The moment of inertia of a thin semicircular | vedclass

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(B) The moment of inertia of a complete circular disc of mass $M'$ and radius $R$ about an axis passing through its center and perpendicular to its pl

Vedclass · Accepted Answer

$\frac{1}{2} MR^2$

Vedclass · Answer

(B) The moment of inertia of a complete circular disc of mass $M'$ and radius $R$ about an axis passing through its center and perpendicular to its plane is $I_{total} = \frac{1}{2} M' R^2$.For a semicircular disc of mass $M$,we can consider it as half of a complete circular disc of mass $2M$ and radius $R$.The moment of inertia of this complete circular disc (mass $2M$) about the axis through its center $O$ is $I_{total} = \frac{1}{2} (2M) R^2 = MR^2$.Since the semicircular disc is exactly half of this complete disc,its moment of inertia about the same axis is half of the total moment of inertia.Therefore,$I = \frac{1}{2} I_{total} = \frac{1}{2} (MR^2) = \frac{1}{2} MR^2$.

The moment of inertia of a thin semicircular disc (mass $= M$ and radius $= R$) about an axis passing through point $O$ and perpendicular to the plane of the disc is given by:

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