Consider a rod of
mass m, and suppose
I apply a force to this rod.
Let's say a force like that.
So we know that as a result of
that applied force, the center
of mass of the rod, which we can
imagine is right at the center
the rod, will translate
with some acceleration, such
that the vector f
is equal to the mass
of the rod times
the acceleration
of the center of mass.
Now recall that
for a rigid body,
this equation will be true
regardless of where on the rod
I apply the force.
So for example, if I draw
the rod again over here,
and I apply the same force,
vector f, but I apply it,
let's say, on the
right hand side,
I'll still have the
same f equals ma.
All I specify with f is its
magnitude and direction.
But for a rigid body, no
matter where on the body
I apply that force,
the acceleration
of the center of mass
will be the same.
Now, we know from
experience, however,
that the motion of
the rod is different
if I push it at the center at
one end or at the other end.
The point is, though, that the
motion, the overall motion,
the overall translation
of the object
doesn't just involve
the translation
of the center of mass, but it
also involves some rotation,
in general.
So in fact, there is a theorem
called Chasles' theorem, which
tells us that the most general
displacement of a rigid body
can be split into
the translation
of the center of mass and
a rotation about the center
of mass.
Our f equals ma
relation tells us
how an applied force affects the
first part of the translation
of the center of mass.
But how does that
applied force affect
the rotation of the rigid body?
We'll see that this depends
not only on the magnitude
and direction of the
force, but also on where
the force is applied, OK?
That's different
than the translation
of the center of
mass, which only
depends upon what the force is.
The rotation about
the center of mass
depends both upon the force
itself and where on the object
that it's applied.
In fact, in the
next few lessons,
we'll see that there's a
special equation of motion
for describing
rotational motion.
It's analogous to Newton's
Second Law of Motion,
analogous to f equals ma,
applying to rotational motion.
And we write that relationship.
Its rotational
equation of motion
is written as tau,
which is a vector,
is equal to i times
alpha, which is a vector.
So tau is a new quantity
we haven't talked
about before called the torque.
It's a vector.
i is the moment of inertia
that we've already encountered,
and alpha is the
angular acceleration
of the rotational motion.
Now, tau is a sort of
rotational analog to the force,
and it depends on not
just the force itself,
but on where the
force is applied.
Computing the torque, tau,
depends upon something
called a vector product
or a cross product.
It's a way of multiplying
two vectors together
to produce a third vector.
In that way, it's different
than the dot product or scalar
product that we
discussed earlier,
where we multiply two
vectors together and get
a scalar or a pure number.
This is a different
operation, the vector product
or cross-product.
Torque is the first physical
quantity we've encountered,
but it won't be the last,
that involves a cross-product
in its definition.
And so, before giving you a
formal definition of torque,
we'll first review the
mathematics of cross-products,
and we'll do that
in the next lesson.
Now, the angular
acceleration, alpha,
tells us about how the
rotational motion changes.
Again, just like in
f equals ma, where
we could divide the two
sides of this equation
into dynamics and kinematics.
Kinematics is telling us about
a geometrical description
of the motion, or the
translation of motion,
and the dynamics f is
telling us about how
the applied forces cause
changes in the motion, changes
in the kinematics.
Likewise, for rotational motion,
the angular acceleration,
alpha, tells us
about the geometry
of the rotational
motion, and specifically
how the rotational
motion is changing,
and the torque is telling us
about the application of forces
and how that causes changes
in the rotation of motion.
So that leaves one
other quantity,
which is the moment of inertia.
And I want to talk about what
that means for just a moment.
So the term "inertia," the
term "inertia" in physics
represents a resistance
to an applied force.
It tells us how difficult it is
to change an object's motion.
So for example, suppose I
have two blocks, one of mass m
and another that's
10 times as massive.
So 10m.
Newton's Second
Law, f equals ma,
tells us that if I want
to accelerate the heavier
mass to the same rate that
I do with a smaller mass,
I'll need to apply a force
that's 10 times larger.
So for translation
of motion, the mass m
represents the notion of
inertia, the resistance
to a force.
It tells us how
much force we have
to apply to achieve a
certain change in motion.
Now, for rotational
motion, that role
is played by the
moment of inertia i.
That tells us how
difficult it is to change
the rotation of an object.
If I increase the moment of
inertia by a factor of 10,
then I'll need a torque that
is 10 times larger in order
to achieve the same change
in rotational motion,
the same angular acceleration.
But recall that our definition
of moment of inertia
for some rigid body
is, if I break up
the rigid body into a
bunch of little pieces,
and for each piece
I take the product
of the mass of that piece, delta
mj, times the perpendicular
distance of that piece
from the rotation
axis-- I call that
rj-- and if I sum
that over the entire object,
the entire body, that
gives me my moment of inertia.
Well, so-- sorry,
this is r squared.
So it's the mass of each
element times the distance
of the axis squared,
delta mj times rj squared.
Now, I can increase the moment
of inertia by a factor of 10
by increasing the total
mass of the object,
but you can see
from this equation
that I can also do
it with the same mass
by changing the
location of the mass.
If I increase the r's, if I
move the mass, the same mass,
but I move it to be farther
away from the rotation axis,
that also achieves an increase
in the moment of inertia.
So what that tells is that
for rotational motion,
it's not just the amount
of mass that matters,
but also how that
mass is distributed.
OK, so in that sense,
rotational inertia
is different than
translational inertia.
For translational motion,
all that matters is the mass.
If I increase the mass
by a factor of 10,
then it will become
a factor of 10
more difficult to get
that object to accelerate.
I need a factor of
10 larger force.
But with rotational
motion, I can
increase the notion of inertia.
I can make it more difficult to
change the rotation of motion
either by changing the
mass, by increasing it,
or by making the
distribution of mass
be further away from
the rotation axis.
So as an example, imagine if you
were rolling a wheel up a hill.
It's different if I have a
wheel whose mass is distributed
evenly over the whole
disk, or if I have
all of the mass in the rim.
If all of the mass is in the
rim, then from this equation,
we see the moment of inertia
is larger for the same amount
of total mass.
And so it's much more difficult
to roll a wheel up the hill
if all of the mass
is in the rim than it
is if the mass is distributed
evenly over the wheel.
So our rotational equivalent
to Newton's Second Law
is, tau equals i
alpha, the torque
equals the moment of inertia
times the angular acceleration.
In the next few lessons, we'll
see how torque is defined,
and we'll derive this expression
for the dynamics of rotation
of motion and see
how to apply it.
