1
00:00:03,480 --> 00:00:06,920
Let's consider a very famous
problem the, Atwood machine.

2
00:00:06,920 --> 00:00:11,780
We have a pulley, A,
suspended from a ceiling.

3
00:00:11,780 --> 00:00:14,330
And a rope is wrapped
around the pulley.

4
00:00:14,330 --> 00:00:18,200
And on each side of the rope,
there's different masses.

5
00:00:18,200 --> 00:00:22,820
So here is block 1,
and block 2, and we

6
00:00:22,820 --> 00:00:25,160
can say here-- it
doesn't matter--

7
00:00:25,160 --> 00:00:28,040
but we'll say that
M2 is bigger than M1.

8
00:00:28,040 --> 00:00:29,840
And that gives us
some intuition that we

9
00:00:29,840 --> 00:00:33,440
expect block 2 to go down
and block 1 to go up.

10
00:00:33,440 --> 00:00:39,750
Now in this problem
there is friction

11
00:00:39,750 --> 00:00:45,690
between the rope and the pulley,
so the rope is not sliding.

12
00:00:45,690 --> 00:00:48,510
And what that means is that
the pulley will rotate.

13
00:00:48,510 --> 00:00:51,840
And also the mass of
the pulley is not 0.

14
00:00:51,840 --> 00:00:54,180
So these were all assumptions
we made way back when

15
00:00:54,180 --> 00:00:56,260
were analyzing
Newton's second law,

16
00:00:56,260 --> 00:00:58,110
but now we have to
take into effect

17
00:00:58,110 --> 00:01:00,990
that there some
rotational inertia to make

18
00:01:00,990 --> 00:01:04,360
the pulley start to have
angular acceleration.

19
00:01:04,360 --> 00:01:08,350
So what we'd like to do is to
identify our three objects--

20
00:01:08,350 --> 00:01:10,920
mass 1, the pulley, and mass 2.

21
00:01:10,920 --> 00:01:14,130
And for mass 1 and mass 2,
use Newton's second law,

22
00:01:14,130 --> 00:01:17,260
for the pulley, we'll use
our torque relationships.

23
00:01:17,260 --> 00:01:19,410
So let's begin by
drawing our free body

24
00:01:19,410 --> 00:01:22,500
diagrams for object 1.

25
00:01:22,500 --> 00:01:27,420
So we have tension in the
rope pulling object 1 up,

26
00:01:27,420 --> 00:01:29,970
we have the
gravitational force down.

27
00:01:29,970 --> 00:01:32,759
And here it doesn't
matter which way

28
00:01:32,759 --> 00:01:35,670
I'm going to choose my
unit vectors, because I

29
00:01:35,670 --> 00:01:38,460
have this idea that
M2 is bigger than M1.

30
00:01:38,460 --> 00:01:41,820
I'm just going to
choose j hat 1 up.

31
00:01:41,820 --> 00:01:46,680
Now for block 2, I have M2g.

32
00:01:46,680 --> 00:01:49,590
Now here's the place where
lots of people get tripped up.

33
00:01:49,590 --> 00:01:53,729
In the past we've been assuming
that the tension in the rope

34
00:01:53,729 --> 00:01:56,280
is uniform everywhere.

35
00:01:56,280 --> 00:02:00,090
In this problem, because
the rope is not slipping

36
00:02:00,090 --> 00:02:03,720
and the pulley is not
massless, the tension

37
00:02:03,720 --> 00:02:07,380
is not constant
everywhere in the rope.

38
00:02:07,380 --> 00:02:10,590
And we'll see more reasons why
that can't be the case, so I'll

39
00:02:10,590 --> 00:02:14,160
have to identify a different
tension on the other side T2

40
00:02:14,160 --> 00:02:17,579
pulling the rope,
pulling the block up.

41
00:02:17,579 --> 00:02:19,890
And I'm going to
choose j hat 2 down.

42
00:02:19,890 --> 00:02:23,920
Notice my unit vectors are
chosen in opposite directions.

43
00:02:23,920 --> 00:02:27,840
The reason for that is that
there's a constraint here

44
00:02:27,840 --> 00:02:31,620
that as block 2 goes
down, block 1 goes up,

45
00:02:31,620 --> 00:02:34,170
if the acceleration of
block 2 is positive,

46
00:02:34,170 --> 00:02:36,120
the acceleration of
block 1 will also

47
00:02:36,120 --> 00:02:39,760
be positive if I choose
unit vector pointing up.

48
00:02:39,760 --> 00:02:43,360
So now I can write Newton's
second law for both of these.

49
00:02:43,360 --> 00:02:51,690
So for block 2, positive down,
and 2g minus T2 equals M2a.

50
00:02:51,690 --> 00:02:54,390
It's the same
acceleration in the rope.

51
00:02:54,390 --> 00:02:58,930
a is equal to a1, equal to
a2, they're all positive.

52
00:02:58,930 --> 00:03:03,480
And for 2, notice
I have positive up

53
00:03:03,480 --> 00:03:07,740
minus M1g equals m1a.

54
00:03:07,740 --> 00:03:10,260
So so far, these are
my two equations.

55
00:03:10,260 --> 00:03:14,700
I have three unknowns-- T1,
T2, and the accelerations,

56
00:03:14,700 --> 00:03:16,660
and only two equations.

57
00:03:16,660 --> 00:03:18,970
Now let's analyze the pulley.

58
00:03:18,970 --> 00:03:21,990
So we have our
pulley A. And what

59
00:03:21,990 --> 00:03:24,570
are the forces on the pulley?

60
00:03:24,570 --> 00:03:27,310
Well forces and torques
are a little bit different,

61
00:03:27,310 --> 00:03:29,640
but let's just draw
our forces first.

62
00:03:29,640 --> 00:03:31,680
There is a tension
holding it up,

63
00:03:31,680 --> 00:03:35,710
we'll call that T3 from
this rope pulling it up.

64
00:03:35,710 --> 00:03:39,390
Now here's where we
have to be careful.

65
00:03:39,390 --> 00:03:44,970
Rope 1 is pulling the pulley
down, that's what we called T1,

66
00:03:44,970 --> 00:03:47,220
so we draw a T1 on this side.

67
00:03:47,220 --> 00:03:49,410
And the same rope
on the other side

68
00:03:49,410 --> 00:03:51,000
is pulling the pulley down.

69
00:03:51,000 --> 00:03:54,840
Notice that it is
a different, T2.

70
00:03:54,840 --> 00:03:58,740
Now this brings us to what
we called our rotational

71
00:03:58,740 --> 00:04:00,750
coordinate system.

72
00:04:00,750 --> 00:04:04,440
I do know expect
that the angular

73
00:04:04,440 --> 00:04:06,690
acceleration of the pulley.

74
00:04:06,690 --> 00:04:09,930
So as the pulley
rotates in this problem,

75
00:04:09,930 --> 00:04:14,100
I expect that it's
rotating in the direction

76
00:04:14,100 --> 00:04:18,899
like by the right
hand rule, this way.

77
00:04:18,899 --> 00:04:28,350
And so I expect to see the
alpha pointing like that.

78
00:04:28,350 --> 00:04:33,659
What I'll do is I'll choose a
coordinate system for an angle

79
00:04:33,659 --> 00:04:35,409
theta.

80
00:04:35,409 --> 00:04:39,490
And that will make
positive direction

81
00:04:39,490 --> 00:04:43,180
like that for my rotational
coordinate system.

82
00:04:43,180 --> 00:04:48,450
Now when we write torque
equals the moment of inertia

83
00:04:48,450 --> 00:04:52,500
of the pulley, times the
angular acceleration,

84
00:04:52,500 --> 00:04:54,360
we have two torques.

85
00:04:54,360 --> 00:05:00,630
The radius is R, radiuses
of R. And if we're

86
00:05:00,630 --> 00:05:04,050
calculating the torque about
the center of the pulley,

87
00:05:04,050 --> 00:05:08,220
we'll call that point 0,
then each of these torques

88
00:05:08,220 --> 00:05:10,180
are in different directions.

89
00:05:10,180 --> 00:05:13,560
So for instance, we
would have to take for T1

90
00:05:13,560 --> 00:05:20,190
going down and this vector from
origin to where T1 is acting,

91
00:05:20,190 --> 00:05:23,910
we can extend that force.

92
00:05:23,910 --> 00:05:29,370
And we see that that torque
is pointing in the direction

93
00:05:29,370 --> 00:05:32,940
in-- direction like that.

94
00:05:32,940 --> 00:05:35,190
And that's opposite
or sign here.

95
00:05:35,190 --> 00:05:39,640
So this torque, minus
T1 R is negative.

96
00:05:39,640 --> 00:05:43,470
What about the torque
from the other force, T2?

97
00:05:43,470 --> 00:05:45,659
Again, we would
draw that vector.

98
00:05:45,659 --> 00:05:47,490
Let's draw them over here.

99
00:05:47,490 --> 00:05:49,890
We have T2 pointing down.

100
00:05:49,890 --> 00:05:52,400
The point 0 is there.

101
00:05:52,400 --> 00:05:58,540
The vector from 0 to 2
is pointing like that.

102
00:05:58,540 --> 00:06:00,150
We extend that vector.

103
00:06:00,150 --> 00:06:02,140
We put the arrow like that.

104
00:06:02,140 --> 00:06:06,540
And we see that
this one is pointing

105
00:06:06,540 --> 00:06:08,080
in our positive direction.

106
00:06:08,080 --> 00:06:14,130
So we have a plus
T2R equals IA alpha.

107
00:06:14,130 --> 00:06:17,250
Now right away we can see why
the tension in the strings

108
00:06:17,250 --> 00:06:18,734
is not the same.

109
00:06:18,734 --> 00:06:20,400
Because if these
tensions were the same,

110
00:06:20,400 --> 00:06:22,740
this quantity would
be 0, but the tensions

111
00:06:22,740 --> 00:06:27,420
can't be the same because
the pulley is rotating.

112
00:06:27,420 --> 00:06:30,360
And that rotational
inertia of the pulley

113
00:06:30,360 --> 00:06:33,210
is coming from the fact
that the two torques are not

114
00:06:33,210 --> 00:06:35,010
the same on both sides.

115
00:06:35,010 --> 00:06:37,530
So we now have our
last equation here.

116
00:06:37,530 --> 00:06:44,190
T2R minus T1R equals IA alpha.

117
00:06:44,190 --> 00:06:48,540
But notice that I've introduced
another variable here,

118
00:06:48,540 --> 00:06:50,820
so I guess again, I have
four equations and only

119
00:06:50,820 --> 00:06:52,450
three unknowns.

120
00:06:52,450 --> 00:06:55,290
We still have one
last constraint.

121
00:06:55,290 --> 00:06:58,440
Because the rope is
sliding along the pulley

122
00:06:58,440 --> 00:07:01,680
and the radius is R,
we know that a point

123
00:07:01,680 --> 00:07:07,900
on the rim of the pulley has
acceleration A of the rope.

124
00:07:07,900 --> 00:07:11,220
But the pulley has an
alpha angular acceleration,

125
00:07:11,220 --> 00:07:15,480
so our constraint conditioned
here is plus R alpha.

126
00:07:15,480 --> 00:07:18,180
Now why did I put a plus sign?

127
00:07:18,180 --> 00:07:21,510
Because when the pulley is
rotating the direction shown,

128
00:07:21,510 --> 00:07:22,800
alpha will be positive.

129
00:07:22,800 --> 00:07:25,670
This quantity, torque, will
be bigger than this one,

130
00:07:25,670 --> 00:07:27,930
and so we'll have
a positive alpha.

131
00:07:27,930 --> 00:07:32,040
When I chose j hat 1
up and j hat 2 down,

132
00:07:32,040 --> 00:07:34,810
that made all my
accelerations positive,

133
00:07:34,810 --> 00:07:37,330
and so I with the plus side.

134
00:07:37,330 --> 00:07:40,930
If I had reversed my
choice of unit vectors,

135
00:07:40,930 --> 00:07:43,600
then this would be a minus sign.

136
00:07:43,600 --> 00:07:47,430
So you have to be extremely
careful by making sure

137
00:07:47,430 --> 00:07:50,520
that the directions you
chose for the linear force

138
00:07:50,520 --> 00:07:53,652
diagrams that give us A,
and the direction we chose

139
00:07:53,652 --> 00:07:55,110
in our rotational
coordinate system

140
00:07:55,110 --> 00:07:58,260
are consistent when
we choose to relate

141
00:07:58,260 --> 00:07:59,980
the constraints between them.

142
00:07:59,980 --> 00:08:06,150
So that's our last condition
that a equals our R alpha.

143
00:08:06,150 --> 00:08:08,920
So I now have four
equations and four unknowns.

144
00:08:08,920 --> 00:08:11,160
And if I want to solve
for the acceleration a,

145
00:08:11,160 --> 00:08:14,650
then I have to need
a strategy here.

146
00:08:14,650 --> 00:08:18,570
And what I'll do to find a
is I'll use this equation

147
00:08:18,570 --> 00:08:19,950
as my backbone.

148
00:08:19,950 --> 00:08:21,210
Why did I do that?

149
00:08:21,210 --> 00:08:25,080
Because I have the
unknowns T2, T1, and alpha.

150
00:08:25,080 --> 00:08:27,735
And I have separate equations
that relate alpha to A,

151
00:08:27,735 --> 00:08:33,440
and T1 to a and T2 to a
And so I can solve for T2.

152
00:08:33,440 --> 00:08:36,480
And substitution here
I'll need a little room.

153
00:08:36,480 --> 00:08:45,960
So I get that T2
is M2g plus M2a.

154
00:08:45,960 --> 00:09:04,260
I'm sorry, minus M2a times R.
Now T1 is M1a plus M1g times R,

155
00:09:04,260 --> 00:09:12,000
and that's equal to times
AI alpha, which is a over R.

156
00:09:12,000 --> 00:09:16,560
So I'll now collect all
of my a terms over here.

157
00:09:16,560 --> 00:09:30,360
And so I get to M2gR minus
M1gR, this term and this term,

158
00:09:30,360 --> 00:09:36,870
is equal to a times
IA over R. Notice

159
00:09:36,870 --> 00:09:45,720
I have 2 plus signs here,
so I get M1 plus M2 times R.

160
00:09:45,720 --> 00:09:50,520
And I therefore conclude,
we'll put it over here,

161
00:09:50,520 --> 00:10:03,440
that the acceleration a of is
equal to M2 minus m1gR, divided

162
00:10:03,440 --> 00:10:11,840
by IA over R, plus M1 plus M2R.

163
00:10:14,540 --> 00:10:16,820
Now, when I have this
result, let's just

164
00:10:16,820 --> 00:10:19,040
check a number of things first.

165
00:10:19,040 --> 00:10:21,420
Notice that if M2
is bigger than M1,

166
00:10:21,420 --> 00:10:25,810
a will be positive,
which is what I expected.

167
00:10:25,810 --> 00:10:32,060
Dimensionally, we have
M2gR, down here, M2R,

168
00:10:32,060 --> 00:10:36,380
so this first term will have
dimensions of acceleration g.

169
00:10:36,380 --> 00:10:38,840
Now over here, we
have IA over R,

170
00:10:38,840 --> 00:10:41,210
but remember moment of
inertia is M times R

171
00:10:41,210 --> 00:10:46,760
square, so this also have the
dimensions of mass and radius.

172
00:10:46,760 --> 00:10:50,660
And so I have confidence
dimensionally.

173
00:10:50,660 --> 00:10:54,780
And the sine of A that
this is the correct answer.

174
00:10:54,780 --> 00:10:58,360
And that's how we solve
the Atwood machine.