moment of inertia of an Annulus proof

fuinneamh

Just after finding out that it has previously come up on the exam but can't find a proof of it. Could anyone help me with it please?

ZorbaTehZ

I would be very surprised if this came up since it has a trivial solution. (This can be somewhat difficult to get your head around)

If you're using Ollie Murphy's book consider the diagram on page 322.
The mass is located at distances x.
Consider an annulus or ringed lamina, of maximum radius B, and inner radius b.
But all the mass is located at a common distance b from the axis in an annulus, so you need only consider that one value for x.

Therefore:

Definition of I (moment of inertia) is:

I = E delta(m) x r^2 { E means sum of all

where delta(m) is the mass of each piece of the ring at a distance r

But since all the pieces are a common distance b:

I = E delta(m) x b^2

E becomes an integral

I = S delta(m) x b^2 dm { S means integral
I = Mb^2

(All that can be condensed down to about 6 lines methinks)