We've already talked about kinetic energy
for an object that's rotating and translating.
Let's take, again, a simple object like a wheel.
Center of mass is moving, and we want
to consider this is a bunch of point-like particles,
th Jth particle with mass MJ, and a position vector RCMJ.
And now what we'd like to do is, what is the angular momentum
of this about some point P?
And we're going to do the same type of decomposition
that we've been doing right along.
Let's draw the vector RJ.
This is the vector from the point P to the object.
Here's the vector capital R, and there's the vector RCMJ.
And our same vector addition, R plus RCMJ.
j Now let's calculate the angular momentum
about the point P. Now just let's
recall two facts about the center of mass reference frame
that we've already used.
The first is that the sum of the velocities, the momentum
in the center of mass reference frame is zero,
and if you integrate this equation in just exactly
the same way, if you added up the mass times each
of the position vectors, that's MJ,
MJ, in the center of mass frame, that's also zero.
So we've used this when we talked about kinetic energy.
Now we're going to consider both of these results.
So recall that angular momentum about a point
is the sum of the vector from the point, cross-product MJ,
times the velocity BJ.
Now we can also use our law of addition velocities,
VJ is V plus the VCMJ, and so now
we're going to have to do both substitutions.
That's why this calculation is a little bit more complicated.
We have R plus RCMJ, cross MJ times capital
D. That is the velocity of the center of mass
with respect to the ground frame plus VCMJ.
Now we have four terms in the cross-product,
and this time we'll write them all out.
So the first term is the sum over J of R cross MJ capital D.
The second term is, we'll take this one,
and we'll cross with that one.
But because MJ is just a scalar, I'm going to pull it in front.
So that's MJ, RCMJ, and V is the same for every single particle
in the object.
Remember, V is just the velocity of one reference frame
with respect to the other.
So I can pull that out, cross V. And you're already noticing
that this term will be zero.
The next term is when I take R cross
MJVJ, V center of mass, J, summed over J,
that's JR cross MJ, VCMJ.
But again, remember that R is the same vector.
It's just from P to the center of mass,
so I didn't need to put the sum here.
I can pull R out of the sum and I have my sum like that.
And again, you're probably noticing that this term is zero
and our final term is RCMJ cross MJ VCMJ.
So our final term is sum R, RCMJ cross MJ VCMJ.
Now from these two conditions, one and two,
both of these two terms are zero.
And that's the power of using the center of mass reference
frame.
So LP, now let's look at the first term.
Here, R is the same for every particle,
V is the same for every particle,
when we're summing over the mass,
we're just getting the total mass,
so the first piece is R cross M total V.
And this last piece is precisely the definition of the angular
momentum of the Jth particle in the center of mass frame.
So that's the sum over J of LCMJ,
because RCM cross MVJ R cross, say whatever speed this
is in the center of mass frame, it could be that sum VCMJ.
BJ minus capital V.
This is exactly the angular momentum
of that Jth mass in the center of mass frame,
and when we total all of this up, we have R cross
and total V plus LCM.
Now what this first term is, if you treated the whole object
as a point mass, M total, moving with V, and here's our point P
and there's R, this is just what we
call the translational angular momentum.
It's treating the whole system as a point-like mass,
and just calculate moving with speed V,
and that's just the angular momentum.
We've called this V.
And this piece is the angular momentum about center of mass.
So sometimes, we can think of the angular momentum
as corresponding to a translational,
it has to do with the actual orbit of the object.
You could call this orbital angular momentum.
And this is the fact that the object is rotating.
You could think of this as the spin angular momentum
about the center of mass.
This is the same type of decomposition
that we saw with kinetic energy.
There was a translational component of kinetic energy
and a rotational component of kinetic energy.
This is the analogous case for angular momentum about a point
P.