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This is the proof up the parallel axis theorem.
On page 254 we have this
diagram. It's just some object
and it has a center of mass which we have placed
at the origin. So the center of mass is at the origin,
and we're going to rotate it through some
axis located at point P. We know that the moment of inertia of an object
is the sum of mr^2 which is
many many pieces, each with
incremental mass dm times
its distance to the axis of rotation squared.
So mathematically we write it like this - it's the integral
of r^2 dm where r is the distance
from each piece of mass to the axis of rotation,
and dm is the incremental mass of each piece.
Now we look at this diagram over here
we see that dm
is located at coordinates X,Y
so the rest of the diagram -
if we let the distance in the x-direction
from the center mass to the axis of rotation in the x-direction be a,
and in the y-direction be b,
then the distance from the center of mass to the axis of rotation is h,
and using Pythagorean Theorem we know that
h^2 = a^2 + b^2
and we also see over here that r^2
is equal to, right here, this equation,
r^2 = (x-a)^2 + (y-b)^2
In this equation here, I'm going to substitute
for the r^2 here the
(x-a)^2 + (y-b)^2 substituted in there
and now I'm going to multiply out the square terms
and rearrange it and look at each term
individually, take the integral of each term individually
the first term x^2 + y^2
I recognize that as capital R squared
That's not drawn in this diagram on the right, but that is the distance from
the center of mass
to the point of mass I'm looking at
... so that's R^2.
These next two terms, if you compare them
to equations that we saw chapter 9
This is the
term for the X coordinate of the center of mass and this is the term
for the y coordinate of the center of mass. So since my
center of mass in this diagram is at the origin
these terms must be zero because
x dm is 0 and y dm is zero, and that's what I've got up here.
So those terms drop out, and the
this last term (a^2 = B^2) we know is the same as h^2,
so that becomes
the integral of h^2 dm
the integral of dm is M - the mass of the object,
and h^2 is a constant, so
the last term is equal to Mh^2.
So therefore, the moment of inertia is equal to
the moment of inertia the center mass which is just
integral of R^2 dm
plus Mh^2! And there you have it, the
parallel axis theorem