First slide

Moment of intertia

Question

Two identical rings each of mass m with their planes mutually perpendicular, radius R are welded at their point of contact O. If the system is free to rotate about an axis passing through the point P perpendicular to the plane of the paper the moment of inertia of the system about this axis is equal to :

Moderate

A

6.5 mR²
B

12 mR²
C

6 mR²
D

11.5 mR²

Solution

Required moment of inertia :
$I = ({mR}^{2} + {mR}^{2}) + [\frac{{mR}^{2}}{2} + m (3 R)^{2}] = 11 .5 {mR}^{2}$

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