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PROFESSOR: So we were doing
velocities and accelerations.
00:00:23.840 --> 00:00:33.070
We came up with-- I guess I
ought to continue to remind us.
00:00:33.070 --> 00:00:36.140
We're talking about
velocities and accelerations
00:00:36.140 --> 00:00:40.630
of a point with respect to
another point, to which we've
00:00:40.630 --> 00:00:44.480
attached-- I'll make
this point here--
00:00:44.480 --> 00:00:49.760
a reference frame, x
prime, y prime, z prime.
00:00:49.760 --> 00:00:52.750
And there's a vector that
goes between these two
00:00:52.750 --> 00:00:56.420
and a vector that goes there
so that we can say r of B
00:00:56.420 --> 00:01:01.660
with respect to O, my
inertial frame, is r of A
00:01:01.660 --> 00:01:06.050
with respect to O plus
r of B with respect
00:01:06.050 --> 00:01:07.550
to A. These are all vectors.
00:01:07.550 --> 00:01:10.070
And then from these, we derive
velocities and acceleration
00:01:10.070 --> 00:01:10.570
formulas.
00:01:10.570 --> 00:01:15.040
And so we've come up with a
couple of very handy formulas.
00:01:15.040 --> 00:01:21.920
The velocity formula, velocity
of B with respect to O,
00:01:21.920 --> 00:01:23.920
is the velocity
of A with respect
00:01:23.920 --> 00:01:31.247
to O plus you have to take
a time derivative of this.
00:01:31.247 --> 00:01:33.580
And so I'm going to give you
just the general expression
00:01:33.580 --> 00:01:35.870
here just as a brief reminder.
00:01:35.870 --> 00:01:39.760
It's the derivative
of B with respect
00:01:39.760 --> 00:01:46.490
to A as seen in the Axyz frame
plus omega with respect to O
00:01:46.490 --> 00:01:49.279
cross rBA.
00:01:49.279 --> 00:01:50.195
These are all vectors.
00:01:52.760 --> 00:01:55.030
So that's the velocity
formula, remember?
00:01:55.030 --> 00:01:57.640
This is if you're in
the frame rotating
00:01:57.640 --> 00:01:59.650
and translating with
it, this is the change
00:01:59.650 --> 00:02:02.860
of length of the vector.
00:02:02.860 --> 00:02:05.430
And this, then, is the
contribution to the velocity
00:02:05.430 --> 00:02:10.199
that you see in the fixed frame
that comes from the rotation.
00:02:10.199 --> 00:02:12.820
And we then got into
polar coordinates.
00:02:12.820 --> 00:02:15.750
And we found out that if
you use polar coordinates,
00:02:15.750 --> 00:02:20.080
then you can express
this as the velocity of A
00:02:20.080 --> 00:02:27.780
with respect to O
plus r dot r hat.
00:02:27.780 --> 00:02:30.530
And I should really say
cylindrical coordinates,
00:02:30.530 --> 00:02:43.250
z dot k hat plus r
theta dot theta hat.
00:02:43.250 --> 00:02:45.070
So that's exactly
the same thing.
00:02:45.070 --> 00:02:47.610
This is full 3D vector notation.
00:02:47.610 --> 00:02:50.680
This is a special case of
a coordinate system which
00:02:50.680 --> 00:02:53.730
we call polar coordinates.
00:02:53.730 --> 00:02:58.010
And we came up with another
formula for accelerations,
00:02:58.010 --> 00:03:04.030
the full 3D vector version
of that, A with respect
00:03:04.030 --> 00:03:10.610
to O plus-- this is
the acceleration of B
00:03:10.610 --> 00:03:14.620
with respect to A, but as
seen in the Axyz frame.
00:03:19.000 --> 00:03:27.320
2 omega cross-- and
this is v, the velocity,
00:03:27.320 --> 00:03:30.700
as seen in the xyz frame.
00:03:30.700 --> 00:03:34.100
So these things, this is no
contribution from acceleration.
00:03:34.100 --> 00:03:36.000
This is no contribution
from acceleration.
00:03:39.020 --> 00:03:47.810
Plus omega naught cross
rBA plus omega cross
00:03:47.810 --> 00:03:55.750
omega cross rBA-- all vectors.
00:03:59.190 --> 00:04:03.130
This is a movement of the
frame, the acceleration of it
00:04:03.130 --> 00:04:04.010
with respect to this.
00:04:04.010 --> 00:04:06.330
This is pure translation.
00:04:06.330 --> 00:04:10.800
This is the acceleration
of the dog on the merry go
00:04:10.800 --> 00:04:14.535
round with respect to the center
of the coordinate system there.
00:04:14.535 --> 00:04:17.170
It does not involve rotation.
00:04:17.170 --> 00:04:21.050
This is this term
that we call Coriolis.
00:04:21.050 --> 00:04:22.650
This is a term we call Euler.
00:04:22.650 --> 00:04:26.080
This is the angular speed up
of the system, the angular
00:04:26.080 --> 00:04:27.280
acceleration.
00:04:27.280 --> 00:04:28.450
And this is centripetal.
00:04:32.100 --> 00:04:33.900
And if you do these,
if you want to go
00:04:33.900 --> 00:04:37.240
to cylindrical coordinates--
and what we're going to do next
00:04:37.240 --> 00:04:39.610
is just do some
applications here.
00:04:39.610 --> 00:04:42.490
The acceleration of B with
respect to O and cylindrical
00:04:42.490 --> 00:04:46.670
coordinates-- it's
a translation piece.
00:04:46.670 --> 00:04:49.970
Plus now in cylindrical,
it's useful to group these.
00:04:56.550 --> 00:05:02.650
So I'm going to put together
this term and this term.
00:05:02.650 --> 00:05:05.024
Because they're both
in the r hat direction.
00:05:05.024 --> 00:05:06.940
And I'm going to put
together these two terms.
00:05:06.940 --> 00:05:11.090
Because they're both in
the theta hat direction.
00:05:11.090 --> 00:05:16.990
And I need a z double dot
k, because it's cylindrical.
00:05:16.990 --> 00:05:20.000
And then I have the
theta hat piece,
00:05:20.000 --> 00:05:29.180
r theta double dot plus 2r
dot theta dot theta hat.
00:05:29.180 --> 00:05:32.880
This is the same thing as this.
00:05:32.880 --> 00:05:37.587
Except this is expressed
in cylindrical coordinates.
00:05:37.587 --> 00:05:39.420
And cylindrical coordinates
are particularly
00:05:39.420 --> 00:05:42.740
good for doing the kind of
problems that are mostly
00:05:42.740 --> 00:05:45.780
done at the level of this
course, which we call planar
00:05:45.780 --> 00:05:50.190
motion problems, confined
to an x, y plane,
00:05:50.190 --> 00:05:54.650
and confined to single
axis rotation in z.
00:05:54.650 --> 00:05:57.620
This is a coordinate system
that's ideally suited
00:05:57.620 --> 00:05:58.810
to do problems like that.
00:05:58.810 --> 00:06:00.410
And that's why we use it.
00:06:00.410 --> 00:06:04.550
So now let's do some examples.
00:06:04.550 --> 00:06:11.550
This is really
quite a powerful--
00:06:11.550 --> 00:06:13.260
now that you have
these two equations,
00:06:13.260 --> 00:06:15.625
you can do a lot of
kinematics and dynamics.
00:06:25.270 --> 00:06:31.010
So we started last time
right at the very end.
00:06:31.010 --> 00:06:36.100
I said, OK, let's do this
problem really quickly, right?
00:06:36.100 --> 00:06:39.650
Constant rotation
rate, constant radius,
00:06:39.650 --> 00:06:42.030
no angular
acceleration, no change
00:06:42.030 --> 00:06:44.340
in length of the thing--
pretty simple problem.
00:06:44.340 --> 00:06:46.630
And we zipped it
off really fast.
00:06:46.630 --> 00:06:50.870
And I wanted to start there,
do that really quickly.
00:06:50.870 --> 00:06:53.760
So this is my case one.
00:06:53.760 --> 00:06:55.495
It's my ball on the string.
00:07:01.260 --> 00:07:08.060
Here's point A. Here's point
B. Here's the theta hat
00:07:08.060 --> 00:07:12.930
direction, r hat direction.
00:07:12.930 --> 00:07:17.880
And it has some
constant length R here.
00:07:17.880 --> 00:07:24.140
So r dot r double dot are 0.
00:07:24.140 --> 00:07:33.370
There's no also z, z dot, z
double dot, no z motion at all.
00:07:33.370 --> 00:07:36.090
Those are all 0.
00:07:36.090 --> 00:07:39.980
Theta dot is a constant.
00:07:39.980 --> 00:07:45.460
I'll call it cap omega in the k
hat direction, right hand rule.
00:07:45.460 --> 00:07:50.080
Theta double dot is 0.
00:07:50.080 --> 00:07:53.177
So the easy way to
use these formulas
00:07:53.177 --> 00:07:55.010
is you just start
knocking out all the terms
00:07:55.010 --> 00:07:56.490
that you don't need.
00:07:56.490 --> 00:08:00.800
But let's, just to give you
a little quick review of how
00:08:00.800 --> 00:08:05.560
to use these things,
use the vector full
00:08:05.560 --> 00:08:08.770
3D version of the velocity
equation for a second.
00:08:08.770 --> 00:08:11.930
The full 3D version
says the velocity of A
00:08:11.930 --> 00:08:16.250
with respect to O, what's
that in this problem?
00:08:16.250 --> 00:08:19.630
That's the translating term.
00:08:19.630 --> 00:08:22.060
So I was standing still,
not going anywhere.
00:08:22.060 --> 00:08:25.070
But if I were walking along,
I'd still spin that thing.
00:08:25.070 --> 00:08:28.070
OK, so this problem,
this term is 0.
00:08:28.070 --> 00:08:33.409
This problem, r
dot-- let's go here.
00:08:33.409 --> 00:08:35.580
This is the rate of
change of the length
00:08:35.580 --> 00:08:40.321
of the string in the coordinate
system of the string walking
00:08:40.321 --> 00:08:40.820
along.
00:08:40.820 --> 00:08:43.110
So what's r dot?
00:08:43.110 --> 00:08:45.180
OK, so that term is 0.
00:08:45.180 --> 00:08:51.170
And omega cross rBA-- well, r is
in the R hat, capital R R hat,
00:08:51.170 --> 00:08:56.030
cross with omega k.
00:08:56.030 --> 00:09:01.883
k cross R is-- k cross of R hat?
00:09:01.883 --> 00:09:02.630
STUDENT: Theta.
00:09:02.630 --> 00:09:03.505
PROFESSOR: Theta hat.
00:09:03.505 --> 00:09:07.230
So we get our familiar--
this first term is 0.
00:09:07.230 --> 00:09:08.860
The second term is 0.
00:09:08.860 --> 00:09:12.830
The third term, we
just get our R omega
00:09:12.830 --> 00:09:14.920
in the theta hat direction.
00:09:14.920 --> 00:09:16.050
We know that to be true.
00:09:16.050 --> 00:09:17.633
The point I'm just
trying to make here
00:09:17.633 --> 00:09:21.530
is you can always just fall
back and use the vector formula,
00:09:21.530 --> 00:09:23.230
full 3D formulas of both.
00:09:23.230 --> 00:09:26.370
Just plug it in, and
everything will just drop out.
00:09:26.370 --> 00:09:30.270
You can also then go to the
cylindrical coordinate terms
00:09:30.270 --> 00:09:32.750
when you happen to
have it all expressed
00:09:32.750 --> 00:09:37.710
in coordinates of that kind
to make it simpler for you.
00:09:37.710 --> 00:09:43.400
OK, so now let's quickly
then do the acceleration of B
00:09:43.400 --> 00:09:49.300
with respect to O. And let's
use the formulation already
00:09:49.300 --> 00:09:50.890
in cylindrical coordinates.
00:09:50.890 --> 00:09:53.760
What's A with respect to O?
00:09:53.760 --> 00:09:54.720
0.
00:09:54.720 --> 00:09:58.200
What's r double dot?
00:09:58.200 --> 00:10:01.470
How about the r theta dot
squared term, 0 or not?
00:10:01.470 --> 00:10:02.290
Nope.
00:10:02.290 --> 00:10:04.243
Let's go on-- z double dot?
00:10:04.243 --> 00:10:04.861
STUDENT: 0.
00:10:04.861 --> 00:10:06.110
PROFESSOR: r theta double dot?
00:10:06.110 --> 00:10:07.082
STUDENT: 0.
00:10:07.082 --> 00:10:08.311
PROFESSOR: r dot theta dot?
00:10:08.311 --> 00:10:08.810
STUDENT: 0
00:10:08.810 --> 00:10:10.540
PROFESSOR: OK, we only
end up with one term.
00:10:10.540 --> 00:10:12.770
Actually, I'm going to keep
this one for a second-- A
00:10:12.770 --> 00:10:16.030
with respect to O minus.
00:10:16.030 --> 00:10:25.360
And we came up with our R
theta dot squared r hat term.
00:10:25.360 --> 00:10:28.390
I leave this in here because
I actually don't have to say.
00:10:28.390 --> 00:10:33.000
If I wanted now to
do this problem,
00:10:33.000 --> 00:10:35.880
if I asked you to do a
problem where I'm doing this,
00:10:35.880 --> 00:10:40.039
and I start accelerating,
do you know how to do it?
00:10:40.039 --> 00:10:41.705
There's the answer--
still there, right?
00:10:46.437 --> 00:11:06.155
All right, let's do a
quick free body diagram.
00:11:11.660 --> 00:11:18.920
Here's our mass, my master
coordinate system out here.
00:11:18.920 --> 00:11:23.735
Let's draw-- let's say it's
right here at 90 degrees.
00:11:26.570 --> 00:11:29.300
What are the external forces
on the mass in this problem?
00:11:33.196 --> 00:11:34.660
STUDENT: [INAUDIBLE].
00:11:34.660 --> 00:11:37.120
PROFESSOR: So there's
tension in the string, right?
00:11:37.120 --> 00:11:41.700
OK, so now this is a really
trivially simple problem.
00:11:41.700 --> 00:11:45.660
So the emphasize here
is on the concept.
00:11:45.660 --> 00:11:50.100
So now when you're asked to come
up with an equation of motion
00:11:50.100 --> 00:11:55.520
or compute the forces on a
mass, use Newton's second law,
00:11:55.520 --> 00:11:58.040
F equals mass
times acceleration.
00:11:58.040 --> 00:12:01.870
You now have the complete
3D vector formulation
00:12:01.870 --> 00:12:08.290
for acceleration of a particle
in a translating rotating
00:12:08.290 --> 00:12:09.870
coordinate system.
00:12:09.870 --> 00:12:12.780
That's all you need to
compute accelerations for lots
00:12:12.780 --> 00:12:14.850
and lots of difficult problems.
00:12:14.850 --> 00:12:18.670
And so if you can write
down the acceleration,
00:12:18.670 --> 00:12:23.960
you say it's equal to what?
00:12:23.960 --> 00:12:26.870
Mass-- if you multiply mass
times that acceleration,
00:12:26.870 --> 00:12:28.269
what's that equal to?
00:12:28.269 --> 00:12:29.060
STUDENT: The force.
00:12:29.060 --> 00:12:31.393
PROFESSOR: The forces that
must be acting on the system.
00:12:31.393 --> 00:12:32.860
And that's the point here.
00:12:32.860 --> 00:12:37.090
So now I want to know the forces
on the system, the summation
00:12:37.090 --> 00:12:38.650
of the external forces.
00:12:38.650 --> 00:12:39.940
And then these are vectors.
00:12:39.940 --> 00:12:42.100
And you can do them
component by component.
00:12:42.100 --> 00:12:47.010
Some of the external forces
in the r hat direction
00:12:47.010 --> 00:12:49.810
must be equal to the
mass-- in this case, just
00:12:49.810 --> 00:12:56.360
a particle-- times the
acceleration of that particle.
00:12:56.360 --> 00:12:58.050
And in this problem,
then that would
00:12:58.050 --> 00:13:01.670
be the mass times the
acceleration of A with respect
00:13:01.670 --> 00:13:10.540
to O minus R theta
dot squared r hat.
00:13:10.540 --> 00:13:15.690
And this would have to be the
r hat component of this thing.
00:13:15.690 --> 00:13:17.990
I said I just want
the R component.
00:13:17.990 --> 00:13:20.180
I'd have to figure out
if I was running along,
00:13:20.180 --> 00:13:23.110
if I had it in the same
direction as r hat.
00:13:23.110 --> 00:13:25.490
What part of that acceleration
is in that direction?
00:13:25.490 --> 00:13:26.700
That would come here.
00:13:26.700 --> 00:13:29.040
If there's 0, you
just make it 0.
00:13:29.040 --> 00:13:34.610
So let's just let
the acceleration of A
00:13:34.610 --> 00:13:36.830
with respect to O be 0.
00:13:36.830 --> 00:13:41.690
Then that says the sum of
the forces in the r direction
00:13:41.690 --> 00:13:49.120
is equal to the mass minus
mR theta dot squared.
00:13:49.120 --> 00:13:52.210
And if we were to draw
a free body diagram,
00:13:52.210 --> 00:13:53.730
we would find out
that, ahh, there
00:13:53.730 --> 00:13:55.980
must be a tension on
the string pulling
00:13:55.980 --> 00:14:00.120
in on the mass
sufficient to give it
00:14:00.120 --> 00:14:03.000
the acceleration
that you've computed.
00:14:03.000 --> 00:14:04.460
So every problem,
when you're asked
00:14:04.460 --> 00:14:06.790
to compute the force,
or the next step up,
00:14:06.790 --> 00:14:09.250
find the equation of motion,
the equation of motion
00:14:09.250 --> 00:14:13.115
is just writing this thing out.
00:14:20.250 --> 00:14:30.600
OK, now I want to move on to
a more interesting problem.
00:14:55.510 --> 00:15:02.260
All right, I'm going
to do this problem.
00:15:02.260 --> 00:15:05.540
So it's a hollow tube.
00:15:05.540 --> 00:15:09.230
And we're going to look at
things like, I put a ping pong
00:15:09.230 --> 00:15:11.010
ball in it.
00:15:11.010 --> 00:15:15.345
And if I swing this tube
around, the ping pong ball
00:15:15.345 --> 00:15:16.220
is going to come out.
00:15:23.290 --> 00:15:25.710
OK, there must be some
forces on that thing
00:15:25.710 --> 00:15:27.405
to cause it to come out.
00:15:27.405 --> 00:15:29.640
There must be some
accelerations on them.
00:15:29.640 --> 00:15:31.960
And so I could
conceivably have an R,
00:15:31.960 --> 00:15:35.070
a theta double dot acceleration.
00:15:35.070 --> 00:15:39.130
It certainly is going to have
theta dot rotation rates.
00:15:39.130 --> 00:15:40.520
The ball is allowed to move.
00:15:40.520 --> 00:15:44.375
So there can be nonzero
r dots, r double dots--
00:15:44.375 --> 00:15:47.390
a lot going on inside of
this simple little tube.
00:15:47.390 --> 00:15:49.500
So that's what I
want to figure out.
00:15:49.500 --> 00:15:52.990
Let's see if we can come up
with a model for this problem.
00:16:08.090 --> 00:16:09.620
So here's my z-axis.
00:16:09.620 --> 00:16:14.980
I have a rotation around it,
some theta dot k hat direction.
00:16:14.980 --> 00:16:17.950
Here's my tube.
00:16:17.950 --> 00:16:20.020
It's rotating around.
00:16:20.020 --> 00:16:22.980
So this is sort of a side
view, your view of the tube.
00:16:22.980 --> 00:16:26.940
Here's that ping
pong ball in there.
00:16:26.940 --> 00:16:35.110
And I'm going to
idealize that ping pong
00:16:35.110 --> 00:16:38.970
ball for a minute, a little
more general problem.
00:16:38.970 --> 00:16:42.570
Let's say I have kind of a
nut on this thing, a disk,
00:16:42.570 --> 00:16:45.280
something hanging
onto the outside.
00:16:45.280 --> 00:16:49.380
And I can control
the rate, the speed
00:16:49.380 --> 00:16:51.650
at which this thing goes out.
00:16:51.650 --> 00:16:54.330
The ping pong
ball, this is going
00:16:54.330 --> 00:16:55.610
to be an application of this.
00:16:55.610 --> 00:16:57.735
But I want to be able to
do several other versions,
00:16:57.735 --> 00:17:02.120
like make the speed
constant for a second.
00:17:02.120 --> 00:17:06.690
OK, and looking
down on this thing,
00:17:06.690 --> 00:17:13.000
top view, here's
our inertial frame,
00:17:13.000 --> 00:17:14.579
maybe out here like this.
00:17:14.579 --> 00:17:16.800
Here's my mass.
00:17:16.800 --> 00:17:20.000
So in polar coordinates,
here's your theta.
00:17:20.000 --> 00:17:24.192
The r is this.
00:17:24.192 --> 00:17:27.150
This is your r hat.
00:17:27.150 --> 00:17:28.810
This is your theta
hat directions.
00:17:33.750 --> 00:17:39.650
I'm going to let the velocity
of A with respect to O be 0,
00:17:39.650 --> 00:17:43.380
so there's no translational
of this system.
00:17:43.380 --> 00:17:49.510
And z, z dot, z double
dot, those are all 0.
00:17:49.510 --> 00:17:53.280
So nothing's happening
in the z direction.
00:17:53.280 --> 00:17:57.090
So I want to first
compute the velocity of B
00:17:57.090 --> 00:17:59.740
with respect to O.
00:17:59.740 --> 00:18:04.970
And you ought to be able to
do that sort of by inspection.
00:18:04.970 --> 00:18:10.830
It comes only from-- oh,
I haven't told you enough.
00:18:10.830 --> 00:18:12.790
What do I want to make
happen in this problem?
00:18:12.790 --> 00:18:20.993
I want to let r dot--
it's going to be some vr,
00:18:20.993 --> 00:18:21.743
and it's constant.
00:18:24.660 --> 00:18:27.259
I'm not going to go quite to
my ping pong shooter here yet.
00:18:27.259 --> 00:18:29.300
I'm going to do a slightly
simpler problem first.
00:18:29.300 --> 00:18:30.260
So this is a constant.
00:18:30.260 --> 00:18:32.980
That means r double dot is 0.
00:18:32.980 --> 00:18:34.860
So this thing is
just-- let's say
00:18:34.860 --> 00:18:37.490
you had threads on this
thing, and it's a screw,
00:18:37.490 --> 00:18:41.950
and it's just moving its
way out at a constant rate.
00:18:41.950 --> 00:18:47.980
And I'm going to have constant
angular, so theta dot.
00:18:47.980 --> 00:18:51.350
I'll call that cap omega k hat.
00:18:51.350 --> 00:18:55.450
So the angular rate
is also constant.
00:18:55.450 --> 00:18:58.070
All right, if that's the
case, can you tell me,
00:18:58.070 --> 00:19:04.030
what's v of B with respect
to my fixed frame O?
00:19:07.800 --> 00:19:11.480
Well, any time you're not sure,
you go back to this formula,
00:19:11.480 --> 00:19:12.560
throw out terms.
00:19:12.560 --> 00:19:14.640
There's no z dot term.
00:19:14.640 --> 00:19:15.670
This is constant.
00:19:15.670 --> 00:19:18.390
This is 0, 0, 0.
00:19:18.390 --> 00:19:19.850
You have this term.
00:19:19.850 --> 00:19:22.560
It's some vr in the
r hat direction.
00:19:22.560 --> 00:19:26.520
You have this term, wherever it
happens to be in the theta hat
00:19:26.520 --> 00:19:27.020
direction.
00:19:39.372 --> 00:19:43.950
r dot in the r hat
direction plus r theta
00:19:43.950 --> 00:19:46.450
dot in the theta
hat direction-- OK,
00:19:46.450 --> 00:19:50.850
we need acceleration
next, B with respect to O.
00:19:50.850 --> 00:19:53.110
And now you can crank
through the terms again.
00:19:53.110 --> 00:19:57.290
This time, the first term is 0.
00:19:57.290 --> 00:19:59.880
The second term is 0,
because it's constant.
00:19:59.880 --> 00:20:02.560
The third term is
definitely not 0.
00:20:02.560 --> 00:20:03.660
The fourth term is 0.
00:20:03.660 --> 00:20:04.280
Fifth term?
00:20:08.040 --> 00:20:09.270
0.
00:20:09.270 --> 00:20:11.140
This term?
00:20:11.140 --> 00:20:13.020
Not 0.
00:20:13.020 --> 00:20:14.160
Let me just write them.
00:20:18.830 --> 00:20:23.600
So you get a minus r theta
dot squared r hat-- that's
00:20:23.600 --> 00:20:30.904
the radial direction term--
plus 2r dot theta dot theta hat.
00:20:30.904 --> 00:20:31.945
That's the accelerations.
00:20:36.000 --> 00:20:38.530
So you have an acceleration
now in the r hat direction
00:20:38.530 --> 00:20:40.640
and in the theta hat direction.
00:20:40.640 --> 00:20:42.610
If there's acceleration
to those directions,
00:20:42.610 --> 00:20:44.665
there must be forces.
00:20:47.680 --> 00:20:51.720
Newton's second law now
says, again, the force
00:20:51.720 --> 00:20:55.200
is the mass times acceleration.
00:20:55.200 --> 00:20:56.930
And this is a vector.
00:20:56.930 --> 00:20:58.390
This is a vector.
00:20:58.390 --> 00:20:59.470
It has two components.
00:20:59.470 --> 00:21:02.585
And the nice thing about these
Newton's laws and vectors
00:21:02.585 --> 00:21:05.180
is you can break the problems
down into their vector
00:21:05.180 --> 00:21:07.070
components and treat
the r direction
00:21:07.070 --> 00:21:09.300
as one equation of
motion, and the theta
00:21:09.300 --> 00:21:11.310
direction as a separate one.
00:21:11.310 --> 00:21:15.240
So we might want to draw
a free body diagram.
00:21:15.240 --> 00:21:20.260
Here's this block
working its way out.
00:21:20.260 --> 00:21:22.660
We know that there's
probably some axial force.
00:21:22.660 --> 00:21:24.700
I'm just going to
call it T. And there's
00:21:24.700 --> 00:21:27.700
some other unknown force here
in the theta hat direction.
00:21:27.700 --> 00:21:30.704
So this is my theta hat.
00:21:30.704 --> 00:21:32.870
I've drawn this just
intentionally in the positive r
00:21:32.870 --> 00:21:33.490
hat direction.
00:21:33.490 --> 00:21:35.210
The sign that comes
out will tell us
00:21:35.210 --> 00:21:38.490
which direction it really
is if you're not certain.
00:21:38.490 --> 00:21:42.250
Just draw it positive, in
the positive r hat direction.
00:21:42.250 --> 00:21:45.520
And that's your
free body diagram.
00:21:45.520 --> 00:21:48.180
If I wanted to put gravity
in there, I might have.
00:21:48.180 --> 00:21:51.200
But we're doing this in
the horizontal plane.
00:21:51.200 --> 00:21:52.560
Gravity is in and out this way.
00:21:52.560 --> 00:21:53.670
It's in the z direction.
00:21:53.670 --> 00:21:57.740
And we know it's constrained,
can't move. z double dot is 0.
00:21:57.740 --> 00:22:01.710
So there's certainly a support
force that picks up the weight.
00:22:01.710 --> 00:22:04.170
But this is in our
horizontal plane.
00:22:04.170 --> 00:22:06.170
There's your free body diagram.
00:22:06.170 --> 00:22:09.310
And we can write two equations
to solve for these things.
00:22:09.310 --> 00:22:16.050
So the sum of the forces in
the r hat direction is T.
00:22:16.050 --> 00:22:19.640
And that must be equal to the
mass times the acceleration
00:22:19.640 --> 00:22:27.690
in the r, minus mr theta dot
squared in the r hat direction.
00:22:27.690 --> 00:22:31.890
Sure enough, the tension
has to pull inwards
00:22:31.890 --> 00:22:33.820
in the minus r hat direction.
00:22:33.820 --> 00:22:39.115
And that's the full result.
00:22:39.115 --> 00:22:44.350
STUDENT: Where are the
T and F forces exactly?
00:22:44.350 --> 00:22:45.620
PROFESSOR: Where are they?
00:22:45.620 --> 00:22:53.450
OK, so I'm going to bring
the ball to the outside
00:22:53.450 --> 00:22:54.590
where you can see it.
00:22:54.590 --> 00:22:58.260
This thing is going in this
direction, horizontal plane, x,
00:22:58.260 --> 00:22:59.380
y plane.
00:22:59.380 --> 00:23:03.520
It's moving its fixed rate out.
00:23:03.520 --> 00:23:05.550
So its speed when
it's in here is r
00:23:05.550 --> 00:23:07.370
over 2 omega in that direction.
00:23:07.370 --> 00:23:10.730
And the speed when it's out here
is r omega in that direction.
00:23:10.730 --> 00:23:12.630
So clearly it's
picking up speed.
00:23:12.630 --> 00:23:17.320
If it's picking up speed, is
it picking up kinetic energy?
00:23:17.320 --> 00:23:19.200
Is there work being
done on it somehow
00:23:19.200 --> 00:23:20.570
to build up that energy?
00:23:20.570 --> 00:23:22.970
So there must be
some forces at play.
00:23:22.970 --> 00:23:26.090
So there's a normal
force from the wall
00:23:26.090 --> 00:23:29.670
of this thing pushing this
ball sideways to speed it up,
00:23:29.670 --> 00:23:30.250
for sure.
00:23:30.250 --> 00:23:31.650
That's one force.
00:23:31.650 --> 00:23:35.810
And the other force is because
I'm not allowing this thing
00:23:35.810 --> 00:23:38.180
just to go freely out.
00:23:38.180 --> 00:23:41.310
I'm constraining it
to constant speed out.
00:23:41.310 --> 00:23:43.900
It would really like to
go a lot faster than that.
00:23:43.900 --> 00:23:46.600
So what's holding it back?
00:23:46.600 --> 00:23:49.510
At any instance in time, it
has centripetal acceleration.
00:23:49.510 --> 00:23:53.290
And what's making
it go in the circle
00:23:53.290 --> 00:23:56.060
is a force that
is-- in this case,
00:23:56.060 --> 00:24:00.570
if that were a nut with threads,
in the threads are applying
00:24:00.570 --> 00:24:03.368
to the nut to keep
it from running away.
00:24:03.368 --> 00:24:05.360
STUDENT: In that
free body diagram,
00:24:05.360 --> 00:24:08.965
F, doesn't F act similar
between theta and r hat?
00:24:08.965 --> 00:24:10.340
And then there's
like a component
00:24:10.340 --> 00:24:13.830
of theta hat in there?
00:24:13.830 --> 00:24:15.700
PROFESSOR: Well,
I've broken down.
00:24:15.700 --> 00:24:18.080
I've chosen.
00:24:18.080 --> 00:24:21.200
The total force
acting on this thing
00:24:21.200 --> 00:24:23.890
is some combination of a
force in that direction
00:24:23.890 --> 00:24:25.760
and a combination in
the axial direction.
00:24:25.760 --> 00:24:31.810
So it has some net direction
that's neither this nor that.
00:24:31.810 --> 00:24:34.735
STUDENT: From that equation,
it looks like what you drew,
00:24:34.735 --> 00:24:39.640
F has an r hat component.
00:24:39.640 --> 00:24:42.830
PROFESSOR: So this
thing is rotating
00:24:42.830 --> 00:24:45.560
about some center over here.
00:24:45.560 --> 00:24:47.460
So this is theta dot.
00:24:47.460 --> 00:24:50.690
Theta is going in this
direction, theta dot.
00:24:50.690 --> 00:24:55.050
And we've chosen a coordinate
system that has unit vectors r
00:24:55.050 --> 00:24:56.510
hat and theta hat.
00:24:56.510 --> 00:24:57.560
And so it makes sense.
00:24:57.560 --> 00:24:59.840
We can express the
acceleration in terms
00:24:59.840 --> 00:25:01.100
of those two components.
00:25:01.100 --> 00:25:05.270
It makes sense to express the
forces in the same direction.
00:25:05.270 --> 00:25:08.620
So I've just arbitrarily said,
I have some unknown force
00:25:08.620 --> 00:25:09.690
that's in this direction.
00:25:09.690 --> 00:25:11.814
And I have another unknown
force in that direction.
00:25:14.340 --> 00:25:17.230
Then I'm saying, they
account for all forces
00:25:17.230 --> 00:25:19.770
in this direction,
whatever their source.
00:25:19.770 --> 00:25:23.000
That tells me that the
sum of the external forces
00:25:23.000 --> 00:25:29.090
in the theta hat direction in
this case is this unknown F.
00:25:29.090 --> 00:25:31.610
But I know from
Newton that that's
00:25:31.610 --> 00:25:33.930
got to be equal to the
mass times the acceleration
00:25:33.930 --> 00:25:35.380
in that direction.
00:25:35.380 --> 00:25:42.520
In this case, then, that is
2mr dot theta dot theta hat.
00:25:45.630 --> 00:25:50.250
So just from
applying the equation
00:25:50.250 --> 00:25:52.580
and applying
Newton's second law,
00:25:52.580 --> 00:25:56.359
I can find out what
that force must be.
00:25:56.359 --> 00:25:57.900
It would've been a
lot more work if I
00:25:57.900 --> 00:26:00.286
had drawn it in some arbitrary
in between direction.
00:26:00.286 --> 00:26:02.660
Because then I'd have to break
it down into its compounds
00:26:02.660 --> 00:26:03.690
to write this.
00:26:03.690 --> 00:26:07.060
So I've made it as easy
as possible for myself.
00:26:07.060 --> 00:26:08.850
So there's a force like this.
00:26:08.850 --> 00:26:10.840
And there is another
force like that.
00:26:10.840 --> 00:26:14.550
This one is caused by the
centripetal acceleration.
00:26:14.550 --> 00:26:19.820
Or this force causes the
centripetal acceleration.
00:26:19.820 --> 00:26:22.080
In order to make something
go in a circular path,
00:26:22.080 --> 00:26:24.940
you have to exert a force on it.
00:26:24.940 --> 00:26:26.900
That's the force.
00:26:26.900 --> 00:26:30.160
In order to accelerate
something angularly,
00:26:30.160 --> 00:26:31.500
you have to apply a force.
00:26:31.500 --> 00:26:33.760
That comes from the Euler term.
00:26:33.760 --> 00:26:36.830
And this is a curious term.
00:26:36.830 --> 00:26:38.250
This is the Coriolis term.
00:26:41.290 --> 00:26:42.580
So where does it come from?
00:26:45.870 --> 00:26:47.900
That's the crux of
the matter here.
00:26:47.900 --> 00:26:48.900
Where does it come from?
00:26:56.600 --> 00:27:01.420
So part of the reading that you
need to do now is Chapter 15.
00:27:01.420 --> 00:27:04.700
Chapter 15, most of it is
going to be complete review.
00:27:04.700 --> 00:27:07.810
It just says the conservation
of a linear momentum, impulse
00:27:07.810 --> 00:27:08.355
and momentum.
00:27:08.355 --> 00:27:10.900
But it also gets into
angular momentum.
00:27:10.900 --> 00:27:14.040
So we're going to talk quite
a lot about angular momentum.
00:27:17.660 --> 00:27:28.240
And I want to do a very brief
little review right now so
00:27:28.240 --> 00:27:29.700
that it applies to this problem.
00:27:32.320 --> 00:27:34.110
So we've come up
with an expression
00:27:34.110 --> 00:27:36.025
that the force in
the theta direction
00:27:36.025 --> 00:27:43.420
here comes-- there's got to
be that, 2mr dot theta dot.
00:27:52.820 --> 00:27:56.400
So here's my point O,
and A for that matter.
00:27:56.400 --> 00:28:00.200
But looking down on
our problem, here's
00:28:00.200 --> 00:28:03.160
my mass at some instant in time.
00:28:03.160 --> 00:28:09.400
My rotation rate is theta dot--
or actually it's constant.
00:28:09.400 --> 00:28:15.170
So it's k hat like
that, cap omega k hat.
00:28:15.170 --> 00:28:20.520
This is my r hat
direction, theta hat.
00:28:20.520 --> 00:28:23.710
Now, I'm going to treat
this as a particle.
00:28:23.710 --> 00:28:26.530
Not long-- we're going to be
talking about the dynamics
00:28:26.530 --> 00:28:27.560
of rigid bodies.
00:28:27.560 --> 00:28:29.510
We're just doing
particles for the moment.
00:28:29.510 --> 00:28:31.860
We think of them
just as point masses
00:28:31.860 --> 00:28:34.660
and don't deal with
their finite extent.
00:28:34.660 --> 00:28:37.410
So we're still thinking
of this as a particle.
00:28:37.410 --> 00:28:40.310
And I'm going to write down
the definition of the angular
00:28:40.310 --> 00:28:42.390
momentum of a particle.
00:28:42.390 --> 00:28:47.590
This is B out here with respect
to my fixed frame here at O.
00:28:47.590 --> 00:28:49.600
And I'm going to use a
lowercase h to describe
00:28:49.600 --> 00:28:51.410
angular momentum of particles.
00:28:51.410 --> 00:28:54.640
And I'll use capital H
to describe the angular
00:28:54.640 --> 00:28:57.670
momentum of rigid bodies.
00:28:57.670 --> 00:29:00.920
So this is a particle,
the definition
00:29:00.920 --> 00:29:03.940
of the angular momentum
of a particle with respect
00:29:03.940 --> 00:29:07.166
to a fixed point.
00:29:07.166 --> 00:29:08.540
We're going to
come back to that.
00:29:08.540 --> 00:29:10.670
That'll turn out to
have some significance.
00:29:10.670 --> 00:29:17.960
The definition of this is
it's RBO, the position vector,
00:29:17.960 --> 00:29:25.535
crossed with the linear momentum
evaluated in the fixed frame.
00:29:25.535 --> 00:29:27.660
So that's the definition
of angular momentum, which
00:29:27.660 --> 00:29:29.470
is you have linear
momentum, and it's
00:29:29.470 --> 00:29:32.810
the cross product with the
position vector out to it.
00:29:32.810 --> 00:29:38.845
So in this case, the
RBO is capital R R hat.
00:29:44.560 --> 00:29:47.300
Well, it's not-- this
varies, excuse me.
00:29:47.300 --> 00:29:50.750
So I'm not going to use--
this I use as a constant.
00:29:50.750 --> 00:29:53.090
I better keep it
as the variable.
00:29:53.090 --> 00:29:57.140
Pardon that, so it's r,
whatever the local position is,
00:29:57.140 --> 00:30:01.970
in the r hat direction crossed
with the linear momentum.
00:30:01.970 --> 00:30:06.270
What's the linear
momentum of that particle?
00:30:06.270 --> 00:30:09.050
Mass times velocity.
00:30:09.050 --> 00:30:09.983
What's the velocity?
00:30:09.983 --> 00:30:10.858
STUDENT: [INAUDIBLE].
00:30:17.182 --> 00:30:19.390
PROFESSOR: Theta dot-- where
did we write it up here?
00:30:19.390 --> 00:30:31.790
Someplace-- the total velocity
is r dot r hat plus r theta
00:30:31.790 --> 00:30:33.300
dot theta hat.
00:30:36.780 --> 00:30:38.640
And we need a mass in here.
00:30:38.640 --> 00:30:44.530
So mass times velocity
would be the momentum.
00:30:44.530 --> 00:30:46.670
And we need the cross
products of those.
00:30:46.670 --> 00:30:49.310
r hat cross r hat, you
get nothing from that.
00:30:49.310 --> 00:30:56.500
And r hat cross theta
hat-- positive or negative?
00:30:56.500 --> 00:30:59.040
Positive k, right?
00:30:59.040 --> 00:31:13.280
So this becomes mr squared theta
dot in the k hat direction.
00:31:19.200 --> 00:31:21.685
And this one is a constant.
00:31:21.685 --> 00:31:23.491
We'll write this as cap omega.
00:31:27.340 --> 00:31:32.970
So this is my angular momentum
of my particle with respect
00:31:32.970 --> 00:31:35.480
to this fixed reference frame.
00:31:35.480 --> 00:31:38.354
And so one final
step that you know
00:31:38.354 --> 00:31:43.220
from your previous
physics is-- how's torque
00:31:43.220 --> 00:31:44.400
related to angular momentum?
00:31:49.710 --> 00:31:50.505
Take a guess.
00:31:50.505 --> 00:31:52.600
What do you remember?
00:31:52.600 --> 00:32:01.870
Time derivative-- so
d by dt of hBO here.
00:32:01.870 --> 00:32:04.380
The time rate of
change of this vector
00:32:04.380 --> 00:32:11.190
is the torque with respect to
this point, with respect to O.
00:32:11.190 --> 00:32:12.920
So what are constants
in this thing?
00:32:12.920 --> 00:32:15.590
We don't have to deal
with their derivatives.
00:32:15.590 --> 00:32:22.640
k hat is-- does the direction
of the angular momentum change?
00:32:22.640 --> 00:32:23.960
It's upwards.
00:32:23.960 --> 00:32:26.290
The derivative of k,
the unit vector k,
00:32:26.290 --> 00:32:28.430
does it change in this problem?
00:32:28.430 --> 00:32:30.930
No, so its time derivative is 0.
00:32:30.930 --> 00:32:32.490
Its time derivative is 0.
00:32:32.490 --> 00:32:33.999
The m time derivative is 0.
00:32:33.999 --> 00:32:36.040
The only thing you have
to take the derivative of
00:32:36.040 --> 00:32:37.605
is r, so 2rr dot.
00:32:54.350 --> 00:32:59.380
So the torque is
2mrr dot cap omega.
00:32:59.380 --> 00:33:01.410
And it's all in the k direction.
00:33:01.410 --> 00:33:11.750
And that had better be
rBO cross F. Torque comes
00:33:11.750 --> 00:33:14.290
as a force times a
moment arm, right?
00:33:14.290 --> 00:33:18.000
And we computed here a force
in the theta direction.
00:33:18.000 --> 00:33:20.610
So what would give us a
torque in the k direction?
00:33:20.610 --> 00:33:23.090
An r cross a theta.
00:33:23.090 --> 00:33:26.440
r hat cross theta
hat gives you a k.
00:33:26.440 --> 00:33:29.800
r cross-- and this force,
the part of the force.
00:33:29.800 --> 00:33:32.450
We have two forces,
one in the r direction
00:33:32.450 --> 00:33:34.470
and one in the theta direction.
00:33:34.470 --> 00:33:38.540
The cross product, this term up
here, the T gives you nothing,
00:33:38.540 --> 00:33:41.620
r cross r hat cross r hat.
00:33:41.620 --> 00:33:46.420
So the only force that matters
here is this Coriolis one.
00:33:46.420 --> 00:33:55.800
And so that force is
2mr dot theta dot.
00:33:55.800 --> 00:33:59.280
And we have the r cross.
00:33:59.280 --> 00:34:01.800
We know this comes out
in the k direction.
00:34:01.800 --> 00:34:05.150
And we get an r in here when
we compute this product.
00:34:05.150 --> 00:34:09.880
And this, to me, looks
an awful lot like this.
00:34:09.880 --> 00:34:11.719
Except I missed an r squared.
00:34:11.719 --> 00:34:12.469
How did I do that?
00:34:19.049 --> 00:34:19.924
STUDENT: [INAUDIBLE].
00:34:25.900 --> 00:34:27.460
PROFESSOR: Oh no,
it's not r squared.
00:34:27.460 --> 00:34:29.650
It started off over here as
the angular momentum has an r
00:34:29.650 --> 00:34:30.858
squared, took the derivative.
00:34:30.858 --> 00:34:33.109
It dropped down to 2rr dot.
00:34:33.109 --> 00:34:36.150
And that's right.
00:34:36.150 --> 00:34:39.650
So the Coriolis
force, what's it got
00:34:39.650 --> 00:34:41.489
to do with the angular momentum?
00:34:46.460 --> 00:34:48.520
That's kind of the point
of the exercise here.
00:34:48.520 --> 00:34:57.700
In order for this ping pong
ball to be accelerated,
00:34:57.700 --> 00:35:00.760
as it goes slowly out this
tube, the angular momentum
00:35:00.760 --> 00:35:02.630
is increasing.
00:35:02.630 --> 00:35:05.030
In order to change angular
momentum with time,
00:35:05.030 --> 00:35:06.910
you have to apply a torque.
00:35:06.910 --> 00:35:09.060
The torque that
you have to apply
00:35:09.060 --> 00:35:12.700
is 2mrr dot in the k direction.
00:35:12.700 --> 00:35:16.670
And that is r cross
the Coriolis force.
00:35:16.670 --> 00:35:19.240
So the Coriolis
force, in this case,
00:35:19.240 --> 00:35:21.990
is the force that's necessary
to increase the angular
00:35:21.990 --> 00:35:24.630
momentum of a system.
00:35:24.630 --> 00:35:26.760
That's very often
the reason-- that's
00:35:26.760 --> 00:35:30.110
what Coriolis force is about.
00:35:30.110 --> 00:35:36.100
So when you shoot an
artillery piece on the Earth,
00:35:36.100 --> 00:35:38.410
you've got that projectile
going out there.
00:35:38.410 --> 00:35:41.760
It has angular momentum
with respect to the Earth.
00:35:41.760 --> 00:35:44.960
And you'll find out that
this little term pops up.
00:35:44.960 --> 00:35:47.920
And in fact, the projectile
doesn't go straight.
00:35:47.920 --> 00:35:48.590
It curves.
00:35:51.550 --> 00:35:54.020
There's lots of things that
because of conservation
00:35:54.020 --> 00:35:59.810
of angular momentum, you end
up with this term popping up.
00:35:59.810 --> 00:36:01.810
In this case, angular
momentum is not conserved.
00:36:01.810 --> 00:36:02.750
It's increasing.
00:36:02.750 --> 00:36:04.860
So you have this force
to make it happen.
00:36:04.860 --> 00:36:06.260
Any questions about this?
00:36:06.260 --> 00:36:08.090
You're going to
use this one a lot.
00:36:08.090 --> 00:36:09.548
You're going to
work with it a lot.
00:36:17.200 --> 00:36:20.390
So anytime you see changes
in angular momentum happening
00:36:20.390 --> 00:36:25.990
in a problem, in these
problems with circular motion,
00:36:25.990 --> 00:36:30.060
velocities of parts
increasing in radius,
00:36:30.060 --> 00:36:32.860
you'll almost always
see this term pop up.
00:36:32.860 --> 00:36:35.200
Any time you see these
changes in angular momentum,
00:36:35.200 --> 00:36:38.030
you'll often see
the Coriolis term.
00:36:38.030 --> 00:36:46.070
All right, now we're going to
do another interesting problem--
00:36:46.070 --> 00:36:48.040
simple but interesting.
00:36:48.040 --> 00:36:52.790
And that is really to go--
let's go do this problem
00:36:52.790 --> 00:36:55.170
where the thing is really
allowed to come freely out.
00:37:02.079 --> 00:37:03.745
All right, you ready
to defend yourself?
00:37:12.050 --> 00:37:15.040
A little short stick is pretty
effective at throwing candy.
00:37:19.640 --> 00:37:20.950
You got your safety glasses on?
00:37:20.950 --> 00:37:24.400
You want me to see if
I can get one up there?
00:37:24.400 --> 00:37:27.600
Aww, also, I obviously
haven't practiced this.
00:37:31.210 --> 00:37:32.550
All right, last one.
00:37:38.380 --> 00:37:40.840
Actually, there's two more.
00:37:48.400 --> 00:37:51.300
What makes this work?
00:37:51.300 --> 00:37:52.880
Let's do this problem.
00:37:56.490 --> 00:37:58.505
So let's look at
our candy shooter.
00:38:17.870 --> 00:38:20.460
So I'm whipping
this thing around.
00:38:20.460 --> 00:38:21.690
Candy is coming out of here.
00:38:21.690 --> 00:38:24.670
It's at B.
00:38:24.670 --> 00:38:28.970
Again, I get no z, not
mess with the z part.
00:38:28.970 --> 00:38:31.170
The z part is trivial
to usually deal with.
00:38:31.170 --> 00:38:33.636
Because it's
totally independent,
00:38:33.636 --> 00:38:34.635
just separate equations.
00:38:34.635 --> 00:38:36.630
It doesn't complicate
things much at all,
00:38:36.630 --> 00:38:38.542
even when you have z.
00:38:38.542 --> 00:38:42.340
Now, in this case, the
velocity-- and here's
00:38:42.340 --> 00:38:45.230
my O frame here.
00:38:45.230 --> 00:38:47.110
The velocity B
with respect to O,
00:38:47.110 --> 00:38:52.520
well, now it's got--
I'm not going to move.
00:38:52.520 --> 00:38:55.080
I'm standing still when I do it.
00:38:55.080 --> 00:38:57.660
So the first term is 0.
00:38:57.660 --> 00:39:02.350
It's got an r term, r dot
in the r hat direction.
00:39:02.350 --> 00:39:08.570
And it has an r theta
dot theta hat term.
00:39:13.810 --> 00:39:17.330
And these can now--
this might be changing.
00:39:17.330 --> 00:39:19.380
This might have a time
derivative, r double dot.
00:39:19.380 --> 00:39:20.130
It certainly does.
00:39:20.130 --> 00:39:22.000
It's accelerating
coming out of that tube.
00:39:22.000 --> 00:39:25.670
And my theta, angular motion,
it can be accelerating, too.
00:39:25.670 --> 00:39:28.985
So we're going to have
to deal with those.
00:39:28.985 --> 00:39:30.360
I need to know
the accelerations.
00:39:36.100 --> 00:39:39.060
Now, I haven't
emphasized it till now,
00:39:39.060 --> 00:39:42.990
but I find it conceptually
useful to-- when
00:39:42.990 --> 00:39:45.750
you work with polar
coordinates, you
00:39:45.750 --> 00:39:48.810
can have this
ability to aggregate
00:39:48.810 --> 00:39:57.910
the terms in these two
component directions.
00:39:57.910 --> 00:39:59.240
So you have this.
00:39:59.240 --> 00:40:01.640
All the r terms go together.
00:40:01.640 --> 00:40:06.870
And we've let the
z double dot term--
00:40:06.870 --> 00:40:08.270
it's just separate by itself.
00:40:08.270 --> 00:40:10.380
It drops out easily.
00:40:10.380 --> 00:40:12.040
And you have the theta hat term.
00:40:12.040 --> 00:40:25.890
And that's the r theta double
dot plus 2r dot theta dot.
00:40:25.890 --> 00:40:27.900
And these are in the
theta hat direction.
00:40:27.900 --> 00:40:30.530
There's your Coriolis term,
your Euler acceleration
00:40:30.530 --> 00:40:33.500
term, centripetal term.
00:40:33.500 --> 00:40:34.039
Yeah?
00:40:34.039 --> 00:40:34.914
STUDENT: [INAUDIBLE].
00:40:38.964 --> 00:40:40.630
PROFESSOR: Well, I
don't know if I ought
00:40:40.630 --> 00:40:42.692
to tell you secrets about me.
00:40:42.692 --> 00:40:45.025
Because it's going to give
you an advantage on the quiz.
00:40:48.300 --> 00:40:50.840
But I've almost
never, ever been known
00:40:50.840 --> 00:40:55.630
to ask a question
that says, "derive."
00:40:55.630 --> 00:40:59.920
But I'll sure ask you
concept questions.
00:40:59.920 --> 00:41:01.930
I really want you to
understand the principles.
00:41:01.930 --> 00:41:04.625
I don't get real hung
up on having you do
00:41:04.625 --> 00:41:07.029
the grungy grind it out things.
00:41:07.029 --> 00:41:08.570
Do I want you to
remember the formula
00:41:08.570 --> 00:41:13.130
for how to take the derivative
of a vector in rotating frame?
00:41:13.130 --> 00:41:15.150
Yeah, that's where
these have come from.
00:41:18.250 --> 00:41:21.830
You had better remember this.
00:41:21.830 --> 00:41:24.850
These two formulas, the
velocity formula and this,
00:41:24.850 --> 00:41:27.820
the acceleration formula,
are just core to this course.
00:41:27.820 --> 00:41:29.670
Now, the way quizzes
are done-- first quiz,
00:41:29.670 --> 00:41:31.900
you come in, one sheet of paper.
00:41:31.900 --> 00:41:35.580
What had better
be on your paper?
00:41:35.580 --> 00:41:37.030
OK?
00:41:37.030 --> 00:41:39.470
And second quiz, two
sheets, final, three sheets,
00:41:39.470 --> 00:41:40.870
that kind of thing.
00:41:40.870 --> 00:41:43.290
But conceptually, you've
got forces in the r,
00:41:43.290 --> 00:41:47.000
forces in the z,
accelerations in r theta
00:41:47.000 --> 00:41:49.455
and z, forces r theta and z.
00:41:52.210 --> 00:41:53.890
And for these planar
motion problems,
00:41:53.890 --> 00:41:57.190
this one is sure easy to use.
00:41:57.190 --> 00:42:00.980
So let's think
about this problem.
00:42:00.980 --> 00:42:06.650
I'm going to let this
be frictionless just
00:42:06.650 --> 00:42:09.410
to make it easy.
00:42:09.410 --> 00:42:12.130
All right, so what
possible forces act
00:42:12.130 --> 00:42:14.870
on the hunk of candy?
00:42:14.870 --> 00:42:17.210
Let's do a free body
diagram of the hunk of candy
00:42:17.210 --> 00:42:19.680
coming out of here.
00:42:19.680 --> 00:42:22.010
What are the forces?
00:42:22.010 --> 00:42:23.640
And let's keep it planar.
00:42:23.640 --> 00:42:25.080
We can get gravity into this.
00:42:25.080 --> 00:42:26.770
But let's just do it in a plane.
00:42:26.770 --> 00:42:29.500
So I'm just going horizontal
and slinging this thing.
00:42:29.500 --> 00:42:31.160
Gravity's in the z
direction, and I've
00:42:31.160 --> 00:42:32.430
constrained it in the z.
00:42:32.430 --> 00:42:37.140
So it's something supporting
the gravity in the tube.
00:42:37.140 --> 00:42:40.540
So definitely there's an
mg on this thing downwards.
00:42:40.540 --> 00:42:42.990
But it's in the z
direction, and we're not
00:42:42.990 --> 00:42:45.200
letting it move in the z.
00:42:45.200 --> 00:42:49.290
What about the horizontal,
this direction?
00:42:49.290 --> 00:42:53.750
The r-- this is my
r hat direction.
00:42:53.750 --> 00:42:56.960
This is my theta hat direction.
00:42:56.960 --> 00:42:59.516
Whoops, not either--
theta is going this way.
00:42:59.516 --> 00:43:03.260
So I've drawn this
as a side view.
00:43:03.260 --> 00:43:06.430
What's the force in
the r hat direction?
00:43:06.430 --> 00:43:07.305
STUDENT: [INAUDIBLE].
00:43:10.600 --> 00:43:12.350
PROFESSOR: OK, in order
to do the problem,
00:43:12.350 --> 00:43:14.980
you have to figure that out.
00:43:14.980 --> 00:43:15.850
What are the forces?
00:43:19.030 --> 00:43:22.150
What are the source of
forces in the r direction?
00:43:22.150 --> 00:43:24.016
Here's the r direction.
00:43:24.016 --> 00:43:24.890
STUDENT: [INAUDIBLE].
00:43:24.890 --> 00:43:25.723
PROFESSOR: Say what?
00:43:28.230 --> 00:43:29.140
Come on, you guys.
00:43:29.140 --> 00:43:31.158
Somebody be-- yeah.
00:43:31.158 --> 00:43:33.650
STUDENT: [INAUDIBLE].
00:43:33.650 --> 00:43:34.860
PROFESSOR: There are not any.
00:43:34.860 --> 00:43:35.776
Is that what you said?
00:43:35.776 --> 00:43:36.710
There aren't any.
00:43:36.710 --> 00:43:38.025
And why's that?
00:43:38.025 --> 00:43:38.900
STUDENT: [INAUDIBLE].
00:43:38.900 --> 00:43:41.780
PROFESSOR: Right, so there's
no forces in the r direction.
00:43:41.780 --> 00:43:44.980
So there's no forces
on this thing in the r.
00:43:44.980 --> 00:43:48.260
And so then this is a side view.
00:43:48.260 --> 00:43:50.260
We could do the top
view looking down.
00:43:50.260 --> 00:43:53.260
Top view, you've got your x, y.
00:43:53.260 --> 00:43:56.580
You also have your--
here's the ball.
00:43:56.580 --> 00:43:58.310
Here's the r hat.
00:43:58.310 --> 00:44:00.040
Here's the theta hat.
00:44:00.040 --> 00:44:04.800
Now, free body diagram
in the top view--
00:44:04.800 --> 00:44:10.937
well, there's some
force here probably.
00:44:10.937 --> 00:44:14.290
STUDENT: I was going to
ask that the fact that we
00:44:14.290 --> 00:44:16.685
have no forces in the r
hat direction, but we do
00:44:16.685 --> 00:44:18.529
have acceleration.
00:44:18.529 --> 00:44:19.445
PROFESSOR: Absolutely.
00:44:19.445 --> 00:44:22.030
She's commenting that
we do have accelerations
00:44:22.030 --> 00:44:23.240
in the r direction, right?
00:44:23.240 --> 00:44:24.490
STUDENT: And we have no force.
00:44:24.490 --> 00:44:25.910
PROFESSOR: And no force.
00:44:25.910 --> 00:44:28.765
So that's the conundrum
of this problem.
00:44:28.765 --> 00:44:30.140
That's the point
of this problem.
00:44:30.140 --> 00:44:30.931
So let me continue.
00:44:34.070 --> 00:44:36.830
Free body diagram--
I'm looking down on it.
00:44:36.830 --> 00:44:38.080
I'm allowing for some force.
00:44:38.080 --> 00:44:42.590
It's the normal force that
comes from the pipe exerting
00:44:42.590 --> 00:44:44.420
the force on the candy.
00:44:44.420 --> 00:44:47.410
And since it's frictionless, it
can only be normal to the pipe.
00:44:47.410 --> 00:44:54.270
OK, so there's the free body
diagram in the top view.
00:44:54.270 --> 00:44:56.466
So we can write two equations.
00:44:56.466 --> 00:44:58.510
We can write three
questions-- this one
00:44:58.510 --> 00:45:01.286
equal to 0, this one
in the r hat direction,
00:45:01.286 --> 00:45:02.785
this one in the
theta hat direction.
00:45:22.600 --> 00:45:27.350
Sum of the forces r hat
direction must be 0.
00:45:27.350 --> 00:45:28.667
And we have a mass.
00:45:28.667 --> 00:45:29.666
We have an acceleration.
00:45:40.170 --> 00:45:42.120
Solve for r double dot.
00:45:50.860 --> 00:45:58.930
Remarkable-- there's no
force in the r hat direction.
00:45:58.930 --> 00:46:05.220
The position of
the object, the r
00:46:05.220 --> 00:46:08.490
coordinate, it has a velocity.
00:46:08.490 --> 00:46:10.250
It has an acceleration.
00:46:14.380 --> 00:46:18.045
But the total acceleration
in the r hat directions
00:46:18.045 --> 00:46:19.590
are actually 0.
00:46:19.590 --> 00:46:25.790
The rate of change
of the velocity
00:46:25.790 --> 00:46:28.860
of this thing in the radial
direction, r double dot,
00:46:28.860 --> 00:46:30.870
is nonzero.
00:46:30.870 --> 00:46:32.151
But there are no forces on it.
00:46:37.340 --> 00:46:42.150
I have pondered another
way to explain this.
00:46:42.150 --> 00:46:43.390
I'm still thinking about it.
00:46:43.390 --> 00:46:45.010
And you think about this, too.
00:46:45.010 --> 00:46:48.300
How do you explain this
in the absence of forces?
00:46:48.300 --> 00:46:49.960
And it's partly
where the concept
00:46:49.960 --> 00:46:53.760
comes from of fictitious forces,
that centrifugal force is
00:46:53.760 --> 00:46:54.580
a force.
00:46:54.580 --> 00:46:55.140
It's not.
00:46:55.140 --> 00:46:56.600
It's an acceleration.
00:46:56.600 --> 00:47:01.330
This is just saying that--
let's go back to this problem.
00:47:01.330 --> 00:47:05.170
In order to make
something go in a circle,
00:47:05.170 --> 00:47:08.200
you have to put a
force on it to cause
00:47:08.200 --> 00:47:11.280
the centripetal acceleration.
00:47:11.280 --> 00:47:16.090
You have to allow the thing
to go out if you're not
00:47:16.090 --> 00:47:18.190
forcing it to go in a circle.
00:47:18.190 --> 00:47:20.940
You have to allow it to go
out at that rate in order
00:47:20.940 --> 00:47:25.020
for there to be no centripetal
accelerations on the object.
00:47:25.020 --> 00:47:25.521
Yeah?
00:47:25.521 --> 00:47:27.603
STUDENT: So I was going
to say, when you start it,
00:47:27.603 --> 00:47:29.026
you push it in the y direction.
00:47:29.026 --> 00:47:29.930
So that's a force.
00:47:29.930 --> 00:47:32.620
It's not in the r hat,
but it's in the y.
00:47:32.620 --> 00:47:33.830
PROFESSOR: To get it started.
00:47:33.830 --> 00:47:35.958
STUDENT: Right, and there's
no force opposing it
00:47:35.958 --> 00:47:38.090
in that direction [INAUDIBLE].
00:47:38.090 --> 00:47:45.990
PROFESSOR: Does it experience
centripetal acceleration?
00:47:45.990 --> 00:47:47.550
So there's no rotation.
00:47:47.550 --> 00:47:50.125
Does it experience
centripetal acceleration?
00:47:53.100 --> 00:47:54.560
What do you think?
00:47:54.560 --> 00:47:58.450
Yeah, because it goes
through curved motion.
00:47:58.450 --> 00:48:00.360
At any instance in time
when it's doing that,
00:48:00.360 --> 00:48:02.279
there is a radius of curvature.
00:48:02.279 --> 00:48:04.070
You can at that instant
in time think of it
00:48:04.070 --> 00:48:07.050
as being in a circular path.
00:48:07.050 --> 00:48:09.840
And sometimes it's
just really easy
00:48:09.840 --> 00:48:13.510
to do these kind of problems
with normal and tangential
00:48:13.510 --> 00:48:14.970
coordinates.
00:48:14.970 --> 00:48:23.670
So I'm looking down
on the x, y plane.
00:48:23.670 --> 00:48:25.330
Gravity is into the Earth.
00:48:25.330 --> 00:48:27.790
So I'm looking down
on a vehicle, a car.
00:48:27.790 --> 00:48:31.940
And that car, the
guy is kind of drunk.
00:48:31.940 --> 00:48:35.100
He's going down
the road like this.
00:48:35.100 --> 00:48:36.440
So here's my y.
00:48:36.440 --> 00:48:37.770
Here's my x.
00:48:37.770 --> 00:48:49.270
And y equals some A sine
2 pi over the wavelength,
00:48:49.270 --> 00:48:58.150
2 pi over lambda, times
x at some A sine kx.
00:48:58.150 --> 00:49:00.380
And as he drives down--
if you're in a car,
00:49:00.380 --> 00:49:03.490
and you're doing that, you get
thrown side to side in the car.
00:49:03.490 --> 00:49:04.997
So you are being accelerated.
00:49:04.997 --> 00:49:06.580
And so we want to
be able to calculate
00:49:06.580 --> 00:49:09.850
the acceleration due to
the fact that you're going
00:49:09.850 --> 00:49:12.480
down and doing a curved path.
00:49:12.480 --> 00:49:16.390
And we deal with
these things sometimes
00:49:16.390 --> 00:49:19.260
with a convenient little
set of coordinates
00:49:19.260 --> 00:49:23.810
that are our normal
and tangential unit
00:49:23.810 --> 00:49:30.140
vectors, u normal
and u tangential,
00:49:30.140 --> 00:49:31.350
at any instant in time.
00:49:34.610 --> 00:49:38.990
And we know that if
this is along the path,
00:49:38.990 --> 00:49:41.700
at any instant in time
you're right here,
00:49:41.700 --> 00:49:45.070
what direction is your velocity?
00:49:45.070 --> 00:49:52.170
Just definition of velocity--
tangent to the path, right?
00:49:52.170 --> 00:49:54.170
So at any instant in
time, the velocity
00:49:54.170 --> 00:49:58.000
has got to be tangent to
the path at that moment.
00:49:58.000 --> 00:50:09.170
So velocity, the vector, has
a magnitude and a unit vector
00:50:09.170 --> 00:50:12.606
uT here I'll call it, tangent.
00:50:12.606 --> 00:50:15.020
That's all there is to it.
00:50:15.020 --> 00:50:20.320
And the acceleration
of this thing
00:50:20.320 --> 00:50:22.435
is your time derivative of this.
00:50:26.330 --> 00:50:34.000
And that's going to give
you a v dot uT hat plus a v.
00:50:34.000 --> 00:50:39.290
And now you need a time
derivative of this guy.
00:50:39.290 --> 00:50:41.904
But this is a unit
length vector.
00:50:41.904 --> 00:50:43.320
You can plug it
into that equation
00:50:43.320 --> 00:50:46.700
for the derivative
of a rotating vector
00:50:46.700 --> 00:50:48.440
and calculate what
this should be.
00:50:48.440 --> 00:50:50.210
You could also just draw it out.
00:50:50.210 --> 00:50:53.060
So I'll draw this for you.
00:50:58.880 --> 00:51:01.750
How are we doing on time?
00:51:01.750 --> 00:51:04.070
I should just be
able to finish this.
00:51:11.500 --> 00:51:16.500
Here's my unit vector in
the tangential direction.
00:51:16.500 --> 00:51:23.320
As I'm going around
this curve, this
00:51:23.320 --> 00:51:27.650
is my tangential direction.
00:51:27.650 --> 00:51:29.280
There's some instant
I have a radius.
00:51:29.280 --> 00:51:31.050
We call that rho.
00:51:31.050 --> 00:51:33.655
And in the little
time, delta t, I
00:51:33.655 --> 00:51:38.300
go through an angle
delta theta in delta t.
00:51:38.300 --> 00:51:42.070
And this is my uT vector here.
00:51:42.070 --> 00:51:46.120
It changes by a little bit.
00:51:46.120 --> 00:51:53.290
That's the change in the uT unit
vector in this time, delta t.
00:51:53.290 --> 00:51:55.870
And it goes perpendicular.
00:51:55.870 --> 00:51:58.652
And it goes in the
positive un direction.
00:52:01.700 --> 00:52:06.050
So delta-- what's the easiest
way to write this one?
00:52:12.030 --> 00:52:26.600
Delta uT equals some
theta dot delta t-- that's
00:52:26.600 --> 00:52:35.390
the angle-- times the length
of the unit vector, 1.
00:52:35.390 --> 00:52:38.410
That's the distance
it goes, so 1.
00:52:38.410 --> 00:52:42.990
So r omega, 1 theta
dot is the distance
00:52:42.990 --> 00:52:48.500
that this unit vector
goes through in delta t.
00:52:48.500 --> 00:52:54.940
And the direction it
goes in is u normal hat.
00:52:54.940 --> 00:53:01.820
So delta uT over
delta t limit as t
00:53:01.820 --> 00:53:11.280
goes to 0, you get theta
dot un, just like before.
00:53:11.280 --> 00:53:13.700
So the time derivative
of this unit vector
00:53:13.700 --> 00:53:17.080
in the tangential
direction is just theta dot
00:53:17.080 --> 00:53:19.770
in the normal direction.
00:53:19.770 --> 00:53:24.090
And then from that,
we can very quickly
00:53:24.090 --> 00:53:27.630
derive the rest of
this acceleration.
00:53:27.630 --> 00:53:30.820
The acceleration then
is that plus this.
00:53:30.820 --> 00:53:41.119
We now know an expression
for-- this is my uT dot term.
00:53:41.119 --> 00:53:42.410
I'm going to plug that in here.
00:53:42.410 --> 00:53:45.510
This then, we need an
expression for v. What's v?
00:53:48.770 --> 00:53:52.000
Well, at that instant in
time, it has some radius rho.
00:53:52.000 --> 00:53:55.100
It has an angular
velocity theta dot.
00:53:55.100 --> 00:53:57.930
So rho theta dot
would be the v here.
00:54:07.980 --> 00:54:13.240
So I'm looking for an
expression for vuT dot.
00:54:18.300 --> 00:54:24.928
So that's v theta dot un.
00:54:24.928 --> 00:54:28.550
But theta dot is v over rho.
00:54:31.150 --> 00:54:37.210
v squared-- sorry about
this-- over rho un.
00:54:37.210 --> 00:54:45.160
So this guy up here, this
is v dot uT plus v squared--
00:54:45.160 --> 00:54:47.150
I'll rewrite it-- over rho un.
00:54:50.082 --> 00:54:51.540
So if you're speeding
up, if you're
00:54:51.540 --> 00:54:54.250
going from 30 miles an
hour to 40 miles an hour,
00:54:54.250 --> 00:54:55.520
that's your tangential.
00:54:55.520 --> 00:54:57.360
That's your speed
along the path.
00:54:57.360 --> 00:54:58.901
That's this term.
00:54:58.901 --> 00:55:00.650
But because you're
going around the curve,
00:55:00.650 --> 00:55:02.720
you have an acceleration
of v squared over rho.
00:55:02.720 --> 00:55:05.010
You've run into this
before in physics.
00:55:05.010 --> 00:55:07.760
This is where it comes from.
00:55:07.760 --> 00:55:12.490
This is a centripetal
acceleration like term.
00:55:12.490 --> 00:55:15.210
If you replace v
with rho theta dot,
00:55:15.210 --> 00:55:17.557
you'd get r theta dot squared.
00:55:17.557 --> 00:55:19.890
So you can either put it in
terms of v squared over rho,
00:55:19.890 --> 00:55:22.820
or you can put it in terms
of rho theta dot squared.
00:55:22.820 --> 00:55:28.350
And rho theta dot squared sounds
a lot like my acceleration term
00:55:28.350 --> 00:55:28.890
right here.
00:55:32.280 --> 00:55:35.500
So with that simple
little formula,
00:55:35.500 --> 00:55:38.580
you can do-- you
need one other thing.
00:55:41.160 --> 00:55:44.960
And you just go look
it up in the book.
00:55:44.960 --> 00:55:48.680
There is an expression from
calculus for the radius
00:55:48.680 --> 00:55:52.570
of curvature of a path.
00:55:52.570 --> 00:55:56.480
And it has first and second
derivatives of y with respect
00:55:56.480 --> 00:55:56.980
to x.
00:55:56.980 --> 00:56:00.710
So you need a dy/dx
and a d2y dx squared.
00:56:00.710 --> 00:56:03.650
From a sine function,
you can calculate that.
00:56:03.650 --> 00:56:05.960
So you calculate
rho from a formula
00:56:05.960 --> 00:56:07.910
that's in the book
for calculating
00:56:07.910 --> 00:56:09.240
radius of curvature.
00:56:09.240 --> 00:56:11.100
And then you're done, all right?
00:56:11.100 --> 00:56:15.520
So see you on Thursday.