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**Moments of Inertia, Centres of Mass... help plz!**

## Homework Statement

A rod of length l rotates about an axis that runs perpendicularly to the rod through one end. The linear density p(x) of the rod in terms of the distance x from the axis is given by:

p(x) = p0(1 + (x^2/l^2))

where p0 is a constant. In terms of l and p0, find the total mass M of the rod, the distance R of the centre of mass of the rod from x=0, and the moment of inertia I and radius of gyration k of the rod about the axis.

## The Attempt at a Solution

For the mass M, I guessed what you needed to do was integrate the p(x) expression between the limits of 0 and l with respect to x. I did this and got (4/3)lp0. Is this the right method?

For the centre of mass, I really don't know! I don't know how to do it for this type of problem, only simple (uniform density) problems.

For the moment of inertia, based on my limited understanding, I decided to take the expression for p(x) and perform the same integration as above but with an expression x^2 in there, so I did this:

p0 * [integral between limits 0 and l] (x^2 + (x^4/l^2) dx

Please don't ask me to explain why I did this, it's based on guesswork and trying to copy the method of other problems I've seen, which were a lot more simple! Anyway, the answer I got was (8/15)(l^3)p0. Am I anywhere close?

For the radius of gyration I know the formula I = mk^2 (where k = radius of gyration) so that should be easy if I have the correct mass and inertia. If I find it based on the answers I've got for M and I here (which I expect are wrong), the answer is sqrt((2l^2)/5).

Sorry, I would try to figure out how to type all the symbols and notation etc properly, but I'm under extreme exam pressure and don't have time! Your help would be greatly welcomed!