1
00:00:02,752 --> 00:00:04,210
DR. PETER DOURMASHKIN:
I would like
2
00:00:04,210 --> 00:00:06,790
to now analyze the motion
of a system of particles
3
00:00:06,790 --> 00:00:09,820
that has both translational
and rotational motion.
4
00:00:09,820 --> 00:00:12,760
So I'm going to
consider a pulley,
5
00:00:12,760 --> 00:00:15,550
and the pulley has
radius R. And there
6
00:00:15,550 --> 00:00:22,960
is a string wrapped around THE
pulley and a block of object 1
7
00:00:22,960 --> 00:00:28,050
that's on a plane, and
another block of object 2.
8
00:00:28,050 --> 00:00:31,720
And as object 2 falls
down, the pulley
9
00:00:31,720 --> 00:00:34,780
rotates and object 1
moves to the right.
10
00:00:34,780 --> 00:00:38,980
And there's a coefficient
of friction between block
11
00:00:38,980 --> 00:00:40,790
1 and the surface.
12
00:00:40,790 --> 00:00:45,030
Now, in order to analyze this
problem, I'm going to apply,
13
00:00:45,030 --> 00:00:50,400
for the pulley, our
torque equals I cm alpha,
14
00:00:50,400 --> 00:00:52,300
and for each of the
blocks, I'll apply
15
00:00:52,300 --> 00:00:58,780
F1 equals m1 a1 and
F2 equals m2 a2.
16
00:00:58,780 --> 00:01:00,670
But the important
thing to realize
17
00:01:00,670 --> 00:01:03,520
is that these three
quantities, the acceleration
18
00:01:03,520 --> 00:01:05,920
of block 1, the
acceleration of block 2,
19
00:01:05,920 --> 00:01:08,140
and the angular
acceleration of the pulley,
20
00:01:08,140 --> 00:01:11,140
are constrained because
this string is not
21
00:01:11,140 --> 00:01:15,540
slipping around the pulley.
22
00:01:15,540 --> 00:01:18,700
And so let's begin to
analyze this type of problem.
23
00:01:18,700 --> 00:01:22,270
So we'll start with
the torque principle.
24
00:01:22,270 --> 00:01:25,240
Now, what's crucial in
all of these problems
25
00:01:25,240 --> 00:01:27,940
is that we're relating
two different quantities,
26
00:01:27,940 --> 00:01:29,140
vectors on both sides.
27
00:01:29,140 --> 00:01:32,620
The physics quantities
have definite direction,
28
00:01:32,620 --> 00:01:35,860
and our alpha a1
and a2 as vectors
29
00:01:35,860 --> 00:01:38,200
are determined by our
choice of coordinates.
30
00:01:38,200 --> 00:01:41,500
So what I'd like to do is
draw a coordinate system,
31
00:01:41,500 --> 00:01:43,370
a rotational coordinate system.
32
00:01:43,370 --> 00:01:47,810
Now, the way I'll do it is
I'll draw an angle theta.
33
00:01:47,810 --> 00:01:50,410
And now I have to draw
a right-hand move,
34
00:01:50,410 --> 00:01:55,480
so my angle theta will
look as though it's going
35
00:01:55,480 --> 00:01:57,550
into the plane of the figure.
36
00:01:57,550 --> 00:02:03,160
And so I write cross
n hat right-hand move,
37
00:02:03,160 --> 00:02:06,490
and I'm going to just to
define that to be k hat.
38
00:02:06,490 --> 00:02:10,770
Now, what that allows me to
do, when I write my point
39
00:02:10,770 --> 00:02:13,030
s here will be cm.
40
00:02:13,030 --> 00:02:16,570
So I'm going to calculate
this about the center of mass,
41
00:02:16,570 --> 00:02:18,370
and I get I cm alpha.
42
00:02:18,370 --> 00:02:21,070
As soon as I draw the
coordinate system, then
43
00:02:21,070 --> 00:02:26,210
this side becomes the
vector alpha z k hat,
44
00:02:26,210 --> 00:02:34,660
where alpha z is the z component
of the angular acceleration.
45
00:02:34,660 --> 00:02:36,490
And technically, the
reason this angle
46
00:02:36,490 --> 00:02:40,150
is there is because this
is the second derivative
47
00:02:40,150 --> 00:02:41,590
of that angle.
48
00:02:41,590 --> 00:02:43,910
And that's well-defined now.
49
00:02:43,910 --> 00:02:47,470
So the next step is
to define the force,
50
00:02:47,470 --> 00:02:50,560
to do what we call
a torque diagram.
51
00:02:50,560 --> 00:02:57,630
So this is my rotational
coordinate system.
52
00:02:57,630 --> 00:03:01,090
The next step is to
construct a torque diagram,
53
00:03:01,090 --> 00:03:04,780
and the way we do that
is we draw the object.
54
00:03:04,780 --> 00:03:08,120
We indicate our rotational
coordinate system.
55
00:03:08,120 --> 00:03:09,980
I don't have to put
the theta anymore.
56
00:03:09,980 --> 00:03:12,700
Now, here's a subtle point.
57
00:03:12,700 --> 00:03:16,750
I'm going to draw
the rope that is
58
00:03:16,750 --> 00:03:22,510
connected to the pulley
as part of my system.
59
00:03:22,510 --> 00:03:28,030
This is the part where
the tension here, I'll
60
00:03:28,030 --> 00:03:33,550
call that T2, and over here,
this is the tension T1.
61
00:03:33,550 --> 00:03:37,180
Now, on the pulley, there
is a gravitational force,
62
00:03:37,180 --> 00:03:41,800
and there's some pivot
force on this pulley.
63
00:03:41,800 --> 00:03:50,320
And now what I want to consider
is the torque about the Cm.
64
00:03:50,320 --> 00:03:56,620
Now, the pivot force, f pivot,
and the gravitational force,
65
00:03:56,620 --> 00:03:59,410
produce no torque
about the pivot,
66
00:03:59,410 --> 00:04:02,740
so I'm just going to eliminate
those for the moment,
67
00:04:02,740 --> 00:04:09,070
and just focus on the
torque due to T1 and T2.
68
00:04:09,070 --> 00:04:16,480
So I draw my vector Rs
T1 and my vector Rs T2.
69
00:04:16,480 --> 00:04:19,269
So this is what a torque
diagram consists of.
70
00:04:19,269 --> 00:04:20,290
Let's summarize it.
71
00:04:20,290 --> 00:04:23,110
It's our system,
the relevant forces
72
00:04:23,110 --> 00:04:26,350
that are producing torque,
vectors from the point we're
73
00:04:26,350 --> 00:04:27,550
calculating the torque.
74
00:04:27,550 --> 00:04:32,950
Our S is the center of mass.
75
00:04:32,950 --> 00:04:35,740
And the vector from where
we're calculating the torque
76
00:04:35,740 --> 00:04:38,300
to where the force is acting.
77
00:04:38,300 --> 00:04:42,830
And now, when we calculate
the cross-product of Rs and T,
78
00:04:42,830 --> 00:04:45,400
we put these two
vectors tail to tail,
79
00:04:45,400 --> 00:04:48,820
and notice that this
vector is giving us
80
00:04:48,820 --> 00:04:52,240
a torque out of the board,
our positive direction
81
00:04:52,240 --> 00:04:54,310
is into the board,
so over here we
82
00:04:54,310 --> 00:04:58,510
have minus T1 R.
Whereas T2, when
83
00:04:58,510 --> 00:05:04,300
we put these two vectors, Rs T2,
and we calculate that torque,
84
00:05:04,300 --> 00:05:07,750
that torque is into
the board, which
85
00:05:07,750 --> 00:05:13,390
is our positive direction,
and so that's plus T2 R2.
86
00:05:13,390 --> 00:05:17,830
And now in our torque principle,
we set these two sides equal,
87
00:05:17,830 --> 00:05:25,810
and we have minus T1 R plus
T2 R equals Icm alpha z.
88
00:05:25,810 --> 00:05:27,820
Now, this is our
first equation, but it
89
00:05:27,820 --> 00:05:30,840
requires some type of thought.
90
00:05:30,840 --> 00:05:34,510
For the first thing, we
see that the tension T2
91
00:05:34,510 --> 00:05:39,490
is equal to Icm over
R alpha z plus T1.
92
00:05:39,490 --> 00:05:43,010
So the tensions on the
side are not equal.
93
00:05:43,010 --> 00:05:46,390
Now, when we studied pulleys
earlier in the semester,
94
00:05:46,390 --> 00:05:50,020
we imposed a condition that the
pulley was frictionless, which
95
00:05:50,020 --> 00:05:53,290
meant that the rope was
sliding along the pulley,
96
00:05:53,290 --> 00:05:56,920
and there was no
rotation in the pulley,
97
00:05:56,920 --> 00:06:00,000
so there was no
contribution to alpha.
98
00:06:00,000 --> 00:06:04,180
And in that case, T2
would be equal to T2.
99
00:06:04,180 --> 00:06:07,330
We also could make a
slightly different statement.
100
00:06:07,330 --> 00:06:10,180
We could say suppose
the mass of the pulley
101
00:06:10,180 --> 00:06:13,090
was very, very small, an
extremely light pulley,
102
00:06:13,090 --> 00:06:18,850
then Icm would be 0, and
again, T2 would be equal to T1.
103
00:06:18,850 --> 00:06:22,390
So when we were dealing
with either massless pulleys
104
00:06:22,390 --> 00:06:26,260
or ropes that were slipping
frictionlessly along a pulley,
105
00:06:26,260 --> 00:06:28,870
the tension on both
sides was equal.
106
00:06:28,870 --> 00:06:31,660
Now, something
different is happening.
107
00:06:31,660 --> 00:06:36,700
We need to apply a greater
torque here than T1
108
00:06:36,700 --> 00:06:39,980
because there is
rotational inertia.
109
00:06:39,980 --> 00:06:42,650
We're causing the
pulley to accelerate.
110
00:06:42,650 --> 00:06:46,600
So this torque from T2 has to be
bigger than the torque from T1,
111
00:06:46,600 --> 00:06:50,860
and therefore T2
is bigger than T1,
112
00:06:50,860 --> 00:06:55,990
so that is a very important way
to apply the torque principle.
113
00:06:55,990 --> 00:06:59,770
When T2 is bigger than T1,
alpha will be positive,
114
00:06:59,770 --> 00:07:02,200
and a positive
angular acceleration
115
00:07:02,200 --> 00:07:05,020
is giving a rotation in which
our angle theta is not only
116
00:07:05,020 --> 00:07:08,900
increasing, but its second
derivative is positive.
117
00:07:08,900 --> 00:07:13,360
So that's crucial for beginning
the analysis of this problem.
118
00:07:13,360 --> 00:07:17,620
The next step is to
analyze Newton's second law
119
00:07:17,620 --> 00:07:21,070
on both objects, M1 and M2.
120
00:07:21,070 --> 00:07:23,560
So I'll save our
result here, I'll
121
00:07:23,560 --> 00:07:28,180
erase what we don't need, and
then continue the analysis.
122
00:07:28,180 --> 00:07:30,280
So returning to our
analysis of a pulley
123
00:07:30,280 --> 00:07:33,070
with two masses
and a string that's
124
00:07:33,070 --> 00:07:35,110
not slipping around
the pulley, I now
125
00:07:35,110 --> 00:07:38,520
want to begin analysis of
F equals Ma on object 1.
126
00:07:38,520 --> 00:07:41,200
So as usual, I draw object 1.
127
00:07:41,200 --> 00:07:44,900
I'll choose i hat 1 to
point in the direction
128
00:07:44,900 --> 00:07:46,840
because I know it's
going to go that way
129
00:07:46,840 --> 00:07:49,500
so my component of
acceleration will be positive.
130
00:07:49,500 --> 00:07:54,820
In my force diagrams, I have a
normal force, I have gravity.
131
00:07:54,820 --> 00:07:58,680
The string is pulling T1,
that's the same tension
132
00:07:58,680 --> 00:08:00,400
at the end of the string.
133
00:08:00,400 --> 00:08:02,520
The tension in the
string is not changing.
134
00:08:02,520 --> 00:08:05,200
We're assuming it's
a massless string.
135
00:08:05,200 --> 00:08:08,790
And I have a friction
force on object 1,
136
00:08:08,790 --> 00:08:10,570
which is kinetic friction.
137
00:08:10,570 --> 00:08:12,900
And now I can write
down Newton's second law
138
00:08:12,900 --> 00:08:14,790
in the horizontal direction.
139
00:08:14,790 --> 00:08:19,020
I could also call j hat 1
up, and my two equations
140
00:08:19,020 --> 00:08:24,770
for Newton's second law
are T1 minus Fk is M1 A1,
141
00:08:24,770 --> 00:08:29,640
and N1 minus M1 g is zero.
142
00:08:29,640 --> 00:08:34,110
Now, I also know that
the kinetic friction, Fk,
143
00:08:34,110 --> 00:08:37,830
is the coefficient of
friction mu times N1.
144
00:08:37,830 --> 00:08:43,730
So my next equation for
F equals M A on object T1
145
00:08:43,730 --> 00:08:50,370
is T1 minus mu N1 equals M1 A1.
146
00:08:50,370 --> 00:08:53,190
Now I have to apply
the same analysis to 2.
147
00:08:53,190 --> 00:08:56,310
Notice I'm not drawing my
force diagram on my sketch.
148
00:08:56,310 --> 00:08:59,040
I do a separate
force diagram on 2.
149
00:08:59,040 --> 00:09:00,860
So here's 2.
150
00:09:00,860 --> 00:09:04,990
I have tension T2
and gravity M2g.
151
00:09:04,990 --> 00:09:08,580
Now, even though I chose
a unit vector up here,
152
00:09:08,580 --> 00:09:12,000
this choice of unit vectors is
completely independent of how
153
00:09:12,000 --> 00:09:14,130
I choose unit vectors for 2.
154
00:09:14,130 --> 00:09:16,050
Because object is
moving down, I would
155
00:09:16,050 --> 00:09:18,780
prefer to choose j hat 2 down.
156
00:09:18,780 --> 00:09:22,180
My acceleration for this
object will be positive.
157
00:09:22,180 --> 00:09:25,540
And then when I apply
f equals MA object 2,
158
00:09:25,540 --> 00:09:31,800
I get M2g minus T2 equals M2 A2.
159
00:09:31,800 --> 00:09:39,812
So that's now my third equation,
that M2g minus T2 equals M2 A2.
160
00:09:39,812 --> 00:09:40,770
And now I look at this.
161
00:09:40,770 --> 00:09:41,980
System of equations.
162
00:09:41,980 --> 00:09:44,430
And what are my unknowns?
163
00:09:44,430 --> 00:09:48,810
T1, T2, alpha, A1, A2.
164
00:09:48,810 --> 00:09:50,640
Five unknowns.
165
00:09:50,640 --> 00:09:55,620
I'm treating properties of the
system, the radius mu, Icm,
166
00:09:55,620 --> 00:09:59,490
actually the N1,
because it's in M1g,
167
00:09:59,490 --> 00:10:05,850
I can simplify this
equation and replace this
168
00:10:05,850 --> 00:10:12,930
with M1g, where
I'm already using
169
00:10:12,930 --> 00:10:14,370
the other Newton's second law.
170
00:10:14,370 --> 00:10:18,670
So I have three equations
and five unknowns.
171
00:10:18,670 --> 00:10:20,790
I cannot solve this system.
172
00:10:20,790 --> 00:10:23,990
But in all of these problems,
there's constraint conditions.
173
00:10:23,990 --> 00:10:28,020
There's constraints between
how the masses are moving
174
00:10:28,020 --> 00:10:30,160
and how the angular
acceleration pulley
175
00:10:30,160 --> 00:10:33,120
is related to the linear
acceleration of the masses.
176
00:10:33,120 --> 00:10:36,060
Let's consider mass 1 and 2.
177
00:10:36,060 --> 00:10:37,940
They're attached by a string.
178
00:10:37,940 --> 00:10:42,070
As mass 2 goes down,
mass 1 goes to the right.
179
00:10:42,070 --> 00:10:44,040
The string is not
stretching, so they're
180
00:10:44,040 --> 00:10:48,030
moving at the same rate, so
they have the same acceleration.
181
00:10:48,030 --> 00:10:55,910
So my first constraint
is that A1 equals A2.
182
00:10:55,910 --> 00:10:58,250
Now in general, I
have to be careful.
183
00:10:58,250 --> 00:10:59,630
Plus or minus.
184
00:10:59,630 --> 00:11:03,590
Why is it a plus sign and
not a minus sign here?
185
00:11:03,590 --> 00:11:08,510
It's a plus sign because I've
chosen i hat to the right
186
00:11:08,510 --> 00:11:11,000
and I've chosen j hat downwards.
187
00:11:11,000 --> 00:11:14,630
If I'd chosen them differently,
that sign could have varied.
188
00:11:14,630 --> 00:11:16,730
Now, let's focus
on the relationship
189
00:11:16,730 --> 00:11:20,750
between the angular acceleration
of the pulley and M2.
190
00:11:20,750 --> 00:11:23,070
Think about the strength.
191
00:11:23,070 --> 00:11:26,940
Here we're on a
point on the rim.
192
00:11:26,940 --> 00:11:31,580
This is a distance R, and
the pulley and the string
193
00:11:31,580 --> 00:11:33,060
are moving together.
194
00:11:33,060 --> 00:11:34,940
So there's a
tangential acceleration
195
00:11:34,940 --> 00:11:38,330
of the pulley
equal to R alpha Z.
196
00:11:38,330 --> 00:11:46,190
This is the tangential
acceleration of pulley
197
00:11:46,190 --> 00:11:48,020
and string.
198
00:11:48,020 --> 00:11:53,300
But the same string has a linear
acceleration, either A1 or A2.
199
00:11:53,300 --> 00:11:55,850
So this has to be
equal to A2, this
200
00:11:55,850 --> 00:12:00,190
is the linear acceleration
of the string.
201
00:12:00,190 --> 00:12:03,050
And so that's our
last constraint, five,
202
00:12:03,050 --> 00:12:07,820
that A1 equals R alpha.
203
00:12:07,820 --> 00:12:12,050
And now I have a system of five
equations and five unknowns.
204
00:12:12,050 --> 00:12:16,920
And the question is, how
can I find the acceleration?
205
00:12:16,920 --> 00:12:19,580
So in general, when opposed
to a system like that,
206
00:12:19,580 --> 00:12:21,260
I want to have some strategy.
207
00:12:21,260 --> 00:12:28,840
Let's make a little space
to clear for our algebra.
208
00:12:28,840 --> 00:12:34,700
OK, now, I look at this
system, and I say to myself,
209
00:12:34,700 --> 00:12:37,490
which equation do I want
to use as a background?
210
00:12:37,490 --> 00:12:41,570
My target is to find A1.
211
00:12:41,570 --> 00:12:43,550
A1 is equal to A2.
212
00:12:43,550 --> 00:12:49,160
Now, when I look at these
equations, T1 depends on A1,
213
00:12:49,160 --> 00:12:53,510
T2 depends on A2,
which is equal to A1,
214
00:12:53,510 --> 00:12:57,260
and alpha is also related to A1.
215
00:12:57,260 --> 00:13:01,490
So I can use this
equation 1 as my backbone,
216
00:13:01,490 --> 00:13:07,850
and substitute in T1, T2,
and alpha into that equation.
217
00:13:07,850 --> 00:13:09,590
And now let's do that.
218
00:13:09,590 --> 00:13:11,870
So when I solve
this equation for T1
219
00:13:11,870 --> 00:13:15,380
equal to M1 A1 plus mu
M1g with the minus sign,
220
00:13:15,380 --> 00:13:24,680
I get minus M1 A1 plus mu M1g
times R, that's my first piece.
221
00:13:24,680 --> 00:13:30,260
I solve for T2, which
is M2g minus M2 A,
222
00:13:30,260 --> 00:13:34,180
so I get M2g minus M2.
223
00:13:34,180 --> 00:13:38,540
Now, A2 is equal to A1, so I
make my second substitution,
224
00:13:38,540 --> 00:13:42,040
multiply it by R, and
that's equal to Icm.
225
00:13:42,040 --> 00:13:44,390
And now I make my
final substitution,
226
00:13:44,390 --> 00:13:50,060
that alpha z is
equal to A1 over R.
227
00:13:50,060 --> 00:13:56,180
So if I now can collect
terms, minus, minus over here,
228
00:13:56,180 --> 00:13:59,060
but there's an R
there, an R there, I'll
229
00:13:59,060 --> 00:14:02,930
divide through by R, and
bring my A1 terms over
230
00:14:02,930 --> 00:14:11,770
to the other side, and I'm
left with minus mu M1g plus M2g
231
00:14:11,770 --> 00:14:17,570
equals Icm over R squared that
has the dimensions of mass,
232
00:14:17,570 --> 00:14:20,395
because moment of inertia,
M R squared divided by R
233
00:14:20,395 --> 00:14:27,770
squared, plus M1 A1 plus M2 A1.
234
00:14:27,770 --> 00:14:31,550
And finally, as a
conclusion, I now
235
00:14:31,550 --> 00:14:34,790
can solve for the
acceleration of my system
236
00:14:34,790 --> 00:14:37,220
in terms of all of
these quantities.
237
00:14:37,220 --> 00:14:40,640
And let's just put it all the
way down here at the bottom
238
00:14:40,640 --> 00:14:51,500
that A1 equals M2g minus
mu M1g over Icm R squared
239
00:14:51,500 --> 00:14:55,100
plus M1 plus M2.
240
00:14:55,100 --> 00:14:57,680
Often, in types of problems
like these when there's
241
00:14:57,680 --> 00:15:02,630
a lot of signers, you might end
up with a minus or a minus sign
242
00:15:02,630 --> 00:15:05,780
down here, and if you looked
at that, that would imply that
243
00:15:05,780 --> 00:15:08,862
with the right choice of
parameters, this could be zero
244
00:15:08,862 --> 00:15:10,570
and that would be an
impossible solution.
245
00:15:10,570 --> 00:15:14,250
So that's always a sign that
there could be something wrong.
246
00:15:14,250 --> 00:15:16,790
The other thing we want to
check is, when does it actually
247
00:15:16,790 --> 00:15:17,480
accelerate?
248
00:15:17,480 --> 00:15:19,040
We have a condition.
249
00:15:19,040 --> 00:15:25,400
So we can conclude
that if M2g is bigger
250
00:15:25,400 --> 00:15:30,890
than Mu M1g then 2 will
start to go downwards.
251
00:15:30,890 --> 00:15:33,770
If M2g were less than
M1g, then the problem
252
00:15:33,770 --> 00:15:37,580
would be very different, because
two would not go downwards.
253
00:15:37,580 --> 00:15:40,700
The friction would not be
kinetic, but would be static.
254
00:15:40,700 --> 00:15:44,960
And that would vary, depending
on how much weight were here.
255
00:15:44,960 --> 00:15:50,060
So if you went from 0 to
mu M1g, the static friction
256
00:15:50,060 --> 00:15:52,380
would depend on how much
weight that's there.
257
00:15:52,380 --> 00:15:57,620
So here is a full analysis of
rotational and translational
258
00:15:57,620 --> 00:15:58,700
motion.
259
00:15:58,700 --> 00:16:01,490
Takes a little bit of time
and a little bit of care,
260
00:16:01,490 --> 00:16:03,920
but we've done it.