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PROFESSOR: OK.
9
00:00:22,930 --> 00:00:26,820
I've been giving
out the money cards
10
00:00:26,820 --> 00:00:30,610
for a few of the lectures,
and two or three questions
11
00:00:30,610 --> 00:00:32,980
came up in those that I
haven't addressed so far.
12
00:00:32,980 --> 00:00:35,179
I'm calling them loose ends.
13
00:00:35,179 --> 00:00:37,220
And I'm going to pick up
a couple of those today,
14
00:00:37,220 --> 00:00:40,080
I think they'll
help you consolidate
15
00:00:40,080 --> 00:00:43,010
the knowledge around the quiz.
16
00:00:43,010 --> 00:00:45,180
So I'm going to tie up a
little loose ends there.
17
00:00:45,180 --> 00:00:48,710
And then the lecture topic
I started last time, which
18
00:00:48,710 --> 00:00:51,650
is making this transition
from thinking about angular
19
00:00:51,650 --> 00:00:56,150
momentum of particles to using
the full angular momentum
20
00:00:56,150 --> 00:00:58,080
equations for
rigid bodies, where
21
00:00:58,080 --> 00:01:00,565
we talk about mass
moments of inertia
22
00:01:00,565 --> 00:01:01,740
and products of inertia.
23
00:01:01,740 --> 00:01:04,190
And that's where we'll
pick up, there again today.
24
00:01:04,190 --> 00:01:07,215
Because that's where we're
going for the next few lectures.
25
00:01:07,215 --> 00:01:07,715
OK.
26
00:01:18,270 --> 00:01:22,190
Let's pick up with a
first example here.
27
00:01:22,190 --> 00:01:29,290
This is on the topic
of basically finding
28
00:01:29,290 --> 00:01:31,310
equations of motion.
29
00:01:31,310 --> 00:01:32,970
And there's been
a little confusion
30
00:01:32,970 --> 00:01:38,120
with people, who have asked
me what do you mean find?
31
00:01:38,120 --> 00:01:39,437
Do you mean solve, et cetera.
32
00:01:39,437 --> 00:01:42,020
And so I'm going to go through,
and just a real quick example,
33
00:01:42,020 --> 00:01:47,030
skipping some of the
steps because my purpose
34
00:01:47,030 --> 00:01:51,810
is emphasizing the steps, not
working out all the details.
35
00:01:51,810 --> 00:01:55,950
So finding equations of motion.
36
00:01:55,950 --> 00:01:58,240
Where does it begin?
37
00:01:58,240 --> 00:01:59,940
One of the really
important steps
38
00:01:59,940 --> 00:02:07,660
is this, determine the
number of independent
39
00:02:07,660 --> 00:02:08,770
coordinates you need.
40
00:02:18,330 --> 00:02:20,980
Because when you've done
that, that tells you,
41
00:02:20,980 --> 00:02:27,890
basically-- it really
starts finding the number
42
00:02:27,890 --> 00:02:28,980
of degrees of freedom.
43
00:02:28,980 --> 00:02:30,730
Should have put this
in a different order.
44
00:02:30,730 --> 00:02:33,020
Degrees of freedom
tells you the number
45
00:02:33,020 --> 00:02:35,110
of independent
coordinates you need.
46
00:02:35,110 --> 00:02:39,590
This is 1, 2, and
then 3, that leads you
47
00:02:39,590 --> 00:02:43,130
to the number of equations
of motion that you need.
48
00:02:43,130 --> 00:02:46,520
So this is really
an important step.
49
00:02:46,520 --> 00:02:52,640
Secondly, draw a
free body diagram.
50
00:02:55,150 --> 00:03:01,350
And third, apply
summation of forces.
51
00:03:05,110 --> 00:03:10,320
External vector equations gives
you mass times acceleration,
52
00:03:10,320 --> 00:03:27,730
and summation of torques gives
you DHDT plus this V cross
53
00:03:27,730 --> 00:03:30,470
P term.
54
00:03:30,470 --> 00:03:32,980
So this is just kind
of the step by step.
55
00:03:32,980 --> 00:03:36,670
So let's apply it briefly.
56
00:03:46,480 --> 00:03:49,870
We've talked a lot about things
on hills, so here's a cart.
57
00:03:49,870 --> 00:03:53,680
It's got wheels
attached by a cord
58
00:03:53,680 --> 00:03:58,630
to a second mass that's sliding.
59
00:03:58,630 --> 00:04:02,685
m1, m2, doesn't stretch
the cord in between.
60
00:04:11,240 --> 00:04:14,320
Let's think of these
things as rigid bodies.
61
00:04:14,320 --> 00:04:15,670
So how many degrees of freedom?
62
00:04:15,670 --> 00:04:18,890
How many possible
degrees of freedom?
63
00:04:18,890 --> 00:04:21,529
For the maximum possible, you
have how many rigid bodies?
64
00:04:21,529 --> 00:04:22,990
2.
65
00:04:22,990 --> 00:04:26,030
How many degrees of freedom
for rigid bodies possible?
66
00:04:26,030 --> 00:04:27,860
6 each.
67
00:04:27,860 --> 00:04:31,750
So we're at 6 times
m plus 3 times
68
00:04:31,750 --> 00:04:33,750
n minus the number
of constraints
69
00:04:33,750 --> 00:04:36,300
is the number of independent
degrees of freedom.
70
00:04:36,300 --> 00:04:39,390
This is the number of rigid
bodies, number of particles,
71
00:04:39,390 --> 00:04:44,010
so we have 6 times 2, 3
times 0 minus constraints,
72
00:04:44,010 --> 00:04:47,505
so this one comes out 12
minus the constraints.
73
00:04:50,032 --> 00:04:51,990
You have to figure out
the constraints quickly.
74
00:04:51,990 --> 00:04:55,830
We're not going to allow
rotation in any of the three
75
00:04:55,830 --> 00:04:56,740
directions on either.
76
00:04:56,740 --> 00:04:58,960
They're on carts, they're
big, they're sliding,
77
00:04:58,960 --> 00:05:00,680
they're not rolling
or any of that.
78
00:05:00,680 --> 00:05:04,540
So no rotation for 3,
no rotation for 3 more.
79
00:05:04,540 --> 00:05:06,390
That's minus 6.
80
00:05:06,390 --> 00:05:14,000
So c equals minus 6, or
c equals 6 for rotation.
81
00:05:14,000 --> 00:05:15,650
And then what else can we say?
82
00:05:19,330 --> 00:05:24,280
There's no-- I'll designate
this the y direction so we
83
00:05:24,280 --> 00:05:25,670
can talk about directions here.
84
00:05:25,670 --> 00:05:30,020
And I'll designate
this x in general.
85
00:05:30,020 --> 00:05:34,620
No acceleration at all in
the y direction, right.
86
00:05:34,620 --> 00:05:36,010
Can't move in the y.
87
00:05:36,010 --> 00:05:39,260
So that gives us
1, 2 for each mass,
88
00:05:39,260 --> 00:05:42,050
plus 2 more, that's
8 constraints
89
00:05:42,050 --> 00:05:43,630
that we've come up with.
90
00:05:43,630 --> 00:05:51,340
Now a 9th constraint is
the fact that these two
91
00:05:51,340 --> 00:05:52,530
are tied together.
92
00:05:52,530 --> 00:05:54,730
And so if you had
just temporarily
93
00:05:54,730 --> 00:06:00,510
assigned a coordinate here,
x1, and another one here, x2,
94
00:06:00,510 --> 00:06:04,284
we know for a fact that x1
has got to be equal to x2,
95
00:06:04,284 --> 00:06:06,075
and that gives you yet
one more constraint.
96
00:06:10,800 --> 00:06:13,900
So that's 9.
97
00:06:13,900 --> 00:06:15,250
So we might just stop there.
98
00:06:15,250 --> 00:06:18,010
We say OK, that's a
total of 9, so the number
99
00:06:18,010 --> 00:06:24,980
of degrees of freedom,
12 minus 9 is 3,
100
00:06:24,980 --> 00:06:29,100
and that implies that you need
three equations of motion.
101
00:06:29,100 --> 00:06:33,810
Some confusion comes, you
know, if something's not--
102
00:06:33,810 --> 00:06:37,430
let me rephrase that.
103
00:06:37,430 --> 00:06:40,810
We haven't talked anything
about the z direction.
104
00:06:40,810 --> 00:06:44,960
I haven't described any
constraints in the z direction.
105
00:06:44,960 --> 00:06:50,040
If this is me in a car and I'm
dragging a sled down a hill,
106
00:06:50,040 --> 00:06:53,460
or I'm in a vehicle and I'm
dragging a sled down a hill.
107
00:06:53,460 --> 00:06:57,990
I don't know if you've ever
been in a vehicle with a trailer
108
00:06:57,990 --> 00:07:00,690
on an icy road in the
winter time, that's
109
00:07:00,690 --> 00:07:05,640
a dicey maneuver, going down a
hill trying to put brakes on.
110
00:07:05,640 --> 00:07:09,280
So this thing could conceivably
move in this direction.
111
00:07:09,280 --> 00:07:12,020
And you can either
constrain it to be that way
112
00:07:12,020 --> 00:07:15,910
to make the problem simple, or
you can just say it's possible,
113
00:07:15,910 --> 00:07:18,740
so we have three
equations of motion,
114
00:07:18,740 --> 00:07:23,460
of which two, for now, the
summation of the forces,
115
00:07:23,460 --> 00:07:27,980
external in the--
since I've got x and y,
116
00:07:27,980 --> 00:07:32,550
z must be this way--
in the y direction.
117
00:07:32,550 --> 00:07:35,300
And we'll make it
in z direction,
118
00:07:35,300 --> 00:07:39,240
this is coordinate system 1,
so this would be z1 or z1.
119
00:07:39,240 --> 00:07:53,610
The summation of forces,
z1, is m1 z1 double dot.
120
00:07:53,610 --> 00:07:55,261
But I'm going to
set that equal to 0,
121
00:07:55,261 --> 00:07:56,510
I just know there's no forces.
122
00:07:56,510 --> 00:07:59,740
So this becomes a trivial
equation of motion.
123
00:07:59,740 --> 00:08:03,620
And I add another one, summation
of forces on the second mass.
124
00:08:03,620 --> 00:08:11,880
This is on m2 in the z2
direction, is m2 z2 double dot,
125
00:08:11,880 --> 00:08:14,020
and we set that equal to 0 also.
126
00:08:14,020 --> 00:08:16,410
So what it boils down
to, I had 3 degrees
127
00:08:16,410 --> 00:08:19,860
of freedom, 2 trivial
equations of motion,
128
00:08:19,860 --> 00:08:22,350
leaving me with just
1 equation of motion
129
00:08:22,350 --> 00:08:24,560
that's going to be meaningful.
130
00:08:24,560 --> 00:08:25,060
Yeah.
131
00:08:25,060 --> 00:08:31,431
AUDIENCE: If x1 is
equal to x2, would that
132
00:08:31,431 --> 00:08:36,124
mean that the rod has to be
entirely along the x-axis?
133
00:08:36,124 --> 00:08:37,606
So that would mean that--
134
00:08:37,606 --> 00:08:39,971
PROFESSOR: So he
asked if x1 equals x2,
135
00:08:39,971 --> 00:08:42,179
does that mean they both
have to be along the x-axis?
136
00:08:42,179 --> 00:08:43,350
I'm assuming that.
137
00:08:43,350 --> 00:08:46,110
So I really am assuming this
thing's going down the hill.
138
00:08:46,110 --> 00:08:49,250
I'm making a point about
this z direction thing
139
00:08:49,250 --> 00:08:51,170
because it's just
a subtlety that you
140
00:08:51,170 --> 00:08:56,070
have to decide on when you're
figuring out how to actually
141
00:08:56,070 --> 00:08:57,710
analyze the situation.
142
00:08:57,710 --> 00:09:00,040
If you really were
thinking about what
143
00:09:00,040 --> 00:09:04,460
happens when a vehicle is going
down a steep, icy hill towing
144
00:09:04,460 --> 00:09:08,777
a trailer, maybe put the
brakes on, maybe not.
145
00:09:08,777 --> 00:09:10,610
You could probably say
well, unless I really
146
00:09:10,610 --> 00:09:15,690
have a disaster, we're not going
to get rollovers and things
147
00:09:15,690 --> 00:09:16,200
like that.
148
00:09:16,200 --> 00:09:18,600
But you could imagine
that it can get out
149
00:09:18,600 --> 00:09:19,840
of the x direction, right.
150
00:09:19,840 --> 00:09:22,167
It could start sliding into z.
151
00:09:22,167 --> 00:09:23,750
It's probably not
going to go anywhere
152
00:09:23,750 --> 00:09:25,500
in the y, that's still
a good assumption.
153
00:09:25,500 --> 00:09:28,860
So this is about modeling,
and how complicated
154
00:09:28,860 --> 00:09:30,790
do you make the modeling.
155
00:09:30,790 --> 00:09:33,610
You actually have to make quite
a few modeling decisions when
156
00:09:33,610 --> 00:09:37,510
you go to do this, and we
tend, in class and examples,
157
00:09:37,510 --> 00:09:39,390
tend to really
oversimplify problems
158
00:09:39,390 --> 00:09:42,680
so that we can do them.
159
00:09:42,680 --> 00:09:44,170
So I've boiled
this down to where
160
00:09:44,170 --> 00:09:46,720
I'm going to end up with one
significant equation of motion.
161
00:10:04,840 --> 00:10:11,100
And that one's going to say
that the summation of the forces
162
00:10:11,100 --> 00:10:14,520
in the x direction--
and I'm just
163
00:10:14,520 --> 00:10:17,050
going to write mx
double dot here,
164
00:10:17,050 --> 00:10:19,780
not putting down m1s
or m2s because we're
165
00:10:19,780 --> 00:10:21,750
going to have to go
to free body diagrams
166
00:10:21,750 --> 00:10:25,310
to figure out how to apply this.
167
00:10:28,544 --> 00:10:29,335
Free body diagrams.
168
00:10:32,800 --> 00:10:33,400
First mass.
169
00:10:39,920 --> 00:10:41,470
So this is m1 here.
170
00:10:48,420 --> 00:10:53,100
m1g, a normal force, no
doubt a friction force,
171
00:10:53,100 --> 00:10:58,310
I'll call that f1, and
a tension in the cord.
172
00:11:01,330 --> 00:11:03,870
And I'm assuming that tension's
going to always be there.
173
00:11:03,870 --> 00:11:05,530
So if, again, in
a situation where
174
00:11:05,530 --> 00:11:07,350
the trailer starts
overtaking the car
175
00:11:07,350 --> 00:11:09,220
and the rope goes
slack, we're not
176
00:11:09,220 --> 00:11:11,490
going to consider
that one today.
177
00:11:11,490 --> 00:11:14,020
So there's your free body
diagram for the first one
178
00:11:14,020 --> 00:11:19,790
and we're going to need to break
mg into a couple of components.
179
00:11:19,790 --> 00:11:24,690
We're going to need
the slope here,
180
00:11:24,690 --> 00:11:28,490
and that translates
into an angle here.
181
00:11:28,490 --> 00:11:31,260
So there's our first
free body diagram.
182
00:11:31,260 --> 00:11:40,010
And our second free body
diagram, second mass, tension,
183
00:11:40,010 --> 00:11:46,838
normal force, another m2g.
184
00:11:49,782 --> 00:11:51,240
And this one's on
wheels, and we're
185
00:11:51,240 --> 00:11:54,030
going to consider
this one frictionless.
186
00:11:54,030 --> 00:11:56,280
So we don't have any friction
force holding it back.
187
00:12:08,950 --> 00:12:10,454
So the reason I've
kind of-- this
188
00:12:10,454 --> 00:12:12,120
seemed like a ridiculous
simple problem,
189
00:12:12,120 --> 00:12:13,880
but the point I
want to make is not
190
00:12:13,880 --> 00:12:17,240
the solution, not the
particular problem,
191
00:12:17,240 --> 00:12:19,390
but an issue that crops up.
192
00:12:19,390 --> 00:12:21,280
How many unknowns are
there in this problem.
193
00:12:24,064 --> 00:12:25,456
AUDIENCE: 2.
194
00:12:25,456 --> 00:12:29,560
PROFESSOR: Well, you
know n1 immediately,
195
00:12:29,560 --> 00:12:36,060
or n2, or the friction
force, or the tension,
196
00:12:36,060 --> 00:12:38,616
or, for that matter, the x
double dot we're looking for.
197
00:12:38,616 --> 00:12:40,490
There's actually-- you
start off this problem
198
00:12:40,490 --> 00:12:43,440
with five unknowns.
199
00:12:43,440 --> 00:12:48,070
But you're only looking to
derive one equation of motion.
200
00:12:48,070 --> 00:12:50,914
So the fact that we're looking
for one equation of motion
201
00:12:50,914 --> 00:12:52,330
doesn't say you
don't have to deal
202
00:12:52,330 --> 00:12:54,610
with several intermediate
equations to get there.
203
00:12:54,610 --> 00:12:57,860
That's just part of the work.
204
00:12:57,860 --> 00:13:11,700
So this is five unknowns to
start with, n1, n2, f1, t,
205
00:13:11,700 --> 00:13:12,670
and x double dot.
206
00:13:17,340 --> 00:13:21,850
So if you'll find out that the
summation of the forces on y
207
00:13:21,850 --> 00:13:33,100
and the y1 on the first mass,
this one, this gives you n1.
208
00:13:33,100 --> 00:13:36,970
Summation of the forces
in the y direction on m2,
209
00:13:36,970 --> 00:13:41,270
this one gives you m2 directly.
210
00:13:41,270 --> 00:13:45,400
From this flows directly, you
know that the friction force
211
00:13:45,400 --> 00:13:47,710
is mu n1.
212
00:13:47,710 --> 00:13:49,504
See, that's a third equation.
213
00:13:49,504 --> 00:13:51,420
So this gives you one
equation, this gives you
214
00:13:51,420 --> 00:13:53,503
another equation, this
gives you a third equation.
215
00:13:53,503 --> 00:13:55,630
You have 5 unknowns,
you need 5 equations.
216
00:13:55,630 --> 00:13:58,760
So there's 3 of them
right off the top.
217
00:13:58,760 --> 00:14:07,790
So this leaves-- you solve
for the those, this leaves t
218
00:14:07,790 --> 00:14:10,530
and x double dot to solve for.
219
00:14:18,990 --> 00:14:26,190
So now the sum of the forces
on mass 1 in the x direction
220
00:14:26,190 --> 00:14:29,280
says m1 x double dot.
221
00:14:33,040 --> 00:14:37,390
And the sum of the forces on the
second mass in the x direction
222
00:14:37,390 --> 00:14:42,080
gives you m2 x double dot.
223
00:14:42,080 --> 00:14:48,060
And I explicitly haven't
said this yet, what I've done
224
00:14:48,060 --> 00:14:52,780
is, one of my requirements
is x1 equals x2,
225
00:14:52,780 --> 00:14:56,070
and I'm just going
to call them x.
226
00:14:56,070 --> 00:14:57,720
Both of these are
exactly the same,
227
00:14:57,720 --> 00:14:59,870
both masses have to move
with the same motion,
228
00:14:59,870 --> 00:15:01,030
is the assumption.
229
00:15:01,030 --> 00:15:05,935
So that means that x1 double
dot equals x2 double dot equals
230
00:15:05,935 --> 00:15:07,310
x double dot, and
that's what I'm
231
00:15:07,310 --> 00:15:10,590
assuming when I'm writing
down these two equations.
232
00:15:10,590 --> 00:15:14,490
I can write those two equations,
one from each free body
233
00:15:14,490 --> 00:15:15,140
diagram.
234
00:15:15,140 --> 00:15:18,130
I've already eliminated
three of the unknowns.
235
00:15:18,130 --> 00:15:23,460
And now because I have two
equations, each have t in them.
236
00:15:23,460 --> 00:15:35,150
Essentially, you eliminate t
and solve for x double dot.
237
00:15:35,150 --> 00:15:39,360
And if you do that
in this problem,
238
00:15:39,360 --> 00:15:40,815
you get your equation of motion.
239
00:16:14,550 --> 00:16:17,992
You eliminate t, solve
for x1 double dot.
240
00:16:17,992 --> 00:16:19,700
Look at this, and you
just look at things
241
00:16:19,700 --> 00:16:21,060
that doesn't make sense.
242
00:16:21,060 --> 00:16:25,120
This says the total mass times
the acceleration is a system,
243
00:16:25,120 --> 00:16:27,640
it's one system.
244
00:16:27,640 --> 00:16:30,040
Mass times the acceleration
of the center of gravity
245
00:16:30,040 --> 00:16:32,090
of the system, if
you will, has got
246
00:16:32,090 --> 00:16:33,930
to be equal to the sums
of the forces on it.
247
00:16:33,930 --> 00:16:39,870
Well, it's got m1 plus m2g sine
theta pulling it down the hill,
248
00:16:39,870 --> 00:16:47,160
and it has minus m
mu m1g cosine theta
249
00:16:47,160 --> 00:16:48,780
dragging it back up the hill.
250
00:16:48,780 --> 00:16:52,180
And that's the entire equation
of motion, it makes sense.
251
00:16:52,180 --> 00:16:58,650
But the equation of motion,
the thing you're looking for,
252
00:16:58,650 --> 00:17:01,410
is the one that ends up with
this acceleration term in it.
253
00:17:01,410 --> 00:17:03,340
If you have multiple
degrees of freedom,
254
00:17:03,340 --> 00:17:05,240
multiple coordinates--
if you have,
255
00:17:05,240 --> 00:17:08,470
let's say, three significant
equations of motion
256
00:17:08,470 --> 00:17:11,000
that result, there
won't necessarily
257
00:17:11,000 --> 00:17:13,180
be one in terms of
each coordinate.
258
00:17:13,180 --> 00:17:16,036
They'll have the
coordinates mixed in them.
259
00:17:16,036 --> 00:17:18,119
Like we did that that two
mass system with springs
260
00:17:18,119 --> 00:17:22,200
the other day, each equation
of motion had x1 and x2.
261
00:17:22,200 --> 00:17:24,200
They don't necessarily separate.
262
00:17:24,200 --> 00:17:26,730
They're coupled through
their coordinates.
263
00:17:26,730 --> 00:17:29,670
This one, it's one equation
of motion for the system,
264
00:17:29,670 --> 00:17:31,640
therefore you have
only one coordinate.
265
00:17:31,640 --> 00:17:34,314
But that's not generally
true of multiple degree
266
00:17:34,314 --> 00:17:35,105
of freedom systems.
267
00:17:38,100 --> 00:17:43,020
OK, that's your method,
though, for a simple problem.
268
00:17:43,020 --> 00:17:47,900
I want to do a little more
difficult problem that
269
00:17:47,900 --> 00:17:50,450
involves rotation.
270
00:18:00,964 --> 00:18:02,630
And this is the
problem, I'm sure you've
271
00:18:02,630 --> 00:18:04,706
done this problem in physics.
272
00:18:04,706 --> 00:18:07,150
It's a classic problem
that people do.
273
00:18:09,922 --> 00:18:26,300
A disk, a pulley, really,
supporting two masses
274
00:18:26,300 --> 00:18:31,390
rotates about this point,
which I'll call a here.
275
00:18:38,150 --> 00:18:43,760
So at some theta there's
no slip, so theta is going
276
00:18:43,760 --> 00:18:47,120
to be related to movement, x.
277
00:18:47,120 --> 00:18:52,820
And I'm going to assign this
one a coordinate, x1 going down.
278
00:18:52,820 --> 00:18:56,100
This one a coordinate,
x2 going up.
279
00:18:56,100 --> 00:18:58,674
A little foreknowledge
here, because you've
280
00:18:58,674 --> 00:18:59,465
worked the problem.
281
00:19:02,240 --> 00:19:05,020
So I want to solve for
the motion of this system.
282
00:19:08,120 --> 00:19:13,700
Now again, we need to know the
number of degrees of freedom.
283
00:19:13,700 --> 00:19:18,890
So it's the maximum possible,
which is our 6m plus
284
00:19:18,890 --> 00:19:22,040
3n minus the constraints.
285
00:19:22,040 --> 00:19:27,092
And let's think about how
we want to model this again.
286
00:19:27,092 --> 00:19:29,300
This time I'm just going to
model these as particles,
287
00:19:29,300 --> 00:19:31,420
doesn't matter how big they are.
288
00:19:31,420 --> 00:19:33,780
My problem, really, they
only go up and down.
289
00:19:33,780 --> 00:19:35,660
So I'm going to model
them as particles.
290
00:19:35,660 --> 00:19:41,000
This is 6 times 0 plus 3
times 2 minus constraints.
291
00:19:41,000 --> 00:19:42,620
So 6 minus the constraints.
292
00:19:42,620 --> 00:19:44,980
So the issue is really
how many constraints.
293
00:19:47,570 --> 00:19:55,230
Well, we're going to
require x1 equal x2.
294
00:19:55,230 --> 00:19:57,756
Cord's taught, doesn't stretch.
295
00:19:57,756 --> 00:19:59,130
If this thing goes
down, that has
296
00:19:59,130 --> 00:20:01,787
to go up exactly
an equal amount,
297
00:20:01,787 --> 00:20:02,870
and that's one constraint.
298
00:20:11,710 --> 00:20:14,690
In that, leaving us 6
minus 1 is 5, leaving us
299
00:20:14,690 --> 00:20:17,654
with a lot of degrees
of freedom here.
300
00:20:17,654 --> 00:20:19,570
Kind of back to the issue
I was making before,
301
00:20:19,570 --> 00:20:24,190
are there any constraints
in the-- I'll call it
302
00:20:24,190 --> 00:20:29,740
x, y, z directions here.
303
00:20:29,740 --> 00:20:33,030
Are there any constraints
in the y or z directions
304
00:20:33,030 --> 00:20:36,280
on either of those masses?
305
00:20:36,280 --> 00:20:39,520
No, I haven't shown any, no
tracks, no guides, no anything.
306
00:20:39,520 --> 00:20:42,500
So technically, there are
no additional constraints
307
00:20:42,500 --> 00:20:43,710
in this problem.
308
00:20:43,710 --> 00:20:49,560
But if there's no forces in
the y, x, I guess y this way,
309
00:20:49,560 --> 00:20:51,820
and no forces in
the z, I'm going
310
00:20:51,820 --> 00:20:54,290
to end up with two trivial
equations of motion
311
00:20:54,290 --> 00:20:56,980
for this one, 1 in
the y, 1 in the z,
312
00:20:56,980 --> 00:20:59,460
for this one, 1 in
the y, 1 in the z.
313
00:20:59,460 --> 00:21:01,880
So back to this
issue of, there's
314
00:21:01,880 --> 00:21:05,480
a difference between constraints
and trivial equations
315
00:21:05,480 --> 00:21:05,980
of motion.
316
00:21:05,980 --> 00:21:08,614
We're going have 4 to
reveal equations of motion.
317
00:21:08,614 --> 00:21:10,280
So really, again, I'm
going to come down
318
00:21:10,280 --> 00:21:13,860
to one significant
equation of motion.
319
00:21:13,860 --> 00:21:23,110
So I have 1 constraint,
4 trivial EOMs,
320
00:21:23,110 --> 00:21:32,390
and 1 significant
equation of motion.
321
00:21:32,390 --> 00:21:32,890
OK.
322
00:21:35,440 --> 00:21:39,700
Now, because I want to
talk about rotation,
323
00:21:39,700 --> 00:21:41,500
we need to pick a coordinate.
324
00:21:41,500 --> 00:21:49,020
Now I can pick-- I can
either let x1 equal x2
325
00:21:49,020 --> 00:21:52,000
and just let it be the sum
x, a single coordinate,
326
00:21:52,000 --> 00:21:55,880
or theta, the rotation.
327
00:21:55,880 --> 00:21:57,960
I'm actually going to
use x for a second.
328
00:22:05,250 --> 00:22:09,990
But there's 2 obvious ways
to approach this problem.
329
00:22:09,990 --> 00:22:15,140
One is to draw a
free body diagrams
330
00:22:15,140 --> 00:22:22,160
of each of these masses, sum the
forces on each, and how many--
331
00:22:22,160 --> 00:22:25,110
if I do that, how many
unknowns do I end up with?
332
00:22:29,120 --> 00:22:30,990
We can draw the free--
here's the free body
333
00:22:30,990 --> 00:22:32,280
diagram for mass 1.
334
00:22:32,280 --> 00:22:33,185
What's it got on it?
335
00:22:33,185 --> 00:22:34,950
Well, m1g.
336
00:22:34,950 --> 00:22:37,064
What else is acting on it?
337
00:22:37,064 --> 00:22:38,972
AUDIENCE: [? Tension. ?]
338
00:22:38,972 --> 00:22:44,590
PROFESSOR: And here's
the second one, m2g,
339
00:22:44,590 --> 00:22:47,100
and tension acting on it.
340
00:22:47,100 --> 00:22:50,090
So now we can sum,
sum of the forces
341
00:22:50,090 --> 00:22:52,520
equals mass times the
acceleration of each one,
342
00:22:52,520 --> 00:22:54,840
and the external forces
are going to involve t.
343
00:22:54,840 --> 00:22:57,810
So you're going to end up
with how many unknowns?
344
00:23:02,960 --> 00:23:04,290
How many unknowns?
345
00:23:04,290 --> 00:23:07,380
I can write two equations
for the sum of the forces
346
00:23:07,380 --> 00:23:08,660
in the x direction.
347
00:23:08,660 --> 00:23:11,520
x double dot is
certainly an unknown.
348
00:23:11,520 --> 00:23:12,020
What else?
349
00:23:12,020 --> 00:23:12,500
AUDIENCE: t.
350
00:23:12,500 --> 00:23:13,041
PROFESSOR: t.
351
00:23:13,041 --> 00:23:15,269
So I end up with
this other [? end. ?]
352
00:23:15,269 --> 00:23:17,560
So that means I'm going to
have to write two equations,
353
00:23:17,560 --> 00:23:18,900
I'm going to have
to eliminate t,
354
00:23:18,900 --> 00:23:20,608
going to go through
the same thing there.
355
00:23:20,608 --> 00:23:22,670
So I don't want to
bother with that.
356
00:23:22,670 --> 00:23:25,485
Is there another way
to do this problem?
357
00:23:32,270 --> 00:23:35,980
This is a problem where you
can use angular momentum
358
00:23:35,980 --> 00:23:38,440
and not have to
deal with t at all.
359
00:23:38,440 --> 00:23:40,029
So let's set that problem up.
360
00:23:44,620 --> 00:23:49,120
You know that the sum of
the torques about that point
361
00:23:49,120 --> 00:23:55,380
a, with respect to point a,
it's going to be derivative,
362
00:23:55,380 --> 00:23:57,910
and since we're dealing
with particles here,
363
00:23:57,910 --> 00:24:01,422
of the angular momentum
with respect to-- I'll
364
00:24:01,422 --> 00:24:04,540
just call it lowercase
h for particles.
365
00:24:04,540 --> 00:24:09,480
Plus this velocity
of a with respect
366
00:24:09,480 --> 00:24:16,920
to an inertial frame across a
linear momentum with respect
367
00:24:16,920 --> 00:24:18,440
to an inertial frame.
368
00:24:18,440 --> 00:24:21,530
That's the full equation
for sum of torques.
369
00:24:21,530 --> 00:24:23,830
What's velocity of a with
respect to o in this problem?
370
00:24:26,930 --> 00:24:27,670
0.
371
00:24:27,670 --> 00:24:29,420
Fortunately, this is
one of those problems
372
00:24:29,420 --> 00:24:32,770
where you can get rid of
this difficult second term.
373
00:24:32,770 --> 00:24:35,990
So it's just torques as the
time derivative of the angular
374
00:24:35,990 --> 00:24:37,100
momentum.
375
00:24:37,100 --> 00:24:44,080
So we need an expression, then,
for both the sum of the torques
376
00:24:44,080 --> 00:24:45,970
with respect to a.
377
00:24:45,970 --> 00:24:47,390
Let's see, what would that be?
378
00:24:50,950 --> 00:24:57,290
So now the external torques
with respect to-- I'll
379
00:24:57,290 --> 00:25:03,220
finish my-- oops, come here
you-- free body diagram.
380
00:25:03,220 --> 00:25:06,870
So now what I really want
is a free body diagram
381
00:25:06,870 --> 00:25:07,795
of the whole system.
382
00:25:10,990 --> 00:25:13,790
So here's the whole system
created as one thing.
383
00:25:13,790 --> 00:25:20,850
You have a force down, m1g,
another force down, m2g.
384
00:25:20,850 --> 00:25:25,630
Up here you have
some normal force up,
385
00:25:25,630 --> 00:25:28,350
that's the support of the pin.
386
00:25:28,350 --> 00:25:38,820
You have tensions in these,
but now this equation
387
00:25:38,820 --> 00:25:40,550
applies to the system.
388
00:25:40,550 --> 00:25:43,910
The ts are internal to the
system, they are irrelevant.
389
00:25:46,640 --> 00:25:49,360
So I'm talking about this whole
thing treated as a system,
390
00:25:49,360 --> 00:25:50,860
and I'm going to
compute the moments
391
00:25:50,860 --> 00:25:53,620
about point a, which is right
there, where that axle is.
392
00:25:53,620 --> 00:25:56,830
Does n create a
moment at the axle?
393
00:25:56,830 --> 00:26:03,260
Nope, but the m1 and m2
times g create moments.
394
00:26:03,260 --> 00:26:04,160
Sure, OK.
395
00:26:07,467 --> 00:26:09,300
I'm going to have
positive out of the board,
396
00:26:09,300 --> 00:26:13,370
the positive moment,
positive angular direction.
397
00:26:13,370 --> 00:26:16,090
So the torques
applied to this system
398
00:26:16,090 --> 00:26:22,630
are r cross t, so you're
going to end up with-- I
399
00:26:22,630 --> 00:26:26,350
want to summarize these.
400
00:26:26,350 --> 00:26:33,734
An m1g, and I didn't write
the radius on this problem,
401
00:26:33,734 --> 00:26:39,345
but at some radius, capital
R. So the torques that are m1g
402
00:26:39,345 --> 00:26:49,190
are positive minus
m2gR, k hat direction,
403
00:26:49,190 --> 00:26:53,460
and that must be equal to the
time derivative of the angular
404
00:26:53,460 --> 00:26:57,570
momentum about a.
405
00:26:57,570 --> 00:27:04,470
Now we need an expression
for the angular
406
00:27:04,470 --> 00:27:06,221
momentum with respect to a.
407
00:27:09,590 --> 00:27:16,260
Angular momentum
is, in general, this
408
00:27:16,260 --> 00:27:19,940
is a r cross linear
momentum, right.
409
00:27:19,940 --> 00:27:28,940
So R for mass 1 with
respect to a cross
410
00:27:28,940 --> 00:27:34,650
the momentum of that
second mass with respect
411
00:27:34,650 --> 00:27:35,880
to an inertial frame.
412
00:27:35,880 --> 00:27:38,340
And a and the inertial
frame are the same thing,
413
00:27:38,340 --> 00:27:40,340
a sticks in the inertial frame.
414
00:27:40,340 --> 00:27:42,970
But angular momentum
is always with respect
415
00:27:42,970 --> 00:27:44,690
to the inertial frame.
416
00:27:44,690 --> 00:27:50,070
Plus the second piece, which
is R of m2 with respect to a
417
00:27:50,070 --> 00:27:54,760
crossed with p for mass 2 with
respect to some inertial frame.
418
00:28:01,230 --> 00:28:03,230
I'm just going to give
you the results for this.
419
00:28:07,020 --> 00:28:18,516
m1 plus m2 R x dot k.
420
00:28:18,516 --> 00:28:25,680
So x dot is this velocity,
R cross and mass times
421
00:28:25,680 --> 00:28:30,360
velocity is momentum, so
the perpendicular radius
422
00:28:30,360 --> 00:28:33,460
to that is the
radius, R. So it's
423
00:28:33,460 --> 00:28:35,830
Rx dot times m,
shouldn't surprise you,
424
00:28:35,830 --> 00:28:36,850
in the k hat direction.
425
00:28:36,850 --> 00:28:38,940
That's the total
angular momentum
426
00:28:38,940 --> 00:28:43,970
that comes from these two
particles with respect to a.
427
00:28:43,970 --> 00:28:45,785
And taking their
time derivative.
428
00:28:50,000 --> 00:28:53,230
These are constant, that's
a constant, this is not,
429
00:28:53,230 --> 00:28:55,660
this is a constant, but it
doesn't change direction,
430
00:28:55,660 --> 00:28:57,420
so this one is pretty simple.
431
00:29:04,970 --> 00:29:12,050
Now I can set equal the sum
of the external torques, that,
432
00:29:12,050 --> 00:29:14,990
to the time derivative
of the angular momentum,
433
00:29:14,990 --> 00:29:16,845
just to fulfill this expression.
434
00:29:19,450 --> 00:29:26,170
And in so doing, I end up with
a solution for x double dot.
435
00:29:26,170 --> 00:29:36,520
m1 minus m2, m1
plus m2, times g.
436
00:29:36,520 --> 00:29:37,935
Turns out the R goes away.
437
00:29:41,950 --> 00:29:46,130
So one equation, never
had to mess with tension.
438
00:29:46,130 --> 00:29:52,070
This is a pretty nice, direct
way of solving this problem.
439
00:29:52,070 --> 00:29:56,674
If you solve for g here, and
you measure x double dot,
440
00:29:56,674 --> 00:29:59,380
this actually gives
you an experimental way
441
00:29:59,380 --> 00:30:02,140
of determining
acceleration of gravity.
442
00:30:02,140 --> 00:30:04,790
It's actually what this thing
was used for a long time
443
00:30:04,790 --> 00:30:07,930
ago, before they had a lot
of the measurement techniques
444
00:30:07,930 --> 00:30:09,170
and things that we do today.
445
00:30:09,170 --> 00:30:12,630
This is a way of determining
the acceleration of gravity.
446
00:30:12,630 --> 00:30:14,580
So these two masses are
quite close together.
447
00:30:14,580 --> 00:30:18,514
This number is pretty
small, you can, however
448
00:30:18,514 --> 00:30:19,805
accurate your timing device is.
449
00:30:23,970 --> 00:30:27,520
Now, just to mention it, I
neglected something in this.
450
00:30:27,520 --> 00:30:29,440
I assumed something and
I didn't even say it.
451
00:30:29,440 --> 00:30:29,970
What was it?
452
00:30:29,970 --> 00:30:31,511
What would screw up
this measurement?
453
00:30:31,511 --> 00:30:33,760
I'm trying to measure the
acceleration of gravity,
454
00:30:33,760 --> 00:30:38,780
if I built this apparatus, would
I get a very good measurement?
455
00:30:38,780 --> 00:30:41,160
AUDIENCE: The pulley
would have to be massless.
456
00:30:41,160 --> 00:30:43,510
PROFESSOR: Yeah, the pulley
would have to be massless.
457
00:30:43,510 --> 00:30:45,260
I've made an assumption
about that, right.
458
00:30:45,260 --> 00:30:47,040
So how would you
fix up this equation
459
00:30:47,040 --> 00:30:48,642
to account for the pulley?
460
00:30:51,946 --> 00:30:54,306
AUDIENCE: You'd have
to take into account
461
00:30:54,306 --> 00:30:55,260
its moment of inertia.
462
00:30:55,260 --> 00:30:56,926
PROFESSOR: Yeah, you'd
put in something.
463
00:30:56,926 --> 00:30:59,750
And where would that
go into the problem?
464
00:30:59,750 --> 00:31:02,765
How would you account its
inertia, moment of inertia
465
00:31:02,765 --> 00:31:03,410
in the problem?
466
00:31:03,410 --> 00:31:04,150
AUDIENCE: ha.
467
00:31:04,150 --> 00:31:06,250
PROFESSOR: Yeah, you'd
just put it into ha.
468
00:31:06,250 --> 00:31:11,080
So this expression for h would
end up with one more term,
469
00:31:11,080 --> 00:31:14,834
it's going to look like--
well, when you take the time
470
00:31:14,834 --> 00:31:16,250
derivative, you're
going to end up
471
00:31:16,250 --> 00:31:17,620
with another piece over here.
472
00:31:17,620 --> 00:31:21,750
Some i about a theta double dot.
473
00:31:21,750 --> 00:31:24,260
You're going to have to
relate theta double dot to x
474
00:31:24,260 --> 00:31:28,930
double dot, which you can,
because x equals R theta.
475
00:31:28,930 --> 00:31:30,660
x double dot is R
theta double dot.
476
00:31:30,660 --> 00:31:33,890
You could fix that and you'd
have an equation of motion.
477
00:31:33,890 --> 00:31:37,340
But that means we need
to know about i about a,
478
00:31:37,340 --> 00:31:42,720
that's where we're going to
at the end of this lecture
479
00:31:42,720 --> 00:31:45,470
and for the next
several lectures.
480
00:31:45,470 --> 00:31:46,900
OK.
481
00:31:46,900 --> 00:31:52,070
That's that example, and
I've got two more brief ones
482
00:31:52,070 --> 00:31:54,050
that I wanted to talk about.
483
00:31:54,050 --> 00:31:55,841
Any last questions?
484
00:31:55,841 --> 00:31:56,340
Yeah.
485
00:31:56,340 --> 00:31:58,200
AUDIENCE: Can you
explain again why
486
00:31:58,200 --> 00:31:59,991
you didn't take the
tensions into account
487
00:31:59,991 --> 00:32:00,990
for your sum of torques?
488
00:32:00,990 --> 00:32:05,680
PROFESSOR: OK, so why did I not
take the tensions into account?
489
00:32:12,520 --> 00:32:21,830
So I can write the
equation of motion
490
00:32:21,830 --> 00:32:26,470
for this thing as
a complete system.
491
00:32:26,470 --> 00:32:29,170
One, the masses and
the pulley are all
492
00:32:29,170 --> 00:32:35,300
the same thing, the summation
of the external torques on that,
493
00:32:35,300 --> 00:32:39,100
they're going to amount up to
taking into account the time
494
00:32:39,100 --> 00:32:41,870
rate of change of the angular
momentum of the system.
495
00:32:41,870 --> 00:32:43,550
Now, if I didn't
understand that,
496
00:32:43,550 --> 00:32:46,614
I could have blindly gone ahead
and put the ts in there, right,
497
00:32:46,614 --> 00:32:49,030
they would've been exactly
equal and opposite with respect
498
00:32:49,030 --> 00:32:50,860
to a and it would
have cancelled out.
499
00:32:50,860 --> 00:32:55,092
So either way, if you're not
sure about that assumption,
500
00:32:55,092 --> 00:32:56,550
you could just put
them in and they
501
00:32:56,550 --> 00:33:00,570
would appear in the torque
equation, but as a minus
502
00:33:00,570 --> 00:33:02,800
tR and a plus tR
and they'd cancel.
503
00:33:05,420 --> 00:33:11,590
I want to move on
to a third example,
504
00:33:11,590 --> 00:33:17,850
and this is the third item
that I want to clear up,
505
00:33:17,850 --> 00:33:19,190
loose ends I'm calling them.
506
00:33:19,190 --> 00:33:21,370
The muddy cards
are really useful.
507
00:33:21,370 --> 00:33:25,100
I get questions in those
that spark something.
508
00:33:25,100 --> 00:33:27,670
And this is a question that
came up two or three times
509
00:33:27,670 --> 00:33:30,030
in the muddy cards and
I haven't addressed it,
510
00:33:30,030 --> 00:33:34,540
and that is, we were
working with rotor problems.
511
00:33:34,540 --> 00:33:37,580
And remember this problem.
512
00:33:37,580 --> 00:33:39,860
You have the rotor,
it had an arm,
513
00:33:39,860 --> 00:33:42,770
I did it this way to
make some things obvious.
514
00:33:42,770 --> 00:33:49,780
But this is the z direction,
it's rotating about that axis.
515
00:33:49,780 --> 00:33:53,680
I've got a point mass up here.
516
00:33:53,680 --> 00:33:57,485
r hat, so this is
R-- actually, I'm
517
00:33:57,485 --> 00:33:59,580
going to make it a
capital R so it's easier
518
00:33:59,580 --> 00:34:01,200
to distinguish from the r hat.
519
00:34:01,200 --> 00:34:02,210
And this is z.
520
00:34:08,620 --> 00:34:18,929
This thing's rotating, it's got
bearings here to keep it going.
521
00:34:18,929 --> 00:34:24,449
And we talked about
torques, so this
522
00:34:24,449 --> 00:34:38,030
is my point A. I want to
write the sum of the torques
523
00:34:38,030 --> 00:34:44,810
about A, time derivative
of the angular momentum.
524
00:34:44,810 --> 00:34:47,060
We've done this problem
before, so I'm just putting up
525
00:34:47,060 --> 00:34:51,860
a couple of points
for review to clear up
526
00:34:51,860 --> 00:34:54,066
some possible misconceptions.
527
00:34:56,989 --> 00:35:01,140
That's this term, so what about
point A now in this problem?
528
00:35:01,140 --> 00:35:03,101
What's the velocity at point A?
529
00:35:03,101 --> 00:35:06,600
0, so again we can
get rid of this guy.
530
00:35:06,600 --> 00:35:08,600
I'm going to come back
to this and do an example
531
00:35:08,600 --> 00:35:10,220
one of these days
where this isn't 0,
532
00:35:10,220 --> 00:35:14,920
where it's really handy
to be able to do a problem
533
00:35:14,920 --> 00:35:17,710
where that's not 0.
534
00:35:17,710 --> 00:35:18,210
OK.
535
00:35:21,100 --> 00:35:22,580
This is true.
536
00:35:22,580 --> 00:35:25,870
I need a free body diagram
of our little mass,
537
00:35:25,870 --> 00:35:30,240
so here's my free body diagram.
538
00:35:30,240 --> 00:35:35,220
And it has possibly a
force in the z direction.
539
00:35:35,220 --> 00:35:37,410
That comes from the
rod, there's rods that's
540
00:35:37,410 --> 00:35:39,090
supporting this thing, right.
541
00:35:39,090 --> 00:35:41,670
There's possibly a force
in the z direction.
542
00:35:41,670 --> 00:35:45,750
There's a force in the r hat
direction, in the R direction.
543
00:35:45,750 --> 00:35:54,950
There's a force in theta
direction going into the board.
544
00:35:54,950 --> 00:35:55,975
And there's mg.
545
00:36:01,110 --> 00:36:04,660
All sorts of forces
on this thing.
546
00:36:04,660 --> 00:36:09,420
And the question was asked,
when we did this problem before
547
00:36:09,420 --> 00:36:13,850
and did the time derivatives
of the angular momentum,
548
00:36:13,850 --> 00:36:16,660
we found that we got--
there's three terms
549
00:36:16,660 --> 00:36:19,540
and I'll write them
down here for you.
550
00:36:19,540 --> 00:36:22,750
I'm just saying in advance
what we're going to do.
551
00:36:22,750 --> 00:36:24,570
When you solve this
problem, you find out
552
00:36:24,570 --> 00:36:27,520
that it takes to torque to
accelerate this shaft and spin.
553
00:36:27,520 --> 00:36:29,840
That the driving one,
that's what makes it happen,
554
00:36:29,840 --> 00:36:30,910
makes it accelerate.
555
00:36:30,910 --> 00:36:34,810
We had two more terms that were
torques at this point, that
556
00:36:34,810 --> 00:36:36,590
is what it takes to
support this system.
557
00:36:36,590 --> 00:36:39,800
It's trying to bend out,
it's trying to bend back,
558
00:36:39,800 --> 00:36:41,300
those are torques
that show up here.
559
00:36:41,300 --> 00:36:43,910
And we actually get them
when we work through this.
560
00:36:43,910 --> 00:36:48,180
But we don't get
something that tells us
561
00:36:48,180 --> 00:36:51,062
about the moment
the torque created
562
00:36:51,062 --> 00:36:52,270
this point caused by gravity.
563
00:36:55,890 --> 00:36:58,870
The question was, why don't we
get the torque about this point
564
00:36:58,870 --> 00:36:59,680
caused by gravity.
565
00:36:59,680 --> 00:37:02,920
There's clearly mg down,
there's clearly a moment arm.
566
00:37:02,920 --> 00:37:05,490
So mgR is the torque
about this point.
567
00:37:05,490 --> 00:37:08,770
And if you were doing the
statics problem in 2.001,
568
00:37:08,770 --> 00:37:10,890
there'd be a torque
around this point caused
569
00:37:10,890 --> 00:37:13,015
by the weight of this thing
just sitting there, not
570
00:37:13,015 --> 00:37:14,170
even spinning.
571
00:37:14,170 --> 00:37:19,430
And what we're doing here
gives you no help with that.
572
00:37:19,430 --> 00:37:24,730
But just for the quick
review of this problem,
573
00:37:24,730 --> 00:37:29,030
more in the line of helping
you think about the quiz.
574
00:37:29,030 --> 00:37:33,660
This then is R, we'll
call this point B
575
00:37:33,660 --> 00:37:37,270
and this is point A, remember
this is RB with respect
576
00:37:37,270 --> 00:37:43,850
to A cross p with respect to o.
577
00:37:43,850 --> 00:37:46,610
And that's where our
angular momentum comes from.
578
00:37:46,610 --> 00:37:57,460
In this problem that is r hat
plus z k hat cross m times
579
00:37:57,460 --> 00:38:03,290
the velocity,
which is R omega z.
580
00:38:03,290 --> 00:38:08,030
And that must be in the
theta hat direction.
581
00:38:08,030 --> 00:38:19,070
When you multiply these
out, omega z is theta dot.
582
00:38:19,070 --> 00:38:22,850
They're kind of interchangeable
in this problem.
583
00:38:22,850 --> 00:38:26,010
So when you multiply this
out, you get two terms.
584
00:38:26,010 --> 00:38:41,230
mR squared theta dot k hat
minus mRz theta dot r hat.
585
00:38:41,230 --> 00:38:42,560
Two terms from this.
586
00:38:45,360 --> 00:38:53,330
And when you do the time
derivative of the dhdt,
587
00:38:53,330 --> 00:38:54,200
you get three terms.
588
00:38:56,710 --> 00:39:02,017
mR squared theta
double dot and the k.
589
00:39:02,017 --> 00:39:03,350
Now, why do you get three terms?
590
00:39:03,350 --> 00:39:06,970
Because this term has
two variables in it that
591
00:39:06,970 --> 00:39:10,490
are functions of time,
theta dot has a derivative,
592
00:39:10,490 --> 00:39:14,550
and r hat has a derivative,
because it rotates.
593
00:39:14,550 --> 00:39:16,320
So one of the key
bits of mathematics
594
00:39:16,320 --> 00:39:18,529
you have to learn
in this course,
595
00:39:18,529 --> 00:39:20,570
I'm kind of giving you a
little quiz review here,
596
00:39:20,570 --> 00:39:23,030
you need to know how to take
the derivative of a rotating
597
00:39:23,030 --> 00:39:24,210
vector.
598
00:39:24,210 --> 00:39:26,250
And that's what we
do here, gives us
599
00:39:26,250 --> 00:39:34,970
two terms minus mR z
theta double dot r hat
600
00:39:34,970 --> 00:39:43,500
minus mRz theta dot
squared theta hat.
601
00:39:43,500 --> 00:39:47,260
So three terms in this time
derivative of the angular
602
00:39:47,260 --> 00:39:52,400
momentum, and they have to be
equal to the external torques.
603
00:39:52,400 --> 00:39:56,470
This is equal to the summation
of the torques about A,
604
00:39:56,470 --> 00:39:58,610
the external torques.
605
00:39:58,610 --> 00:40:01,980
Well, you'll need a
torque in the k direction.
606
00:40:01,980 --> 00:40:04,320
That's what it takes to
accelerate the thing,
607
00:40:04,320 --> 00:40:05,110
make it go faster.
608
00:40:09,590 --> 00:40:14,890
This mass has a
force on it to make
609
00:40:14,890 --> 00:40:18,500
it go faster, that's this f
in the theta hat direction.
610
00:40:18,500 --> 00:40:21,650
And that rods have
a push on that mass,
611
00:40:21,650 --> 00:40:23,825
the mass pushes back on the rod.
612
00:40:23,825 --> 00:40:25,200
So if in the theta
direction it's
613
00:40:25,200 --> 00:40:28,290
like that, the mass pushes back
on the rod, it twists the rod,
614
00:40:28,290 --> 00:40:29,060
or tries to.
615
00:40:29,060 --> 00:40:33,936
That's a torque about this
in the r hat direction.
616
00:40:33,936 --> 00:40:35,435
So there's centripetal
acceleration,
617
00:40:35,435 --> 00:40:38,590
it takes force to cause
centripetal acceleration.
618
00:40:38,590 --> 00:40:41,270
It's that force is inward.
619
00:40:41,270 --> 00:40:44,020
It's about a moment arm
z, and so this gives you
620
00:40:44,020 --> 00:40:47,000
a torque about the point A
in the theta hat direction.
621
00:40:47,000 --> 00:40:50,500
So these are three different
terms, each one has a purpose.
622
00:40:50,500 --> 00:40:52,450
No work is done here,
no work is done here,
623
00:40:52,450 --> 00:40:53,616
because there's no movement.
624
00:40:56,990 --> 00:40:59,830
Now, but gravity, we
started this question
625
00:40:59,830 --> 00:41:01,620
as why doesn't gravity
pop out of this.
626
00:41:04,220 --> 00:41:07,220
Because this only tells
you about the time rate
627
00:41:07,220 --> 00:41:09,820
of change of angular momentum.
628
00:41:09,820 --> 00:41:13,810
Gravity has nothing to
do with angular momentum.
629
00:41:13,810 --> 00:41:17,870
r cross p is all that
angular momentum is.
630
00:41:17,870 --> 00:41:19,630
The linear momentum of little
631
00:41:19,630 --> 00:41:24,060
clumps of mass times the
radius from the point you're
632
00:41:24,060 --> 00:41:26,000
computing the angular momentum.
633
00:41:26,000 --> 00:41:28,060
Has nothing to do
with g, never will.
634
00:41:30,760 --> 00:41:35,970
You'll never get the g
related static moment
635
00:41:35,970 --> 00:41:38,200
out of this equation.
636
00:41:38,200 --> 00:41:41,292
It's there, though, and if
you were designing the system,
637
00:41:41,292 --> 00:41:42,750
you'd have to take
it into account.
638
00:42:08,530 --> 00:42:12,270
So remember, I didn't bring it
today, but I have my shaker.
639
00:42:12,270 --> 00:42:13,635
I've bolted it to the floor.
640
00:42:18,120 --> 00:42:22,720
Inside of that shaker is
a little rotating mass.
641
00:42:22,720 --> 00:42:25,570
It has a little arm
and eccentricity, it
642
00:42:25,570 --> 00:42:29,916
has some mass that
I'm going to make m,
643
00:42:29,916 --> 00:42:35,250
it's rotating theta direction.
644
00:42:35,250 --> 00:42:39,000
And it rotates a constant speed.
645
00:42:39,000 --> 00:42:44,940
So it's some constant omega,
theta double dot equals 0.
646
00:42:44,940 --> 00:42:47,040
So I just got my shaker
bolted to the floors,
647
00:42:47,040 --> 00:42:49,650
it's putting a lot of
vibration into the floor.
648
00:42:49,650 --> 00:42:51,920
And the question
that someone came up
649
00:42:51,920 --> 00:42:55,660
with on a muddy card that was
a really inside insightful
650
00:42:55,660 --> 00:42:59,960
question, why-- or
they didn't say why--
651
00:42:59,960 --> 00:43:03,810
they said, shouldn't
the torque required
652
00:43:03,810 --> 00:43:07,960
to drive this thing somehow
be affected by gravity?
653
00:43:11,710 --> 00:43:14,320
So does the torque
that it takes to run
654
00:43:14,320 --> 00:43:17,710
this around and around
depend on gravity,
655
00:43:17,710 --> 00:43:19,980
was the question that was asked.
656
00:43:19,980 --> 00:43:21,910
Let's take a quick look at that.
657
00:43:21,910 --> 00:43:25,000
We just discovered that dhdt
doesn't tell you anything
658
00:43:25,000 --> 00:43:28,030
about torque from
gravity, right?
659
00:43:28,030 --> 00:43:30,170
Well, let's see
what happens then.
660
00:43:30,170 --> 00:43:35,000
So the summation of the external
torques-- I'll call this now
661
00:43:35,000 --> 00:43:37,080
point A, where it's
rotating about.
662
00:43:37,080 --> 00:43:39,720
This point now doesn't
move in this problem.
663
00:43:39,720 --> 00:43:41,430
It's an inertial point.
664
00:43:41,430 --> 00:43:44,460
Summation of the torques
with respect to A
665
00:43:44,460 --> 00:43:49,200
is dh with respect to A, dt,
and there's no additional terms
666
00:43:49,200 --> 00:43:53,370
because that
velocity point is 0.
667
00:43:53,370 --> 00:44:01,100
And that's d by dt, the
torque is just r cross p,
668
00:44:01,100 --> 00:44:05,200
so that is me theta dot.
669
00:44:05,200 --> 00:44:07,435
That's the velocity,
that's the momentum.
670
00:44:21,100 --> 00:44:22,805
I've left out something.
671
00:44:31,650 --> 00:44:36,670
So r cross p, I need
an e squared in here.
672
00:44:36,670 --> 00:44:44,730
me squared theta dot
in the k hat direction.
673
00:44:44,730 --> 00:44:46,590
I need to take the time
derivative of that.
674
00:44:46,590 --> 00:44:48,923
That's a constant, that's a
constant, that's a constant,
675
00:44:48,923 --> 00:44:50,940
it only comes from this term.
676
00:44:53,470 --> 00:44:58,980
And that gives me me squared
theta double dot k hat
677
00:44:58,980 --> 00:45:01,960
direction, and that's got to be
equal to the sum of the torques
678
00:45:01,960 --> 00:45:04,640
in the system, the
external torques.
679
00:45:04,640 --> 00:45:05,390
And what are they?
680
00:45:09,790 --> 00:45:11,400
So torques about this point.
681
00:45:11,400 --> 00:45:16,250
So axial forces in this
thing contribute no torques,
682
00:45:16,250 --> 00:45:19,290
transverse forces,
external forces only
683
00:45:19,290 --> 00:45:25,100
come from the mg on this thing.
684
00:45:25,100 --> 00:45:28,510
So my torques on
this system, there
685
00:45:28,510 --> 00:45:30,550
is some mechanical
torque being applied.
686
00:45:30,550 --> 00:45:33,830
That's what I'm looking for.
687
00:45:33,830 --> 00:45:35,460
I've got a motor
driving this thing,
688
00:45:35,460 --> 00:45:47,290
so there's some t of t in there,
some torque, minus mge cosine
689
00:45:47,290 --> 00:45:49,270
theta is the moment arm.
690
00:45:49,270 --> 00:45:51,700
So there's this force,
there's this moment arm
691
00:45:51,700 --> 00:45:54,570
is e cosine theta.
692
00:45:54,570 --> 00:45:57,260
So this is the external
torque caused by gravity,
693
00:45:57,260 --> 00:46:01,766
but all of this equals what?
694
00:46:01,766 --> 00:46:03,665
What's theta double dot?
695
00:46:03,665 --> 00:46:04,165
0.
696
00:46:08,713 --> 00:46:17,340
The external torque
is mge cosine omega t,
697
00:46:17,340 --> 00:46:19,950
theta is omega t.
698
00:46:19,950 --> 00:46:23,560
And so indeed, as this
thing goes around,
699
00:46:23,560 --> 00:46:27,170
when it's coming up, you've
got to apply enough torque
700
00:46:27,170 --> 00:46:29,020
to lift it against gravity.
701
00:46:29,020 --> 00:46:31,269
When it clears the top,
gravity is helping it,
702
00:46:31,269 --> 00:46:32,560
it's going down the other side.
703
00:46:32,560 --> 00:46:35,640
So in fact, if you plotted the
torque as a function of time
704
00:46:35,640 --> 00:46:40,040
for this system, it's like this.
705
00:46:40,040 --> 00:46:42,220
It's just lifting
that mass up and down.
706
00:46:42,220 --> 00:46:43,970
Then, of course, if
there's any friction
707
00:46:43,970 --> 00:46:45,320
in this thing, et
cetera, it's going
708
00:46:45,320 --> 00:46:47,570
to have to apply a little
bit of torque for that, too.
709
00:46:47,570 --> 00:46:50,640
But indeed, this is
an insightful question
710
00:46:50,640 --> 00:46:55,850
that someone asked, is
that the gravity does
711
00:46:55,850 --> 00:46:57,510
have to enter into this thing.
712
00:46:57,510 --> 00:47:00,220
So there will be a torque
that the motor has to supply
713
00:47:00,220 --> 00:47:02,420
to drive this thing in gravity.
714
00:47:02,420 --> 00:47:02,920
Yeah.
715
00:47:02,920 --> 00:47:06,798
AUDIENCE: Should that
expression also have--
716
00:47:06,798 --> 00:47:10,150
the expression for torque also
me squared theta double dot--
717
00:47:10,150 --> 00:47:12,980
PROFESSOR: Ah, now
theta double dot is?
718
00:47:12,980 --> 00:47:14,830
Yeah, see, it would.
719
00:47:14,830 --> 00:47:17,450
If this thing was
spinning up and I
720
00:47:17,450 --> 00:47:21,180
was trying to account for the
torque required to spin it up,
721
00:47:21,180 --> 00:47:23,070
then here is.
722
00:47:23,070 --> 00:47:26,440
Then I would include that, this
would be an equation of motion
723
00:47:26,440 --> 00:47:28,630
that says all these
things are true,
724
00:47:28,630 --> 00:47:31,200
and I can solve
for torque again.
725
00:47:31,200 --> 00:47:34,480
And it will allow me to decide
how fast I could spin it up.
726
00:47:34,480 --> 00:47:36,230
If I have a dinky
little motor, it
727
00:47:36,230 --> 00:47:37,770
doesn't spin up
very fast, if I had
728
00:47:37,770 --> 00:47:40,061
a really powerful motor that
could really put it to it,
729
00:47:40,061 --> 00:47:41,170
spin it up quickly.
730
00:47:41,170 --> 00:47:41,670
OK.
731
00:47:57,410 --> 00:48:00,490
So now I want to move on
to the third topic, which
732
00:48:00,490 --> 00:48:03,640
is to kind of go back to where
I left off last time, talking
733
00:48:03,640 --> 00:48:07,270
about we need to move on from
particles to rigid bodies
734
00:48:07,270 --> 00:48:09,640
so we can do more
interesting problems.
735
00:48:09,640 --> 00:48:12,530
So I want to pick up with
the subject of angular
736
00:48:12,530 --> 00:48:13,865
momentum for rigid bodies.
737
00:48:32,580 --> 00:48:35,990
Now last time I just barely
scratched the surface of this.
738
00:48:35,990 --> 00:48:38,060
And lots of muddy cards
said I don't get it.
739
00:48:38,060 --> 00:48:41,010
I didn't expect you to get
it with it being half baked
740
00:48:41,010 --> 00:48:42,980
and the first time
you've seen it.
741
00:48:42,980 --> 00:48:46,840
So we're going to continue
and we won't finish today.
742
00:48:46,840 --> 00:48:53,620
So let's think about
a general rigid body.
743
00:48:53,620 --> 00:49:05,020
Here's my inertial system,
got a body out here
744
00:49:05,020 --> 00:49:06,570
that's rotating
about some point,
745
00:49:06,570 --> 00:49:08,880
A. A could even be
outside the body
746
00:49:08,880 --> 00:49:10,135
and have it rotate about it.
747
00:49:13,520 --> 00:49:15,620
And attached to A is
a reference frame.
748
00:49:21,070 --> 00:49:23,280
Little x, little y, little z.
749
00:49:23,280 --> 00:49:25,490
So my Axy frame.
750
00:49:25,490 --> 00:49:32,650
Now, I put up last time, there's
two pages out of Williams
751
00:49:32,650 --> 00:49:37,190
which gives the
equations for the moment
752
00:49:37,190 --> 00:49:43,920
and products of inertia in terms
of summations of masses times
753
00:49:43,920 --> 00:49:45,750
particle locations.
754
00:49:45,750 --> 00:49:48,090
And in order to do
that, Williams defines
755
00:49:48,090 --> 00:49:50,250
a coordinate system
on this body,
756
00:49:50,250 --> 00:49:53,310
and that coordinate system
is fixed to the body,
757
00:49:53,310 --> 00:49:55,470
rotates to the body,
and Williams calls that
758
00:49:55,470 --> 00:49:57,050
coordinate system little oxyz.
759
00:50:00,160 --> 00:50:04,660
In his book, he calls the
inertial frame big Oxyz.
760
00:50:04,660 --> 00:50:06,670
It's really hard to do
that on a blackboard
761
00:50:06,670 --> 00:50:09,170
and how you'd be able
to tell it apart, OK.
762
00:50:09,170 --> 00:50:15,630
So I'm going to depart, and my
frame in here is an Axyz frame.
763
00:50:15,630 --> 00:50:17,860
But A and o, if you're
reading that handout,
764
00:50:17,860 --> 00:50:19,220
are the same thing.
765
00:50:19,220 --> 00:50:20,370
A and little o.
766
00:50:20,370 --> 00:50:22,960
It's a frame fixed to the
body that's rotating with it.
767
00:50:34,260 --> 00:50:39,680
We can write angular
momentum for rigid bodies
768
00:50:39,680 --> 00:50:43,820
as a vector hx having a
component in the i direction, j
769
00:50:43,820 --> 00:50:54,140
direction, and these
coordinates as the product
770
00:50:54,140 --> 00:50:58,070
of a matrix of constants.
771
00:50:58,070 --> 00:51:02,930
And these constants are these
moments of inertia and products
772
00:51:02,930 --> 00:51:03,765
of inertia terms.
773
00:51:18,410 --> 00:51:19,580
And so forth.
774
00:51:19,580 --> 00:51:21,410
I'll write out a couple
more of these, iy.
775
00:51:31,060 --> 00:51:34,900
It's a symmetric matrix,
and you multiply it
776
00:51:34,900 --> 00:51:44,000
by the components
of the rotation
777
00:51:44,000 --> 00:51:48,990
that you are
rotating this object,
778
00:51:48,990 --> 00:51:52,710
so here's a vector omega.
779
00:51:52,710 --> 00:51:57,220
This object is rotating about
A, the direction, the axis
780
00:51:57,220 --> 00:51:59,270
of rotation is like that.
781
00:51:59,270 --> 00:52:02,060
And you can break
this rotation rate
782
00:52:02,060 --> 00:52:04,960
into components
in the xyz system.
783
00:52:04,960 --> 00:52:07,620
And that's what these are,
these are the components of it.
784
00:52:07,620 --> 00:52:11,290
So you multiply out
this matrix in a vector,
785
00:52:11,290 --> 00:52:14,030
you will get
individual equations
786
00:52:14,030 --> 00:52:17,670
for the hx, hy, and hz
components of the angular
787
00:52:17,670 --> 00:52:19,415
momentum of that object.
788
00:53:02,550 --> 00:53:04,230
Now let's consider,
let's just do
789
00:53:04,230 --> 00:53:17,100
a case where the spin is
only about the z-axis.
790
00:53:17,100 --> 00:53:19,315
We do lots of these
problems, the book
791
00:53:19,315 --> 00:53:20,856
has a whole chapter
on it and they're
792
00:53:20,856 --> 00:53:22,150
called planar motion problems.
793
00:53:22,150 --> 00:53:26,970
We just typically pick the spin
around the z as a convention.
794
00:53:26,970 --> 00:53:35,350
And if you have a case like
that, then h here is i times
795
00:53:35,350 --> 00:53:39,170
0, 0, omega z.
796
00:53:39,170 --> 00:53:53,830
Then you multiply that out, you
get ixz omega z, iyz omega z,
797
00:53:53,830 --> 00:53:59,090
and izz omega z.
798
00:53:59,090 --> 00:54:03,560
Vector times the square matrix
gives you back a vector.
799
00:54:03,560 --> 00:54:06,700
That's what you get back.
800
00:54:06,700 --> 00:54:13,450
And if you want to write h as
a vector, which we frequently
801
00:54:13,450 --> 00:54:19,370
do, h, now this is
with respect to A,
802
00:54:19,370 --> 00:54:23,460
and we'll find that i here
as also with respect to A.
803
00:54:23,460 --> 00:54:29,140
Have to be very careful in your
construction of this matrix.
804
00:54:29,140 --> 00:54:33,770
It has to do with the point
about which you are computing
805
00:54:33,770 --> 00:54:35,410
your angular momentum.
806
00:54:35,410 --> 00:54:37,590
OK, if you want to
write this as a vector,
807
00:54:37,590 --> 00:54:47,539
then this becomes hx i hat
plus hy j hat plus hz k hat.
808
00:54:47,539 --> 00:54:49,330
That's just where the
unit vectors come in.
809
00:54:49,330 --> 00:54:51,700
When you want to express
this as a vector,
810
00:54:51,700 --> 00:54:57,740
you take these three components,
and these are hx, hy, hz.
811
00:55:07,340 --> 00:55:09,650
This little double
subscript, the first one
812
00:55:09,650 --> 00:55:13,770
tells you the component
of h, this is hx, hy, hz.
813
00:55:13,770 --> 00:55:17,210
The second one tells
you the axis of rotation
814
00:55:17,210 --> 00:55:19,720
about which the object
is spinning to give you
815
00:55:19,720 --> 00:55:22,800
this piece of angular momentum.
816
00:55:22,800 --> 00:55:30,720
So ixz is hx spinning
at rate omega z.
817
00:55:33,730 --> 00:55:39,045
Now, the direction of spin was?
818
00:55:39,045 --> 00:55:40,670
What's the unit vector
in the direction
819
00:55:40,670 --> 00:55:42,706
of rotation for this problem?
820
00:55:48,020 --> 00:55:50,520
What's omega?
821
00:55:50,520 --> 00:55:53,950
We said we're going to start
off with just-- direction,
822
00:55:53,950 --> 00:55:56,700
it's only spinning
in z direction.
823
00:55:56,700 --> 00:55:59,420
So it's just spinning
in z direction.
824
00:55:59,420 --> 00:56:04,000
But I multiply this thing
out, I get three terms.
825
00:56:04,000 --> 00:56:06,660
And I get a term in
the i, a j, and a k.
826
00:56:09,230 --> 00:56:19,920
Now these two terms,
so this is i, x is z,
827
00:56:19,920 --> 00:56:34,210
omega is zi plus iyz omega
zj plus izz omega zk.
828
00:56:34,210 --> 00:56:38,200
That's these three terms.
829
00:56:38,200 --> 00:56:44,980
These two terms
exist because I've
830
00:56:44,980 --> 00:56:50,720
assumed that these off
diagonal terms are not 0.
831
00:56:50,720 --> 00:56:53,650
The problem we started with,
we started with an example
832
00:56:53,650 --> 00:56:54,150
last time.
833
00:56:54,150 --> 00:56:56,850
Our bicycle wheel thing with
the unbalanced masses on it,
834
00:56:56,850 --> 00:57:02,440
we use the Williams formulas to
compute these different terms.
835
00:57:02,440 --> 00:57:09,840
If the off diagonal
terms here are not 0,
836
00:57:09,840 --> 00:57:13,130
then when you write the
angular momentum expression,
837
00:57:13,130 --> 00:57:15,660
you get parts of
the angular momentum
838
00:57:15,660 --> 00:57:19,290
that are not in the
direction of spin.
839
00:57:19,290 --> 00:57:22,000
That's a really
important conclusion.
840
00:57:22,000 --> 00:57:26,290
So the off diagonal terms
lead to angular momentum
841
00:57:26,290 --> 00:57:27,997
not in the direction of spin.
842
00:57:27,997 --> 00:57:29,580
And when you take
the time derivative,
843
00:57:29,580 --> 00:57:32,280
you end up with torques,
and they're right back
844
00:57:32,280 --> 00:57:33,520
to this problem up here.
845
00:57:36,650 --> 00:57:41,280
If you have off diagonal
terms in this matrix,
846
00:57:41,280 --> 00:57:44,310
when you spin it
around one of its axes,
847
00:57:44,310 --> 00:57:46,225
it is dynamically unbalanced.
848
00:57:52,370 --> 00:57:54,890
If these are not 0,
you spin it around one
849
00:57:54,890 --> 00:57:58,230
of the axes of the system
for which these are defining,
850
00:57:58,230 --> 00:58:00,810
in which these are
defined, you find out
851
00:58:00,810 --> 00:58:04,800
that you get unbalanced
torques in the system.
852
00:58:04,800 --> 00:58:07,290
So those two go together.
853
00:58:07,290 --> 00:58:30,610
Now, another way of
saying that is any time
854
00:58:30,610 --> 00:58:34,220
you end up with the angular
momentum vector not pointing
855
00:58:34,220 --> 00:58:37,790
in the same direction
as the rotation,
856
00:58:37,790 --> 00:58:40,558
then the system is going to
be dynamically unbalanced.
857
00:58:57,700 --> 00:59:01,500
Actually, I kind of want to
keep Atwood's machine here.
858
01:00:15,130 --> 01:00:19,740
So this was our unbalanced
bicycle wheel problem
859
01:00:19,740 --> 01:00:22,200
we had talked about last time.
860
01:00:22,200 --> 01:00:25,794
I can simulate that with this.
861
01:00:31,710 --> 01:00:35,190
I basically have
drawn it like this.
862
01:00:35,190 --> 01:00:36,230
So this is the problem.
863
01:00:38,950 --> 01:00:42,140
This thing is
definitely unbalanced,
864
01:00:42,140 --> 01:00:46,380
it's trying to do this
as it goes around.
865
01:00:46,380 --> 01:00:48,890
And last time we
actually worked up,
866
01:00:48,890 --> 01:00:54,830
from the William formulas, what
the moment of inertia matrix
867
01:00:54,830 --> 01:00:55,350
looked like.
868
01:00:55,350 --> 01:01:04,190
So now this xyz system
are attached and rotating
869
01:01:04,190 --> 01:01:06,500
with that frame.
870
01:01:06,500 --> 01:01:22,400
So my axis of spin is-- this
one's a little exaggerated.
871
01:01:22,400 --> 01:01:25,260
That drawing is like this.
872
01:01:25,260 --> 01:01:30,580
The x is like that.
873
01:01:30,580 --> 01:01:36,240
So x is like this, z is
like that, minus x minus z.
874
01:01:36,240 --> 01:01:39,100
So when this thing spins, that's
the problem that's drawn there.
875
01:01:41,810 --> 01:01:46,600
And if I compute with those
with Williams formulas,
876
01:01:46,600 --> 01:01:52,830
the various quantities--
so i with respect
877
01:01:52,830 --> 01:01:55,220
to A for this system.
878
01:02:01,070 --> 01:02:08,280
The first term, the ixx
term, is summation miyi
879
01:02:08,280 --> 01:02:15,260
squared plus zi
squared, and so forth.
880
01:02:15,260 --> 01:02:19,990
You get a bunch of terms, and
I will write out one other one
881
01:02:19,990 --> 01:02:20,490
here.
882
01:02:20,490 --> 01:02:30,010
This term over in
the corner is ixz,
883
01:02:30,010 --> 01:02:40,270
and that's minus summation
of the mixizi and so forth.
884
01:02:40,270 --> 01:02:44,430
And if we went through and
worked up each of these things,
885
01:02:44,430 --> 01:02:49,290
i with respect to
A for this problem,
886
01:02:49,290 --> 01:03:04,406
comes out mz1 squared 0
minus mx1z1 0 minus mx1z1.
887
01:03:07,970 --> 01:03:18,844
The middle term mx1 squared
plus z1 squared, 0, 0,
888
01:03:18,844 --> 01:03:24,160
and mx1 squared.
889
01:03:24,160 --> 01:03:27,020
So that's what this mass
moment of inertia matrix
890
01:03:27,020 --> 01:03:29,090
looks like for
these two particles.
891
01:04:06,190 --> 01:04:08,230
So now if I want to
write the angular
892
01:04:08,230 --> 01:04:14,170
momentum of this system
using this new notation,
893
01:04:14,170 --> 01:04:18,010
I would say that it's
i computed with respect
894
01:04:18,010 --> 01:04:24,970
to A times my omega, and
our case is 0, 0, omega z.
895
01:04:28,670 --> 01:04:33,940
And if we write that out, we
do that, multiply that out, we
896
01:04:33,940 --> 01:04:48,403
end up with a minus mx1z1 omega
z, 0, and mx1 squared omega z.
897
01:04:51,790 --> 01:04:57,850
These are our three
components, hx, hy, hz.
898
01:05:08,690 --> 01:05:13,300
And if you wanted to
write it as a vector,
899
01:05:13,300 --> 01:05:15,420
then you'd add the unit vectors.
900
01:05:15,420 --> 01:05:22,260
So the hx and the i plus 0
for the hy plus hz in the k.
901
01:05:30,970 --> 01:05:35,010
So now if you went and took the
time derivative of those terms,
902
01:05:35,010 --> 01:05:38,010
what do you get?
903
01:05:38,010 --> 01:05:38,780
AUDIENCE: Torques.
904
01:05:38,780 --> 01:05:40,560
PROFESSOR: Torques.
905
01:05:40,560 --> 01:05:43,070
And you'll get 3 terms.
906
01:05:43,070 --> 01:05:46,000
When we did the example a minute
ago, what we're doing here
907
01:05:46,000 --> 01:05:48,530
is not very different from that.
908
01:05:48,530 --> 01:05:51,590
You're going to get the torque
that it takes to accelerate it
909
01:05:51,590 --> 01:05:53,830
around the spin
axis, but you're also
910
01:05:53,830 --> 01:05:57,020
going to get the torque two
derivatives of this one, which
911
01:05:57,020 --> 01:05:58,330
gives you two terms.
912
01:05:58,330 --> 01:06:00,000
And these are the
moments of torques
913
01:06:00,000 --> 01:06:03,030
about that center of
the axle, in this case,
914
01:06:03,030 --> 01:06:05,050
trying to twist
the system around.
915
01:06:07,970 --> 01:06:12,080
Now, reach some closure here.
916
01:06:12,080 --> 01:06:13,570
We've got a good stopping point.
917
01:06:29,090 --> 01:06:34,073
Here's our system one last time.
918
01:06:34,073 --> 01:06:34,823
Here's the z-axis.
919
01:06:37,650 --> 01:06:40,550
The angular momentum
that comes out of this,
920
01:06:40,550 --> 01:06:46,120
you have a component
hz in the z direction,
921
01:06:46,120 --> 01:06:47,600
and you end up
with a component--
922
01:06:47,600 --> 01:06:57,150
it's got a minus in it-- in
the x direction like this,
923
01:06:57,150 --> 01:07:01,910
so that the total h vector with
respect to A looks like that.
924
01:07:01,910 --> 01:07:04,075
And it's not in the
direction of spin,
925
01:07:04,075 --> 01:07:10,030
it's actually perpendicular
to our bar here.
926
01:07:10,030 --> 01:07:13,050
And it's dynamically unbalanced.
927
01:07:13,050 --> 01:07:19,940
So just to-- how do
we make the transition
928
01:07:19,940 --> 01:07:27,060
from that to rigid bodies?
929
01:07:27,060 --> 01:07:31,520
The Williams formulas,
that are these,
930
01:07:31,520 --> 01:07:34,780
say that if you want the mass
moment of inertia of a body,
931
01:07:34,780 --> 01:07:36,950
all you have to do is sum
up all the little mass
932
01:07:36,950 --> 01:07:41,470
bits at the correct
distances off of axes,
933
01:07:41,470 --> 01:07:42,360
and you will get it.
934
01:07:42,360 --> 01:07:46,410
So when you have particles,
you can just add them up.
935
01:07:46,410 --> 01:07:49,130
When you have a rigid
body, those summations
936
01:07:49,130 --> 01:07:50,980
become integrals.
937
01:07:50,980 --> 01:08:02,546
And, for example,
izz is the integral
938
01:08:02,546 --> 01:08:03,670
of-- how should I say this.
939
01:08:08,340 --> 01:08:13,595
x squared plus y squared
dm, every little mass bit.
940
01:08:20,250 --> 01:08:22,696
It looks like-- is there
an exam some distance away?
941
01:08:22,696 --> 01:08:24,029
I see a lot of people vanishing.
942
01:08:24,029 --> 01:08:26,120
OK, so let me--
I'll tell you what.
943
01:08:26,120 --> 01:08:29,250
I'll just make it easy
for it and let you go.
944
01:08:29,250 --> 01:08:33,399
Let me just say one thing
to where we're going.
945
01:08:33,399 --> 01:08:40,399
For every rigid body,
there is a different set
946
01:08:40,399 --> 01:08:48,810
of axes for which, when you
go to make up this matrix,
947
01:08:48,810 --> 01:08:51,790
you can make a diagonal.
948
01:08:51,790 --> 01:08:53,910
And those are called
the principal axes,
949
01:08:53,910 --> 01:08:55,520
and that's where
we're going next,
950
01:08:55,520 --> 01:08:59,189
those play a really important
role in what we want to do.
951
01:08:59,189 --> 01:09:00,920
OK.