The moment of inertia of a semicircular ring | vedclass

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(B) complete ring of mass $M_{total}$ and radius $R$ has a moment of inertia $I = M_{total}R^2$ about an axis passing through its center and perpendic

Vedclass · Accepted Answer

$1$

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(B) complete ring of mass $M_{total}$ and radius $R$ has a moment of inertia $I = M_{total}R^2$ about an axis passing through its center and perpendicular to its plane.For a semicircular ring of mass $M$,the mass is distributed such that the distance of every mass element from the center is exactly $R$.The moment of inertia $I$ is defined as $\int r^2 dm$.Since every mass element $dm$ of the semicircular ring is at a constant distance $R$ from the center,we have $I = \int R^2 dm = R^2 \int dm = MR^2$.Comparing this with the given expression $\frac{1}{x} MR^2$,we get $\frac{1}{x} = 1$,which implies $x = 1$.

The moment of inertia of a semicircular ring about an axis passing through the center and perpendicular to the plane of the ring is $\frac{1}{x} MR^2$,where $R$ is the radius and $M$ is the mass of the semicircular ring. The value of $x$ will be $...........$

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