WEBVTT

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Let's consider a very famous
problem the, Atwood machine.

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We have a pulley, A,
suspended from a ceiling.

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And a rope is wrapped
around the pulley.

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And on each side of the rope,
there's different masses.

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So here is block 1,
and block 2, and we

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can say here-- it
doesn't matter--

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but we'll say that
M2 is bigger than M1.

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And that gives us
some intuition that we

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expect block 2 to go down
and block 1 to go up.

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Now in this problem
there is friction

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between the rope and the pulley,
so the rope is not sliding.

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And what that means is that
the pulley will rotate.

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And also the mass of
the pulley is not 0.

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So these were all assumptions
we made way back when

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were analyzing
Newton's second law,

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but now we have to
take into effect

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that there some
rotational inertia to make

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the pulley start to have
angular acceleration.

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So what we'd like to do is to
identify our three objects--

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mass 1, the pulley, and mass 2.

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And for mass 1 and mass 2,
use Newton's second law,

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for the pulley, we'll use
our torque relationships.

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So let's begin by
drawing our free body

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diagrams for object 1.

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So we have tension in the
rope pulling object 1 up,

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we have the
gravitational force down.

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And here it doesn't
matter which way

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I'm going to choose my
unit vectors, because I

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have this idea that
M2 is bigger than M1.

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I'm just going to
choose j hat 1 up.

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Now for block 2, I have M2g.

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Now here's the place where
lots of people get tripped up.

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In the past we've been assuming
that the tension in the rope

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is uniform everywhere.

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In this problem, because
the rope is not slipping

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and the pulley is not
massless, the tension

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is not constant
everywhere in the rope.

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And we'll see more reasons why
that can't be the case, so I'll

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have to identify a different
tension on the other side T2

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pulling the rope,
pulling the block up.

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And I'm going to
choose j hat 2 down.

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Notice my unit vectors are
chosen in opposite directions.

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The reason for that is that
there's a constraint here

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that as block 2 goes
down, block 1 goes up,

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if the acceleration of
block 2 is positive,

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the acceleration of
block 1 will also

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be positive if I choose
unit vector pointing up.

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So now I can write Newton's
second law for both of these.

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So for block 2, positive down,
and 2g minus T2 equals M2a.

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It's the same
acceleration in the rope.

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a is equal to a1, equal to
a2, they're all positive.

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And for 2, notice
I have positive up

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minus M1g equals m1a.

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So so far, these are
my two equations.

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I have three unknowns-- T1,
T2, and the accelerations,

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and only two equations.

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Now let's analyze the pulley.

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So we have our
pulley A. And what

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are the forces on the pulley?

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Well forces and torques
are a little bit different,

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but let's just draw
our forces first.

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There is a tension
holding it up,

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we'll call that T3 from
this rope pulling it up.

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Now here's where we
have to be careful.

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Rope 1 is pulling the pulley
down, that's what we called T1,

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so we draw a T1 on this side.

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And the same rope
on the other side

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is pulling the pulley down.

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Notice that it is
a different, T2.

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Now this brings us to what
we called our rotational

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coordinate system.

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I do know expect
that the angular

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acceleration of the pulley.

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So as the pulley
rotates in this problem,

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I expect that it's
rotating in the direction

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like by the right
hand rule, this way.

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And so I expect to see the
alpha pointing like that.

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What I'll do is I'll choose a
coordinate system for an angle

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theta.

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And that will make
positive direction

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like that for my rotational
coordinate system.

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Now when we write torque
equals the moment of inertia

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of the pulley, times the
angular acceleration,

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we have two torques.

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The radius is R, radiuses
of R. And if we're

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calculating the torque about
the center of the pulley,

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we'll call that point 0,
then each of these torques

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are in different directions.

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So for instance, we
would have to take for T1

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going down and this vector from
origin to where T1 is acting,

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we can extend that force.

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And we see that that torque
is pointing in the direction

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in-- direction like that.

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And that's opposite
or sign here.

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So this torque, minus
T1 R is negative.

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What about the torque
from the other force, T2?

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Again, we would
draw that vector.

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Let's draw them over here.

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We have T2 pointing down.

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The point 0 is there.

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The vector from 0 to 2
is pointing like that.

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We extend that vector.

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We put the arrow like that.

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And we see that
this one is pointing

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in our positive direction.

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So we have a plus
T2R equals IA alpha.

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Now right away we can see why
the tension in the strings

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is not the same.

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Because if these
tensions were the same,

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this quantity would
be 0, but the tensions

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can't be the same because
the pulley is rotating.

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And that rotational
inertia of the pulley

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is coming from the fact
that the two torques are not

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the same on both sides.

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So we now have our
last equation here.

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T2R minus T1R equals IA alpha.

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But notice that I've introduced
another variable here,

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so I guess again, I have
four equations and only

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three unknowns.

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We still have one
last constraint.

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Because the rope is
sliding along the pulley

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and the radius is R,
we know that a point

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on the rim of the pulley has
acceleration A of the rope.

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But the pulley has an
alpha angular acceleration,

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so our constraint conditioned
here is plus R alpha.

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Now why did I put a plus sign?

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Because when the pulley is
rotating the direction shown,

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alpha will be positive.

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This quantity, torque, will
be bigger than this one,

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and so we'll have
a positive alpha.

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When I chose j hat 1
up and j hat 2 down,

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that made all my
accelerations positive,

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and so I with the plus side.

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If I had reversed my
choice of unit vectors,

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then this would be a minus sign.

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So you have to be extremely
careful by making sure

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that the directions you
chose for the linear force

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diagrams that give us A,
and the direction we chose

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in our rotational
coordinate system

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are consistent when
we choose to relate

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the constraints between them.

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So that's our last condition
that a equals our R alpha.

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So I now have four
equations and four unknowns.

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And if I want to solve
for the acceleration a,

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then I have to need
a strategy here.

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And what I'll do to find a
is I'll use this equation

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as my backbone.

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Why did I do that?

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Because I have the
unknowns T2, T1, and alpha.

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And I have separate equations
that relate alpha to A,

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and T1 to a and T2 to a
And so I can solve for T2.

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And substitution here
I'll need a little room.

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So I get that T2
is M2g plus M2a.

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I'm sorry, minus M2a times R.
Now T1 is M1a plus M1g times R,

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and that's equal to times
AI alpha, which is a over R.

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So I'll now collect all
of my a terms over here.

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And so I get to M2gR minus
M1gR, this term and this term,

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is equal to a times
IA over R. Notice

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I have 2 plus signs here,
so I get M1 plus M2 times R.

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And I therefore conclude,
we'll put it over here,

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that the acceleration a of is
equal to M2 minus m1gR, divided

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by IA over R, plus M1 plus M2R.

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Now, when I have this
result, let's just

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check a number of things first.

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Notice that if M2
is bigger than M1,

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a will be positive,
which is what I expected.

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Dimensionally, we have
M2gR, down here, M2R,

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so this first term will have
dimensions of acceleration g.

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Now over here, we
have IA over R,

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but remember moment of
inertia is M times R

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square, so this also have the
dimensions of mass and radius.

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And so I have confidence
dimensionally.

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And the sine of A that
this is the correct answer.

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And that's how we solve
the Atwood machine.