.
00:55:50.588 --> 00:55:53.940
So the same statement
for a rigid body
00:55:53.940 --> 00:55:56.810
is that the torque with
respect to some point A, which
00:55:56.810 --> 00:56:01.020
can be moving now--
even accelerating--
00:56:01.020 --> 00:56:03.520
is the time rate of
change of the angular
00:56:03.520 --> 00:56:07.240
momentum with respect to A
of that rigid body plus v A/o
00:56:07.240 --> 00:56:10.220
cross P B/o.
00:56:10.220 --> 00:56:15.070
And again, when would this
messy second term go to 0?
00:56:15.070 --> 00:56:21.100
When the velocity of A is 0,
fixed axis rotation, or when
00:56:21.100 --> 00:56:24.450
this is parallel to that, which
is true for the center of mass
00:56:24.450 --> 00:56:25.960
always.
00:56:25.960 --> 00:56:31.650
So the same two special
cases apply for rigid bodies
00:56:31.650 --> 00:56:37.920
when velocity of A with
respect to 0 equals 0
00:56:37.920 --> 00:56:52.580
or when the velocity of A with
respect to o is parallel the P.
00:56:52.580 --> 00:56:56.895
And the most important case
of that is always true.
00:57:12.850 --> 00:57:13.890
It's always the case.
00:57:13.890 --> 00:57:15.560
And in these you can say this.
00:57:15.560 --> 00:57:19.590
Then, the torque,
with respect to A,
00:57:19.590 --> 00:57:25.840
is dH, with respect to A dt.
00:57:25.840 --> 00:57:27.017
No second terms.
00:57:31.310 --> 00:57:35.610
And it's those kinds
of-- you've applied,
00:57:35.610 --> 00:57:39.270
generally in your physics like
in 801, you've used formulas
00:57:39.270 --> 00:57:40.120
like this a lot.
00:57:42.730 --> 00:57:44.220
Done problems
either with respect
00:57:44.220 --> 00:57:51.500
to the center of mass so objects
were doing things like that.
00:57:51.500 --> 00:57:54.740
You'll do the torque
formulas with respect
00:57:54.740 --> 00:57:55.760
to the center of mass.
00:57:55.760 --> 00:58:00.900
Or when you have things that
are pinned to points and rotate
00:58:00.900 --> 00:58:05.230
about fixed axes, then
use the other formulation
00:58:05.230 --> 00:58:10.730
where it is with respect
to a non-moving point.
00:58:10.730 --> 00:58:12.230
So this is fixed axis rotation.
00:58:27.660 --> 00:58:30.370
So that's the dry
derivation part of it.
00:58:30.370 --> 00:58:33.100
I'm going to see if I
can find an example here.
00:58:42.430 --> 00:58:44.029
This is quite a
bit like a number
00:58:44.029 --> 00:58:45.070
of the homework problems.
00:59:18.620 --> 00:59:20.210
So I've got a carnival ride.
00:59:24.210 --> 00:59:26.470
I got a bar.
00:59:26.470 --> 00:59:29.890
And it's got a seat out here.
00:59:29.890 --> 00:59:32.240
You're riding in it.
00:59:32.240 --> 00:59:33.660
So you're taking this ride.
00:59:38.510 --> 00:59:40.010
It's rotating.
00:59:40.010 --> 00:59:42.330
This is some fixed axis here.
00:59:47.666 --> 00:59:53.120
You have a fixed coordinate
system, your o x, y, z.
00:59:53.120 --> 00:59:55.690
And then you'd probably have
some rotating coordinate system
00:59:55.690 --> 01:00:03.210
with the point A fixed here
at the axis of rotation.
01:00:03.210 --> 01:00:09.580
And this is going-- what's
unusual about this ride,
01:00:09.580 --> 01:00:11.860
not only can it go
around and round,
01:00:11.860 --> 01:00:14.220
but the arm can go in and out.
01:00:17.230 --> 01:00:19.790
So it might take a
path inwards, like you
01:00:19.790 --> 01:00:22.660
pull the arm in as
you're going around.
01:00:22.660 --> 01:00:25.050
What are the forces that
you feel in the ride?
01:00:28.040 --> 01:00:31.426
So this is my point B out here.
01:00:31.426 --> 01:00:32.550
It's where the person's at.
01:00:32.550 --> 01:00:36.165
You have some mass, m.
01:00:38.770 --> 01:00:49.010
And let's let, for now,
the first case-- theta
01:00:49.010 --> 01:00:59.510
dot, let that be constant, the
constant angular rate here.
01:00:59.510 --> 01:01:04.460
Here's theta and r dot.
01:01:13.220 --> 01:01:17.020
Now, you already know quite
a bit about things like this.
01:01:17.020 --> 01:01:21.510
If your riding in that
bucket, what forces
01:01:21.510 --> 01:01:23.920
do you think you would feel?
01:01:23.920 --> 01:01:26.600
Or I should say
it more carefully.
01:01:26.600 --> 01:01:29.920
There will be accelerations
that you feel in that bucket.
01:01:29.920 --> 01:01:31.700
You'll feel like forces on you.
01:01:31.700 --> 01:01:33.575
But what are the
accelerations that you
01:01:33.575 --> 01:01:36.100
will feel-- you expect to
be if you're riding in it?
01:01:39.812 --> 01:01:41.210
AUDIENCE: [INAUDIBLE].
01:01:41.210 --> 01:01:42.460
PROFESSOR: I hear one here.
01:01:42.460 --> 01:01:43.690
AUDIENCE: Centripetal
acceleration.
01:01:43.690 --> 01:01:45.960
PROFESSOR: So there will be
centripetal acceleration,
01:01:45.960 --> 01:01:46.880
right.
01:01:46.880 --> 01:01:48.150
Everybody agree with that?
01:01:48.150 --> 01:01:49.070
Anything else?
01:01:49.070 --> 01:01:49.974
AUDIENCE: Coriolis acceleration.
01:01:49.974 --> 01:01:51.807
PROFESSOR: There's going
to be some Coriolis
01:01:51.807 --> 01:01:54.762
because r dot is not 0.
01:01:54.762 --> 01:01:57.710
It's changing in
position, but when
01:01:57.710 --> 01:01:59.740
the length of the
arm of something
01:01:59.740 --> 01:02:03.054
rotating at constant speed
changes, what momentum changes?
01:02:09.090 --> 01:02:09.840
AUDIENCE: Angular.
01:02:09.840 --> 01:02:11.048
PROFESSOR: Angular, for sure.
01:02:11.048 --> 01:02:12.980
How about linear momentum?
01:02:12.980 --> 01:02:16.260
Yeah, it's changing too because
r cross P is angular momentum.
01:02:16.260 --> 01:02:17.810
So both linear and
angular momentum
01:02:17.810 --> 01:02:22.320
are changing as this radius
gets longer or shorter.
01:02:22.320 --> 01:02:25.500
And if that angular
momentum changes,
01:02:25.500 --> 01:02:27.506
it takes torque to drive it.
01:02:27.506 --> 01:02:29.880
And so we ought to be able to
use the formulas that we've
01:02:29.880 --> 01:02:32.280
just derived to
calculate something
01:02:32.280 --> 01:02:35.430
about the torques required
to make this happen
01:02:35.430 --> 01:02:36.750
and the forces on the rider.
01:02:44.580 --> 01:02:47.710
So we'll treat this
one as a particle.
01:02:47.710 --> 01:02:52.420
So H, the angular momentum
of the rider out here,
01:02:52.420 --> 01:03:07.480
at B with respect to o, is
going to be r B/o cross P.
01:03:07.480 --> 01:03:10.780
And in this problem, I'll
use polar coordinates.
01:03:10.780 --> 01:03:11.550
Pretty easy.
01:03:11.550 --> 01:03:13.470
This is a planar motion problem.
01:03:13.470 --> 01:03:17.040
It's confined to the x,y
plane and rotation z,
01:03:17.040 --> 01:03:19.640
so polar coordinates
are pretty convenient.
01:03:19.640 --> 01:03:26.020
So this should look like
radius r r hat cross--
01:03:26.020 --> 01:03:29.880
and the linear
momentum of this is
01:03:29.880 --> 01:03:36.190
the mass-- times r dot r hat.
01:03:36.190 --> 01:03:41.090
That's the extension
rate, but it also
01:03:41.090 --> 01:03:49.720
has velocity in this theta
hat direction, r theta hat.
01:03:49.720 --> 01:03:56.450
So this is P was respected
o, m, v. And this
01:03:56.450 --> 01:03:59.810
is the radius-- this is
the r crossed into it.
01:03:59.810 --> 01:04:02.920
Now r hat cross r
hat gives you 0,
01:04:02.920 --> 01:04:05.690
so you only get a
single term out of this.
01:04:05.690 --> 01:04:10.970
And you get an r cross-- r hat
cross, theta hat, positive k.
01:04:10.970 --> 01:04:25.750
So this looks like plus m r
squared, theta dot, k hat.
01:04:25.750 --> 01:04:27.820
That's my H B/o.
01:04:34.910 --> 01:04:37.880
Now, I would-- that's my first
piece that I wanted to get.
01:04:37.880 --> 01:04:38.790
That's the a part.
01:04:38.790 --> 01:04:41.200
That's find the
angular momentum.
01:04:41.200 --> 01:04:43.688
So this is a.
01:04:43.688 --> 01:04:45.150
b, I want to know the torque.
01:04:54.950 --> 01:04:59.010
Well, which formula can we use?
01:04:59.010 --> 01:05:01.900
Are we allowed to use--
do we have to account
01:05:01.900 --> 01:05:04.202
for that second term?
01:05:04.202 --> 01:05:04.827
AUDIENCE: No.
01:05:04.827 --> 01:05:05.410
PROFESSOR: No.
01:05:05.410 --> 01:05:05.959
Why?
01:05:05.959 --> 01:05:06.750
AUDIENCE: Velocity.
01:05:06.750 --> 01:05:07.180
PROFESSOR: Yeah.
01:05:07.180 --> 01:05:08.971
The velocity of the
point about which we're
01:05:08.971 --> 01:05:12.550
computing, the angular momentum,
and therefore, the torques,
01:05:12.550 --> 01:05:13.380
is not moving.
01:05:13.380 --> 01:05:15.650
So you only have to deal
with the first term.
01:05:15.650 --> 01:05:20.520
So the torque required to
move that particle at B
01:05:20.520 --> 01:05:24.040
with respect to o is just
a time rate of change
01:05:24.040 --> 01:05:28.870
d by dt of h B
with respect to o.
01:05:28.870 --> 01:05:44.030
And that's d by dt of m r
squared theta dot k hat.
01:05:44.030 --> 01:05:46.977
So what are the constants
in this expression
01:05:46.977 --> 01:05:49.060
so we don't have to worry
about their derivatives?
01:05:49.060 --> 01:05:51.160
Does k hat change length?
01:05:51.160 --> 01:05:51.660
No.
01:05:51.660 --> 01:05:53.010
Change direction?
01:05:53.010 --> 01:05:54.360
No.
01:05:54.360 --> 01:05:59.170
Theta dot, does it
change in this problem?
01:05:59.170 --> 01:06:00.100
No, we fixed it.
01:06:00.100 --> 01:06:02.620
We arbitrarily started
off saying that we'll just
01:06:02.620 --> 01:06:04.810
let theta dot be constant.
01:06:04.810 --> 01:06:06.390
r, though, is changing.
01:06:06.390 --> 01:06:08.420
So the time derivative
of this particular one,
01:06:08.420 --> 01:06:10.910
we only have to deal with the r.
01:06:10.910 --> 01:06:13.175
Actually, I'll forget
that for a second
01:06:13.175 --> 01:06:14.550
because I want to
get both terms,
01:06:14.550 --> 01:06:16.120
and then we'll let it be 0.
01:06:16.120 --> 01:06:19.000
So this time derivative
then, when you work it
01:06:19.000 --> 01:06:47.080
out-- the derivative
of the r term
01:06:47.080 --> 01:06:56.730
gives you 2mr r dot
theta dot k hat.
01:06:56.730 --> 01:06:58.764
And I'll just go ahead
and forget for a minute
01:06:58.764 --> 01:06:59.930
that I said that's constant.
01:06:59.930 --> 01:07:03.870
Let's get the other term
that might be there.
01:07:03.870 --> 01:07:16.785
That term gives us m r squared
theta double dot k hat.
01:07:19.950 --> 01:07:24.580
This term comes from what
we'll find is the Coriolis one.
01:07:24.580 --> 01:07:28.585
And this is from what we call
the Eulerian acceleration.
01:07:31.445 --> 01:07:35.305
The torque-- let me get
this up a little higher.
01:07:39.640 --> 01:07:48.270
We should be able to
write as some r cross f.
01:07:48.270 --> 01:07:52.930
And so the r cross
f terms, this will
01:07:52.930 --> 01:07:56.790
be from the Coriolis force.
01:07:56.790 --> 01:08:02.460
And this would be from
that Eulerian force.
01:08:02.460 --> 01:08:06.700
So this will look
like r r hat cross--
01:08:06.700 --> 01:08:10.620
I'm just factoring this back
out into its cross products--
01:08:10.620 --> 01:08:17.819
r r hat cross 2m
r dot theta dot,
01:08:17.819 --> 01:08:26.939
in the theta hat direction,
plus r theta double dot,
01:08:26.939 --> 01:08:30.120
in the theta hat direction.
01:08:30.120 --> 01:08:32.689
So there's two terms in
this torque expression.
01:08:32.689 --> 01:08:37.970
They come from r
cross, two force terms.
01:08:37.970 --> 01:08:41.380
The first force term is what we
know to be the Coriolis force.
01:08:41.380 --> 01:08:44.816
And the second force term--
I am missing an m here.
01:08:48.560 --> 01:08:50.880
r m theta double dot
theta hat, that's
01:08:50.880 --> 01:08:57.140
the force that it takes to--
if the thing were accelerating,
01:08:57.140 --> 01:09:01.370
just to accelerate that ball,
theta double dot takes a force.
01:09:01.370 --> 01:09:04.420
You do these cross products.
r cross theta hat give you k.
01:09:04.420 --> 01:09:06.380
R cross theta hat give you k.
01:09:06.380 --> 01:09:09.200
It all comes out in
the right directions.
01:09:09.200 --> 01:09:11.380
And now to do the
problem we had,
01:09:11.380 --> 01:09:13.229
we said, oh yeah,
let this one be 0.
01:09:13.229 --> 01:09:14.590
So we'll let this term go to 0.
01:09:17.220 --> 01:09:18.590
We're left with a single term.
01:10:14.430 --> 01:10:16.070
And if we plugged
in some numbers--
01:10:16.070 --> 01:10:17.670
let's just see how
this works out.
01:10:17.670 --> 01:10:19.610
And this we know is just
r cross the Coriolis.
01:10:27.570 --> 01:10:33.310
Well, let's let m-- if
you let m be 100 kilograms
01:10:33.310 --> 01:10:44.580
and r be 5 meters
and r dot, 0.4 meters
01:10:44.580 --> 01:10:48.380
per second-- they're
all perfectly reasonable
01:10:48.380 --> 01:10:50.320
dimensions.
01:10:50.320 --> 01:10:59.020
And theta dot equals
3 radians per second.
01:10:59.020 --> 01:11:02.664
So 2 pi radians per second means
it goes around once a second.
01:11:02.664 --> 01:11:04.830
So this is a little less
than half a rev per second.
01:11:04.830 --> 01:11:05.442
Yeah?
01:11:05.442 --> 01:11:07.233
AUDIENCE: Is this still
[INAUDIBLE] though,
01:11:07.233 --> 01:11:08.420
or does it--
01:11:08.420 --> 01:11:10.310
PROFESSOR: Probably.
01:11:10.310 --> 01:11:11.350
Yeah.
01:11:11.350 --> 01:11:12.650
Good catch!
01:11:12.650 --> 01:11:15.870
Got to have on r in it because
that r comes from here, right?
01:11:15.870 --> 01:11:18.180
So I had the Coriolis
force written down,
01:11:18.180 --> 01:11:21.660
but not the r you have
to multiply it by.
01:11:21.660 --> 01:11:24.020
So now if you plug in
all of these numbers,
01:11:24.020 --> 01:11:26.510
let's see if it's a
ride you could survive.
01:11:34.220 --> 01:11:38.069
I actually computed the
Coriolis force first.
01:11:38.069 --> 01:11:40.360
I think it's just interesting
to get a physical feeling
01:11:40.360 --> 01:11:43.570
for how-- whether or not
you can feel these forces.
01:11:43.570 --> 01:11:45.910
So the Coriolis force
is just everything
01:11:45.910 --> 01:11:49.860
but the r, 2m r dot theta dot.
01:11:49.860 --> 01:11:56.540
So if you calculate that,
you get 240 newtons.
01:11:56.540 --> 01:12:00.262
And r times that, the torque.
01:12:03.650 --> 01:12:16.520
5 times that, about
1,200 Newton meters.
01:12:16.520 --> 01:12:21.960
And the acceleration, how
do we get acceleration?
01:12:21.960 --> 01:12:26.650
Well, f cor, Coriolis
force, is some mass
01:12:26.650 --> 01:12:27.930
times an acceleration.
01:12:27.930 --> 01:12:32.320
So we can solve, from
that 240 newtons,
01:12:32.320 --> 01:12:34.040
the acceleration of the system.
01:12:34.040 --> 01:12:45.500
Our acceleration of B with
respect to o, 240 newtons
01:12:45.500 --> 01:12:56.890
divided by 100 kilograms, 2.4
meters per second squared.
01:12:56.890 --> 01:13:00.134
And what's that in g's.
01:13:00.134 --> 01:13:01.070
AUDIENCE: About 1/4.
01:13:01.070 --> 01:13:03.082
PROFESSOR: Yeah, about
a quarter of a g.
01:13:06.250 --> 01:13:08.230
And again, would you feel that?
01:13:08.230 --> 01:13:12.080
So now you riding in the bucket.
01:13:12.080 --> 01:13:13.730
Let's say it's just
spinning around.
01:13:13.730 --> 01:13:15.760
The arm is not going
in and out at all.
01:13:15.760 --> 01:13:16.790
So you're riding in it.
01:13:16.790 --> 01:13:19.840
What force would you feel?
01:13:19.840 --> 01:13:24.110
Would you feel any forces
pushing you into your seat?
01:13:24.110 --> 01:13:26.150
Constant speed, r dot's 0.
01:13:36.810 --> 01:13:38.680
Somebody out there,
would you feel any force
01:13:38.680 --> 01:13:41.120
if you're going around and
around in this thing sitting
01:13:41.120 --> 01:13:41.680
in a seat?
01:13:41.680 --> 01:13:42.010
AUDIENCE: Yes.
01:13:42.010 --> 01:13:42.593
PROFESSOR: OK.
01:13:42.593 --> 01:13:44.000
What's it come from?
01:13:44.000 --> 01:13:44.310
AUDIENCE: Centripetal
acceleration.
01:13:44.310 --> 01:13:45.813
PROFESSOR: Centripetal
acceleration.
01:13:48.820 --> 01:13:51.264
And that is in what direction?
01:13:51.264 --> 01:13:52.180
AUDIENCE: It's inward.
01:13:52.180 --> 01:13:53.638
PROFESSOR:
Accelerations is inward.
01:13:53.638 --> 01:13:55.560
What would you actually feel?
01:13:55.560 --> 01:13:59.350
You'd feel-- if this
seat could swing out
01:13:59.350 --> 01:14:02.090
so your facing in, you'd feel
like you're being thrown out
01:14:02.090 --> 01:14:04.800
in your seat, right?
01:14:04.800 --> 01:14:07.180
Because you have to
have a force on you
01:14:07.180 --> 01:14:09.460
to make that
acceleration happen.
01:14:09.460 --> 01:14:10.665
Acceleration is inward.
01:14:10.665 --> 01:14:13.360
The force had better
be pushing you inward.
01:14:13.360 --> 01:14:16.460
So it's pushing on your back if
you're look in, going around.
01:14:16.460 --> 01:14:22.750
So that's the-- I've
forgotten the term for it.
01:14:22.750 --> 01:14:26.020
But put the astronauts
in a centrifuge,
01:14:26.020 --> 01:14:28.710
spin around to see if
they can take high g's.
01:14:28.710 --> 01:14:33.080
That's the inward
high g acceleration
01:14:33.080 --> 01:14:35.200
due to the centripetal
acceleration,
01:14:35.200 --> 01:14:36.800
makes you go in a circle.
01:14:36.800 --> 01:14:40.520
But now not only are you going
to feel that, but now you start
01:14:40.520 --> 01:14:42.430
changing the length of the arm.
01:14:42.430 --> 01:14:46.010
And if you change r dot
and make it positive
01:14:46.010 --> 01:14:47.820
so the arm is getting
longer, and we made
01:14:47.820 --> 01:14:49.340
this 0.4 meters per second.
01:14:49.340 --> 01:14:51.350
So it's moving out
about like that.
01:14:51.350 --> 01:14:53.190
It's not real fast.
01:14:53.190 --> 01:14:56.660
But it's going to create a
quarter of a g acceleration
01:14:56.660 --> 01:14:57.160
on you.
01:14:57.160 --> 01:15:00.325
And if r dot's positive, which
direction is that acceleration?
01:15:03.500 --> 01:15:04.190
I was careful.
01:15:04.190 --> 01:15:05.670
We walked all the
way through this.
01:15:05.670 --> 01:15:09.480
The acceleration is positive,
but in what direction?
01:15:09.480 --> 01:15:11.292
It's the Coriolis.
01:15:11.292 --> 01:15:12.750
AUDIENCE: Theta hat.
01:15:12.750 --> 01:15:14.760
PROFESSOR: Theta hat,
positive theta hat,
01:15:14.760 --> 01:15:16.800
so it's in the direction
of increasing theta.
01:15:16.800 --> 01:15:20.460
So this one is
perpendicular to the arm.
01:15:20.460 --> 01:15:24.190
And this is the--
one of my ping pong
01:15:24.190 --> 01:15:28.510
balls in here, the force that
actually causes it to speed up
01:15:28.510 --> 01:15:30.940
is partly Coriolis and
partly or Eulerian.
01:15:30.940 --> 01:15:34.270
If I can make this go at
constant speed, the force that
01:15:34.270 --> 01:15:38.050
drives that-- the force
that actually speeds that up
01:15:38.050 --> 01:15:41.090
as they're going out is the
Coriolis force, the normal one.
01:15:41.090 --> 01:15:42.730
So if you were on
this ride, you'd
01:15:42.730 --> 01:15:45.760
be feel about a quarter of a
g perpendicular to the arm.
01:15:45.760 --> 01:15:47.780
You'd be filling
another force inward
01:15:47.780 --> 01:15:49.837
that's the
centripetal-- caused by
01:15:49.837 --> 01:15:51.045
the centripetal acceleration.
01:15:51.045 --> 01:15:52.510
So you'd be feeling
both of them.
01:15:52.510 --> 01:15:54.176
How big is the
centripetal acceleration?
01:15:56.810 --> 01:16:02.274
Like r omega squared, right?
01:16:02.274 --> 01:16:03.030
AUDIENCE: Right.
01:16:03.030 --> 01:16:04.380
PROFESSOR: r is 5.
01:16:04.380 --> 01:16:06.065
Omega is 3.
01:16:06.065 --> 01:16:18.007
3 Squared is 9, times 5-- 9
times 5, 45, divided by 10,
01:16:18.007 --> 01:16:18.590
4 and 1/2 g's.
01:16:22.740 --> 01:16:24.815
You wouldn't notice
the Coriolis very much.
01:16:24.815 --> 01:16:29.470
It would be a tough ride
on that side of things.
01:16:32.980 --> 01:16:37.592
Now, we've done that-- you've
seen a simple application,
01:16:37.592 --> 01:16:39.050
pretty straightforward
application,
01:16:39.050 --> 01:16:44.350
of using time derivative
of angular momentum
01:16:44.350 --> 01:16:46.640
to calculate torques.
01:16:46.640 --> 01:16:49.710
So on homework, again,
there's a couple problems
01:16:49.710 --> 01:16:52.400
very similar to the one I
just did, things going around,
01:16:52.400 --> 01:16:54.610
circus rides, that
kind of stuff.
01:16:54.610 --> 01:16:58.850
So have a good weekend.
01:16:58.850 --> 01:17:01.460
See you on Tuesday.