×

# Moments of Inertia (2)

This note is a continuation of a previous note.

The process for this will be similar to the spherical shell, that is, we will take a limit as the inner radius approaches the outer. This time, however, we will transfer to cylindrical coordinates with bounds:

$$r_1 \leq r \leq r_2$$

$$0 \leq \theta \leq 2\pi$$

$$0 \leq z \leq l$$

Our integral, with the Jacobian, is:

$$I = \displaystyle \lim_{r_1 \to r_2} \frac{m \int_{0}^{2\pi} \int_{0}^{l} \int_{r_1}^{r_2} r^3 dr dz d\theta}{\int_{0}^{2\pi} \int_{0}^{l} \int_{r_1}^{r_2} r dr dz d\theta}$$

Evaluating this, we obtain:

$$I = \displaystyle \lim_{r_1 \to r_2} \frac{m(r_2^4 -r_1^4)}{2(r_2^2 -r_1^2)} = \frac{0}{0}$$

Differentiating the numerator and denominator with respect to $$r_1$$ and taking the limit, we have:

$$I = \displaystyle \lim_{r_1 \to r_2} \frac{m(-4r_1^3)}{-4r_1} = mr^2$$

Which is the intended result.

(4) Solid Disk:

This is actually not, strictly speaking, a three dimensional moment of inertia, as we have no height. Hence, we change not to cylindrical coordinates, but to polar coordinates (which are essentially cylindrical coordinates without the z coordinate). We have bounds:

$$0 \leq r \leq r_0$$

$$0 \leq \theta \leq 2\pi$$

And our integral is (note that we must still include the Jacobian):

$$I = \frac{m \int_{0}^{2\pi} \int_{0}^{r_0} r^3 dr d\theta}{\int_{0}^{2\pi} \int_{0}^{r_0} r dr d\theta}$$

$$I = \frac{mr^2}{2}$$

This integral is exactly the same as (4), except we add:

$$0 \leq z \leq l$$

Then we have:

$$I = \frac{m \int_{0}^{2\pi} \int_{0}^{l} \int_{0}^{r_0} r^3 dr dl d\theta}{\int_{0}^{2\pi} \int_{0}^{l} \int_{0}^{r_0} r dr dl d\theta}$$

And immediately we have:

$$I= \frac{mr^2}{2}$$

Now, note that (4) and (5) have the same moment of inertia. This should make sense, as (4) is just the special case of (5), when $$l=0$$.

Note by Ethan Robinett
2 years, 2 months ago