I will do an example to apply that parallel axis theorem.
In a separate segment, we calculated the moment of
inertia all of a uniform rod, rotating about its center,
perpendicular to axis, and perpendicular to the length
direction of the rod.
I will now offset that axis, and use the
parallel access theorem.
Here is that rod: length l, mass m, with uniform mass
distribution.
This was the center, and therefore the center of mass.
We calculated earlier that the moment of inertia about this
axis of rotation equals 1/12 ml squared, but now I would
like to know what the moment of inertia is about an axis of
rotation which is also perpendicular to the rod.
Let it go through point P, and this separation between these
two axes equals d.
That moment of inertia equals 1/12 ml squared, which is the
moment of inertia--
the axis through the center of mass, plus m times d squared.
We derived this one, but you could have looked
it up in the tables--
whatever you prefer.
When you want to know what it is about this axis, you apply
the parallel axis theorem.
If you want to rotate this rod about point Q, about an axis
perpendicular to the direction of the rod, then you would
find that I q equals 1/12 ml squared plus m--
d is now 1/2 l--
so you get plus m times 1/4 l squared.
That equals 1/3 ml squared.
This result you can also find in many tables, but if somehow
you don't have the tables and you have the value of moment
of inertia about the center of mass, then you can evaluate
all the others, so it comes in very handy.