Question 37 of 280

Moment of inertia of a disc of mass $M$ and radius '$R$' about any of its diameter is $\dfrac{M{R}^{2}}{4}$. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, $\dfrac{x}{2}M{R}^{2}$. The value of $x$ is ______.

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Answer: 3

Given here,
Moment of inertia of a disc of mass $M$ and radius '$R$' about any of its diameter is $\dfrac{M{R}^{2}}{4}$.
Moment of inertia of disc about centre is ${I}_{cm}=\dfrac{M{R}^{2}}{4}+\dfrac{M{R}^{2}}{4}=\dfrac{M{R}^{2}}{2}$
Using parallel axis theorem, we can write
$I={I}_{cm}+M{d}^{2}$
$=\dfrac{M{R}^{2}}{2}+M{R}^{2}$
$=\dfrac{3}{2}M{R}^{2}$
Hence, the value of $x=3$.