First slide

Moment of intertia

Question

The moment of inertia of a circular loop of of mass M and radius radius R, about an axis at a distance of R/2 parallel to horizontal diameter of loop is [AIIMS 2012]

Moderate

A

MR²
B

$\frac{1}{2}$ MR²
C

2MR²
D

$\frac{3}{4}$ MR²

Solution

According to theorem of parallel axes,
$I = I_{CM} + M {(\frac{R}{2})}^{2}$
$\Rightarrow I = \frac{1}{2} M R^{2} + \frac{M R^{2}}{4}$
Moment of inertia, $I = \frac{3}{4} M R^{2}$

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