First slide

Moment of intertia

Question

From a circular disc of radius R and mass 9M, a small disc of mass M and radius $\frac{R}{3}$ is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is :

Moderate

A

MR²
B

2MR²
C

$\frac{4}{9} {MR}^{2}$
D

$\frac{40}{9} {MR}^{2}$

Solution

Use parallel axis theorem
$I_{0} = \frac{1}{2} (9 M) R^{2} - \frac{1}{2} (M) {(\frac{R}{3})}^{2} = \frac{40}{9} {MR}^{2}$

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