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?

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Answer 1 · 2018-07-04T00:32:15

$3/10 MR^2$

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.

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

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