Moment of Inertia on Semi-Hollow Cylinder

AI Thread Summary
The discussion focuses on deriving the moment of inertia for a hollow cylinder using integration. The initial setup involves defining mass, outer radius, and inner radius, with the assumption of uniform composition. The integration process leads to an expression for moment of inertia, but the user initially struggles to arrive at the expected formula. A key insight is provided regarding the factorization of the numerator, which simplifies the expression to the correct form. Ultimately, the user acknowledges the mistake in handling exponents and appreciates the guidance towards the correct solution.
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Homework Statement



A hollow cylinder has mass m, and outside radius r_{2}, and an inside radius r_{1}. Use intergration to show that the moment of inertia about its axis is give by I=\frac{1}{2}m(r^2_2+r_1^2).

Uniform composition is assumed.

Homework Equations



I=\int r^2dm
dm=\sigma dA

and

A=\pi r^2
(note, later on I constrict the lower and upper bounds of integration to account for the now entire area of the circular side, which sounds as though it is assumed here)

The Attempt at a Solution



Note: Using equations above.

I=\int r^2dm

I=\int r^2\sigma dA

dA=2\pi rdr

I=\sigma 2\pi\int r^3dr

(here I insert my upper and lower bounds)
I=\sigma 2\pi\int_{r_1}^{r_2} r^3dr

I=\sigma 2\pi \left(\frac{r^4_2}{4}-\frac{r^4_1}{4} \right )

I=\sigma \pi \left(\frac{r^4_2}{2}-\frac{r^4_1}{2} \right )

I=\frac {1}{2}\sigma \pi \left(r^4_2-r^4_1 \right )

\sigma=\frac{m}{A}

and since

A=\pi r^2_1-\pi r^2_2
then

I=\frac{m}{2(\pi r^2_1-\pi r^2_2)} \pi \left(r^4_2-r^4_1 \right )I=\frac{\left(r^4_2-r^4_1 \right )m}{2 (r^2_1-r^2_2)}This is as far as I venture into the problem before I realize that I've done something wrong if I'm supposed to get the answer they assume at the beginning of the problem. Where did I go wrong?

Thank you very much for the help, it's greatly appreciated.
 
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That is correct so far. Remember a2-b2=(a+b)(a-b). So if you factorize the numerator what do you get?
 
Given:

I=\frac{\left(r^4_2-r^4_1 \right )m}{2 (r^2_1-r^2_2)}

Then:

I=\frac{1}{2}m(r^2_1+r^2_2)

Great! Thank you so much! I actually forgot how those exponents would go into each other, that was my issue. Thanks for leading me toward the right answer there!
 
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