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DR. PETER DOURMASHKIN:
I would like

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to now analyze the motion
of a system of particles

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that has both translational
and rotational motion.

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So I'm going to
consider a pulley,

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and the pulley has
radius R. And there

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is a string wrapped around THE
pulley and a block of object 1

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that's on a plane, and
another block of object 2.

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And as object 2 falls
down, the pulley

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rotates and object 1
moves to the right.

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And there's a coefficient
of friction between block

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1 and the surface.

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Now, in order to analyze this
problem, I'm going to apply,

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for the pulley, our
torque equals I cm alpha,

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and for each of the
blocks, I'll apply

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F1 equals m1 a1 and
F2 equals m2 a2.

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But the important
thing to realize

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is that these three
quantities, the acceleration

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of block 1, the
acceleration of block 2,

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and the angular
acceleration of the pulley,

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are constrained because
this string is not

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slipping around the pulley.

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And so let's begin to
analyze this type of problem.

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So we'll start with
the torque principle.

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Now, what's crucial in
all of these problems

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is that we're relating
two different quantities,

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vectors on both sides.

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The physics quantities
have definite direction,

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and our alpha a1
and a2 as vectors

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are determined by our
choice of coordinates.

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So what I'd like to do is
draw a coordinate system,

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a rotational coordinate system.

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Now, the way I'll do it is
I'll draw an angle theta.

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And now I have to draw
a right-hand move,

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so my angle theta will
look as though it's going

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into the plane of the figure.

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And so I write cross
n hat right-hand move,

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and I'm going to just to
define that to be k hat.

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Now, what that allows me to
do, when I write my point

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s here will be cm.

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So I'm going to calculate
this about the center of mass,

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and I get I cm alpha.

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As soon as I draw the
coordinate system, then

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this side becomes the
vector alpha z k hat,

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where alpha z is the z component
of the angular acceleration.

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And technically, the
reason this angle

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is there is because this
is the second derivative

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of that angle.

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And that's well-defined now.

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So the next step is
to define the force,

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to do what we call
a torque diagram.

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So this is my rotational
coordinate system.

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The next step is to
construct a torque diagram,

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and the way we do that
is we draw the object.

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We indicate our rotational
coordinate system.

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I don't have to put
the theta anymore.

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Now, here's a subtle point.

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I'm going to draw
the rope that is

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connected to the pulley
as part of my system.

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This is the part where
the tension here, I'll

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call that T2, and over here,
this is the tension T1.

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Now, on the pulley, there
is a gravitational force,

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and there's some pivot
force on this pulley.

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And now what I want to consider
is the torque about the Cm.

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Now, the pivot force, f pivot,
and the gravitational force,

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produce no torque
about the pivot,

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so I'm just going to eliminate
those for the moment,

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and just focus on the
torque due to T1 and T2.

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So I draw my vector Rs
T1 and my vector Rs T2.

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So this is what a torque
diagram consists of.

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Let's summarize it.

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It's our system,
the relevant forces

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that are producing torque,
vectors from the point we're

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calculating the torque.

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Our S is the center of mass.

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And the vector from where
we're calculating the torque

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to where the force is acting.

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And now, when we calculate
the cross-product of Rs and T,

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we put these two
vectors tail to tail,

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and notice that this
vector is giving us

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a torque out of the board,
our positive direction

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is into the board,
so over here we

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have minus T1 R.
Whereas T2, when

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we put these two vectors, Rs T2,
and we calculate that torque,

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that torque is into
the board, which

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is our positive direction,
and so that's plus T2 R2.

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And now in our torque principle,
we set these two sides equal,

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and we have minus T1 R plus
T2 R equals Icm alpha z.

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Now, this is our
first equation, but it

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requires some type of thought.

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For the first thing, we
see that the tension T2

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is equal to Icm over
R alpha z plus T1.

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So the tensions on the
side are not equal.

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Now, when we studied pulleys
earlier in the semester,

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we imposed a condition that the
pulley was frictionless, which

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meant that the rope was
sliding along the pulley,

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and there was no
rotation in the pulley,

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so there was no
contribution to alpha.

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And in that case, T2
would be equal to T2.

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We also could make a
slightly different statement.

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We could say suppose
the mass of the pulley

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was very, very small, an
extremely light pulley,

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then Icm would be 0, and
again, T2 would be equal to T1.

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So when we were dealing
with either massless pulleys

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or ropes that were slipping
frictionlessly along a pulley,

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the tension on both
sides was equal.

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Now, something
different is happening.

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We need to apply a greater
torque here than T1

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because there is
rotational inertia.

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We're causing the
pulley to accelerate.

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So this torque from T2 has to be
bigger than the torque from T1,

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and therefore T2
is bigger than T1,

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so that is a very important way
to apply the torque principle.

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When T2 is bigger than T1,
alpha will be positive,

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and a positive
angular acceleration

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is giving a rotation in which
our angle theta is not only

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increasing, but its second
derivative is positive.

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So that's crucial for beginning
the analysis of this problem.

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The next step is to
analyze Newton's second law

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on both objects, M1 and M2.

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So I'll save our
result here, I'll

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erase what we don't need, and
then continue the analysis.

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So returning to our
analysis of a pulley

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with two masses
and a string that's

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not slipping around
the pulley, I now

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want to begin analysis of
F equals Ma on object 1.

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So as usual, I draw object 1.

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I'll choose i hat 1 to
point in the direction

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because I know it's
going to go that way

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so my component of
acceleration will be positive.

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In my force diagrams, I have a
normal force, I have gravity.

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The string is pulling T1,
that's the same tension

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at the end of the string.

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The tension in the
string is not changing.

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We're assuming it's
a massless string.

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And I have a friction
force on object 1,

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which is kinetic friction.

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And now I can write
down Newton's second law

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in the horizontal direction.

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I could also call j hat 1
up, and my two equations

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for Newton's second law
are T1 minus Fk is M1 A1,

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and N1 minus M1 g is zero.

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Now, I also know that
the kinetic friction, Fk,

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is the coefficient of
friction mu times N1.

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So my next equation for
F equals M A on object T1

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is T1 minus mu N1 equals M1 A1.

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Now I have to apply
the same analysis to 2.

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Notice I'm not drawing my
force diagram on my sketch.

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I do a separate
force diagram on 2.

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So here's 2.

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I have tension T2
and gravity M2g.

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Now, even though I chose
a unit vector up here,

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this choice of unit vectors is
completely independent of how

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I choose unit vectors for 2.

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Because object is
moving down, I would

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prefer to choose j hat 2 down.

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My acceleration for this
object will be positive.

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And then when I apply
f equals MA object 2,

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I get M2g minus T2 equals M2 A2.

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So that's now my third equation,
that M2g minus T2 equals M2 A2.

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And now I look at this.

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System of equations.

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And what are my unknowns?

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T1, T2, alpha, A1, A2.

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Five unknowns.

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I'm treating properties of the
system, the radius mu, Icm,

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actually the N1,
because it's in M1g,

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I can simplify this
equation and replace this

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with M1g, where
I'm already using

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the other Newton's second law.

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So I have three equations
and five unknowns.

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I cannot solve this system.

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But in all of these problems,
there's constraint conditions.

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There's constraints between
how the masses are moving

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and how the angular
acceleration pulley

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is related to the linear
acceleration of the masses.

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Let's consider mass 1 and 2.

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They're attached by a string.

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As mass 2 goes down,
mass 1 goes to the right.

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The string is not
stretching, so they're

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moving at the same rate, so
they have the same acceleration.

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So my first constraint
is that A1 equals A2.

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Now in general, I
have to be careful.

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Plus or minus.

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Why is it a plus sign and
not a minus sign here?

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It's a plus sign because I've
chosen i hat to the right

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and I've chosen j hat downwards.

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If I'd chosen them differently,
that sign could have varied.

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Now, let's focus
on the relationship

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between the angular acceleration
of the pulley and M2.

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Think about the strength.

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Here we're on a
point on the rim.

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This is a distance R, and
the pulley and the string

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are moving together.

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So there's a
tangential acceleration

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of the pulley
equal to R alpha Z.

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This is the tangential
acceleration of pulley

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and string.

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But the same string has a linear
acceleration, either A1 or A2.

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So this has to be
equal to A2, this

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is the linear acceleration
of the string.

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And so that's our
last constraint, five,

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that A1 equals R alpha.

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And now I have a system of five
equations and five unknowns.

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And the question is, how
can I find the acceleration?

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So in general, when opposed
to a system like that,

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I want to have some strategy.

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Let's make a little space
to clear for our algebra.

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OK, now, I look at this
system, and I say to myself,

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which equation do I want
to use as a background?

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My target is to find A1.

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A1 is equal to A2.

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Now, when I look at these
equations, T1 depends on A1,

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T2 depends on A2,
which is equal to A1,

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and alpha is also related to A1.

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So I can use this
equation 1 as my backbone,

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and substitute in T1, T2,
and alpha into that equation.

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And now let's do that.

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So when I solve
this equation for T1

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equal to M1 A1 plus mu
M1g with the minus sign,

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I get minus M1 A1 plus mu M1g
times R, that's my first piece.

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I solve for T2, which
is M2g minus M2 A,

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so I get M2g minus M2.

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Now, A2 is equal to A1, so I
make my second substitution,

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multiply it by R, and
that's equal to Icm.

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And now I make my
final substitution,

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that alpha z is
equal to A1 over R.

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So if I now can collect
terms, minus, minus over here,

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but there's an R
there, an R there, I'll

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divide through by R, and
bring my A1 terms over

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to the other side, and I'm
left with minus mu M1g plus M2g

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equals Icm over R squared that
has the dimensions of mass,

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because moment of inertia,
M R squared divided by R

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squared, plus M1 A1 plus M2 A1.

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And finally, as a
conclusion, I now

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can solve for the
acceleration of my system

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in terms of all of
these quantities.

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And let's just put it all the
way down here at the bottom

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that A1 equals M2g minus
mu M1g over Icm R squared

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00:14:51,500 --> 00:14:55,100
plus M1 plus M2.

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00:14:55,100 --> 00:14:57,680
Often, in types of problems
like these when there's

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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

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00:15:08,862 --> 00:15:10,570
and that would be an
impossible solution.

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00:15:10,570 --> 00:15:14,250
So that's always a sign that
there could be something wrong.

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00:15:14,250 --> 00:15:16,790
The other thing we want to
check is, when does it actually

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00:15:16,790 --> 00:15:17,480
accelerate?

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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

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00:15:25,400 --> 00:15:30,890
than Mu M1g then 2 will
start to go downwards.

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00:15:30,890 --> 00:15:33,770
If M2g were less than
M1g, then the problem

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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

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00:15:50,060 --> 00:15:52,380
would depend on how much
weight that's there.

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00:15:52,380 --> 00:15:57,620
So here is a full analysis of
rotational and translational

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00:15:57,620 --> 00:15:58,700
motion.

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00:15:58,700 --> 00:16:01,490
Takes a little bit of time
and a little bit of care,

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00:16:01,490 --> 00:16:03,920
but we've done it.