Moments of Inertia (2)

This note is a continuation of a previous note.

(3) Cylindrical Shell about z-axis:

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):

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

Which very readily simplifies to:

\(I = \frac{mr^2}{2}\)

(5) Solid Cylinder about z-axis:

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
3 years, 8 months ago

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