First slide

Moment of intertia

Question

A thin rod OA of length 2R and mass m is attached to a thin ring of mass m and radius R. as shown in the figure. Then moment of inertia of the composite body about an axis passing through O and perpendicular to the plane of the ring is

Easy

A

$23 m R^{2} / 3$
B

$34 m R^{2} / 3$
C

$17 m R^{2} / 3$
D

$12 m R^{2}$

Solution

$I_{0} = \frac{1}{3} m {(2 R)}^{2} + [m R^{2} + m {(3 R)}^{2}] = \frac{34}{3} m R^{2}$

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