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PROFESSOR: So I'm
going to give you
9
00:00:22,330 --> 00:00:30,130
a quick example of what I think
is a good way to do solutions.
10
00:00:30,130 --> 00:00:31,405
Our approach to solutions.
11
00:00:45,090 --> 00:00:45,980
State the problem.
12
00:00:48,910 --> 00:00:51,030
I'm going to give you a
little formulaic here.
13
00:00:51,030 --> 00:00:51,720
Draw figures.
14
00:01:00,870 --> 00:01:03,440
On the first day I said I like
to think of dynamics problem.
15
00:01:03,440 --> 00:01:06,500
I break them down
into three categories.
16
00:01:06,500 --> 00:01:08,185
One is to describe the motion.
17
00:01:14,120 --> 00:01:15,810
What does describing
the motion mean?
18
00:01:15,810 --> 00:01:21,670
Well how many-- the
number of degrees of
19
00:01:21,670 --> 00:01:23,790
freedom in the problem.
20
00:01:23,790 --> 00:01:30,450
That implies the number
of equations of motion
21
00:01:30,450 --> 00:01:33,360
that you're going to need
to solve that problem.
22
00:01:33,360 --> 00:01:35,710
So you got to identify the
number of degrees of freedom.
23
00:01:35,710 --> 00:01:41,370
It also then tells you the
number of coordinates you need.
24
00:01:44,310 --> 00:01:46,730
So this is all part of
describing the motion.
25
00:01:46,730 --> 00:01:50,040
It's figuring out how many
coordinates, assigning them.
26
00:01:58,400 --> 00:02:01,860
So assigning the coordinates.
27
00:02:01,860 --> 00:02:05,520
And then finally, essentially
all underneath this you
28
00:02:05,520 --> 00:02:06,896
essentially do the kinematics.
29
00:02:21,240 --> 00:02:27,097
And that's the velocities,
accelerations, so forth.
30
00:02:27,097 --> 00:02:29,055
So once you have-- you've
explained the motion.
31
00:02:34,420 --> 00:02:35,960
And this I guess is four.
32
00:02:35,960 --> 00:02:43,155
Explain the correct
physical laws.
33
00:02:47,060 --> 00:02:48,460
You know how they apply.
34
00:02:48,460 --> 00:02:50,280
f equals ma.
35
00:02:50,280 --> 00:02:52,925
Newton's first, second, third
law of conservation of momentum
36
00:02:52,925 --> 00:02:56,750
or whatever you want-- you
think is the appropriate thing.
37
00:02:56,750 --> 00:02:59,840
So explain what the physical
laws are and apply them.
38
00:03:02,620 --> 00:03:05,100
And finally, do the math.
39
00:03:10,287 --> 00:03:12,120
So if you could break
problems down that way
40
00:03:12,120 --> 00:03:14,450
it'll give you a
nice, logical flow.
41
00:03:14,450 --> 00:03:16,980
So I'm going to give you a
bit of an example problem.
42
00:03:27,140 --> 00:03:29,640
And I'm also going to kind
of pose a brain teaser to you
43
00:03:29,640 --> 00:03:31,060
at the end of this
problem today.
44
00:03:31,060 --> 00:03:33,660
I want to give you
something to think about.
45
00:03:33,660 --> 00:03:37,710
So I'm going to draw my problem.
46
00:03:37,710 --> 00:03:43,130
This is a block on an incline.
47
00:03:46,840 --> 00:03:50,820
It's got some scales
to measure your weight.
48
00:03:50,820 --> 00:03:55,510
And you're standing
on this thing.
49
00:03:55,510 --> 00:03:56,695
Riding it down the incline.
50
00:04:00,000 --> 00:04:02,870
And the first
question about this
51
00:04:02,870 --> 00:04:08,490
is to find the position
as a function of time.
52
00:04:26,534 --> 00:04:27,450
So that's the problem.
53
00:04:27,450 --> 00:04:28,980
That's part a.
54
00:04:32,650 --> 00:04:34,670
So state the problem,
find the position.
55
00:04:45,024 --> 00:04:46,190
So how-- well what do we do?
56
00:04:46,190 --> 00:04:48,690
Well draw figures,
I've started with that.
57
00:04:48,690 --> 00:04:50,240
Next, describe the motion.
58
00:04:50,240 --> 00:04:56,140
I need a free-- We
have to figure out
59
00:04:56,140 --> 00:04:59,360
how many degrees of
freedom this problem has.
60
00:04:59,360 --> 00:05:01,030
I'm going to just
declare no rotation.
61
00:05:01,030 --> 00:05:04,180
Going to treat it as a particle.
62
00:05:04,180 --> 00:05:06,871
So a particle has how many,
generally how many degrees of
63
00:05:06,871 --> 00:05:07,370
freedom?
64
00:05:07,370 --> 00:05:09,190
How many coordinates
to completely describe
65
00:05:09,190 --> 00:05:10,226
where it's at?
66
00:05:10,226 --> 00:05:11,040
AUDIENCE: Three.
67
00:05:11,040 --> 00:05:12,240
PROFESSOR: Three.
68
00:05:12,240 --> 00:05:14,880
So I may need as many
as three coordinates
69
00:05:14,880 --> 00:05:17,180
to describe the
motion of this thing.
70
00:05:17,180 --> 00:05:19,780
And if I really doing
complete equations of motion
71
00:05:19,780 --> 00:05:21,330
I need three
equations of motion.
72
00:05:21,330 --> 00:05:25,560
So this is two, three--
describing the motion I
73
00:05:25,560 --> 00:05:28,576
have three degrees of freedom.
74
00:05:28,576 --> 00:05:30,075
I'm going to need
three coordinates.
75
00:05:36,820 --> 00:05:42,200
So in this case
here's my picture
76
00:05:42,200 --> 00:05:45,560
I'm going just to set up a
Cartesian coordinate system
77
00:05:45,560 --> 00:05:49,970
aligned in a helpful way.
78
00:05:49,970 --> 00:05:53,390
X, y, z coming out.
79
00:05:53,390 --> 00:05:56,910
And this is my fixed
inertial reference frame.
80
00:05:56,910 --> 00:05:58,820
And here's my center of mass.
81
00:05:58,820 --> 00:06:02,050
And basically my coordinates
are describing the position
82
00:06:02,050 --> 00:06:03,130
of the center of mass.
83
00:06:19,140 --> 00:06:22,230
So that's pretty much the
describing the motion,
84
00:06:22,230 --> 00:06:23,340
what I need for now.
85
00:06:23,340 --> 00:06:25,250
We'll get to the velocities
and accelerations
86
00:06:25,250 --> 00:06:27,666
when we get to the math part.
87
00:06:27,666 --> 00:06:28,415
Apply the physics.
88
00:06:36,570 --> 00:06:38,380
I'm going to use
Newton's second law.
89
00:06:48,570 --> 00:06:55,260
Sum of the external forces,
mass times the acceleration.
90
00:06:55,260 --> 00:06:59,584
That's the physical
law I'm going to apply.
91
00:06:59,584 --> 00:07:00,625
Draw a free body diagram.
92
00:07:16,800 --> 00:07:18,870
And I'm just going
to consider the block
93
00:07:18,870 --> 00:07:22,709
and the person this is the whole
collection, it's one thing.
94
00:07:22,709 --> 00:07:24,500
I'm just not keep
drawing the person on it,
95
00:07:24,500 --> 00:07:26,840
but-- so here's my
object, including
96
00:07:26,840 --> 00:07:29,540
the weight of the person.
97
00:07:29,540 --> 00:07:35,410
And it's going to have--
here's its center of mass.
98
00:07:35,410 --> 00:07:39,040
Obviously in mg,
gravitational force,
99
00:07:39,040 --> 00:07:42,610
it's going to have
a normal force.
100
00:07:42,610 --> 00:07:44,492
Going to have a friction
force and how do I
101
00:07:44,492 --> 00:07:46,950
figure out what the friction--
which direction the friction
102
00:07:46,950 --> 00:07:47,645
is?
103
00:07:47,645 --> 00:07:51,310
I assume motion down the hill.
104
00:07:51,310 --> 00:07:53,070
Friction will oppose it.
105
00:07:53,070 --> 00:07:55,040
I draw in the arrows
in the direction
106
00:07:55,040 --> 00:07:57,590
I expect the forces to act.
107
00:07:57,590 --> 00:07:58,965
Then I'll use the
sign convention
108
00:07:58,965 --> 00:08:01,340
that the arrows
tell me what signs.
109
00:08:01,340 --> 00:08:06,780
So since my x-coordinate
is down the hill.
110
00:08:06,780 --> 00:08:08,080
This is y.
111
00:08:08,080 --> 00:08:11,320
This is x, friction x,
in the minus x direction.
112
00:08:17,820 --> 00:08:23,900
And I haven't-- I've left out
a key piece of information.
113
00:08:23,900 --> 00:08:26,710
Got to have the
angle of the slope.
114
00:08:26,710 --> 00:08:37,270
And once you have the angle
of the slope that's theta,
115
00:08:37,270 --> 00:08:38,840
this is theta.
116
00:08:38,840 --> 00:08:42,500
And I'm going to need to--
this is in a direction of one
117
00:08:42,500 --> 00:08:44,120
of my coordinates
and so is this,
118
00:08:44,120 --> 00:08:47,120
but I need to break
the gravitational piece
119
00:08:47,120 --> 00:08:51,520
into components lined
up with my coordinates.
120
00:08:51,520 --> 00:08:55,420
And so you have a
theta here as well.
121
00:08:55,420 --> 00:08:57,445
And now I can write
my equation to motion.
122
00:09:02,450 --> 00:09:05,350
And the nice thing
about vectors is
123
00:09:05,350 --> 00:09:08,690
that when you have three
equations of motion,
124
00:09:08,690 --> 00:09:12,950
three coordinates, each that
are components of vectors
125
00:09:12,950 --> 00:09:14,770
in the x, y and z
direction, it gives us
126
00:09:14,770 --> 00:09:17,130
three equations immediately.
127
00:09:17,130 --> 00:09:19,590
So for example, the
summation of the forces
128
00:09:19,590 --> 00:09:22,230
in this problem in
the z-direction,
129
00:09:22,230 --> 00:09:25,770
the external forces are sums to?
130
00:09:25,770 --> 00:09:26,469
AUDIENCE: 0.
131
00:09:26,469 --> 00:09:27,010
PROFESSOR: 0.
132
00:09:27,010 --> 00:09:29,620
OK so we get a trivial
solution out of that.
133
00:09:29,620 --> 00:09:34,340
And we don't have
to go much further.
134
00:09:34,340 --> 00:09:36,760
Summation of the forces
in the y-direction
135
00:09:36,760 --> 00:09:38,180
gives us some
useful information.
136
00:09:42,770 --> 00:09:50,860
And then the y-directed forces
I have an n and a minus mg
137
00:09:50,860 --> 00:09:53,910
and I think it's cosine theta.
138
00:09:59,320 --> 00:10:01,580
Which tells me
immediately what n is.
139
00:10:01,580 --> 00:10:09,630
So from statics I get to-- and
from what we know about models
140
00:10:09,630 --> 00:10:16,870
of friction then we know that
we can model the friction as mu
141
00:10:16,870 --> 00:10:25,120
times n for mu mg cosine theta.
142
00:10:25,120 --> 00:10:27,726
So from the statics I
learn a bunch of things
143
00:10:27,726 --> 00:10:29,225
that I need to know
for the problem.
144
00:10:34,760 --> 00:10:43,389
So now I get to the real
heart of the problem,
145
00:10:43,389 --> 00:10:45,430
writing my equation of
motion in the x-direction.
146
00:10:56,370 --> 00:11:00,580
And the forces in
the x-direction mg
147
00:11:00,580 --> 00:11:02,100
sine theta down the hill.
148
00:11:02,100 --> 00:11:04,920
So it's positive.
149
00:11:04,920 --> 00:11:08,540
Minus the friction
is up the hill.
150
00:11:08,540 --> 00:11:09,500
So I get mg.
151
00:11:18,264 --> 00:11:34,280
I'm mixing up my m's here but
there's no other m so-- this
152
00:11:34,280 --> 00:11:40,280
basically says, that x double
dot, the m's all cancel out.
153
00:11:40,280 --> 00:11:52,470
That x double dot is g sine
theta minus mu g cosine theta.
154
00:11:52,470 --> 00:11:57,140
And that just happens to be
a pretty simple to solve,
155
00:11:57,140 --> 00:11:58,920
ordinary differential equation.
156
00:11:58,920 --> 00:12:01,650
This is an equation in which
the acceleration in the problem
157
00:12:01,650 --> 00:12:02,290
is constant.
158
00:12:02,290 --> 00:12:04,000
The data's not changing.
159
00:12:04,000 --> 00:12:05,530
None of these
things are changing
160
00:12:05,530 --> 00:12:07,497
and so you can just
solve this one.
161
00:12:07,497 --> 00:12:09,205
Now we're to the third
part, do the math.
162
00:12:16,110 --> 00:12:18,550
This one you can just integrate.
163
00:12:18,550 --> 00:12:21,525
And so you find out
that well, x dot then
164
00:12:21,525 --> 00:12:25,370
and I'm just going to
call this c some constant.
165
00:12:25,370 --> 00:12:31,340
So x dot ct plus an initial
velocity, if it had one.
166
00:12:31,340 --> 00:12:34,280
And x of t, what
you're looking for.
167
00:12:45,120 --> 00:12:48,370
And now so v0 and x0 are
just your initial conditions.
168
00:12:48,370 --> 00:12:51,410
More than likely 0 if
you set it up cleverly.
169
00:12:51,410 --> 00:12:53,890
So that's just
modeling quickly what
170
00:12:53,890 --> 00:12:56,490
I think a good way to
lay out a problem is.
171
00:12:56,490 --> 00:13:00,230
Describing the motion,
explaining the physics,
172
00:13:00,230 --> 00:13:05,580
doing the math, drawing good
pictures, stating the problem.
173
00:13:05,580 --> 00:13:07,809
All right so now
the brain teaser
174
00:13:07,809 --> 00:13:09,100
that I want you to think about.
175
00:13:33,360 --> 00:13:35,725
So here's the mass of the
block plus the scales.
176
00:13:39,760 --> 00:13:42,290
And here's the-- here
you're standing on it.
177
00:13:45,800 --> 00:13:47,190
So mass of the person.
178
00:13:52,550 --> 00:13:53,930
You're riding this
down the hill.
179
00:14:10,630 --> 00:14:11,530
So this is part b.
180
00:14:22,706 --> 00:14:24,205
Think about that
when in the shower.
181
00:14:27,400 --> 00:14:30,610
If you've got a really
simple way to do it,
182
00:14:30,610 --> 00:14:32,110
great write it up.
183
00:14:32,110 --> 00:14:34,920
It's not terribly hard and it's
mostly-- I'll give you a hint.
184
00:14:34,920 --> 00:14:39,500
Thinking in terms of free
body diagram helps a lot.
185
00:14:39,500 --> 00:14:41,930
And we'll come back to
this kind of fun problem.
186
00:14:56,440 --> 00:14:58,890
OK, want to go--
that was part one.
187
00:14:58,890 --> 00:15:03,210
I want to go onto this recapping
the center of mass quickly.
188
00:15:03,210 --> 00:15:11,519
We learned a couple
of important things.
189
00:15:11,519 --> 00:15:13,310
We got-- we talked
about the center of mass
190
00:15:13,310 --> 00:15:16,730
because we were just talking
about Newton's three laws.
191
00:15:16,730 --> 00:15:18,894
From looking at the
first law, found
192
00:15:18,894 --> 00:15:20,810
that it's useful in
determining whether or not
193
00:15:20,810 --> 00:15:22,530
you're in a inertial
frame, we used it.
194
00:15:22,530 --> 00:15:24,320
Second law we've
just applied it.
195
00:15:24,320 --> 00:15:26,960
We used to do-- get
equations of motion.
196
00:15:26,960 --> 00:15:31,730
Third law was about--
we used it when we're
197
00:15:31,730 --> 00:15:33,150
thinking about center of mass.
198
00:15:33,150 --> 00:15:36,040
We used it to define what
the center of mass was.
199
00:15:36,040 --> 00:15:40,730
So we said the total
mass of a system,
200
00:15:40,730 --> 00:15:41,890
I better draw my picture.
201
00:15:41,890 --> 00:15:43,245
Here's my system of particles.
202
00:15:45,940 --> 00:15:50,910
M1 with position vectors.
203
00:15:50,910 --> 00:15:53,420
So a whole mess of particles
with their position vectors.
204
00:15:53,420 --> 00:15:56,270
This is ri with respect
to O for example.
205
00:15:56,270 --> 00:15:59,480
This is my O x, y, z frame.
206
00:15:59,480 --> 00:16:03,680
We said that this total
mass of the particles
207
00:16:03,680 --> 00:16:11,520
somewhere out here there's a
center of mass with a position
208
00:16:11,520 --> 00:16:22,300
vector rg with respect to O.
So the definition of my center
209
00:16:22,300 --> 00:16:29,140
of mass is this is equal to the
summation of the m, i, r, i, o
210
00:16:29,140 --> 00:16:33,170
and these are position vectors.
211
00:16:33,170 --> 00:16:35,760
So that's the definition
of my center of mass.
212
00:16:35,760 --> 00:16:43,215
If I take two time derivatives
of that we arrived at mt rg
213
00:16:43,215 --> 00:16:46,760
with the respect
to O double dot.
214
00:16:46,760 --> 00:16:53,500
Summation over i of my m,
i, r, i, o double dots.
215
00:16:57,040 --> 00:17:01,730
And then importantly
that's the summation
216
00:17:01,730 --> 00:17:07,883
of all of the external
forces on each
217
00:17:07,883 --> 00:17:09,869
of these-- each one
of these by itself
218
00:17:09,869 --> 00:17:12,780
satisfies Newton's
law, second law, which
219
00:17:12,780 --> 00:17:14,690
he wrote about particles.
220
00:17:14,690 --> 00:17:16,105
Each one has a summation.
221
00:17:16,105 --> 00:17:17,980
I've summed these and
I'm going to sum these,
222
00:17:17,980 --> 00:17:23,920
but each one has a summation of
internal forces acting on it.
223
00:17:23,920 --> 00:17:27,130
These I call the f, i, j's.
224
00:17:27,130 --> 00:17:29,900
And we learn-- something
about the third law
225
00:17:29,900 --> 00:17:31,530
tells us about that summation.
226
00:17:31,530 --> 00:17:34,390
The third law tells us what?
227
00:17:34,390 --> 00:17:38,650
That goes to 0 and that was
that's the really powerful
228
00:17:38,650 --> 00:17:42,530
piece of the third law
that we make great use of.
229
00:17:42,530 --> 00:17:46,190
Because this now essentially
allows us to say,
230
00:17:46,190 --> 00:17:49,060
that the summation
of the external
231
00:17:49,060 --> 00:17:53,110
forces on an assembly
of particles,
232
00:17:53,110 --> 00:18:06,140
on a system of particles,
is equal to the total mass
233
00:18:06,140 --> 00:18:11,820
times the acceleration
of the center of mass.
234
00:18:11,820 --> 00:18:17,080
And that-- what that
does in one stroke
235
00:18:17,080 --> 00:18:22,400
takes you from Newton, who's
laws applied to particles
236
00:18:22,400 --> 00:18:27,487
and allows you to apply Newton's
second law to rigid bodies.
237
00:18:27,487 --> 00:18:29,070
Because rigid bodies
can be thought up
238
00:18:29,070 --> 00:18:33,250
a bunch of particles,
which are represented
239
00:18:33,250 --> 00:18:36,520
in that simple equation.
240
00:18:36,520 --> 00:18:39,430
And that gets you from
particles to rigid bodies.
241
00:18:39,430 --> 00:18:42,190
And we all know from
physics that a summation
242
00:18:42,190 --> 00:18:44,200
of the external
force on this thing
243
00:18:44,200 --> 00:18:45,834
is the mass times
the acceleration
244
00:18:45,834 --> 00:18:46,750
of the center of mass.
245
00:18:50,000 --> 00:18:53,730
So that's actually quite
an important powerful law.
246
00:18:53,730 --> 00:18:57,060
It provides this for us.
247
00:18:57,060 --> 00:19:01,900
Now I said I wanted to
give you a quick, very
248
00:19:01,900 --> 00:19:04,970
useful application of
thinking about center of mass.
249
00:19:04,970 --> 00:19:08,870
I showed you the other day, I
had my carbon fiber tube here.
250
00:19:08,870 --> 00:19:11,120
Showed you that trick for
finding the center of mass
251
00:19:11,120 --> 00:19:12,703
just by sliding your
fingers along it.
252
00:19:12,703 --> 00:19:14,730
And you end up at
the center of mass.
253
00:19:14,730 --> 00:19:17,190
Well as a practical
matter other things--
254
00:19:17,190 --> 00:19:19,900
you really want to be
able find center of mass.
255
00:19:19,900 --> 00:19:23,290
This is a glider, a
sailplane called an LS8.
256
00:19:23,290 --> 00:19:26,330
I happen-- I'm a glider
flight instructor.
257
00:19:26,330 --> 00:19:29,660
Been flying gliders for 35
years or something like that.
258
00:19:29,660 --> 00:19:34,280
And I flew a glider of
this type just recently.
259
00:19:34,280 --> 00:19:38,980
That machine has a 49 to
one let's call it 50 to one
260
00:19:38,980 --> 00:19:41,710
to make it easy, glide ratio.
261
00:19:41,710 --> 00:19:45,030
Means if you're a mile above
the ground in still air,
262
00:19:45,030 --> 00:19:47,850
you will go 50 miles before
you touch the ground.
263
00:19:47,850 --> 00:19:50,840
So they're really amazing
high performance machines.
264
00:19:50,840 --> 00:19:53,930
One of the things about all
aircraft that you actually
265
00:19:53,930 --> 00:19:56,810
need to know is, you
really need to know where
266
00:19:56,810 --> 00:19:59,630
the center of mass of it is.
267
00:19:59,630 --> 00:20:02,140
And if the center of mass
is in the wrong place
268
00:20:02,140 --> 00:20:04,780
the plane will not fly properly.
269
00:20:04,780 --> 00:20:07,475
And so you can't just go
throw on 50 pounds of lead
270
00:20:07,475 --> 00:20:09,300
in the tail of that
plane and expect
271
00:20:09,300 --> 00:20:12,160
to survive the next flight.
272
00:20:12,160 --> 00:20:14,250
So you have to know
where the center of mass
273
00:20:14,250 --> 00:20:16,560
is, and in fact, you
want the center of mass
274
00:20:16,560 --> 00:20:21,040
about 25% of the-- if the
wing is this wide from front
275
00:20:21,040 --> 00:20:22,890
to back it's called the cord.
276
00:20:22,890 --> 00:20:25,600
About 25% back from
the leading edge
277
00:20:25,600 --> 00:20:30,560
is about where the center
of lift of a wing is.
278
00:20:30,560 --> 00:20:33,110
And you want your
center of mass also
279
00:20:33,110 --> 00:20:35,350
called the center of
gravity in these situations,
280
00:20:35,350 --> 00:20:37,890
you want it to be pretty
close to the center of lift
281
00:20:37,890 --> 00:20:41,130
so that their balance in
the plane flies nicely.
282
00:20:41,130 --> 00:20:52,780
So is there a simple way to find
the center of mass of something
283
00:20:52,780 --> 00:20:55,380
like a sailplane?
284
00:20:55,380 --> 00:20:58,060
So I'm going to drew-- this
is exactly how you do it.
285
00:20:58,060 --> 00:21:01,490
So I'll draw a quick picture
of my sailplane here.
286
00:21:11,302 --> 00:21:13,760
You usually have a little skid
or a tail wheel on the back.
287
00:21:13,760 --> 00:21:17,210
And to find the center of mass
you just set them on scales.
288
00:21:21,270 --> 00:21:24,410
You weigh it.
289
00:21:24,410 --> 00:21:26,340
You pick a coordinate system.
290
00:21:26,340 --> 00:21:29,200
Doesn't matter where it
is, as long as it's fixed.
291
00:21:29,200 --> 00:21:32,960
The easiest place is right
at the nose of the sailplane.
292
00:21:32,960 --> 00:21:37,180
So we'll make this
x, make this y.
293
00:21:37,180 --> 00:21:39,080
You take it and you
can take a tape measure
294
00:21:39,080 --> 00:21:45,180
and measure the distance
from your reference point
295
00:21:45,180 --> 00:21:48,200
to the position where the
wheel sits on the scales.
296
00:21:48,200 --> 00:21:50,720
We'll call that L1 here.
297
00:21:50,720 --> 00:21:53,820
And that's typically
about five feet.
298
00:21:53,820 --> 00:21:58,230
And back here you
have another position
299
00:21:58,230 --> 00:22:02,350
to where you have your second
set of scales, that's L2.
300
00:22:02,350 --> 00:22:05,465
And a typical sailplane
that's about 15 feet.
301
00:22:14,420 --> 00:22:16,247
Apply Newton's law.
302
00:22:16,247 --> 00:22:17,580
This thing's not going anywhere.
303
00:22:17,580 --> 00:22:19,910
Sum of the forces in
the vertical direction
304
00:22:19,910 --> 00:22:23,710
is equal to 0.
305
00:22:23,710 --> 00:22:25,390
Free body diagram.
306
00:22:25,390 --> 00:22:30,990
Well you have somewhere about
here, where the wing is,
307
00:22:30,990 --> 00:22:33,170
this is your center of mass.
308
00:22:33,170 --> 00:22:38,880
And you have m total
times g downwards there.
309
00:22:38,880 --> 00:22:42,715
You have two weights
pushing on the sailplane.
310
00:22:42,715 --> 00:22:46,920
A W2 pushing up,
holding up the tail.
311
00:22:46,920 --> 00:22:51,040
And a W1 holding
up the main gear.
312
00:22:51,040 --> 00:22:58,250
And so from the sum of the
forces in the y-direction,
313
00:22:58,250 --> 00:22:59,960
that had better be 0.
314
00:22:59,960 --> 00:23:14,390
So you know that W1
plus W2 minus mtg is 0.
315
00:23:14,390 --> 00:23:18,260
And so you find out that the
total weight of the sailplane
316
00:23:18,260 --> 00:23:23,874
is no surprise, the sum of
the two weights on the scales.
317
00:23:23,874 --> 00:23:24,868
AUDIENCE: [INAUDIBLE].
318
00:23:27,850 --> 00:23:29,530
PROFESSOR: Yeah.
319
00:23:29,530 --> 00:23:33,960
Let's do this, m, t, g.
320
00:23:33,960 --> 00:23:35,480
So the total weight
times gravity
321
00:23:35,480 --> 00:23:38,820
is just the sum of the two
readings on the scales.
322
00:23:38,820 --> 00:23:43,200
And the second piece that you
need to do to do this problem
323
00:23:43,200 --> 00:23:45,440
is, you can have an
equation that says,
324
00:23:45,440 --> 00:23:48,360
the sum of the external
torques with respect
325
00:23:48,360 --> 00:23:52,930
to you're-- through
a fixed point O,
326
00:23:52,930 --> 00:23:56,520
is equal to the mass moment
of inertia times the angular
327
00:23:56,520 --> 00:23:58,530
acceleration, oftentimes
written as alpha.
328
00:23:58,530 --> 00:24:03,560
In this case, that's going
to be 0, it's going nowhere.
329
00:24:03,560 --> 00:24:06,300
So what are the external torques
with respect to this point?
330
00:24:06,300 --> 00:24:08,220
Well we have a right-handed
coordinate system.
331
00:24:08,220 --> 00:24:10,485
You have W1 up, times L1.
332
00:24:15,080 --> 00:24:23,210
W2 up, times L2
minus W1 plus W2,
333
00:24:23,210 --> 00:24:29,750
which is the total weight
of the sailplane times
334
00:24:29,750 --> 00:24:35,760
rg, the location of
the center of mass.
335
00:24:35,760 --> 00:24:37,770
That's this distance
that we're looking for.
336
00:24:46,570 --> 00:24:48,860
So we know everything
here, except rg.
337
00:24:48,860 --> 00:24:52,690
So we can solve for
rg with respect to O
338
00:24:52,690 --> 00:25:03,680
and it's simply W1 L1 plus
W2 L2 over W1 plus W2.
339
00:25:03,680 --> 00:25:15,720
And if you run the numbers,
typically W1 600 pounds,
340
00:25:15,720 --> 00:25:23,560
W2 the numbers I've done
here is-- might be 40 pounds.
341
00:25:26,940 --> 00:25:29,590
And I've already said
five feet and 15 feet.
342
00:25:29,590 --> 00:25:31,580
And you work the
answer and you come up
343
00:25:31,580 --> 00:25:40,830
with 5.62 feet, which puts
this-- here is the wheel
344
00:25:40,830 --> 00:25:44,650
and you're a little bit
aft of the main gear.
345
00:25:44,650 --> 00:25:47,240
And that's where you,
basically where you want to be.
346
00:25:47,240 --> 00:25:53,247
That's a real practical use of
knowing about centers of mass
347
00:25:53,247 --> 00:25:54,330
and how to calculate them.
348
00:26:00,970 --> 00:26:06,080
So that's the second item I
wanted to talk about today.
349
00:26:06,080 --> 00:26:08,370
Essentially a recap
of the center of mass.
350
00:26:08,370 --> 00:26:11,480
And now I want to
move on to talking
351
00:26:11,480 --> 00:26:18,290
about a serious
introduction to-- we've
352
00:26:18,290 --> 00:26:21,380
had the introduction,
velocities and accelerations.
353
00:26:21,380 --> 00:26:24,510
We have to have a
way of writing down
354
00:26:24,510 --> 00:26:35,170
the acceleration of a mass,
a point, a dog in a rotating,
355
00:26:35,170 --> 00:26:38,740
translating, reference
frame with the possibility
356
00:26:38,740 --> 00:26:41,670
that in addition to
that, the dog's moving.
357
00:26:41,670 --> 00:26:44,230
So we want to have equations--
we want to have the ability
358
00:26:44,230 --> 00:26:47,310
to write down expressions
for the velocity
359
00:26:47,310 --> 00:26:54,110
and acceleration of a mass
moving in a translating,
360
00:26:54,110 --> 00:26:57,690
rotating, reference frame.
361
00:26:57,690 --> 00:26:58,680
So we've started this.
362
00:26:58,680 --> 00:27:02,080
We did pretty much did
velocities to begin with.
363
00:27:31,740 --> 00:27:33,080
So here's my inertial frame.
364
00:27:41,030 --> 00:27:43,800
Call it O or O x, y, z.
365
00:27:43,800 --> 00:27:46,470
Here's my rigid body out there.
366
00:27:51,350 --> 00:27:57,200
It has a point a something
else, b might be the dog.
367
00:27:57,200 --> 00:28:00,810
And we've described
the position of this
368
00:28:00,810 --> 00:28:04,940
as the position of this
point a, with respect to O.
369
00:28:04,940 --> 00:28:10,640
And at this point we're going
to locate a reference frame
370
00:28:10,640 --> 00:28:13,700
attached to the rigid body.
371
00:28:13,700 --> 00:28:15,570
And so it's going
to be called a,
372
00:28:15,570 --> 00:28:19,530
and I'll call it x
prime, y prime, z prime.
373
00:28:19,530 --> 00:28:21,350
It's attached to
the rigid body, it
374
00:28:21,350 --> 00:28:24,230
rotates with the rigid
body and its attached
375
00:28:24,230 --> 00:28:26,340
at some fixed point.
376
00:28:26,340 --> 00:28:27,860
Now what would
oftentimes would be
377
00:28:27,860 --> 00:28:30,950
a smart choice for that
fixed point at point A?
378
00:28:35,144 --> 00:28:36,076
AUDIENCE: [INAUDIBLE].
379
00:28:36,076 --> 00:28:38,492
PROFESSOR: If you're going to
write an equation expressing
380
00:28:38,492 --> 00:28:40,280
the motion of this.
381
00:28:40,280 --> 00:28:41,832
Where would you make point a?
382
00:28:41,832 --> 00:28:42,759
[INTERPOSING VOICES]
383
00:28:42,759 --> 00:28:43,800
AUDIENCE: Center of mass.
384
00:28:43,800 --> 00:28:45,350
PROFESSOR: Center of mass.
385
00:28:45,350 --> 00:28:48,320
So very, very often, especially
when objects are free
386
00:28:48,320 --> 00:28:50,820
floating around out there you're
going to make smart choices
387
00:28:50,820 --> 00:28:52,770
and you're going to put
this coordinate system
388
00:28:52,770 --> 00:28:53,936
right on the center of mass.
389
00:28:53,936 --> 00:28:56,040
But it doesn't have
to be, but it can be.
390
00:28:59,420 --> 00:29:04,390
So we were interested
in knowing things
391
00:29:04,390 --> 00:29:09,590
about the motion of this point
in our inertial reference
392
00:29:09,590 --> 00:29:15,080
frame, in terms of positions
of our coordinate system.
393
00:29:15,080 --> 00:29:23,190
And then also this vector
here rd, with respect to a.
394
00:29:23,190 --> 00:29:29,590
Now last time we came up with
expressions for the velocity
395
00:29:29,590 --> 00:29:35,285
of b with respect to O.
396
00:29:35,285 --> 00:29:39,320
We said in general it's the
velocity of your-- where
397
00:29:39,320 --> 00:29:41,320
your coordinate
system's located.
398
00:29:41,320 --> 00:29:46,260
The translating-- the velocity
of the translating frame
399
00:29:46,260 --> 00:29:55,780
plus the derivative of
rba, time derivative
400
00:29:55,780 --> 00:30:00,430
of the position as seen from,
if you were sitting at a.
401
00:30:03,360 --> 00:30:06,520
And another way to
say that, or this
402
00:30:06,520 --> 00:30:12,420
is a derivative taken with the
rotation rate momentarily set
403
00:30:12,420 --> 00:30:14,490
equal to 0.
404
00:30:14,490 --> 00:30:16,370
Another way to think of it.
405
00:30:16,370 --> 00:30:20,180
Plus a piece that
comes from rotation.
406
00:30:20,180 --> 00:30:24,150
So the rotation with respect
to the fixed frame, these
407
00:30:24,150 --> 00:30:30,010
are all vectors, the
rotation with respect
408
00:30:30,010 --> 00:30:36,720
to the fixed frame, cross
product with r, b, a.
409
00:30:36,720 --> 00:30:38,460
And this-- and we
said this is actually
410
00:30:38,460 --> 00:30:41,590
a general formula for the
derivative of-- this piece is
411
00:30:41,590 --> 00:30:48,840
the derivative of a vector
in a frame, in a fixed frame.
412
00:30:48,840 --> 00:30:51,970
You have two pieces,
the derivative as
413
00:30:51,970 --> 00:30:55,197
seen without rotation
plus the contribution that
414
00:30:55,197 --> 00:30:56,030
comes from rotation.
415
00:31:05,430 --> 00:31:08,920
When I did this center of
mass thing a second ago,
416
00:31:08,920 --> 00:31:12,350
I just kind of quickly wrote
down two time derivatives
417
00:31:12,350 --> 00:31:13,820
of the position vector.
418
00:31:13,820 --> 00:31:18,310
There's no omega cross
O's in there right?
419
00:31:18,310 --> 00:31:20,960
Why could I do that?
420
00:31:20,960 --> 00:31:24,520
This is actually kind of an
important distinct point.
421
00:31:24,520 --> 00:31:29,530
I could do that they didn't say
very specifically when I did it
422
00:31:29,530 --> 00:31:31,920
was an assumption I was making.
423
00:31:31,920 --> 00:31:34,610
Except for perhaps they drew it.
424
00:31:34,610 --> 00:31:39,430
This was done in a
Cartesian coordinate system.
425
00:31:39,430 --> 00:31:41,250
And my coordinates
were x, y and z
426
00:31:41,250 --> 00:31:47,471
and the unit vectors were
i, j, k and do they move?
427
00:31:47,471 --> 00:31:47,970
No.
428
00:31:47,970 --> 00:31:49,178
What's their time derivative?
429
00:31:49,178 --> 00:31:50,220
[INTERPOSING VOICES]
430
00:31:50,220 --> 00:31:52,761
PROFESSOR: When you don't-- when
the inner vectors don't have
431
00:31:52,761 --> 00:31:54,811
time derivatives you
don't get these terms.
432
00:31:54,811 --> 00:31:57,060
This is the only term that
contributes so I could just
433
00:31:57,060 --> 00:31:58,390
write that equation.
434
00:31:58,390 --> 00:32:02,350
But we now have a reference
frame attached to a body
435
00:32:02,350 --> 00:32:04,540
and this reference
frame is rotating.
436
00:32:04,540 --> 00:32:07,940
And that means that the
direction of the unit vectors
437
00:32:07,940 --> 00:32:11,710
attach-- the unit vector
attached to x-prime here
438
00:32:11,710 --> 00:32:13,520
is moving, it's rotating.
439
00:32:13,520 --> 00:32:15,700
And it's going to have
a time derivative.
440
00:32:15,700 --> 00:32:17,855
So we have to--
and that is given
441
00:32:17,855 --> 00:32:19,480
and you take those
time derivatives you
442
00:32:19,480 --> 00:32:20,920
get this second piece.
443
00:32:24,140 --> 00:32:26,330
I'm going to give you
the answer in advance.
444
00:32:26,330 --> 00:32:32,460
The acceleration of b
with respect to O I'm
445
00:32:32,460 --> 00:32:37,460
going to give you
the full 3D equation.
446
00:32:37,460 --> 00:32:41,240
Then we'll go back and see
a bit where it comes from.
447
00:32:41,240 --> 00:32:45,660
So here's-- it's the time
derivative of that velocity
448
00:32:45,660 --> 00:32:55,376
expression with respect to time
taken in the inertial frame O,
449
00:32:55,376 --> 00:32:55,875
x, y, z.
450
00:33:02,310 --> 00:33:04,911
And am I going to have
enough room to get this on?
451
00:33:04,911 --> 00:33:05,535
It'll be close.
452
00:33:19,090 --> 00:33:22,280
All right this has
several pieces.
453
00:33:22,280 --> 00:33:28,070
It's got a contribution of the
acceleration of a with respect
454
00:33:28,070 --> 00:33:30,551
to O. That's just the
acceleration of this point.
455
00:33:30,551 --> 00:33:33,050
Has nothing to do with rotation,
so it's just a straight out
456
00:33:33,050 --> 00:33:37,550
acceleration of my translating
frame with respect to O.
457
00:33:37,550 --> 00:33:40,490
That's the first piece.
458
00:33:40,490 --> 00:33:44,690
The second piece is related-- is
the derivative of this guy that
459
00:33:44,690 --> 00:33:46,620
comes from the
derivative of this.
460
00:33:46,620 --> 00:33:54,010
It's the acceleration of b
with respect to a as seen
461
00:33:54,010 --> 00:33:54,895
in this a frame.
462
00:34:00,000 --> 00:34:01,500
If you read the
Williams book, it's
463
00:34:01,500 --> 00:34:03,940
called the relative
acceleration.
464
00:34:03,940 --> 00:34:06,732
It's relative to the-- if
you were sitting at point A,
465
00:34:06,732 --> 00:34:08,565
it's what you would see
as the acceleration.
466
00:34:12,770 --> 00:34:25,239
Plus 2 omega cross
velocity of b with respect
467
00:34:25,239 --> 00:34:36,850
to a as seen from a plus
omega dot, the derivative
468
00:34:36,850 --> 00:34:57,070
of the rotation rate,
cross rba plus omega cross,
469
00:34:57,070 --> 00:35:03,590
omega cross r, b, a.
470
00:35:07,536 --> 00:35:09,110
Kind of daunting right?
471
00:35:09,110 --> 00:35:11,250
A little messy.
472
00:35:11,250 --> 00:35:19,740
Basically one, two, three,
four, five different terms.
473
00:35:19,740 --> 00:35:24,130
And you're going to-- and they
all have names and meanings.
474
00:35:24,130 --> 00:35:27,370
And one of the things
that will really help you
475
00:35:27,370 --> 00:35:30,420
is to get familiar,
you really need
476
00:35:30,420 --> 00:35:36,480
to be familiar with the meaning
of each one of the terms.
477
00:35:36,480 --> 00:35:38,690
And it's not terribly difficult.
478
00:35:38,690 --> 00:35:43,240
This one, just the acceleration
of the translating frame.
479
00:35:43,240 --> 00:35:44,850
So if it's a
merry-go-round sitting
480
00:35:44,850 --> 00:35:47,580
on a train and the train's
heading down the track,
481
00:35:47,580 --> 00:35:50,320
its acceleration of the train.
482
00:35:50,320 --> 00:35:52,370
Rotating frame is attached
to the merry-go-round.
483
00:35:54,990 --> 00:35:58,300
And if you've got the
dog on the merry-go-round
484
00:35:58,300 --> 00:36:02,990
this is then the acceleration
of the dog relative to this,
485
00:36:02,990 --> 00:36:04,460
the merry-go-round.
486
00:36:04,460 --> 00:36:07,040
This position of the
coordinate system
487
00:36:07,040 --> 00:36:11,060
attached to the merry-go-round
has no rotation in it.
488
00:36:11,060 --> 00:36:17,110
This is the velocity
of that point, the dog,
489
00:36:17,110 --> 00:36:19,850
as seen from the A frame.
490
00:36:19,850 --> 00:36:22,100
Again, you have no
sense of rotation.
491
00:36:22,100 --> 00:36:24,250
Rotation is not a part of this.
492
00:36:24,250 --> 00:36:27,050
Cross product with
the rotation rate.
493
00:36:30,430 --> 00:36:31,780
This is the accelerate.
494
00:36:31,780 --> 00:36:35,810
This is the angular acceleration
cross product with rba.
495
00:36:35,810 --> 00:36:38,910
Now that's a term--
what does that mean?
496
00:36:38,910 --> 00:36:44,400
I'm swinging a baseball bat and
I'm accelerating this thing.
497
00:36:44,400 --> 00:36:48,540
Idealize it as just something
on a radius accelerating.
498
00:36:48,540 --> 00:36:52,030
The acceleration
of a point out here
499
00:36:52,030 --> 00:36:55,520
is the radius times the
angular acceleration.
500
00:36:55,520 --> 00:36:57,300
So that's all this term is.
501
00:36:57,300 --> 00:36:59,870
And it's called the
Euler acceleration.
502
00:36:59,870 --> 00:37:01,702
But it's just simply
r theta double dot.
503
00:37:04,866 --> 00:37:08,190
This, if you multiply it out
and just think about units,
504
00:37:08,190 --> 00:37:10,790
this ends up looking
like r omega squared.
505
00:37:10,790 --> 00:37:13,040
Have you run into that before?
506
00:37:13,040 --> 00:37:15,290
What's that?
507
00:37:15,290 --> 00:37:16,618
Common language.
508
00:37:16,618 --> 00:37:18,630
AUDIENCE: [INAUDIBLE].
509
00:37:18,630 --> 00:37:21,010
PROFESSOR: That's as a
centrifugal-- centripetal,
510
00:37:21,010 --> 00:37:23,060
this is centripetal
acceleration.
511
00:37:23,060 --> 00:37:25,600
So this is the centripetal
acceleration term,
512
00:37:25,600 --> 00:37:28,060
that's the Euler
acceleration term,
513
00:37:28,060 --> 00:37:30,025
this is the local acceleration.
514
00:37:30,025 --> 00:37:32,100
This is the acceleration
of your frame.
515
00:37:32,100 --> 00:37:33,910
This is the strange one.
516
00:37:33,910 --> 00:37:35,685
This is the Coriolis
acceleration.
517
00:37:40,390 --> 00:37:42,520
And we'll get
familiar with it too.
518
00:37:42,520 --> 00:37:48,520
So that's the full blown
3D acceleration equation.
519
00:37:48,520 --> 00:37:50,700
And by the way the
vector-- the velocity one
520
00:37:50,700 --> 00:37:52,730
is also perfect 3D.
521
00:37:52,730 --> 00:38:01,170
Now in this course we won't do
much in the way of 3D dynamics
522
00:38:01,170 --> 00:38:01,860
problems.
523
00:38:01,860 --> 00:38:02,793
Yes.
524
00:38:02,793 --> 00:38:08,070
AUDIENCE: Does the point
b on the rigid plane move?
525
00:38:08,070 --> 00:38:10,195
PROFESSOR: Does the-- it may.
526
00:38:10,195 --> 00:38:19,450
It could be this is the-- an
asteroid out there in space
527
00:38:19,450 --> 00:38:22,540
and you've got--
this is home base
528
00:38:22,540 --> 00:38:26,580
and that's a guy out there
in a space suit running.
529
00:38:29,650 --> 00:38:31,910
So we want to be
able to describe
530
00:38:31,910 --> 00:38:38,500
the acceleration of that guy
as seen from a fixed reference
531
00:38:38,500 --> 00:38:40,710
frame.
532
00:38:40,710 --> 00:38:43,720
Now why would we want to
know that acceleration?
533
00:38:43,720 --> 00:38:47,180
Why do we want to know
it in a fixed frame?
534
00:38:47,180 --> 00:38:53,480
Well if you want to calculate
the forces on the person.
535
00:38:53,480 --> 00:38:57,200
Well how much-- what's he
have to do with his feet
536
00:38:57,200 --> 00:38:59,600
to brace himself or whatever?
537
00:38:59,600 --> 00:39:01,570
What are the actual forces?
538
00:39:01,570 --> 00:39:04,970
You have to know the
acceleration on the person.
539
00:39:04,970 --> 00:39:08,030
But Newton's laws, in
order to say f equals ma,
540
00:39:08,030 --> 00:39:13,690
Newton's laws have to be applied
in inertial reference frames.
541
00:39:13,690 --> 00:39:16,590
Is this thing out there doing
this an a inertial reference
542
00:39:16,590 --> 00:39:17,571
frame?
543
00:39:17,571 --> 00:39:18,070
No.
544
00:39:18,070 --> 00:39:20,140
So you can't
calculate the forces
545
00:39:20,140 --> 00:39:23,530
without having some idea
of this inertial frame.
546
00:39:23,530 --> 00:39:26,750
So this is the way of
getting the acceleration
547
00:39:26,750 --> 00:39:33,690
on-- at a location on a moving,
rotating body with respect
548
00:39:33,690 --> 00:39:36,410
to an inertial frame.
549
00:39:36,410 --> 00:39:40,160
And with all the terms present.
550
00:39:40,160 --> 00:39:45,250
Now most discourse has
generally has addressed problems
551
00:39:45,250 --> 00:39:49,740
which are in most textbooks
address only planar motion
552
00:39:49,740 --> 00:39:51,270
problems.
553
00:39:51,270 --> 00:39:55,530
Planar motion basically
means that we can find
554
00:39:55,530 --> 00:39:59,280
the translations to a plane.
555
00:39:59,280 --> 00:40:04,860
So imagine an x, and a y, and
a z upwards coordinate system
556
00:40:04,860 --> 00:40:06,450
attached to the top this table.
557
00:40:06,450 --> 00:40:09,370
And I only allow motions
that are around the table.
558
00:40:09,370 --> 00:40:12,860
And I only allow motions that
have a single axis rotation.
559
00:40:12,860 --> 00:40:14,720
And that's lined up with z.
560
00:40:14,720 --> 00:40:17,210
Those are essentially
planar motion problems.
561
00:40:17,210 --> 00:40:20,650
And most courses in dynamics,
it lists [INAUDIBLE].
562
00:40:20,650 --> 00:40:22,290
That's as far as they get.
563
00:40:22,290 --> 00:40:24,610
And the most of the
problems that you'll do
564
00:40:24,610 --> 00:40:27,670
will be planar motion problems.
565
00:40:27,670 --> 00:40:33,970
But that equation reduces to the
planar motion problem as well.
566
00:40:33,970 --> 00:40:36,270
We will do a little bit
of 3D, because there's
567
00:40:36,270 --> 00:40:37,941
a class of problems
that I really
568
00:40:37,941 --> 00:40:40,440
think it's important for you
to understand that just come up
569
00:40:40,440 --> 00:40:43,997
all the time that
require a little 3D.
570
00:40:43,997 --> 00:40:45,830
And as you want to have
some things going on
571
00:40:45,830 --> 00:40:47,240
out of the plane,
but we'll still
572
00:40:47,240 --> 00:40:53,420
confine the axis of rotation
to a single direction.
573
00:40:53,420 --> 00:40:53,920
Yeah.
574
00:40:53,920 --> 00:40:58,330
AUDIENCE: [INAUDIBLE]
in a planar,
575
00:40:58,330 --> 00:41:02,740
but then would you have three to
view the freedom? [INAUDIBLE].
576
00:41:02,740 --> 00:41:04,430
PROFESSOR: That's
a great question
577
00:41:04,430 --> 00:41:05,880
she said if you
had a dog running
578
00:41:05,880 --> 00:41:08,630
on the merry-go-round how many
degrees of freedom do you have?
579
00:41:08,630 --> 00:41:13,510
So in general rigid
bodies, each rigid body,
580
00:41:13,510 --> 00:41:16,100
each independent
rigid body has--
581
00:41:16,100 --> 00:41:18,580
you have to describe its
location of its center of mass
582
00:41:18,580 --> 00:41:20,270
and that takes how
many coordinates?
583
00:41:20,270 --> 00:41:23,340
How many coordinates
I'll call them.
584
00:41:23,340 --> 00:41:25,172
AUDIENCE: [INAUDIBLE].
585
00:41:25,172 --> 00:41:28,660
PROFESSOR: Well
in general three.
586
00:41:28,660 --> 00:41:35,227
And it can now rotate
around three different axes.
587
00:41:35,227 --> 00:41:36,310
And that takes three more.
588
00:41:36,310 --> 00:41:39,740
So rigid bodies have
six degrees of freedom.
589
00:41:39,740 --> 00:41:42,620
And any problem when you go
to address the problem you
590
00:41:42,620 --> 00:41:45,460
essentially for a rigid
body you start with six.
591
00:41:45,460 --> 00:41:47,270
And you start
applying constraints
592
00:41:47,270 --> 00:41:50,850
to reduce it down to
the number of remaining
593
00:41:50,850 --> 00:41:52,810
degrees of freedom.
594
00:41:52,810 --> 00:41:55,380
So if it's confined to
a plane and no z-motion
595
00:41:55,380 --> 00:41:58,250
is allowed one constraint.
596
00:41:58,250 --> 00:42:01,470
If it is on a
plane and it's only
597
00:42:01,470 --> 00:42:04,640
allowed to rotate
about the z-axis that
598
00:42:04,640 --> 00:42:08,920
means you've constrained its
rotation in around y and x.
599
00:42:08,920 --> 00:42:10,640
So that's two more.
600
00:42:10,640 --> 00:42:16,560
And so now you're down to three
degrees of freedom left, xy
601
00:42:16,560 --> 00:42:19,890
and a rotation about the z-axis.
602
00:42:19,890 --> 00:42:21,680
So planar motion
problems generally
603
00:42:21,680 --> 00:42:26,450
have three degrees of freedom.
604
00:42:26,450 --> 00:42:28,230
But instant-- let's
just say we're
605
00:42:28,230 --> 00:42:30,020
just interested in
just something that
606
00:42:30,020 --> 00:42:33,240
rotates and doesn't translate.
607
00:42:33,240 --> 00:42:35,610
How many degrees of freedom
does that have then?
608
00:42:35,610 --> 00:42:36,770
Just one.
609
00:42:36,770 --> 00:42:40,259
x and y are forced not
to-- no motion, two more
610
00:42:40,259 --> 00:42:41,550
constraints you're down to one.
611
00:42:41,550 --> 00:42:45,712
So lots of problems we do
are in fact single degree
612
00:42:45,712 --> 00:42:46,545
of freedom problems.
613
00:42:57,190 --> 00:43:01,640
So to do planar motion
problems we oftentimes
614
00:43:01,640 --> 00:43:04,010
use polar coordinates.
615
00:43:04,010 --> 00:43:08,420
So I'm going to
introduce r theta.
616
00:43:08,420 --> 00:43:10,980
And I'm actually going to call
it cylindrical coordinates.
617
00:43:21,300 --> 00:43:26,298
And cylindrical coordinates then
you have an r, a theta and z.
618
00:43:34,910 --> 00:43:37,900
And let's think about well
let's see, I have a demo,
619
00:43:37,900 --> 00:43:40,540
a little demo here.
620
00:43:40,540 --> 00:43:45,450
So here's a problem with
a single axis of rotation.
621
00:43:48,740 --> 00:43:54,570
And it's a-- there is a mass out
here and just this is a rhyme.
622
00:43:54,570 --> 00:43:58,260
And so think of this
think of this mass
623
00:43:58,260 --> 00:44:03,350
out here as being a bug
walking out this rod.
624
00:44:03,350 --> 00:44:08,032
And the rod, this thing
goes round and round.
625
00:44:08,032 --> 00:44:09,580
It's not a
merry-go-round but it's
626
00:44:09,580 --> 00:44:12,660
a merry-go-round with
a gang plank on it.
627
00:44:12,660 --> 00:44:14,035
It's going up at
an angle and you
628
00:44:14,035 --> 00:44:16,118
can walk the gang plank
while the merry-go-round's
629
00:44:16,118 --> 00:44:16,690
going around.
630
00:44:16,690 --> 00:44:19,330
So that's what we got here.
631
00:44:19,330 --> 00:44:23,540
So this is actually allowed to
change position of this mass.
632
00:44:23,540 --> 00:44:28,900
So how would I describe that
with cylindrical coordinates?
633
00:44:28,900 --> 00:44:34,900
Let me so it's going to take--
one would be a side view.
634
00:44:38,160 --> 00:44:40,930
So you see your
vertical axis here
635
00:44:40,930 --> 00:44:43,135
and have to have a bearing
to hold it in place.
636
00:44:46,790 --> 00:44:53,970
Here's the arm, here's the
bug walking out the arm,
637
00:44:53,970 --> 00:44:55,480
has some rotation rate.
638
00:44:58,200 --> 00:45:13,440
Theta dot this is the z-axis
and the position of this point
639
00:45:13,440 --> 00:45:22,280
is described by a r
vector, in the r hat
640
00:45:22,280 --> 00:45:23,800
direction, which I
think is the unit
641
00:45:23,800 --> 00:45:25,630
vectors you're used to using.
642
00:45:25,630 --> 00:45:30,375
And then this is the z-component
in the k hat direction.
643
00:45:32,920 --> 00:45:45,730
And this vector here would be r
of v with respect to what shall
644
00:45:45,730 --> 00:45:46,230
I call it?
645
00:45:46,230 --> 00:45:55,800
I'll make this a and
over here someplace
646
00:45:55,800 --> 00:46:07,400
I have a fixed inertial let's
see I got to be careful here.
647
00:46:07,400 --> 00:46:13,620
I want that to be z then I
have a y going out here x, y, z
648
00:46:13,620 --> 00:46:15,180
pointing upwards.
649
00:46:15,180 --> 00:46:17,480
And my-- this fixed
inertial system the unit
650
00:46:17,480 --> 00:46:25,840
vectors here this would
be i hat, j hat and k hat.
651
00:46:25,840 --> 00:46:30,920
But these-- this rotating
system with its unit
652
00:46:30,920 --> 00:46:35,360
vector little k and this
vector they're the same,
653
00:46:35,360 --> 00:46:38,020
they're parallel.
654
00:46:38,020 --> 00:46:42,620
But looking down on this, this
is my polar coordinate system.
655
00:46:42,620 --> 00:46:45,560
Now I'm going to
look at my top view.
656
00:46:45,560 --> 00:46:46,870
I will see a projection.
657
00:46:49,700 --> 00:46:54,340
I'll just see the
r, this is rr hat,
658
00:46:54,340 --> 00:46:57,840
this is my point
B. This is theta.
659
00:47:01,600 --> 00:47:03,640
And I have a unit vector.
660
00:47:03,640 --> 00:47:08,190
So the unit vector r hat is
something-- the unit long
661
00:47:08,190 --> 00:47:10,870
pointing in this direction.
662
00:47:10,870 --> 00:47:14,320
And the unit vector in this
direction is theta hat.
663
00:47:14,320 --> 00:47:18,430
And it's perpendicular
to that radius.
664
00:47:18,430 --> 00:47:20,330
So now I have my
three unit vectors.
665
00:47:20,330 --> 00:47:24,530
One pointing in the
direction of r, here's
666
00:47:24,530 --> 00:47:29,920
also my unit vector is just to
make sure there's no confusion.
667
00:47:29,920 --> 00:47:32,218
This unit vector is
in this direction.
668
00:47:35,360 --> 00:47:36,940
K is in that direction.
669
00:47:36,940 --> 00:47:39,050
Theta is in that direction.
670
00:47:44,450 --> 00:47:47,080
And over here you still
have your-- now here's
671
00:47:47,080 --> 00:47:54,570
my x, y, z out of the
board inertial frame.
672
00:47:54,570 --> 00:47:59,166
And this-- my inertial
frame this might be r, b, o.
673
00:48:02,770 --> 00:48:04,180
So in my inertial frame.
674
00:48:04,180 --> 00:48:07,085
I want to know
what's going on here.
675
00:48:07,085 --> 00:48:08,960
I want to be able to
calculate the velocities
676
00:48:08,960 --> 00:48:09,876
and the accelerations.
677
00:48:16,970 --> 00:48:21,730
So the notation here gets--
can get a little confusing.
678
00:48:21,730 --> 00:48:26,010
The rbo notation that
I've been using all along,
679
00:48:26,010 --> 00:48:30,880
that's the motion-- that's
the position vector describing
680
00:48:30,880 --> 00:48:34,650
that point in my inertial frame.
681
00:48:34,650 --> 00:48:39,310
And my-- just lowercase r here,
no scrub scripts or anything,
682
00:48:39,310 --> 00:48:44,850
that's just going to-- that's
my polar coordinate r theta
683
00:48:44,850 --> 00:48:47,500
and z that happened
to be in this case,
684
00:48:47,500 --> 00:48:50,120
this is a rotating frame.
685
00:48:50,120 --> 00:48:52,190
This is a rotating frame.
686
00:48:52,190 --> 00:48:58,870
It's the center of this
coordinate system's at a.
687
00:48:58,870 --> 00:49:00,720
But this thing rotates.
688
00:49:00,720 --> 00:49:03,400
So this is a pretty
simplified version
689
00:49:03,400 --> 00:49:07,760
of this general problem.
690
00:49:07,760 --> 00:49:11,250
Now because it's simplified,
you can actually--
691
00:49:11,250 --> 00:49:12,680
it's a lot easier to use.
692
00:49:12,680 --> 00:49:14,240
Also has some real limitations.
693
00:49:14,240 --> 00:49:17,230
You can only going just--
there's limited things
694
00:49:17,230 --> 00:49:18,500
that you can describe with it.
695
00:49:42,970 --> 00:49:48,070
So let's start by describing
velocities in cylindrical
696
00:49:48,070 --> 00:49:48,570
coordinates.
697
00:50:23,290 --> 00:50:27,800
Remember this rba is the
length of this guy here.
698
00:50:27,800 --> 00:50:31,710
And it's made up
of rr hat zk hat.
699
00:50:59,900 --> 00:51:01,585
So to express the
velocity we have
700
00:51:01,585 --> 00:51:03,710
to take a time derivative
of this r, b, a
701
00:51:03,710 --> 00:51:07,210
and I'm going to express
it in terms of r theta nz.
702
00:51:07,210 --> 00:51:09,190
And to get acceleration
I have to take two time
703
00:51:09,190 --> 00:51:12,210
derivatives of this, but
this is going to be expressed
704
00:51:12,210 --> 00:51:14,880
in my cylindrical coordinates.
705
00:51:14,880 --> 00:51:17,420
This is where I'm going.
706
00:51:17,420 --> 00:51:24,660
And lots of problems--
many, many of these problems
707
00:51:24,660 --> 00:51:27,240
have fixed axes of
rotations and this velocity
708
00:51:27,240 --> 00:51:29,540
and this acceleration are zero.
709
00:51:29,540 --> 00:51:31,080
You just drop it out.
710
00:51:31,080 --> 00:51:34,000
I'm going to do that
just to keep this-- make
711
00:51:34,000 --> 00:51:35,250
this problem a little simpler.
712
00:51:35,250 --> 00:51:37,890
So we can just focus
on these terms.
713
00:51:37,890 --> 00:51:45,970
So let's just let the there be
no translational of this frame.
714
00:51:45,970 --> 00:51:48,900
And that says that
Va with respect
715
00:51:48,900 --> 00:51:53,679
to O the acceleration of A
with respect to O over zero.
716
00:51:53,679 --> 00:51:55,470
So I want you to just
focus on these terms.
717
00:51:55,470 --> 00:51:57,160
I don't lose anything,
I can put these back in
718
00:51:57,160 --> 00:51:58,200
later if I need them.
719
00:51:58,200 --> 00:52:00,157
I just don't want to
keep carrying them along.
720
00:52:30,530 --> 00:52:33,220
So I have my--
remember my side view.
721
00:52:37,440 --> 00:52:47,140
This is r, r hat zk
hat that's my point.
722
00:52:47,140 --> 00:52:48,210
And my top view.
723
00:53:06,970 --> 00:53:12,990
This is my projection
just looking down on it
724
00:53:12,990 --> 00:53:15,580
what I see is the length r.
725
00:53:15,580 --> 00:53:20,929
And what I see in my unit vector
going that way r direction
726
00:53:20,929 --> 00:53:21,970
and theta that direction.
727
00:53:25,290 --> 00:53:32,870
And this is x and the
i and a y with a j hat
728
00:53:32,870 --> 00:53:36,640
vector looking down on it.
729
00:53:36,640 --> 00:53:42,660
My rotation rate mega with
respect to my inertial frame,
730
00:53:42,660 --> 00:53:46,100
is sum theta dot k hat.
731
00:53:50,850 --> 00:53:54,350
All right so now let's find
the velocity of b with respect
732
00:53:54,350 --> 00:53:59,640
to O. Well it's 0, no
translation, plus--
733
00:53:59,640 --> 00:54:06,850
and now I need a time derivative
of rb with respect to a.
734
00:54:06,850 --> 00:54:20,402
But this is then r, b, a
is r, r hat plus z k hat.
735
00:54:20,402 --> 00:54:22,675
And I need the time
derivative of that.
736
00:54:26,360 --> 00:54:41,980
So I get an r dot r hat plus
an r r hat dot plus a z dot k.
737
00:54:41,980 --> 00:54:44,840
So this is the
product of two things.
738
00:54:44,840 --> 00:54:46,760
They're both time dependent.
739
00:54:46,760 --> 00:54:51,380
So I have to get two pieces, k
does not change in direction.
740
00:54:51,380 --> 00:54:53,330
So it has no time derivative.
741
00:54:53,330 --> 00:54:55,570
So I only have a z dot k.
742
00:54:55,570 --> 00:54:57,300
So this is a result
of doing this,
743
00:54:57,300 --> 00:54:59,740
but I now have to
figure out what
744
00:54:59,740 --> 00:55:03,205
is the time derivative of the
unit vector in the r direction.
745
00:55:17,910 --> 00:55:21,260
So when I told-- when we worked
out this formula the other day
746
00:55:21,260 --> 00:55:24,640
for the time derivative
of a rotating vector,
747
00:55:24,640 --> 00:55:27,300
I mostly did it, it was kind
of an intuitive argument.
748
00:55:27,300 --> 00:55:28,820
So on this one
occasion I'm going
749
00:55:28,820 --> 00:55:32,130
to give you an example
of actually figuring out
750
00:55:32,130 --> 00:55:35,390
what the derivative of
this rotating vector is.
751
00:55:35,390 --> 00:55:39,120
And if you go read
that kinematics handout
752
00:55:39,120 --> 00:55:42,920
and it does this in kind of full
blown form for-- in general.
753
00:55:42,920 --> 00:55:45,780
So I'm just going to
do it as one example.
754
00:55:45,780 --> 00:55:51,880
So here's our looking down on
this, the projection on the xy
755
00:55:51,880 --> 00:55:55,100
plane, here's our r-vector.
756
00:55:55,100 --> 00:55:58,060
And here's this
unit vector and it
757
00:55:58,060 --> 00:56:02,900
starts from-- I have a unit
vector starting from a it's
758
00:56:02,900 --> 00:56:06,370
unit-- it's one long.
759
00:56:06,370 --> 00:56:08,930
And this is r hat.
760
00:56:08,930 --> 00:56:15,900
And it's of unit length and it's
in this particular direction.
761
00:56:15,900 --> 00:56:21,050
Now in a little bit it
time delta t, it moves.
762
00:56:21,050 --> 00:56:23,490
It moves to here.
763
00:56:23,490 --> 00:56:33,480
So this is delta r hat and
what direction does it move?
764
00:56:33,480 --> 00:56:34,920
AUDIENCE: [INAUDIBLE].
765
00:56:34,920 --> 00:56:38,830
PROFESSOR: Yeah, it moves in
the-- moves in the theta hat
766
00:56:38,830 --> 00:56:41,090
direction.
767
00:56:41,090 --> 00:56:45,470
And the amount that it
moves is the rotation rate,
768
00:56:45,470 --> 00:56:49,050
theta dot, delta t.
769
00:56:52,190 --> 00:56:59,640
So delta r hat, if I
solve for this, delta t.
770
00:56:59,640 --> 00:57:04,930
And this is in the
theta hat direction,
771
00:57:04,930 --> 00:57:09,185
is theta dot theta hat.
772
00:57:13,280 --> 00:57:17,810
So this is the
limit as t, delta t
773
00:57:17,810 --> 00:57:24,770
goes to 0 you get the derivative
of r hat with respect to time.
774
00:57:24,770 --> 00:57:27,870
Its direction is in
the theta hat direction
775
00:57:27,870 --> 00:57:31,980
and its magnitude is theta dot.
776
00:57:31,980 --> 00:57:35,740
So that's the time derivative
of the unit vector r hat.
777
00:57:39,390 --> 00:57:40,210
Yeah.
778
00:57:40,210 --> 00:57:43,556
AUDIENCE: How does
that work with units?
779
00:57:43,556 --> 00:57:47,700
PROFESSOR: How does
it work with units?
780
00:57:47,700 --> 00:57:55,640
What's left out of here is
that this is unit length
781
00:57:55,640 --> 00:57:57,426
and has dimensions.
782
00:57:57,426 --> 00:58:02,270
It's unit length, one whatever
unit system you're working.
783
00:58:02,270 --> 00:58:06,440
So that is implicitly in here.
784
00:58:06,440 --> 00:58:11,900
It's one meter theta dot and
that theta dot, the delta t,
785
00:58:11,900 --> 00:58:12,880
the times go away.
786
00:58:12,880 --> 00:58:17,940
You're left with one meter times
the magnitude of theta dot.
787
00:58:17,940 --> 00:58:19,630
So the distance
it actually moves
788
00:58:19,630 --> 00:58:25,370
is r theta, the r delta theta,
delta theta is theta dot delta
789
00:58:25,370 --> 00:58:27,645
t and the radius
happens to be 1.
790
00:58:31,100 --> 00:58:34,830
So whatever unit system you're
working in it's a unit vector.
791
00:58:34,830 --> 00:58:36,130
Has unit length.
792
00:58:36,130 --> 00:58:38,330
So its units are
buried right there.
793
00:58:38,330 --> 00:58:40,890
Good question.
794
00:58:40,890 --> 00:58:45,120
OK so now we know what this is.
795
00:58:45,120 --> 00:58:50,670
So now we can come back finish
our description of the velocity
796
00:58:50,670 --> 00:58:58,820
of b with respect to
a then is r dot r hat
797
00:58:58,820 --> 00:59:10,795
plus z dot k hat
plus theta dot times
798
00:59:10,795 --> 00:59:17,050
so r times the derivative of
the unit vector r, which we just
799
00:59:17,050 --> 00:59:21,292
figured out is theta
dot theta hat times r.
800
00:59:21,292 --> 00:59:25,620
R theta dot theta hat.
801
00:59:25,620 --> 00:59:28,620
So that's my velocity
of b with respect to a.
802
00:59:28,620 --> 00:59:31,640
My velocity of B with respect
to O all you have to add in
803
00:59:31,640 --> 00:59:33,720
is the velocity of A
with perspective to O,
804
00:59:33,720 --> 00:59:35,580
which we've let be 0 for now.
805
00:59:35,580 --> 00:59:40,080
So for the moment this is
also d with respect to O.
806
00:59:40,080 --> 00:59:42,910
But this is the general
piece of the velocity
807
00:59:42,910 --> 00:59:46,294
of b with respect to a in
polar cylindrical coordinates.
808
00:59:50,170 --> 00:59:54,050
Now we could have-- so I've
actually worked it out, just
809
00:59:54,050 --> 00:59:57,090
shown you, just drew the picture
and figured out the derivative.
810
00:59:57,090 --> 00:59:59,850
We could have used
that magic formula.
811
00:59:59,850 --> 01:00:03,920
The formula for the derivative
of a vector in a rotating
812
01:00:03,920 --> 01:00:04,420
frame.
813
01:00:15,720 --> 01:00:18,330
So I'll just do that
quickly to remind you
814
01:00:18,330 --> 01:00:20,020
how we could have done this.
815
01:00:20,020 --> 01:00:27,620
rba with respect to time
as seen in the O frame.
816
01:00:30,570 --> 01:00:36,990
Is the partial derivative
of rba with respect
817
01:00:36,990 --> 01:00:47,577
to time as seen in the rotating
frame, plus omega cross r, b,
818
01:00:47,577 --> 01:00:48,076
a.
819
01:00:55,370 --> 01:01:02,030
This term is that and that.
820
01:01:04,840 --> 01:01:08,760
The derivative of
this rba as seen
821
01:01:08,760 --> 01:01:11,440
from inside of the
rotating frame,
822
01:01:11,440 --> 01:01:17,090
is just the change in length,
this is rba here from the side.
823
01:01:17,090 --> 01:01:21,389
So the change in length of
that vector, the derivative
824
01:01:21,389 --> 01:01:22,680
of-- the time derivative of it.
825
01:01:22,680 --> 01:01:28,290
It's the vector sum of
the r dot plus z dot.
826
01:01:28,290 --> 01:01:36,130
So this piece comes
from this and this.
827
01:01:36,130 --> 01:01:42,840
And this piece should-- this
one here, it better be this.
828
01:01:42,840 --> 01:01:44,650
Well this is--
let's figure it out.
829
01:01:44,650 --> 01:01:49,140
This is omega in
the k hat direction,
830
01:01:49,140 --> 01:01:59,520
cross and rba is r
r hat plus z k hat.
831
01:01:59,520 --> 01:02:03,610
k cross k is 0.
832
01:02:03,610 --> 01:02:09,460
k cross r theta hat.
833
01:02:09,460 --> 01:02:15,530
K cross-- k hat cross r
hat is theta hat positive.
834
01:02:15,530 --> 01:02:21,770
r omega theta hat, same thing
as r theta dot theta hat.
835
01:02:21,770 --> 01:02:25,310
So we could have just applied
this formula for the derivative
836
01:02:25,310 --> 01:02:29,730
of a rotating
vector and we would
837
01:02:29,730 --> 01:02:30,854
have gotten the same thing.
838
01:02:41,230 --> 01:02:43,347
OK just ran out of boards.
839
01:03:20,180 --> 01:03:22,460
Now a quick little exercise
you could do on your own
840
01:03:22,460 --> 01:03:25,510
is, we're going to need
to be able to calculate
841
01:03:25,510 --> 01:03:28,092
the derivative of theta hat.
842
01:03:28,092 --> 01:03:29,800
Well just plug it in
that little formula.
843
01:03:36,650 --> 01:03:41,250
And the first term you'll find
out the derivative of the theta
844
01:03:41,250 --> 01:03:48,800
hat, the length doesn't change
in time, it's a unit vector.
845
01:03:48,800 --> 01:03:50,920
So you only have
the second piece.
846
01:03:50,920 --> 01:03:54,800
So it's sum omega cross
theta and you're going
847
01:03:54,800 --> 01:03:57,859
to get minus theta dot r hat.
848
01:04:17,940 --> 01:04:21,140
So I really want to get here.
849
01:04:21,140 --> 01:04:23,480
The acceleration
of b and O. That's
850
01:04:23,480 --> 01:04:28,420
the real-- that's the
single piece we really
851
01:04:28,420 --> 01:04:30,210
need to finish the kinematics.
852
01:04:30,210 --> 01:04:32,080
So we can do most any problems.
853
01:04:32,080 --> 01:04:35,100
Got to be able to describe
the acceleration of a point
854
01:04:35,100 --> 01:04:37,770
and translating rotating frame.
855
01:04:37,770 --> 01:04:41,960
And that's going to be the
acceleration of a with respect
856
01:04:41,960 --> 01:04:53,970
to O, plus a time derivative of
the velocity of b with respect
857
01:04:53,970 --> 01:04:55,250
to O.
858
01:04:55,250 --> 01:04:58,790
We've calculated
the velocity, we
859
01:04:58,790 --> 01:05:05,300
need to be able to essentially
carry out this derivative.
860
01:05:05,300 --> 01:05:10,250
Two time derivatives of the
r, b, a, or a single time
861
01:05:10,250 --> 01:05:12,870
derivative of bva.
862
01:05:12,870 --> 01:05:18,150
Well we just
computed the velocity
863
01:05:18,150 --> 01:05:20,310
in this-- of this
rotating frame and this
864
01:05:20,310 --> 01:05:22,150
is our final expression.
865
01:05:22,150 --> 01:05:26,860
So we need to compute the
time derivative of that.
866
01:05:26,860 --> 01:05:32,650
I just-- so it's
going to look like r
867
01:05:32,650 --> 01:05:43,520
dot r hat plus over here a
term r theta dot theta hat
868
01:05:43,520 --> 01:05:50,250
and pardon me for
doing this I think
869
01:05:50,250 --> 01:05:51,740
it'll be cleaner in the end.
870
01:05:51,740 --> 01:05:55,180
I'm going to start with my
z dot k, keep it over here,
871
01:05:55,180 --> 01:06:01,500
plus r dot r hat plus this term.
872
01:06:04,440 --> 01:06:06,500
And this is going to
take up a lot room.
873
01:06:06,500 --> 01:06:08,600
Just spread out this way so it--
874
01:06:17,000 --> 01:06:19,410
So let's just-- I'm going to
write down where this comes
875
01:06:19,410 --> 01:06:21,860
out, this is a little
tedious, but then you'll
876
01:06:21,860 --> 01:06:26,190
have seen it once hopefully
believe that it really works.
877
01:06:31,410 --> 01:06:34,350
So these terms, this
first term here just
878
01:06:34,350 --> 01:06:36,650
gives you z double dot.
879
01:06:36,650 --> 01:06:37,770
So let's write her down.
880
01:06:37,770 --> 01:06:44,870
So just z double dot k hat
time derivative of this,
881
01:06:44,870 --> 01:06:50,780
plus an r double
dot r hat, but now I
882
01:06:50,780 --> 01:06:53,740
have to take-- do it, flip it
and do the other side of it.
883
01:06:53,740 --> 01:07:05,460
So I get my-- how do
I want to do this?
884
01:07:15,909 --> 01:07:16,659
Yeah, I like this.
885
01:07:26,730 --> 01:07:29,826
So this term kept--
leads to this.
886
01:07:29,826 --> 01:07:35,270
This term brings you to here.
887
01:07:35,270 --> 01:07:50,410
This term, you get r dot times
theta dot theta hat plus an r
888
01:07:50,410 --> 01:07:56,660
and now you need to take a time
derivative of theta dot theta
889
01:07:56,660 --> 01:07:58,550
hat.
890
01:07:58,550 --> 01:07:59,810
So that's going to expand.
891
01:07:59,810 --> 01:08:02,080
So this brings you to here.
892
01:08:28,680 --> 01:08:33,399
Now we've done this derivative,
so we can put it in.
893
01:08:33,399 --> 01:08:59,930
So this gives us
this term over here,
894
01:08:59,930 --> 01:09:01,449
so let's keep adding these up.
895
01:09:11,520 --> 01:09:15,319
Notice this gives me an r
dot theta dot theta hat.
896
01:09:15,319 --> 01:09:20,399
This gives me an r dot
theta dot theta hat.
897
01:09:23,050 --> 01:09:24,330
Two identical terms.
898
01:09:28,680 --> 01:09:38,649
Now this term gives me an r
theta double dot theta hat.
899
01:09:38,649 --> 01:09:40,340
And now this-- now
we need to take
900
01:09:40,340 --> 01:09:43,020
the derivative of theta
hat and that gives you
901
01:09:43,020 --> 01:09:46,515
minus theta dot r hat.
902
01:09:46,515 --> 01:09:59,980
So you get a minus r but
multiplied by theta dot again
903
01:09:59,980 --> 01:10:04,220
squared r hat.
904
01:10:04,220 --> 01:10:05,820
So I think we're about there.
905
01:10:05,820 --> 01:10:07,820
We're going to start
collecting things together.
906
01:10:23,220 --> 01:10:30,930
Now we have a two
r dot theta dot
907
01:10:30,930 --> 01:10:46,750
theta hat plus an r theta double
dot theta hat, minus r theta
908
01:10:46,750 --> 01:10:49,780
dot squared r half.
909
01:10:49,780 --> 01:10:51,880
So these things,
these derivatives,
910
01:10:51,880 --> 01:10:55,730
just all kind of flowed
down and led to more terms.
911
01:10:58,760 --> 01:11:06,140
But now if we compare-- that
we get one, two, three, four,
912
01:11:06,140 --> 01:11:08,960
five, I clump these together.
913
01:11:08,960 --> 01:11:14,250
This is the change in
length of the r vector.
914
01:11:14,250 --> 01:11:19,500
Stretch in the position has a
z component and a r component.
915
01:11:19,500 --> 01:11:23,630
This is the movement of the
coordinate system if it moves.
916
01:11:23,630 --> 01:11:26,900
This is the Coriolis term.
917
01:11:26,900 --> 01:11:30,470
This is the Euler
acceleration and this is
918
01:11:30,470 --> 01:11:33,350
the centripetal acceleration.
919
01:11:33,350 --> 01:11:36,040
So this is the--
what happens when
920
01:11:36,040 --> 01:11:41,700
you start with that
messy vector thing
921
01:11:41,700 --> 01:11:47,320
and apply it, restrict it to a
cylindrical coordinate problem,
922
01:11:47,320 --> 01:11:50,400
which is basically
planar motion.
923
01:11:50,400 --> 01:11:53,470
But you allow some things
in the z direction only.
924
01:11:53,470 --> 01:11:58,090
So polar coordinates is a
more limited form of that,
925
01:11:58,090 --> 01:12:01,290
but it's-- every
term comes back,
926
01:12:01,290 --> 01:12:03,270
every term is still in it.
927
01:12:03,270 --> 01:12:07,950
So acceleration of
the moving coordinate
928
01:12:07,950 --> 01:12:11,080
system, change of the
length of the position
929
01:12:11,080 --> 01:12:14,580
vector in the moving
coordinate system,
930
01:12:14,580 --> 01:12:19,610
the Coriolis term the Euler
acceleration term that angular
931
01:12:19,610 --> 01:12:25,490
acceleration speedup and
finally the centripetal term.
932
01:12:25,490 --> 01:12:29,030
Now the way you go
about solving problems,
933
01:12:29,030 --> 01:12:32,490
usins-- doing problems
in polar coordinates.
934
01:12:32,490 --> 01:12:36,110
So now you're asked to
find an equation of motion.
935
01:12:36,110 --> 01:12:40,560
This is an expression
for the acceleration
936
01:12:40,560 --> 01:12:42,770
of whatever it is you're
trying to describe
937
01:12:42,770 --> 01:12:46,370
in an inertial frame.
938
01:12:46,370 --> 01:12:51,690
So that when you say-- you
can now say f equals ma.
939
01:12:51,690 --> 01:12:53,700
If you know the--
sometimes you're
940
01:12:53,700 --> 01:12:55,550
given the forces in
problems and your asked
941
01:12:55,550 --> 01:12:59,370
to find the accelerations.
942
01:12:59,370 --> 01:13:01,140
But what if you're
given the acceleration
943
01:13:01,140 --> 01:13:03,610
and you're asked
to find the force?
944
01:13:03,610 --> 01:13:06,566
All right, I give you the
simplest problem of this kind.
945
01:13:09,810 --> 01:13:13,030
What's the tension
in the string?
946
01:13:13,030 --> 01:13:17,517
Well if I know
that-- so I just say,
947
01:13:17,517 --> 01:13:19,850
the way you do these problems
is how many things can you
948
01:13:19,850 --> 01:13:20,480
eliminate?
949
01:13:20,480 --> 01:13:22,710
Well I wasn't moving,
this term goes away.
950
01:13:22,710 --> 01:13:23,930
That's zero.
951
01:13:23,930 --> 01:13:26,370
Z wasn't involved,
it's not changing,
952
01:13:26,370 --> 01:13:30,259
it's just constant angular
rotation, that term is 0.
953
01:13:30,259 --> 01:13:32,300
What's the change in length
of the string while I
954
01:13:32,300 --> 01:13:33,660
was doing it?
955
01:13:33,660 --> 01:13:36,410
Now if that term's 0 we're
getting easier all time.
956
01:13:36,410 --> 01:13:41,860
How far-- how fast was
the length getting longer?
957
01:13:41,860 --> 01:13:43,180
That terms gone away.
958
01:13:43,180 --> 01:13:46,432
Was I speeding up, or slowing
down, or constant speed?
959
01:13:46,432 --> 01:13:47,890
Well we'll say it's
constant speed,
960
01:13:47,890 --> 01:13:50,060
ooh this problem's getting
easier all the time.
961
01:13:50,060 --> 01:13:51,930
I'm down to one term.
962
01:13:51,930 --> 01:13:53,090
f equals ma.
963
01:14:03,730 --> 01:14:10,960
Minus r theta dot squared r hat.
964
01:14:10,960 --> 01:14:15,750
So the force that I must have
been applying to the string
965
01:14:15,750 --> 01:14:18,690
was in the minus r
hat direction and had
966
01:14:18,690 --> 01:14:22,380
magnitude mr omega squared.
967
01:14:25,400 --> 01:14:27,010
So that's actually
all there is to it.
968
01:14:27,010 --> 01:14:29,830
We're using polar coordinates
and cylindrical coordinates
969
01:14:29,830 --> 01:14:33,210
to do second law problems.
970
01:14:33,210 --> 01:14:35,410
So there's a couple of
problems that you're
971
01:14:35,410 --> 01:14:37,770
doing these kind of
things on the homework
972
01:14:37,770 --> 01:14:40,430
set that's being put out today.
973
01:14:40,430 --> 01:14:44,330
So give them a try.
974
01:14:44,330 --> 01:14:45,827
Have a good weekend.