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
. $I = {MR}^{2} = \frac{1}{x} M R^{2}$
$\Rightarrow x = 1$

Aicharya · 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
. $I = {MR}^{2} = \frac{1}{x} M R^{2}$
$\Rightarrow x = 1$

The moment of inertia of a semicircular ring about an axis, passing through the center and perpendicular to…

Solution