Homework Help: Moment of Inertia in slender rod - density varies

A slender rod with length L has a mass per unit length that varies with distance from the left-hand end, where x=0, according to dm/dx = (gamma)x, where (gamma) has units of kg/m^2.

I calculated the total mass of the rod by integrating. I got

M = 0.5(gamma)L^2

In the second part I found the moment of inertia with the axis of rotation to be at x=0, perpendicular to the rod. I don't know how to represent the integration symbol, so I'll use "|." I am integrating from 0 to L

I = |r^2dm
I = |L^2(gamma)L = (gamma)|L^3 = [(gamma)L^4]/4
= [2M/L^2][0.25L^4] = 0.5ML^2

The third part asks me to do the same thing for the rod, this time with the axis of rotation at the opposite end of the rod. I know I have to integrate from L to 0, but shouldn't this just give me the negative of my previous expression? I know it's wrong, but I don't know how the mass expression changes at the other end. Can someone help?

 

Ok, this makes sense. But they want the answer in terms of M and L. When I substitute my expression for gamma after integration, I get

[(4LMX^3)-(6MX^4)]/(6L^2)

There's no way I can see to get rid of X, so now I'm stuck...again.

 

ahhh. thanks a lot. got (1/6)ML^2