# How to calculate moment of inertia of a disc about its one diameter if given mass per unit area is proportional to distance from centre?

Jul 4, 2018

$\frac{3}{10} M {R}^{2}$

#### Explanation:

Let the radius of the disc be $R$ and its total mass be $M$. Let the mass per unit area be given by $\mu r$, where $\mu$ is a constant.

Divide the disc into rings with inner and outer radii $r$ and $r + \mathrm{dr}$ respectively.

• Area of a ring : $2 \pi r \setminus \mathrm{dr}$
• mass of the ring : $2 \pi \mu {r}^{2} \mathrm{dr}$

Thus, net mass :

$M = {\int}_{0}^{R} 2 \pi \mu {r}^{2} \mathrm{dr} = \frac{2 \pi}{3} \setminus \mu {R}^{3}$

The moment of inertia of this ring about an axis passing through the center of the disc and normal to it is $2 \pi \mu {r}^{2} \mathrm{dr} \times {r}^{2}$.

A simple application of the perpendicular axis theorem then shows that the moment of inertia of the ring about a diameter is

$\frac{1}{2} \times 2 \pi \mu {r}^{4} \mathrm{dr} = \pi \mu {r}^{4} \mathrm{dr}$

Hence the moment of inertia of the disc is

$I = {\int}_{0}^{R} \pi \mu {r}^{4} \mathrm{dr} = \frac{\pi}{5} \mu {R}^{5}$

Hence

$\frac{I}{M} = \frac{3}{10} {R}^{2} \quad \implies \quad I = \frac{3}{10} M {R}^{2}$