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Consider a rod of
mass m, and suppose

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I apply a force to this rod.

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Let's say a force like that.

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So we know that as a result of
that applied force, the center

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of mass of the rod, which we can
imagine is right at the center

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the rod, will translate
with some acceleration, such

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that the vector f
is equal to the mass

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of the rod times
the acceleration

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of the center of mass.

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Now recall that
for a rigid body,

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this equation will be true
regardless of where on the rod

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I apply the force.

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So for example, if I draw
the rod again over here,

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and I apply the same force,
vector f, but I apply it,

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let's say, on the
right hand side,

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I'll still have the
same f equals ma.

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All I specify with f is its
magnitude and direction.

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But for a rigid body, no
matter where on the body

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I apply that force,
the acceleration

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of the center of mass
will be the same.

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Now, we know from
experience, however,

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that the motion of
the rod is different

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if I push it at the center at
one end or at the other end.

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The point is, though, that the
motion, the overall motion,

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the overall translation
of the object

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doesn't just involve
the translation

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of the center of mass, but it
also involves some rotation,

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in general.

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So in fact, there is a theorem
called Chasles' theorem, which

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tells us that the most general
displacement of a rigid body

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can be split into
the translation

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of the center of mass and
a rotation about the center

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of mass.

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Our f equals ma
relation tells us

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how an applied force affects the
first part of the translation

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of the center of mass.

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But how does that
applied force affect

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the rotation of the rigid body?

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We'll see that this depends
not only on the magnitude

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and direction of the
force, but also on where

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the force is applied, OK?

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That's different
than the translation

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of the center of
mass, which only

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depends upon what the force is.

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The rotation about
the center of mass

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depends both upon the force
itself and where on the object

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that it's applied.

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In fact, in the
next few lessons,

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we'll see that there's a
special equation of motion

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for describing
rotational motion.

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It's analogous to Newton's
Second Law of Motion,

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analogous to f equals ma,
applying to rotational motion.

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And we write that relationship.

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Its rotational
equation of motion

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is written as tau,
which is a vector,

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is equal to i times
alpha, which is a vector.

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So tau is a new quantity
we haven't talked

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about before called the torque.

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It's a vector.

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i is the moment of inertia
that we've already encountered,

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and alpha is the
angular acceleration

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of the rotational motion.

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Now, tau is a sort of
rotational analog to the force,

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and it depends on not
just the force itself,

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but on where the
force is applied.

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Computing the torque, tau,
depends upon something

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called a vector product
or a cross product.

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It's a way of multiplying
two vectors together

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to produce a third vector.

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In that way, it's different
than the dot product or scalar

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product that we
discussed earlier,

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where we multiply two
vectors together and get

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a scalar or a pure number.

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This is a different
operation, the vector product

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or cross-product.

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Torque is the first physical
quantity we've encountered,

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but it won't be the last,
that involves a cross-product

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in its definition.

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And so, before giving you a
formal definition of torque,

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we'll first review the
mathematics of cross-products,

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and we'll do that
in the next lesson.

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Now, the angular
acceleration, alpha,

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tells us about how the
rotational motion changes.

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Again, just like in
f equals ma, where

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we could divide the two
sides of this equation

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into dynamics and kinematics.

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Kinematics is telling us about
a geometrical description

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of the motion, or the
translation of motion,

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and the dynamics f is
telling us about how

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the applied forces cause
changes in the motion, changes

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in the kinematics.

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Likewise, for rotational motion,
the angular acceleration,

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alpha, tells us
about the geometry

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of the rotational
motion, and specifically

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how the rotational
motion is changing,

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and the torque is telling us
about the application of forces

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and how that causes changes
in the rotation of motion.

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So that leaves one
other quantity,

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which is the moment of inertia.

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And I want to talk about what
that means for just a moment.

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So the term "inertia," the
term "inertia" in physics

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represents a resistance
to an applied force.

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It tells us how difficult it is
to change an object's motion.

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So for example, suppose I
have two blocks, one of mass m

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and another that's
10 times as massive.

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So 10m.

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Newton's Second
Law, f equals ma,

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tells us that if I want
to accelerate the heavier

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mass to the same rate that
I do with a smaller mass,

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I'll need to apply a force
that's 10 times larger.

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So for translation
of motion, the mass m

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represents the notion of
inertia, the resistance

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to a force.

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It tells us how
much force we have

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to apply to achieve a
certain change in motion.

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Now, for rotational
motion, that role

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is played by the
moment of inertia i.

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That tells us how
difficult it is to change

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the rotation of an object.

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If I increase the moment of
inertia by a factor of 10,

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then I'll need a torque that
is 10 times larger in order

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to achieve the same change
in rotational motion,

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the same angular acceleration.

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But recall that our definition
of moment of inertia

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for some rigid body
is, if I break up

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the rigid body into a
bunch of little pieces,

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and for each piece
I take the product

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of the mass of that piece, delta
mj, times the perpendicular

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distance of that piece
from the rotation

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axis-- I call that
rj-- and if I sum

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that over the entire object,
the entire body, that

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gives me my moment of inertia.

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Well, so-- sorry,
this is r squared.

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So it's the mass of each
element times the distance

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of the axis squared,
delta mj times rj squared.

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Now, I can increase the moment
of inertia by a factor of 10

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by increasing the total
mass of the object,

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but you can see
from this equation

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that I can also do
it with the same mass

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by changing the
location of the mass.

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If I increase the r's, if I
move the mass, the same mass,

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but I move it to be farther
away from the rotation axis,

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that also achieves an increase
in the moment of inertia.

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So what that tells is that
for rotational motion,

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it's not just the amount
of mass that matters,

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but also how that
mass is distributed.

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OK, so in that sense,
rotational inertia

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is different than
translational inertia.

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For translational motion,
all that matters is the mass.

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If I increase the mass
by a factor of 10,

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then it will become
a factor of 10

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more difficult to get
that object to accelerate.

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I need a factor of
10 larger force.

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But with rotational
motion, I can

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increase the notion of inertia.

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I can make it more difficult to
change the rotation of motion

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either by changing the
mass, by increasing it,

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or by making the
distribution of mass

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be further away from
the rotation axis.

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So as an example, imagine if you
were rolling a wheel up a hill.

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It's different if I have a
wheel whose mass is distributed

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evenly over the whole
disk, or if I have

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all of the mass in the rim.

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If all of the mass is in the
rim, then from this equation,

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we see the moment of inertia
is larger for the same amount

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of total mass.

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And so it's much more difficult
to roll a wheel up the hill

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if all of the mass
is in the rim than it

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is if the mass is distributed
evenly over the wheel.

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So our rotational equivalent
to Newton's Second Law

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is, tau equals i
alpha, the torque

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equals the moment of inertia
times the angular acceleration.

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In the next few lessons, we'll
see how torque is defined,

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and we'll derive this expression
for the dynamics of rotation

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of motion and see
how to apply it.