First slide

Moment of intertia

Question

Two rings of the same radius r and mass m are placed such that their centres are at a common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of one of the rings is:

Easy

A

(1 /2)mr²
B

mr²
C

(3/2) mr²
D

2mr²

Solution

$I = {mr}^{2} + \frac{{mr}^{2}}{2} = \frac{3}{2} {mr}^{2}$

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