Moment of Inertia on Semi-Hollow Cylinder

[trautlein]

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.

 

[rock.freak667]

That is correct so far. Remember a²-b²=(a+b)(a-b). So if you factorize the numerator what do you get?

 

[trautlein]

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!