The moment of inertia of a semicircular ring about a line perpendicular to the plane of the ring and passing through its centre is given as $I$ $= M R^{2}$ , where $m$ and $R$ are the mass and radius of the ring. Comparing the given value
. $I = {MR}^{2} = \frac{1}{x} M R^{2}$
$\Rightarrow x = 1$

Question

The moment of inertia of a semicircular ring about a line perpendicular to the plane of the ring and passing through its centre is given as $I$ $= M R^{2}$ , where $m$ and $R$ are the mass and radius of the ring. Comparing the given value
. $I = {MR}^{2} = \frac{1}{x} M R^{2}$
$\Rightarrow x = 1$

TestHub · Accepted Answer

The moment of inertia of a semicircular ring about a line perpendicular to the plane of the ring and passing through its centre is given as $I$ $= M R^{2}$ , where $m$ and $R$ are the mass and radius of the ring. Comparing the given value
. $I = {MR}^{2} = \frac{1}{x} M R^{2}$
$\Rightarrow x = 1$

Physics - Rotation Question with Solution | TestHub

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