# How to find the moment of inertia of a cut-out object?

## I've been trying to solve this question for a day now. Can someone point me in the right direction? I'd really appreciate it!

May 22, 2018

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

#### Explanation:

The center of mass of the original triangle (the part that has been cut out) was at the center of the circle, at a distance $R$ from the pivot. The distance of the center of mass of the triangle in its new position from the pivot is the same. Thus, the moment of inertia contributed by the triangle now is the same as its original contribution (by the parallel axis theorem).

The moment of inertia of the whole system is thus the same as that of the complete circular disc about the pivot.

This is

$\frac{1}{2} M {R}^{2} + M {R}^{2} = \frac{3}{2} M {R}^{2}$

(using the parallel axis theorem and the known result that the moment of inertia of a uniform circular disc about an axis passing through its center and normal to it is $\frac{1}{2} M {R}^{2}$