1
00:00:00,080 --> 00:00:02,430
The following content is
provided under a Creative

2
00:00:02,430 --> 00:00:03,810
Commons license.

3
00:00:03,810 --> 00:00:06,050
Your support will help
MIT OpenCourseWare

4
00:00:06,050 --> 00:00:10,140
continue to offer high quality
educational resources for free.

5
00:00:10,140 --> 00:00:12,690
To make a donation or to
view additional materials

6
00:00:12,690 --> 00:00:16,590
from hundreds of MIT courses,
visit MIT OpenCourseWare

7
00:00:16,590 --> 00:00:17,260
at ocw.mit.edu.

8
00:00:21,465 --> 00:00:23,840
PROFESSOR: So we were doing
velocities and accelerations.

9
00:00:23,840 --> 00:00:33,070
We came up with-- I guess I
ought to continue to remind us.

10
00:00:33,070 --> 00:00:36,140
We're talking about
velocities and accelerations

11
00:00:36,140 --> 00:00:40,630
of a point with respect to
another point, to which we've

12
00:00:40,630 --> 00:00:44,480
attached-- I'll make
this point here--

13
00:00:44,480 --> 00:00:49,760
a reference frame, x
prime, y prime, z prime.

14
00:00:49,760 --> 00:00:52,750
And there's a vector that
goes between these two

15
00:00:52,750 --> 00:00:56,420
and a vector that goes there
so that we can say r of B

16
00:00:56,420 --> 00:01:01,660
with respect to O, my
inertial frame, is r of A

17
00:01:01,660 --> 00:01:06,050
with respect to O plus
r of B with respect

18
00:01:06,050 --> 00:01:07,550
to A. These are all vectors.

19
00:01:07,550 --> 00:01:10,070
And then from these, we derive
velocities and acceleration

20
00:01:10,070 --> 00:01:10,570
formulas.

21
00:01:10,570 --> 00:01:15,040
And so we've come up with a
couple of very handy formulas.

22
00:01:15,040 --> 00:01:21,920
The velocity formula, velocity
of B with respect to O,

23
00:01:21,920 --> 00:01:23,920
is the velocity
of A with respect

24
00:01:23,920 --> 00:01:31,247
to O plus you have to take
a time derivative of this.

25
00:01:31,247 --> 00:01:33,580
And so I'm going to give you
just the general expression

26
00:01:33,580 --> 00:01:35,870
here just as a brief reminder.

27
00:01:35,870 --> 00:01:39,760
It's the derivative
of B with respect

28
00:01:39,760 --> 00:01:46,490
to A as seen in the Axyz frame
plus omega with respect to O

29
00:01:46,490 --> 00:01:49,279
cross rBA.

30
00:01:49,279 --> 00:01:50,195
These are all vectors.

31
00:01:52,760 --> 00:01:55,030
So that's the velocity
formula, remember?

32
00:01:55,030 --> 00:01:57,640
This is if you're in
the frame rotating

33
00:01:57,640 --> 00:01:59,650
and translating with
it, this is the change

34
00:01:59,650 --> 00:02:02,860
of length of the vector.

35
00:02:02,860 --> 00:02:05,430
And this, then, is the
contribution to the velocity

36
00:02:05,430 --> 00:02:10,199
that you see in the fixed frame
that comes from the rotation.

37
00:02:10,199 --> 00:02:12,820
And we then got into
polar coordinates.

38
00:02:12,820 --> 00:02:15,750
And we found out that if
you use polar coordinates,

39
00:02:15,750 --> 00:02:20,080
then you can express
this as the velocity of A

40
00:02:20,080 --> 00:02:27,780
with respect to O
plus r dot r hat.

41
00:02:27,780 --> 00:02:30,530
And I should really say
cylindrical coordinates,

42
00:02:30,530 --> 00:02:43,250
z dot k hat plus r
theta dot theta hat.

43
00:02:43,250 --> 00:02:45,070
So that's exactly
the same thing.

44
00:02:45,070 --> 00:02:47,610
This is full 3D vector notation.

45
00:02:47,610 --> 00:02:50,680
This is a special case of
a coordinate system which

46
00:02:50,680 --> 00:02:53,730
we call polar coordinates.

47
00:02:53,730 --> 00:02:58,010
And we came up with another
formula for accelerations,

48
00:02:58,010 --> 00:03:04,030
the full 3D vector version
of that, A with respect

49
00:03:04,030 --> 00:03:10,610
to O plus-- this is
the acceleration of B

50
00:03:10,610 --> 00:03:14,620
with respect to A, but as
seen in the Axyz frame.

51
00:03:19,000 --> 00:03:27,320
2 omega cross-- and
this is v, the velocity,

52
00:03:27,320 --> 00:03:30,700
as seen in the xyz frame.

53
00:03:30,700 --> 00:03:34,100
So these things, this is no
contribution from acceleration.

54
00:03:34,100 --> 00:03:36,000
This is no contribution
from acceleration.

55
00:03:39,020 --> 00:03:47,810
Plus omega naught cross
rBA plus omega cross

56
00:03:47,810 --> 00:03:55,750
omega cross rBA-- all vectors.

57
00:03:59,190 --> 00:04:03,130
This is a movement of the
frame, the acceleration of it

58
00:04:03,130 --> 00:04:04,010
with respect to this.

59
00:04:04,010 --> 00:04:06,330
This is pure translation.

60
00:04:06,330 --> 00:04:10,800
This is the acceleration
of the dog on the merry go

61
00:04:10,800 --> 00:04:14,535
round with respect to the center
of the coordinate system there.

62
00:04:14,535 --> 00:04:17,170
It does not involve rotation.

63
00:04:17,170 --> 00:04:21,050
This is this term
that we call Coriolis.

64
00:04:21,050 --> 00:04:22,650
This is a term we call Euler.

65
00:04:22,650 --> 00:04:26,080
This is the angular speed up
of the system, the angular

66
00:04:26,080 --> 00:04:27,280
acceleration.

67
00:04:27,280 --> 00:04:28,450
And this is centripetal.

68
00:04:32,100 --> 00:04:33,900
And if you do these,
if you want to go

69
00:04:33,900 --> 00:04:37,240
to cylindrical coordinates--
and what we're going to do next

70
00:04:37,240 --> 00:04:39,610
is just do some
applications here.

71
00:04:39,610 --> 00:04:42,490
The acceleration of B with
respect to O and cylindrical

72
00:04:42,490 --> 00:04:46,670
coordinates-- it's
a translation piece.

73
00:04:46,670 --> 00:04:49,970
Plus now in cylindrical,
it's useful to group these.

74
00:04:56,550 --> 00:05:02,650
So I'm going to put together
this term and this term.

75
00:05:02,650 --> 00:05:05,024
Because they're both
in the r hat direction.

76
00:05:05,024 --> 00:05:06,940
And I'm going to put
together these two terms.

77
00:05:06,940 --> 00:05:11,090
Because they're both in
the theta hat direction.

78
00:05:11,090 --> 00:05:16,990
And I need a z double dot
k, because it's cylindrical.

79
00:05:16,990 --> 00:05:20,000
And then I have the
theta hat piece,

80
00:05:20,000 --> 00:05:29,180
r theta double dot plus 2r
dot theta dot theta hat.

81
00:05:29,180 --> 00:05:32,880
This is the same thing as this.

82
00:05:32,880 --> 00:05:37,587
Except this is expressed
in cylindrical coordinates.

83
00:05:37,587 --> 00:05:39,420
And cylindrical coordinates
are particularly

84
00:05:39,420 --> 00:05:42,740
good for doing the kind of
problems that are mostly

85
00:05:42,740 --> 00:05:45,780
done at the level of this
course, which we call planar

86
00:05:45,780 --> 00:05:50,190
motion problems, confined
to an x, y plane,

87
00:05:50,190 --> 00:05:54,650
and confined to single
axis rotation in z.

88
00:05:54,650 --> 00:05:57,620
This is a coordinate system
that's ideally suited

89
00:05:57,620 --> 00:05:58,810
to do problems like that.

90
00:05:58,810 --> 00:06:00,410
And that's why we use it.

91
00:06:00,410 --> 00:06:04,550
So now let's do some examples.

92
00:06:04,550 --> 00:06:11,550
This is really
quite a powerful--

93
00:06:11,550 --> 00:06:13,260
now that you have
these two equations,

94
00:06:13,260 --> 00:06:15,625
you can do a lot of
kinematics and dynamics.

95
00:06:25,270 --> 00:06:31,010
So we started last time
right at the very end.

96
00:06:31,010 --> 00:06:36,100
I said, OK, let's do this
problem really quickly, right?

97
00:06:36,100 --> 00:06:39,650
Constant rotation
rate, constant radius,

98
00:06:39,650 --> 00:06:42,030
no angular
acceleration, no change

99
00:06:42,030 --> 00:06:44,340
in length of the thing--
pretty simple problem.

100
00:06:44,340 --> 00:06:46,630
And we zipped it
off really fast.

101
00:06:46,630 --> 00:06:50,870
And I wanted to start there,
do that really quickly.

102
00:06:50,870 --> 00:06:53,760
So this is my case one.

103
00:06:53,760 --> 00:06:55,495
It's my ball on the string.

104
00:07:01,260 --> 00:07:08,060
Here's point A. Here's point
B. Here's the theta hat

105
00:07:08,060 --> 00:07:12,930
direction, r hat direction.

106
00:07:12,930 --> 00:07:17,880
And it has some
constant length R here.

107
00:07:17,880 --> 00:07:24,140
So r dot r double dot are 0.

108
00:07:24,140 --> 00:07:33,370
There's no also z, z dot, z
double dot, no z motion at all.

109
00:07:33,370 --> 00:07:36,090
Those are all 0.

110
00:07:36,090 --> 00:07:39,980
Theta dot is a constant.

111
00:07:39,980 --> 00:07:45,460
I'll call it cap omega in the k
hat direction, right hand rule.

112
00:07:45,460 --> 00:07:50,080
Theta double dot is 0.

113
00:07:50,080 --> 00:07:53,177
So the easy way to
use these formulas

114
00:07:53,177 --> 00:07:55,010
is you just start
knocking out all the terms

115
00:07:55,010 --> 00:07:56,490
that you don't need.

116
00:07:56,490 --> 00:08:00,800
But let's, just to give you
a little quick review of how

117
00:08:00,800 --> 00:08:05,560
to use these things,
use the vector full

118
00:08:05,560 --> 00:08:08,770
3D version of the velocity
equation for a second.

119
00:08:08,770 --> 00:08:11,930
The full 3D version
says the velocity of A

120
00:08:11,930 --> 00:08:16,250
with respect to O, what's
that in this problem?

121
00:08:16,250 --> 00:08:19,630
That's the translating term.

122
00:08:19,630 --> 00:08:22,060
So I was standing still,
not going anywhere.

123
00:08:22,060 --> 00:08:25,070
But if I were walking along,
I'd still spin that thing.

124
00:08:25,070 --> 00:08:28,070
OK, so this problem,
this term is 0.

125
00:08:28,070 --> 00:08:33,409
This problem, r
dot-- let's go here.

126
00:08:33,409 --> 00:08:35,580
This is the rate of
change of the length

127
00:08:35,580 --> 00:08:40,321
of the string in the coordinate
system of the string walking

128
00:08:40,321 --> 00:08:40,820
along.

129
00:08:40,820 --> 00:08:43,110
So what's r dot?

130
00:08:43,110 --> 00:08:45,180
OK, so that term is 0.

131
00:08:45,180 --> 00:08:51,170
And omega cross rBA-- well, r is
in the R hat, capital R R hat,

132
00:08:51,170 --> 00:08:56,030
cross with omega k.

133
00:08:56,030 --> 00:09:01,883
k cross R is-- k cross of R hat?

134
00:09:01,883 --> 00:09:02,630
STUDENT: Theta.

135
00:09:02,630 --> 00:09:03,505
PROFESSOR: Theta hat.

136
00:09:03,505 --> 00:09:07,230
So we get our familiar--
this first term is 0.

137
00:09:07,230 --> 00:09:08,860
The second term is 0.

138
00:09:08,860 --> 00:09:12,830
The third term, we
just get our R omega

139
00:09:12,830 --> 00:09:14,920
in the theta hat direction.

140
00:09:14,920 --> 00:09:16,050
We know that to be true.

141
00:09:16,050 --> 00:09:17,633
The point I'm just
trying to make here

142
00:09:17,633 --> 00:09:21,530
is you can always just fall
back and use the vector formula,

143
00:09:21,530 --> 00:09:23,230
full 3D formulas of both.

144
00:09:23,230 --> 00:09:26,370
Just plug it in, and
everything will just drop out.

145
00:09:26,370 --> 00:09:30,270
You can also then go to the
cylindrical coordinate terms

146
00:09:30,270 --> 00:09:32,750
when you happen to
have it all expressed

147
00:09:32,750 --> 00:09:37,710
in coordinates of that kind
to make it simpler for you.

148
00:09:37,710 --> 00:09:43,400
OK, so now let's quickly
then do the acceleration of B

149
00:09:43,400 --> 00:09:49,300
with respect to O. And let's
use the formulation already

150
00:09:49,300 --> 00:09:50,890
in cylindrical coordinates.

151
00:09:50,890 --> 00:09:53,760
What's A with respect to O?

152
00:09:53,760 --> 00:09:54,720
0.

153
00:09:54,720 --> 00:09:58,200
What's r double dot?

154
00:09:58,200 --> 00:10:01,470
How about the r theta dot
squared term, 0 or not?

155
00:10:01,470 --> 00:10:02,290
Nope.

156
00:10:02,290 --> 00:10:04,243
Let's go on-- z double dot?

157
00:10:04,243 --> 00:10:04,861
STUDENT: 0.

158
00:10:04,861 --> 00:10:06,110
PROFESSOR: r theta double dot?

159
00:10:06,110 --> 00:10:07,082
STUDENT: 0.

160
00:10:07,082 --> 00:10:08,311
PROFESSOR: r dot theta dot?

161
00:10:08,311 --> 00:10:08,810
STUDENT: 0

162
00:10:08,810 --> 00:10:10,540
PROFESSOR: OK, we only
end up with one term.

163
00:10:10,540 --> 00:10:12,770
Actually, I'm going to keep
this one for a second-- A

164
00:10:12,770 --> 00:10:16,030
with respect to O minus.

165
00:10:16,030 --> 00:10:25,360
And we came up with our R
theta dot squared r hat term.

166
00:10:25,360 --> 00:10:28,390
I leave this in here because
I actually don't have to say.

167
00:10:28,390 --> 00:10:33,000
If I wanted now to
do this problem,

168
00:10:33,000 --> 00:10:35,880
if I asked you to do a
problem where I'm doing this,

169
00:10:35,880 --> 00:10:40,039
and I start accelerating,
do you know how to do it?

170
00:10:40,039 --> 00:10:41,705
There's the answer--
still there, right?

171
00:10:46,437 --> 00:11:06,155
All right, let's do a
quick free body diagram.

172
00:11:11,660 --> 00:11:18,920
Here's our mass, my master
coordinate system out here.

173
00:11:18,920 --> 00:11:23,735
Let's draw-- let's say it's
right here at 90 degrees.

174
00:11:26,570 --> 00:11:29,300
What are the external forces
on the mass in this problem?

175
00:11:33,196 --> 00:11:34,660
STUDENT: [INAUDIBLE].

176
00:11:34,660 --> 00:11:37,120
PROFESSOR: So there's
tension in the string, right?

177
00:11:37,120 --> 00:11:41,700
OK, so now this is a really
trivially simple problem.

178
00:11:41,700 --> 00:11:45,660
So the emphasize here
is on the concept.

179
00:11:45,660 --> 00:11:50,100
So now when you're asked to come
up with an equation of motion

180
00:11:50,100 --> 00:11:55,520
or compute the forces on a
mass, use Newton's second law,

181
00:11:55,520 --> 00:11:58,040
F equals mass
times acceleration.

182
00:11:58,040 --> 00:12:01,870
You now have the complete
3D vector formulation

183
00:12:01,870 --> 00:12:08,290
for acceleration of a particle
in a translating rotating

184
00:12:08,290 --> 00:12:09,870
coordinate system.

185
00:12:09,870 --> 00:12:12,780
That's all you need to
compute accelerations for lots

186
00:12:12,780 --> 00:12:14,850
and lots of difficult problems.

187
00:12:14,850 --> 00:12:18,670
And so if you can write
down the acceleration,

188
00:12:18,670 --> 00:12:23,960
you say it's equal to what?

189
00:12:23,960 --> 00:12:26,870
Mass-- if you multiply mass
times that acceleration,

190
00:12:26,870 --> 00:12:28,269
what's that equal to?

191
00:12:28,269 --> 00:12:29,060
STUDENT: The force.

192
00:12:29,060 --> 00:12:31,393
PROFESSOR: The forces that
must be acting on the system.

193
00:12:31,393 --> 00:12:32,860
And that's the point here.

194
00:12:32,860 --> 00:12:37,090
So now I want to know the forces
on the system, the summation

195
00:12:37,090 --> 00:12:38,650
of the external forces.

196
00:12:38,650 --> 00:12:39,940
And then these are vectors.

197
00:12:39,940 --> 00:12:42,100
And you can do them
component by component.

198
00:12:42,100 --> 00:12:47,010
Some of the external forces
in the r hat direction

199
00:12:47,010 --> 00:12:49,810
must be equal to the
mass-- in this case, just

200
00:12:49,810 --> 00:12:56,360
a particle-- times the
acceleration of that particle.

201
00:12:56,360 --> 00:12:58,050
And in this problem,
then that would

202
00:12:58,050 --> 00:13:01,670
be the mass times the
acceleration of A with respect

203
00:13:01,670 --> 00:13:10,540
to O minus R theta
dot squared r hat.

204
00:13:10,540 --> 00:13:15,690
And this would have to be the
r hat component of this thing.

205
00:13:15,690 --> 00:13:17,990
I said I just want
the R component.

206
00:13:17,990 --> 00:13:20,180
I'd have to figure out
if I was running along,

207
00:13:20,180 --> 00:13:23,110
if I had it in the same
direction as r hat.

208
00:13:23,110 --> 00:13:25,490
What part of that acceleration
is in that direction?

209
00:13:25,490 --> 00:13:26,700
That would come here.

210
00:13:26,700 --> 00:13:29,040
If there's 0, you
just make it 0.

211
00:13:29,040 --> 00:13:34,610
So let's just let
the acceleration of A

212
00:13:34,610 --> 00:13:36,830
with respect to O be 0.

213
00:13:36,830 --> 00:13:41,690
Then that says the sum of
the forces in the r direction

214
00:13:41,690 --> 00:13:49,120
is equal to the mass minus
mR theta dot squared.

215
00:13:49,120 --> 00:13:52,210
And if we were to draw
a free body diagram,

216
00:13:52,210 --> 00:13:53,730
we would find out
that, ahh, there

217
00:13:53,730 --> 00:13:55,980
must be a tension on
the string pulling

218
00:13:55,980 --> 00:14:00,120
in on the mass
sufficient to give it

219
00:14:00,120 --> 00:14:03,000
the acceleration
that you've computed.

220
00:14:03,000 --> 00:14:04,460
So every problem,
when you're asked

221
00:14:04,460 --> 00:14:06,790
to compute the force,
or the next step up,

222
00:14:06,790 --> 00:14:09,250
find the equation of motion,
the equation of motion

223
00:14:09,250 --> 00:14:13,115
is just writing this thing out.

224
00:14:20,250 --> 00:14:30,600
OK, now I want to move on to
a more interesting problem.

225
00:14:55,510 --> 00:15:02,260
All right, I'm going
to do this problem.

226
00:15:02,260 --> 00:15:05,540
So it's a hollow tube.

227
00:15:05,540 --> 00:15:09,230
And we're going to look at
things like, I put a ping pong

228
00:15:09,230 --> 00:15:11,010
ball in it.

229
00:15:11,010 --> 00:15:15,345
And if I swing this tube
around, the ping pong ball

230
00:15:15,345 --> 00:15:16,220
is going to come out.

231
00:15:23,290 --> 00:15:25,710
OK, there must be some
forces on that thing

232
00:15:25,710 --> 00:15:27,405
to cause it to come out.

233
00:15:27,405 --> 00:15:29,640
There must be some
accelerations on them.

234
00:15:29,640 --> 00:15:31,960
And so I could
conceivably have an R,

235
00:15:31,960 --> 00:15:35,070
a theta double dot acceleration.

236
00:15:35,070 --> 00:15:39,130
It certainly is going to have
theta dot rotation rates.

237
00:15:39,130 --> 00:15:40,520
The ball is allowed to move.

238
00:15:40,520 --> 00:15:44,375
So there can be nonzero
r dots, r double dots--

239
00:15:44,375 --> 00:15:47,390
a lot going on inside of
this simple little tube.

240
00:15:47,390 --> 00:15:49,500
So that's what I
want to figure out.

241
00:15:49,500 --> 00:15:52,990
Let's see if we can come up
with a model for this problem.

242
00:16:08,090 --> 00:16:09,620
So here's my z-axis.

243
00:16:09,620 --> 00:16:14,980
I have a rotation around it,
some theta dot k hat direction.

244
00:16:14,980 --> 00:16:17,950
Here's my tube.

245
00:16:17,950 --> 00:16:20,020
It's rotating around.

246
00:16:20,020 --> 00:16:22,980
So this is sort of a side
view, your view of the tube.

247
00:16:22,980 --> 00:16:26,940
Here's that ping
pong ball in there.

248
00:16:26,940 --> 00:16:35,110
And I'm going to
idealize that ping pong

249
00:16:35,110 --> 00:16:38,970
ball for a minute, a little
more general problem.

250
00:16:38,970 --> 00:16:42,570
Let's say I have kind of a
nut on this thing, a disk,

251
00:16:42,570 --> 00:16:45,280
something hanging
onto the outside.

252
00:16:45,280 --> 00:16:49,380
And I can control
the rate, the speed

253
00:16:49,380 --> 00:16:51,650
at which this thing goes out.

254
00:16:51,650 --> 00:16:54,330
The ping pong
ball, this is going

255
00:16:54,330 --> 00:16:55,610
to be an application of this.

256
00:16:55,610 --> 00:16:57,735
But I want to be able to
do several other versions,

257
00:16:57,735 --> 00:17:02,120
like make the speed
constant for a second.

258
00:17:02,120 --> 00:17:06,690
OK, and looking
down on this thing,

259
00:17:06,690 --> 00:17:13,000
top view, here's
our inertial frame,

260
00:17:13,000 --> 00:17:14,579
maybe out here like this.

261
00:17:14,579 --> 00:17:16,800
Here's my mass.

262
00:17:16,800 --> 00:17:20,000
So in polar coordinates,
here's your theta.

263
00:17:20,000 --> 00:17:24,192
The r is this.

264
00:17:24,192 --> 00:17:27,150
This is your r hat.

265
00:17:27,150 --> 00:17:28,810
This is your theta
hat directions.

266
00:17:33,750 --> 00:17:39,650
I'm going to let the velocity
of A with respect to O be 0,

267
00:17:39,650 --> 00:17:43,380
so there's no translational
of this system.

268
00:17:43,380 --> 00:17:49,510
And z, z dot, z double
dot, those are all 0.

269
00:17:49,510 --> 00:17:53,280
So nothing's happening
in the z direction.

270
00:17:53,280 --> 00:17:57,090
So I want to first
compute the velocity of B

271
00:17:57,090 --> 00:17:59,740
with respect to O.

272
00:17:59,740 --> 00:18:04,970
And you ought to be able to
do that sort of by inspection.

273
00:18:04,970 --> 00:18:10,830
It comes only from-- oh,
I haven't told you enough.

274
00:18:10,830 --> 00:18:12,790
What do I want to make
happen in this problem?

275
00:18:12,790 --> 00:18:20,993
I want to let r dot--
it's going to be some vr,

276
00:18:20,993 --> 00:18:21,743
and it's constant.

277
00:18:24,660 --> 00:18:27,259
I'm not going to go quite to
my ping pong shooter here yet.

278
00:18:27,259 --> 00:18:29,300
I'm going to do a slightly
simpler problem first.

279
00:18:29,300 --> 00:18:30,260
So this is a constant.

280
00:18:30,260 --> 00:18:32,980
That means r double dot is 0.

281
00:18:32,980 --> 00:18:34,860
So this thing is
just-- let's say

282
00:18:34,860 --> 00:18:37,490
you had threads on this
thing, and it's a screw,

283
00:18:37,490 --> 00:18:41,950
and it's just moving its
way out at a constant rate.

284
00:18:41,950 --> 00:18:47,980
And I'm going to have constant
angular, so theta dot.

285
00:18:47,980 --> 00:18:51,350
I'll call that cap omega k hat.

286
00:18:51,350 --> 00:18:55,450
So the angular rate
is also constant.

287
00:18:55,450 --> 00:18:58,070
All right, if that's the
case, can you tell me,

288
00:18:58,070 --> 00:19:04,030
what's v of B with respect
to my fixed frame O?

289
00:19:07,800 --> 00:19:11,480
Well, any time you're not sure,
you go back to this formula,

290
00:19:11,480 --> 00:19:12,560
throw out terms.

291
00:19:12,560 --> 00:19:14,640
There's no z dot term.

292
00:19:14,640 --> 00:19:15,670
This is constant.

293
00:19:15,670 --> 00:19:18,390
This is 0, 0, 0.

294
00:19:18,390 --> 00:19:19,850
You have this term.

295
00:19:19,850 --> 00:19:22,560
It's some vr in the
r hat direction.

296
00:19:22,560 --> 00:19:26,520
You have this term, wherever it
happens to be in the theta hat

297
00:19:26,520 --> 00:19:27,020
direction.

298
00:19:39,372 --> 00:19:43,950
r dot in the r hat
direction plus r theta

299
00:19:43,950 --> 00:19:46,450
dot in the theta
hat direction-- OK,

300
00:19:46,450 --> 00:19:50,850
we need acceleration
next, B with respect to O.

301
00:19:50,850 --> 00:19:53,110
And now you can crank
through the terms again.

302
00:19:53,110 --> 00:19:57,290
This time, the first term is 0.

303
00:19:57,290 --> 00:19:59,880
The second term is 0,
because it's constant.

304
00:19:59,880 --> 00:20:02,560
The third term is
definitely not 0.

305
00:20:02,560 --> 00:20:03,660
The fourth term is 0.

306
00:20:03,660 --> 00:20:04,280
Fifth term?

307
00:20:08,040 --> 00:20:09,270
0.

308
00:20:09,270 --> 00:20:11,140
This term?

309
00:20:11,140 --> 00:20:13,020
Not 0.

310
00:20:13,020 --> 00:20:14,160
Let me just write them.

311
00:20:18,830 --> 00:20:23,600
So you get a minus r theta
dot squared r hat-- that's

312
00:20:23,600 --> 00:20:30,904
the radial direction term--
plus 2r dot theta dot theta hat.

313
00:20:30,904 --> 00:20:31,945
That's the accelerations.

314
00:20:36,000 --> 00:20:38,530
So you have an acceleration
now in the r hat direction

315
00:20:38,530 --> 00:20:40,640
and in the theta hat direction.

316
00:20:40,640 --> 00:20:42,610
If there's acceleration
to those directions,

317
00:20:42,610 --> 00:20:44,665
there must be forces.

318
00:20:47,680 --> 00:20:51,720
Newton's second law now
says, again, the force

319
00:20:51,720 --> 00:20:55,200
is the mass times acceleration.

320
00:20:55,200 --> 00:20:56,930
And this is a vector.

321
00:20:56,930 --> 00:20:58,390
This is a vector.

322
00:20:58,390 --> 00:20:59,470
It has two components.

323
00:20:59,470 --> 00:21:02,585
And the nice thing about these
Newton's laws and vectors

324
00:21:02,585 --> 00:21:05,180
is you can break the problems
down into their vector

325
00:21:05,180 --> 00:21:07,070
components and treat
the r direction

326
00:21:07,070 --> 00:21:09,300
as one equation of
motion, and the theta

327
00:21:09,300 --> 00:21:11,310
direction as a separate one.

328
00:21:11,310 --> 00:21:15,240
So we might want to draw
a free body diagram.

329
00:21:15,240 --> 00:21:20,260
Here's this block
working its way out.

330
00:21:20,260 --> 00:21:22,660
We know that there's
probably some axial force.

331
00:21:22,660 --> 00:21:24,700
I'm just going to
call it T. And there's

332
00:21:24,700 --> 00:21:27,700
some other unknown force here
in the theta hat direction.

333
00:21:27,700 --> 00:21:30,704
So this is my theta hat.

334
00:21:30,704 --> 00:21:32,870
I've drawn this just
intentionally in the positive r

335
00:21:32,870 --> 00:21:33,490
hat direction.

336
00:21:33,490 --> 00:21:35,210
The sign that comes
out will tell us

337
00:21:35,210 --> 00:21:38,490
which direction it really
is if you're not certain.

338
00:21:38,490 --> 00:21:42,250
Just draw it positive, in
the positive r hat direction.

339
00:21:42,250 --> 00:21:45,520
And that's your
free body diagram.

340
00:21:45,520 --> 00:21:48,180
If I wanted to put gravity
in there, I might have.

341
00:21:48,180 --> 00:21:51,200
But we're doing this in
the horizontal plane.

342
00:21:51,200 --> 00:21:52,560
Gravity is in and out this way.

343
00:21:52,560 --> 00:21:53,670
It's in the z direction.

344
00:21:53,670 --> 00:21:57,740
And we know it's constrained,
can't move. z double dot is 0.

345
00:21:57,740 --> 00:22:01,710
So there's certainly a support
force that picks up the weight.

346
00:22:01,710 --> 00:22:04,170
But this is in our
horizontal plane.

347
00:22:04,170 --> 00:22:06,170
There's your free body diagram.

348
00:22:06,170 --> 00:22:09,310
And we can write two equations
to solve for these things.

349
00:22:09,310 --> 00:22:16,050
So the sum of the forces in
the r hat direction is T.

350
00:22:16,050 --> 00:22:19,640
And that must be equal to the
mass times the acceleration

351
00:22:19,640 --> 00:22:27,690
in the r, minus mr theta dot
squared in the r hat direction.

352
00:22:27,690 --> 00:22:31,890
Sure enough, the tension
has to pull inwards

353
00:22:31,890 --> 00:22:33,820
in the minus r hat direction.

354
00:22:33,820 --> 00:22:39,115
And that's the full result.

355
00:22:39,115 --> 00:22:44,350
STUDENT: Where are the
T and F forces exactly?

356
00:22:44,350 --> 00:22:45,620
PROFESSOR: Where are they?

357
00:22:45,620 --> 00:22:53,450
OK, so I'm going to bring
the ball to the outside

358
00:22:53,450 --> 00:22:54,590
where you can see it.

359
00:22:54,590 --> 00:22:58,260
This thing is going in this
direction, horizontal plane, x,

360
00:22:58,260 --> 00:22:59,380
y plane.

361
00:22:59,380 --> 00:23:03,520
It's moving its fixed rate out.

362
00:23:03,520 --> 00:23:05,550
So its speed when
it's in here is r

363
00:23:05,550 --> 00:23:07,370
over 2 omega in that direction.

364
00:23:07,370 --> 00:23:10,730
And the speed when it's out here
is r omega in that direction.

365
00:23:10,730 --> 00:23:12,630
So clearly it's
picking up speed.

366
00:23:12,630 --> 00:23:17,320
If it's picking up speed, is
it picking up kinetic energy?

367
00:23:17,320 --> 00:23:19,200
Is there work being
done on it somehow

368
00:23:19,200 --> 00:23:20,570
to build up that energy?

369
00:23:20,570 --> 00:23:22,970
So there must be
some forces at play.

370
00:23:22,970 --> 00:23:26,090
So there's a normal
force from the wall

371
00:23:26,090 --> 00:23:29,670
of this thing pushing this
ball sideways to speed it up,

372
00:23:29,670 --> 00:23:30,250
for sure.

373
00:23:30,250 --> 00:23:31,650
That's one force.

374
00:23:31,650 --> 00:23:35,810
And the other force is because
I'm not allowing this thing

375
00:23:35,810 --> 00:23:38,180
just to go freely out.

376
00:23:38,180 --> 00:23:41,310
I'm constraining it
to constant speed out.

377
00:23:41,310 --> 00:23:43,900
It would really like to
go a lot faster than that.

378
00:23:43,900 --> 00:23:46,600
So what's holding it back?

379
00:23:46,600 --> 00:23:49,510
At any instance in time, it
has centripetal acceleration.

380
00:23:49,510 --> 00:23:53,290
And what's making
it go in the circle

381
00:23:53,290 --> 00:23:56,060
is a force that
is-- in this case,

382
00:23:56,060 --> 00:24:00,570
if that were a nut with threads,
in the threads are applying

383
00:24:00,570 --> 00:24:03,368
to the nut to keep
it from running away.

384
00:24:03,368 --> 00:24:05,360
STUDENT: In that
free body diagram,

385
00:24:05,360 --> 00:24:08,965
F, doesn't F act similar
between theta and r hat?

386
00:24:08,965 --> 00:24:10,340
And then there's
like a component

387
00:24:10,340 --> 00:24:13,830
of theta hat in there?

388
00:24:13,830 --> 00:24:15,700
PROFESSOR: Well,
I've broken down.

389
00:24:15,700 --> 00:24:18,080
I've chosen.

390
00:24:18,080 --> 00:24:21,200
The total force
acting on this thing

391
00:24:21,200 --> 00:24:23,890
is some combination of a
force in that direction

392
00:24:23,890 --> 00:24:25,760
and a combination in
the axial direction.

393
00:24:25,760 --> 00:24:31,810
So it has some net direction
that's neither this nor that.

394
00:24:31,810 --> 00:24:34,735
STUDENT: From that equation,
it looks like what you drew,

395
00:24:34,735 --> 00:24:39,640
F has an r hat component.

396
00:24:39,640 --> 00:24:42,830
PROFESSOR: So this
thing is rotating

397
00:24:42,830 --> 00:24:45,560
about some center over here.

398
00:24:45,560 --> 00:24:47,460
So this is theta dot.

399
00:24:47,460 --> 00:24:50,690
Theta is going in this
direction, theta dot.

400
00:24:50,690 --> 00:24:55,050
And we've chosen a coordinate
system that has unit vectors r

401
00:24:55,050 --> 00:24:56,510
hat and theta hat.

402
00:24:56,510 --> 00:24:57,560
And so it makes sense.

403
00:24:57,560 --> 00:24:59,840
We can express the
acceleration in terms

404
00:24:59,840 --> 00:25:01,100
of those two components.

405
00:25:01,100 --> 00:25:05,270
It makes sense to express the
forces in the same direction.

406
00:25:05,270 --> 00:25:08,620
So I've just arbitrarily said,
I have some unknown force

407
00:25:08,620 --> 00:25:09,690
that's in this direction.

408
00:25:09,690 --> 00:25:11,814
And I have another unknown
force in that direction.

409
00:25:14,340 --> 00:25:17,230
Then I'm saying, they
account for all forces

410
00:25:17,230 --> 00:25:19,770
in this direction,
whatever their source.

411
00:25:19,770 --> 00:25:23,000
That tells me that the
sum of the external forces

412
00:25:23,000 --> 00:25:29,090
in the theta hat direction in
this case is this unknown F.

413
00:25:29,090 --> 00:25:31,610
But I know from
Newton that that's

414
00:25:31,610 --> 00:25:33,930
got to be equal to the
mass times the acceleration

415
00:25:33,930 --> 00:25:35,380
in that direction.

416
00:25:35,380 --> 00:25:42,520
In this case, then, that is
2mr dot theta dot theta hat.

417
00:25:45,630 --> 00:25:50,250
So just from
applying the equation

418
00:25:50,250 --> 00:25:52,580
and applying
Newton's second law,

419
00:25:52,580 --> 00:25:56,359
I can find out what
that force must be.

420
00:25:56,359 --> 00:25:57,900
It would've been a
lot more work if I

421
00:25:57,900 --> 00:26:00,286
had drawn it in some arbitrary
in between direction.

422
00:26:00,286 --> 00:26:02,660
Because then I'd have to break
it down into its compounds

423
00:26:02,660 --> 00:26:03,690
to write this.

424
00:26:03,690 --> 00:26:07,060
So I've made it as easy
as possible for myself.

425
00:26:07,060 --> 00:26:08,850
So there's a force like this.

426
00:26:08,850 --> 00:26:10,840
And there is another
force like that.

427
00:26:10,840 --> 00:26:14,550
This one is caused by the
centripetal acceleration.

428
00:26:14,550 --> 00:26:19,820
Or this force causes the
centripetal acceleration.

429
00:26:19,820 --> 00:26:22,080
In order to make something
go in a circular path,

430
00:26:22,080 --> 00:26:24,940
you have to exert a force on it.

431
00:26:24,940 --> 00:26:26,900
That's the force.

432
00:26:26,900 --> 00:26:30,160
In order to accelerate
something angularly,

433
00:26:30,160 --> 00:26:31,500
you have to apply a force.

434
00:26:31,500 --> 00:26:33,760
That comes from the Euler term.

435
00:26:33,760 --> 00:26:36,830
And this is a curious term.

436
00:26:36,830 --> 00:26:38,250
This is the Coriolis term.

437
00:26:41,290 --> 00:26:42,580
So where does it come from?

438
00:26:45,870 --> 00:26:47,900
That's the crux of
the matter here.

439
00:26:47,900 --> 00:26:48,900
Where does it come from?

440
00:26:56,600 --> 00:27:01,420
So part of the reading that you
need to do now is Chapter 15.

441
00:27:01,420 --> 00:27:04,700
Chapter 15, most of it is
going to be complete review.

442
00:27:04,700 --> 00:27:07,810
It just says the conservation
of a linear momentum, impulse

443
00:27:07,810 --> 00:27:08,355
and momentum.

444
00:27:08,355 --> 00:27:10,900
But it also gets into
angular momentum.

445
00:27:10,900 --> 00:27:14,040
So we're going to talk quite
a lot about angular momentum.

446
00:27:17,660 --> 00:27:28,240
And I want to do a very brief
little review right now so

447
00:27:28,240 --> 00:27:29,700
that it applies to this problem.

448
00:27:32,320 --> 00:27:34,110
So we've come up
with an expression

449
00:27:34,110 --> 00:27:36,025
that the force in
the theta direction

450
00:27:36,025 --> 00:27:43,420
here comes-- there's got to
be that, 2mr dot theta dot.

451
00:27:52,820 --> 00:27:56,400
So here's my point O,
and A for that matter.

452
00:27:56,400 --> 00:28:00,200
But looking down on
our problem, here's

453
00:28:00,200 --> 00:28:03,160
my mass at some instant in time.

454
00:28:03,160 --> 00:28:09,400
My rotation rate is theta dot--
or actually it's constant.

455
00:28:09,400 --> 00:28:15,170
So it's k hat like
that, cap omega k hat.

456
00:28:15,170 --> 00:28:20,520
This is my r hat
direction, theta hat.

457
00:28:20,520 --> 00:28:23,710
Now, I'm going to treat
this as a particle.

458
00:28:23,710 --> 00:28:26,530
Not long-- we're going to be
talking about the dynamics

459
00:28:26,530 --> 00:28:27,560
of rigid bodies.

460
00:28:27,560 --> 00:28:29,510
We're just doing
particles for the moment.

461
00:28:29,510 --> 00:28:31,860
We think of them
just as point masses

462
00:28:31,860 --> 00:28:34,660
and don't deal with
their finite extent.

463
00:28:34,660 --> 00:28:37,410
So we're still thinking
of this as a particle.

464
00:28:37,410 --> 00:28:40,310
And I'm going to write down
the definition of the angular

465
00:28:40,310 --> 00:28:42,390
momentum of a particle.

466
00:28:42,390 --> 00:28:47,590
This is B out here with respect
to my fixed frame here at O.

467
00:28:47,590 --> 00:28:49,600
And I'm going to use a
lowercase h to describe

468
00:28:49,600 --> 00:28:51,410
angular momentum of particles.

469
00:28:51,410 --> 00:28:54,640
And I'll use capital H
to describe the angular

470
00:28:54,640 --> 00:28:57,670
momentum of rigid bodies.

471
00:28:57,670 --> 00:29:00,920
So this is a particle,
the definition

472
00:29:00,920 --> 00:29:03,940
of the angular momentum
of a particle with respect

473
00:29:03,940 --> 00:29:07,166
to a fixed point.

474
00:29:07,166 --> 00:29:08,540
We're going to
come back to that.

475
00:29:08,540 --> 00:29:10,670
That'll turn out to
have some significance.

476
00:29:10,670 --> 00:29:17,960
The definition of this is
it's RBO, the position vector,

477
00:29:17,960 --> 00:29:25,535
crossed with the linear momentum
evaluated in the fixed frame.

478
00:29:25,535 --> 00:29:27,660
So that's the definition
of angular momentum, which

479
00:29:27,660 --> 00:29:29,470
is you have linear
momentum, and it's

480
00:29:29,470 --> 00:29:32,810
the cross product with the
position vector out to it.

481
00:29:32,810 --> 00:29:38,845
So in this case, the
RBO is capital R R hat.

482
00:29:44,560 --> 00:29:47,300
Well, it's not-- this
varies, excuse me.

483
00:29:47,300 --> 00:29:50,750
So I'm not going to use--
this I use as a constant.

484
00:29:50,750 --> 00:29:53,090
I better keep it
as the variable.

485
00:29:53,090 --> 00:29:57,140
Pardon that, so it's r,
whatever the local position is,

486
00:29:57,140 --> 00:30:01,970
in the r hat direction crossed
with the linear momentum.

487
00:30:01,970 --> 00:30:06,270
What's the linear
momentum of that particle?

488
00:30:06,270 --> 00:30:09,050
Mass times velocity.

489
00:30:09,050 --> 00:30:09,983
What's the velocity?

490
00:30:09,983 --> 00:30:10,858
STUDENT: [INAUDIBLE].

491
00:30:17,182 --> 00:30:19,390
PROFESSOR: Theta dot-- where
did we write it up here?

492
00:30:19,390 --> 00:30:31,790
Someplace-- the total velocity
is r dot r hat plus r theta

493
00:30:31,790 --> 00:30:33,300
dot theta hat.

494
00:30:36,780 --> 00:30:38,640
And we need a mass in here.

495
00:30:38,640 --> 00:30:44,530
So mass times velocity
would be the momentum.

496
00:30:44,530 --> 00:30:46,670
And we need the cross
products of those.

497
00:30:46,670 --> 00:30:49,310
r hat cross r hat, you
get nothing from that.

498
00:30:49,310 --> 00:30:56,500
And r hat cross theta
hat-- positive or negative?

499
00:30:56,500 --> 00:30:59,040
Positive k, right?

500
00:30:59,040 --> 00:31:13,280
So this becomes mr squared theta
dot in the k hat direction.

501
00:31:19,200 --> 00:31:21,685
And this one is a constant.

502
00:31:21,685 --> 00:31:23,491
We'll write this as cap omega.

503
00:31:27,340 --> 00:31:32,970
So this is my angular momentum
of my particle with respect

504
00:31:32,970 --> 00:31:35,480
to this fixed reference frame.

505
00:31:35,480 --> 00:31:38,354
And so one final
step that you know

506
00:31:38,354 --> 00:31:43,220
from your previous
physics is-- how's torque

507
00:31:43,220 --> 00:31:44,400
related to angular momentum?

508
00:31:49,710 --> 00:31:50,505
Take a guess.

509
00:31:50,505 --> 00:31:52,600
What do you remember?

510
00:31:52,600 --> 00:32:01,870
Time derivative-- so
d by dt of hBO here.

511
00:32:01,870 --> 00:32:04,380
The time rate of
change of this vector

512
00:32:04,380 --> 00:32:11,190
is the torque with respect to
this point, with respect to O.

513
00:32:11,190 --> 00:32:12,920
So what are constants
in this thing?

514
00:32:12,920 --> 00:32:15,590
We don't have to deal
with their derivatives.

515
00:32:15,590 --> 00:32:22,640
k hat is-- does the direction
of the angular momentum change?

516
00:32:22,640 --> 00:32:23,960
It's upwards.

517
00:32:23,960 --> 00:32:26,290
The derivative of k,
the unit vector k,

518
00:32:26,290 --> 00:32:28,430
does it change in this problem?

519
00:32:28,430 --> 00:32:30,930
No, so its time derivative is 0.

520
00:32:30,930 --> 00:32:32,490
Its time derivative is 0.

521
00:32:32,490 --> 00:32:33,999
The m time derivative is 0.

522
00:32:33,999 --> 00:32:36,040
The only thing you have
to take the derivative of

523
00:32:36,040 --> 00:32:37,605
is r, so 2rr dot.

524
00:32:54,350 --> 00:32:59,380
So the torque is
2mrr dot cap omega.

525
00:32:59,380 --> 00:33:01,410
And it's all in the k direction.

526
00:33:01,410 --> 00:33:11,750
And that had better be
rBO cross F. Torque comes

527
00:33:11,750 --> 00:33:14,290
as a force times a
moment arm, right?

528
00:33:14,290 --> 00:33:18,000
And we computed here a force
in the theta direction.

529
00:33:18,000 --> 00:33:20,610
So what would give us a
torque in the k direction?

530
00:33:20,610 --> 00:33:23,090
An r cross a theta.

531
00:33:23,090 --> 00:33:26,440
r hat cross theta
hat gives you a k.

532
00:33:26,440 --> 00:33:29,800
r cross-- and this force,
the part of the force.

533
00:33:29,800 --> 00:33:32,450
We have two forces,
one in the r direction

534
00:33:32,450 --> 00:33:34,470
and one in the theta direction.

535
00:33:34,470 --> 00:33:38,540
The cross product, this term up
here, the T gives you nothing,

536
00:33:38,540 --> 00:33:41,620
r cross r hat cross r hat.

537
00:33:41,620 --> 00:33:46,420
So the only force that matters
here is this Coriolis one.

538
00:33:46,420 --> 00:33:55,800
And so that force is
2mr dot theta dot.

539
00:33:55,800 --> 00:33:59,280
And we have the r cross.

540
00:33:59,280 --> 00:34:01,800
We know this comes out
in the k direction.

541
00:34:01,800 --> 00:34:05,150
And we get an r in here when
we compute this product.

542
00:34:05,150 --> 00:34:09,880
And this, to me, looks
an awful lot like this.

543
00:34:09,880 --> 00:34:11,719
Except I missed an r squared.

544
00:34:11,719 --> 00:34:12,469
How did I do that?

545
00:34:19,049 --> 00:34:19,924
STUDENT: [INAUDIBLE].

546
00:34:25,900 --> 00:34:27,460
PROFESSOR: Oh no,
it's not r squared.

547
00:34:27,460 --> 00:34:29,650
It started off over here as
the angular momentum has an r

548
00:34:29,650 --> 00:34:30,858
squared, took the derivative.

549
00:34:30,858 --> 00:34:33,109
It dropped down to 2rr dot.

550
00:34:33,109 --> 00:34:36,150
And that's right.

551
00:34:36,150 --> 00:34:39,650
So the Coriolis
force, what's it got

552
00:34:39,650 --> 00:34:41,489
to do with the angular momentum?

553
00:34:46,460 --> 00:34:48,520
That's kind of the point
of the exercise here.

554
00:34:48,520 --> 00:34:57,700
In order for this ping pong
ball to be accelerated,

555
00:34:57,700 --> 00:35:00,760
as it goes slowly out this
tube, the angular momentum

556
00:35:00,760 --> 00:35:02,630
is increasing.

557
00:35:02,630 --> 00:35:05,030
In order to change angular
momentum with time,

558
00:35:05,030 --> 00:35:06,910
you have to apply a torque.

559
00:35:06,910 --> 00:35:09,060
The torque that
you have to apply

560
00:35:09,060 --> 00:35:12,700
is 2mrr dot in the k direction.

561
00:35:12,700 --> 00:35:16,670
And that is r cross
the Coriolis force.

562
00:35:16,670 --> 00:35:19,240
So the Coriolis
force, in this case,

563
00:35:19,240 --> 00:35:21,990
is the force that's necessary
to increase the angular

564
00:35:21,990 --> 00:35:24,630
momentum of a system.

565
00:35:24,630 --> 00:35:26,760
That's very often
the reason-- that's

566
00:35:26,760 --> 00:35:30,110
what Coriolis force is about.

567
00:35:30,110 --> 00:35:36,100
So when you shoot an
artillery piece on the Earth,

568
00:35:36,100 --> 00:35:38,410
you've got that projectile
going out there.

569
00:35:38,410 --> 00:35:41,760
It has angular momentum
with respect to the Earth.

570
00:35:41,760 --> 00:35:44,960
And you'll find out that
this little term pops up.

571
00:35:44,960 --> 00:35:47,920
And in fact, the projectile
doesn't go straight.

572
00:35:47,920 --> 00:35:48,590
It curves.

573
00:35:51,550 --> 00:35:54,020
There's lots of things that
because of conservation

574
00:35:54,020 --> 00:35:59,810
of angular momentum, you end
up with this term popping up.

575
00:35:59,810 --> 00:36:01,810
In this case, angular
momentum is not conserved.

576
00:36:01,810 --> 00:36:02,750
It's increasing.

577
00:36:02,750 --> 00:36:04,860
So you have this force
to make it happen.

578
00:36:04,860 --> 00:36:06,260
Any questions about this?

579
00:36:06,260 --> 00:36:08,090
You're going to
use this one a lot.

580
00:36:08,090 --> 00:36:09,548
You're going to
work with it a lot.

581
00:36:17,200 --> 00:36:20,390
So anytime you see changes
in angular momentum happening

582
00:36:20,390 --> 00:36:25,990
in a problem, in these
problems with circular motion,

583
00:36:25,990 --> 00:36:30,060
velocities of parts
increasing in radius,

584
00:36:30,060 --> 00:36:32,860
you'll almost always
see this term pop up.

585
00:36:32,860 --> 00:36:35,200
Any time you see these
changes in angular momentum,

586
00:36:35,200 --> 00:36:38,030
you'll often see
the Coriolis term.

587
00:36:38,030 --> 00:36:46,070
All right, now we're going to
do another interesting problem--

588
00:36:46,070 --> 00:36:48,040
simple but interesting.

589
00:36:48,040 --> 00:36:52,790
And that is really to go--
let's go do this problem

590
00:36:52,790 --> 00:36:55,170
where the thing is really
allowed to come freely out.

591
00:37:02,079 --> 00:37:03,745
All right, you ready
to defend yourself?

592
00:37:12,050 --> 00:37:15,040
A little short stick is pretty
effective at throwing candy.

593
00:37:19,640 --> 00:37:20,950
You got your safety glasses on?

594
00:37:20,950 --> 00:37:24,400
You want me to see if
I can get one up there?

595
00:37:24,400 --> 00:37:27,600
Aww, also, I obviously
haven't practiced this.

596
00:37:31,210 --> 00:37:32,550
All right, last one.

597
00:37:38,380 --> 00:37:40,840
Actually, there's two more.

598
00:37:48,400 --> 00:37:51,300
What makes this work?

599
00:37:51,300 --> 00:37:52,880
Let's do this problem.

600
00:37:56,490 --> 00:37:58,505
So let's look at
our candy shooter.

601
00:38:17,870 --> 00:38:20,460
So I'm whipping
this thing around.

602
00:38:20,460 --> 00:38:21,690
Candy is coming out of here.

603
00:38:21,690 --> 00:38:24,670
It's at B.

604
00:38:24,670 --> 00:38:28,970
Again, I get no z, not
mess with the z part.

605
00:38:28,970 --> 00:38:31,170
The z part is trivial
to usually deal with.

606
00:38:31,170 --> 00:38:33,636
Because it's
totally independent,

607
00:38:33,636 --> 00:38:34,635
just separate equations.

608
00:38:34,635 --> 00:38:36,630
It doesn't complicate
things much at all,

609
00:38:36,630 --> 00:38:38,542
even when you have z.

610
00:38:38,542 --> 00:38:42,340
Now, in this case, the
velocity-- and here's

611
00:38:42,340 --> 00:38:45,230
my O frame here.

612
00:38:45,230 --> 00:38:47,110
The velocity B
with respect to O,

613
00:38:47,110 --> 00:38:52,520
well, now it's got--
I'm not going to move.

614
00:38:52,520 --> 00:38:55,080
I'm standing still when I do it.

615
00:38:55,080 --> 00:38:57,660
So the first term is 0.

616
00:38:57,660 --> 00:39:02,350
It's got an r term, r dot
in the r hat direction.

617
00:39:02,350 --> 00:39:08,570
And it has an r theta
dot theta hat term.

618
00:39:13,810 --> 00:39:17,330
And these can now--
this might be changing.

619
00:39:17,330 --> 00:39:19,380
This might have a time
derivative, r double dot.

620
00:39:19,380 --> 00:39:20,130
It certainly does.

621
00:39:20,130 --> 00:39:22,000
It's accelerating
coming out of that tube.

622
00:39:22,000 --> 00:39:25,670
And my theta, angular motion,
it can be accelerating, too.

623
00:39:25,670 --> 00:39:28,985
So we're going to have
to deal with those.

624
00:39:28,985 --> 00:39:30,360
I need to know
the accelerations.

625
00:39:36,100 --> 00:39:39,060
Now, I haven't
emphasized it till now,

626
00:39:39,060 --> 00:39:42,990
but I find it conceptually
useful to-- when

627
00:39:42,990 --> 00:39:45,750
you work with polar
coordinates, you

628
00:39:45,750 --> 00:39:48,810
can have this
ability to aggregate

629
00:39:48,810 --> 00:39:57,910
the terms in these two
component directions.

630
00:39:57,910 --> 00:39:59,240
So you have this.

631
00:39:59,240 --> 00:40:01,640
All the r terms go together.

632
00:40:01,640 --> 00:40:06,870
And we've let the
z double dot term--

633
00:40:06,870 --> 00:40:08,270
it's just separate by itself.

634
00:40:08,270 --> 00:40:10,380
It drops out easily.

635
00:40:10,380 --> 00:40:12,040
And you have the theta hat term.

636
00:40:12,040 --> 00:40:25,890
And that's the r theta double
dot plus 2r dot theta dot.

637
00:40:25,890 --> 00:40:27,900
And these are in the
theta hat direction.

638
00:40:27,900 --> 00:40:30,530
There's your Coriolis term,
your Euler acceleration

639
00:40:30,530 --> 00:40:33,500
term, centripetal term.

640
00:40:33,500 --> 00:40:34,039
Yeah?

641
00:40:34,039 --> 00:40:34,914
STUDENT: [INAUDIBLE].

642
00:40:38,964 --> 00:40:40,630
PROFESSOR: Well, I
don't know if I ought

643
00:40:40,630 --> 00:40:42,692
to tell you secrets about me.

644
00:40:42,692 --> 00:40:45,025
Because it's going to give
you an advantage on the quiz.

645
00:40:48,300 --> 00:40:50,840
But I've almost
never, ever been known

646
00:40:50,840 --> 00:40:55,630
to ask a question
that says, "derive."

647
00:40:55,630 --> 00:40:59,920
But I'll sure ask you
concept questions.

648
00:40:59,920 --> 00:41:01,930
I really want you to
understand the principles.

649
00:41:01,930 --> 00:41:04,625
I don't get real hung
up on having you do

650
00:41:04,625 --> 00:41:07,029
the grungy grind it out things.

651
00:41:07,029 --> 00:41:08,570
Do I want you to
remember the formula

652
00:41:08,570 --> 00:41:13,130
for how to take the derivative
of a vector in rotating frame?

653
00:41:13,130 --> 00:41:15,150
Yeah, that's where
these have come from.

654
00:41:18,250 --> 00:41:21,830
You had better remember this.

655
00:41:21,830 --> 00:41:24,850
These two formulas, the
velocity formula and this,

656
00:41:24,850 --> 00:41:27,820
the acceleration formula,
are just core to this course.

657
00:41:27,820 --> 00:41:29,670
Now, the way quizzes
are done-- first quiz,

658
00:41:29,670 --> 00:41:31,900
you come in, one sheet of paper.

659
00:41:31,900 --> 00:41:35,580
What had better
be on your paper?

660
00:41:35,580 --> 00:41:37,030
OK?

661
00:41:37,030 --> 00:41:39,470
And second quiz, two
sheets, final, three sheets,

662
00:41:39,470 --> 00:41:40,870
that kind of thing.

663
00:41:40,870 --> 00:41:43,290
But conceptually, you've
got forces in the r,

664
00:41:43,290 --> 00:41:47,000
forces in the z,
accelerations in r theta

665
00:41:47,000 --> 00:41:49,455
and z, forces r theta and z.

666
00:41:52,210 --> 00:41:53,890
And for these planar
motion problems,

667
00:41:53,890 --> 00:41:57,190
this one is sure easy to use.

668
00:41:57,190 --> 00:42:00,980
So let's think
about this problem.

669
00:42:00,980 --> 00:42:06,650
I'm going to let this
be frictionless just

670
00:42:06,650 --> 00:42:09,410
to make it easy.

671
00:42:09,410 --> 00:42:12,130
All right, so what
possible forces act

672
00:42:12,130 --> 00:42:14,870
on the hunk of candy?

673
00:42:14,870 --> 00:42:17,210
Let's do a free body
diagram of the hunk of candy

674
00:42:17,210 --> 00:42:19,680
coming out of here.

675
00:42:19,680 --> 00:42:22,010
What are the forces?

676
00:42:22,010 --> 00:42:23,640
And let's keep it planar.

677
00:42:23,640 --> 00:42:25,080
We can get gravity into this.

678
00:42:25,080 --> 00:42:26,770
But let's just do it in a plane.

679
00:42:26,770 --> 00:42:29,500
So I'm just going horizontal
and slinging this thing.

680
00:42:29,500 --> 00:42:31,160
Gravity's in the z
direction, and I've

681
00:42:31,160 --> 00:42:32,430
constrained it in the z.

682
00:42:32,430 --> 00:42:37,140
So it's something supporting
the gravity in the tube.

683
00:42:37,140 --> 00:42:40,540
So definitely there's an
mg on this thing downwards.

684
00:42:40,540 --> 00:42:42,990
But it's in the z
direction, and we're not

685
00:42:42,990 --> 00:42:45,200
letting it move in the z.

686
00:42:45,200 --> 00:42:49,290
What about the horizontal,
this direction?

687
00:42:49,290 --> 00:42:53,750
The r-- this is my
r hat direction.

688
00:42:53,750 --> 00:42:56,960
This is my theta hat direction.

689
00:42:56,960 --> 00:42:59,516
Whoops, not either--
theta is going this way.

690
00:42:59,516 --> 00:43:03,260
So I've drawn this
as a side view.

691
00:43:03,260 --> 00:43:06,430
What's the force in
the r hat direction?

692
00:43:06,430 --> 00:43:07,305
STUDENT: [INAUDIBLE].

693
00:43:10,600 --> 00:43:12,350
PROFESSOR: OK, in order
to do the problem,

694
00:43:12,350 --> 00:43:14,980
you have to figure that out.

695
00:43:14,980 --> 00:43:15,850
What are the forces?

696
00:43:19,030 --> 00:43:22,150
What are the source of
forces in the r direction?

697
00:43:22,150 --> 00:43:24,016
Here's the r direction.

698
00:43:24,016 --> 00:43:24,890
STUDENT: [INAUDIBLE].

699
00:43:24,890 --> 00:43:25,723
PROFESSOR: Say what?

700
00:43:28,230 --> 00:43:29,140
Come on, you guys.

701
00:43:29,140 --> 00:43:31,158
Somebody be-- yeah.

702
00:43:31,158 --> 00:43:33,650
STUDENT: [INAUDIBLE].

703
00:43:33,650 --> 00:43:34,860
PROFESSOR: There are not any.

704
00:43:34,860 --> 00:43:35,776
Is that what you said?

705
00:43:35,776 --> 00:43:36,710
There aren't any.

706
00:43:36,710 --> 00:43:38,025
And why's that?

707
00:43:38,025 --> 00:43:38,900
STUDENT: [INAUDIBLE].

708
00:43:38,900 --> 00:43:41,780
PROFESSOR: Right, so there's
no forces in the r direction.

709
00:43:41,780 --> 00:43:44,980
So there's no forces
on this thing in the r.

710
00:43:44,980 --> 00:43:48,260
And so then this is a side view.

711
00:43:48,260 --> 00:43:50,260
We could do the top
view looking down.

712
00:43:50,260 --> 00:43:53,260
Top view, you've got your x, y.

713
00:43:53,260 --> 00:43:56,580
You also have your--
here's the ball.

714
00:43:56,580 --> 00:43:58,310
Here's the r hat.

715
00:43:58,310 --> 00:44:00,040
Here's the theta hat.

716
00:44:00,040 --> 00:44:04,800
Now, free body diagram
in the top view--

717
00:44:04,800 --> 00:44:10,937
well, there's some
force here probably.

718
00:44:10,937 --> 00:44:14,290
STUDENT: I was going to
ask that the fact that we

719
00:44:14,290 --> 00:44:16,685
have no forces in the r
hat direction, but we do

720
00:44:16,685 --> 00:44:18,529
have acceleration.

721
00:44:18,529 --> 00:44:19,445
PROFESSOR: Absolutely.

722
00:44:19,445 --> 00:44:22,030
She's commenting that
we do have accelerations

723
00:44:22,030 --> 00:44:23,240
in the r direction, right?

724
00:44:23,240 --> 00:44:24,490
STUDENT: And we have no force.

725
00:44:24,490 --> 00:44:25,910
PROFESSOR: And no force.

726
00:44:25,910 --> 00:44:28,765
So that's the conundrum
of this problem.

727
00:44:28,765 --> 00:44:30,140
That's the point
of this problem.

728
00:44:30,140 --> 00:44:30,931
So let me continue.

729
00:44:34,070 --> 00:44:36,830
Free body diagram--
I'm looking down on it.

730
00:44:36,830 --> 00:44:38,080
I'm allowing for some force.

731
00:44:38,080 --> 00:44:42,590
It's the normal force that
comes from the pipe exerting

732
00:44:42,590 --> 00:44:44,420
the force on the candy.

733
00:44:44,420 --> 00:44:47,410
And since it's frictionless, it
can only be normal to the pipe.

734
00:44:47,410 --> 00:44:54,270
OK, so there's the free body
diagram in the top view.

735
00:44:54,270 --> 00:44:56,466
So we can write two equations.

736
00:44:56,466 --> 00:44:58,510
We can write three
questions-- this one

737
00:44:58,510 --> 00:45:01,286
equal to 0, this one
in the r hat direction,

738
00:45:01,286 --> 00:45:02,785
this one in the
theta hat direction.

739
00:45:22,600 --> 00:45:27,350
Sum of the forces r hat
direction must be 0.

740
00:45:27,350 --> 00:45:28,667
And we have a mass.

741
00:45:28,667 --> 00:45:29,666
We have an acceleration.

742
00:45:40,170 --> 00:45:42,120
Solve for r double dot.

743
00:45:50,860 --> 00:45:58,930
Remarkable-- there's no
force in the r hat direction.

744
00:45:58,930 --> 00:46:05,220
The position of
the object, the r

745
00:46:05,220 --> 00:46:08,490
coordinate, it has a velocity.

746
00:46:08,490 --> 00:46:10,250
It has an acceleration.

747
00:46:14,380 --> 00:46:18,045
But the total acceleration
in the r hat directions

748
00:46:18,045 --> 00:46:19,590
are actually 0.

749
00:46:19,590 --> 00:46:25,790
The rate of change
of the velocity

750
00:46:25,790 --> 00:46:28,860
of this thing in the radial
direction, r double dot,

751
00:46:28,860 --> 00:46:30,870
is nonzero.

752
00:46:30,870 --> 00:46:32,151
But there are no forces on it.

753
00:46:37,340 --> 00:46:42,150
I have pondered another
way to explain this.

754
00:46:42,150 --> 00:46:43,390
I'm still thinking about it.

755
00:46:43,390 --> 00:46:45,010
And you think about this, too.

756
00:46:45,010 --> 00:46:48,300
How do you explain this
in the absence of forces?

757
00:46:48,300 --> 00:46:49,960
And it's partly
where the concept

758
00:46:49,960 --> 00:46:53,760
comes from of fictitious forces,
that centrifugal force is

759
00:46:53,760 --> 00:46:54,580
a force.

760
00:46:54,580 --> 00:46:55,140
It's not.

761
00:46:55,140 --> 00:46:56,600
It's an acceleration.

762
00:46:56,600 --> 00:47:01,330
This is just saying that--
let's go back to this problem.

763
00:47:01,330 --> 00:47:05,170
In order to make
something go in a circle,

764
00:47:05,170 --> 00:47:08,200
you have to put a
force on it to cause

765
00:47:08,200 --> 00:47:11,280
the centripetal acceleration.

766
00:47:11,280 --> 00:47:16,090
You have to allow the thing
to go out if you're not

767
00:47:16,090 --> 00:47:18,190
forcing it to go in a circle.

768
00:47:18,190 --> 00:47:20,940
You have to allow it to go
out at that rate in order

769
00:47:20,940 --> 00:47:25,020
for there to be no centripetal
accelerations on the object.

770
00:47:25,020 --> 00:47:25,521
Yeah?

771
00:47:25,521 --> 00:47:27,603
STUDENT: So I was going
to say, when you start it,

772
00:47:27,603 --> 00:47:29,026
you push it in the y direction.

773
00:47:29,026 --> 00:47:29,930
So that's a force.

774
00:47:29,930 --> 00:47:32,620
It's not in the r hat,
but it's in the y.

775
00:47:32,620 --> 00:47:33,830
PROFESSOR: To get it started.

776
00:47:33,830 --> 00:47:35,958
STUDENT: Right, and there's
no force opposing it

777
00:47:35,958 --> 00:47:38,090
in that direction [INAUDIBLE].

778
00:47:38,090 --> 00:47:45,990
PROFESSOR: Does it experience
centripetal acceleration?

779
00:47:45,990 --> 00:47:47,550
So there's no rotation.

780
00:47:47,550 --> 00:47:50,125
Does it experience
centripetal acceleration?

781
00:47:53,100 --> 00:47:54,560
What do you think?

782
00:47:54,560 --> 00:47:58,450
Yeah, because it goes
through curved motion.

783
00:47:58,450 --> 00:48:00,360
At any instance in time
when it's doing that,

784
00:48:00,360 --> 00:48:02,279
there is a radius of curvature.

785
00:48:02,279 --> 00:48:04,070
You can at that instant
in time think of it

786
00:48:04,070 --> 00:48:07,050
as being in a circular path.

787
00:48:07,050 --> 00:48:09,840
And sometimes it's
just really easy

788
00:48:09,840 --> 00:48:13,510
to do these kind of problems
with normal and tangential

789
00:48:13,510 --> 00:48:14,970
coordinates.

790
00:48:14,970 --> 00:48:23,670
So I'm looking down
on the x, y plane.

791
00:48:23,670 --> 00:48:25,330
Gravity is into the Earth.

792
00:48:25,330 --> 00:48:27,790
So I'm looking down
on a vehicle, a car.

793
00:48:27,790 --> 00:48:31,940
And that car, the
guy is kind of drunk.

794
00:48:31,940 --> 00:48:35,100
He's going down
the road like this.

795
00:48:35,100 --> 00:48:36,440
So here's my y.

796
00:48:36,440 --> 00:48:37,770
Here's my x.

797
00:48:37,770 --> 00:48:49,270
And y equals some A sine
2 pi over the wavelength,

798
00:48:49,270 --> 00:48:58,150
2 pi over lambda, times
x at some A sine kx.

799
00:48:58,150 --> 00:49:00,380
And as he drives down--
if you're in a car,

800
00:49:00,380 --> 00:49:03,490
and you're doing that, you get
thrown side to side in the car.

801
00:49:03,490 --> 00:49:04,997
So you are being accelerated.

802
00:49:04,997 --> 00:49:06,580
And so we want to
be able to calculate

803
00:49:06,580 --> 00:49:09,850
the acceleration due to
the fact that you're going

804
00:49:09,850 --> 00:49:12,480
down and doing a curved path.

805
00:49:12,480 --> 00:49:16,390
And we deal with
these things sometimes

806
00:49:16,390 --> 00:49:19,260
with a convenient little
set of coordinates

807
00:49:19,260 --> 00:49:23,810
that are our normal
and tangential unit

808
00:49:23,810 --> 00:49:30,140
vectors, u normal
and u tangential,

809
00:49:30,140 --> 00:49:31,350
at any instant in time.

810
00:49:34,610 --> 00:49:38,990
And we know that if
this is along the path,

811
00:49:38,990 --> 00:49:41,700
at any instant in time
you're right here,

812
00:49:41,700 --> 00:49:45,070
what direction is your velocity?

813
00:49:45,070 --> 00:49:52,170
Just definition of velocity--
tangent to the path, right?

814
00:49:52,170 --> 00:49:54,170
So at any instant in
time, the velocity

815
00:49:54,170 --> 00:49:58,000
has got to be tangent to
the path at that moment.

816
00:49:58,000 --> 00:50:09,170
So velocity, the vector, has
a magnitude and a unit vector

817
00:50:09,170 --> 00:50:12,606
uT here I'll call it, tangent.

818
00:50:12,606 --> 00:50:15,020
That's all there is to it.

819
00:50:15,020 --> 00:50:20,320
And the acceleration
of this thing

820
00:50:20,320 --> 00:50:22,435
is your time derivative of this.

821
00:50:26,330 --> 00:50:34,000
And that's going to give
you a v dot uT hat plus a v.

822
00:50:34,000 --> 00:50:39,290
And now you need a time
derivative of this guy.

823
00:50:39,290 --> 00:50:41,904
But this is a unit
length vector.

824
00:50:41,904 --> 00:50:43,320
You can plug it
into that equation

825
00:50:43,320 --> 00:50:46,700
for the derivative
of a rotating vector

826
00:50:46,700 --> 00:50:48,440
and calculate what
this should be.

827
00:50:48,440 --> 00:50:50,210
You could also just draw it out.

828
00:50:50,210 --> 00:50:53,060
So I'll draw this for you.

829
00:50:58,880 --> 00:51:01,750
How are we doing on time?

830
00:51:01,750 --> 00:51:04,070
I should just be
able to finish this.

831
00:51:11,500 --> 00:51:16,500
Here's my unit vector in
the tangential direction.

832
00:51:16,500 --> 00:51:23,320
As I'm going around
this curve, this

833
00:51:23,320 --> 00:51:27,650
is my tangential direction.

834
00:51:27,650 --> 00:51:29,280
There's some instant
I have a radius.

835
00:51:29,280 --> 00:51:31,050
We call that rho.

836
00:51:31,050 --> 00:51:33,655
And in the little
time, delta t, I

837
00:51:33,655 --> 00:51:38,300
go through an angle
delta theta in delta t.

838
00:51:38,300 --> 00:51:42,070
And this is my uT vector here.

839
00:51:42,070 --> 00:51:46,120
It changes by a little bit.

840
00:51:46,120 --> 00:51:53,290
That's the change in the uT unit
vector in this time, delta t.

841
00:51:53,290 --> 00:51:55,870
And it goes perpendicular.

842
00:51:55,870 --> 00:51:58,652
And it goes in the
positive un direction.

843
00:52:01,700 --> 00:52:06,050
So delta-- what's the easiest
way to write this one?

844
00:52:12,030 --> 00:52:26,600
Delta uT equals some
theta dot delta t-- that's

845
00:52:26,600 --> 00:52:35,390
the angle-- times the length
of the unit vector, 1.

846
00:52:35,390 --> 00:52:38,410
That's the distance
it goes, so 1.

847
00:52:38,410 --> 00:52:42,990
So r omega, 1 theta
dot is the distance

848
00:52:42,990 --> 00:52:48,500
that this unit vector
goes through in delta t.

849
00:52:48,500 --> 00:52:54,940
And the direction it
goes in is u normal hat.

850
00:52:54,940 --> 00:53:01,820
So delta uT over
delta t limit as t

851
00:53:01,820 --> 00:53:11,280
goes to 0, you get theta
dot un, just like before.

852
00:53:11,280 --> 00:53:13,700
So the time derivative
of this unit vector

853
00:53:13,700 --> 00:53:17,080
in the tangential
direction is just theta dot

854
00:53:17,080 --> 00:53:19,770
in the normal direction.

855
00:53:19,770 --> 00:53:24,090
And then from that,
we can very quickly

856
00:53:24,090 --> 00:53:27,630
derive the rest of
this acceleration.

857
00:53:27,630 --> 00:53:30,820
The acceleration then
is that plus this.

858
00:53:30,820 --> 00:53:41,119
We now know an expression
for-- this is my uT dot term.

859
00:53:41,119 --> 00:53:42,410
I'm going to plug that in here.

860
00:53:42,410 --> 00:53:45,510
This then, we need an
expression for v. What's v?

861
00:53:48,770 --> 00:53:52,000
Well, at that instant in
time, it has some radius rho.

862
00:53:52,000 --> 00:53:55,100
It has an angular
velocity theta dot.

863
00:53:55,100 --> 00:53:57,930
So rho theta dot
would be the v here.

864
00:54:07,980 --> 00:54:13,240
So I'm looking for an
expression for vuT dot.

865
00:54:18,300 --> 00:54:24,928
So that's v theta dot un.

866
00:54:24,928 --> 00:54:28,550
But theta dot is v over rho.

867
00:54:31,150 --> 00:54:37,210
v squared-- sorry about
this-- over rho un.

868
00:54:37,210 --> 00:54:45,160
So this guy up here, this
is v dot uT plus v squared--

869
00:54:45,160 --> 00:54:47,150
I'll rewrite it-- over rho un.

870
00:54:50,082 --> 00:54:51,540
So if you're speeding
up, if you're

871
00:54:51,540 --> 00:54:54,250
going from 30 miles an
hour to 40 miles an hour,

872
00:54:54,250 --> 00:54:55,520
that's your tangential.

873
00:54:55,520 --> 00:54:57,360
That's your speed
along the path.

874
00:54:57,360 --> 00:54:58,901
That's this term.

875
00:54:58,901 --> 00:55:00,650
But because you're
going around the curve,

876
00:55:00,650 --> 00:55:02,720
you have an acceleration
of v squared over rho.

877
00:55:02,720 --> 00:55:05,010
You've run into this
before in physics.

878
00:55:05,010 --> 00:55:07,760
This is where it comes from.

879
00:55:07,760 --> 00:55:12,490
This is a centripetal
acceleration like term.

880
00:55:12,490 --> 00:55:15,210
If you replace v
with rho theta dot,

881
00:55:15,210 --> 00:55:17,557
you'd get r theta dot squared.

882
00:55:17,557 --> 00:55:19,890
So you can either put it in
terms of v squared over rho,

883
00:55:19,890 --> 00:55:22,820
or you can put it in terms
of rho theta dot squared.

884
00:55:22,820 --> 00:55:28,350
And rho theta dot squared sounds
a lot like my acceleration term

885
00:55:28,350 --> 00:55:28,890
right here.

886
00:55:32,280 --> 00:55:35,500
So with that simple
little formula,

887
00:55:35,500 --> 00:55:38,580
you can do-- you
need one other thing.

888
00:55:41,160 --> 00:55:44,960
And you just go look
it up in the book.

889
00:55:44,960 --> 00:55:48,680
There is an expression from
calculus for the radius

890
00:55:48,680 --> 00:55:52,570
of curvature of a path.

891
00:55:52,570 --> 00:55:56,480
And it has first and second
derivatives of y with respect

892
00:55:56,480 --> 00:55:56,980
to x.

893
00:55:56,980 --> 00:56:00,710
So you need a dy/dx
and a d2y dx squared.

894
00:56:00,710 --> 00:56:03,650
From a sine function,
you can calculate that.

895
00:56:03,650 --> 00:56:05,960
So you calculate
rho from a formula

896
00:56:05,960 --> 00:56:07,910
that's in the book
for calculating

897
00:56:07,910 --> 00:56:09,240
radius of curvature.

898
00:56:09,240 --> 00:56:11,100
And then you're done, all right?

899
00:56:11,100 --> 00:56:15,520
So see you on Thursday.