Question

A thin circular disc lying in the \( xy \)-plane has a surface mass density \[ \sigma(r) = \begin{cases} \sigma_0\left(1 - \frac{r^2}{R^2}\right), & \text{if } r \le R \\ 0, & \text{if } r > R \end{cases} \] where \( r \) is the distance from its center. Its moment of inertia about the z-axis passing through its center is:

• A. \( \frac{\sigma_0 R^4}{4} \)
• B. \( \frac{\pi \sigma_0 R^4}{6} \)
• C. \( \sigma_0 R^4 \)
• D. \( 2\pi \sigma_0 R^4 \)

Hint:

For surface mass density varying with radius, always express \( dm = \sigma(r) 2\pi r dr \) and integrate \( r^3 \sigma(r) \) to find moment of inertia.

Answer 1

The Correct Option is B

Step 1: Formula for moment of inertia of a surface mass distribution.
For a thin disc, \[ I = \int r^2 \, dm = \int_0^R r^2 (2\pi r \sigma(r)) dr = 2\pi \int_0^R \sigma(r) r^3 dr. \] Step 2: Substitute given surface density.
\[ I = 2\pi \sigma_0 \int_0^R \left(1 - \frac{r^2}{R^2}\right) r^3 dr = 2\pi \sigma_0 \left[\frac{r^4}{4} - \frac{r^6}{6R^2}\right]_0^R. \] Step 3: Simplify.
\[ I = 2\pi \sigma_0 \left(\frac{R^4}{4} - \frac{R^4}{6}\right) = 2\pi \sigma_0 R^4 \left(\frac{1}{12}\right) = \frac{\pi \sigma_0 R^4}{6}. \] Step 4: Final Answer.
The moment of inertia about the z-axis is \( \frac{\pi \sigma_0 R^4}{6}. \)