I finding my mistake moment of inertia

[dswatson]

How do i prove,using integration, that the moment of inertia of a hollow cylinder that has mass M, an outside radius R2, and an inside radius R1 is given by...
I=.5(M)(R2^2+R1^2)

here is the work I have done...
I am really close but do not see where I made my mistake
Can someone help me find it?

If you look at the cylinder like a bunch of hoops stacked together then

I(hoop)=MR^2

dV=(2*pi*R)(dR)(L)

P=sigma
P=dM/dV
dM=(2*pi*R)(dR)(L)(P)

I=int(R^2*dM)
I=int[a,b](R^2)(2*pi*R)(dR)(L)(P)
I=(2*pi*L*P)*int[R1,R2](R^3dR)
I=(2*pi*L*P)*[(R2^4-R1^4)/4]
I=(pi*L*P)*[(R2^2-R1^2)(R2^2+R1^2)/2]
V=pi*L(R2^2-R1^2)
M=pi*L*P(R2^2-R1^2)
(R2^2-R1^2)=(pi*L*P)/M
I=(R2^2+R1^2)/2M

I have most of the equation correct but M is in the denominator when if needs to be just the opposite

Thank you in advance

 

[wdednam]

Hi there,

Everything is fine up to the last two steps.

From M = pi*L*P(R2^2 - R1^2) you should get M/(pi*L*P) = (R2^2 - R1^2) and not (pi*L*P)/M = (R2^2 - R1^2).

When you change this you should get the right answer.

Don't worry about it, it's just a small mistake. So, you shouldn't get penalized heavily in an exam for it.

I make mistakes like these all the time. The way to avoid them is to always double-check all the steps in your calculations.

Hope this helps,

Wynand.