Question

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?

Answer 1

`3/10 MR^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+dr` respectively.

• Area of a ring : `2pi r\ dr`
• mass of the ring : `2 pi mu r^2 dr`

Thus, net mass :

`M = int_0^R 2 pi mu r^2 dr = (2 pi)/3 \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 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

`1/2 times 2 pi mu r^4 dr = pi mur^4 dr`

Hence the moment of inertia of the disc is

`I = int_0^R pi mu r^4 dr = pi/5 mu R^5`

Hence

`I/M = 3/10 R^2 quad implies quad I = 3/10 MR^2`