The moment of inertia of a semicircular ring | vedclass

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Question

(A) Consider a small element of the semicircular ring of angular width $d	heta$ at an angle $	heta$ from the horizontal.The mass of this small eleme

Vedclass · Accepted Answer

$MR^2$

Vedclass · Answer

(A) Consider a small element of the semicircular ring of angular width $d	heta$ at an angle $	heta$ from the horizontal.The mass of this small element is $dm = \frac{M}{\pi R} \cdot R d	heta = \frac{M}{\pi} d	heta$.The moment of inertia of this small element about the centre is $dI = dm \cdot R^2$.To find the total moment of inertia $I$,we integrate $dI$ from $	heta = 0$ to $	heta = \pi$:$I = \int_{0}^{\pi} \frac{M}{\pi} d	heta \cdot R^2 = \frac{MR^2}{\pi} \int_{0}^{\pi} d	heta$.$I = \frac{MR^2}{\pi} [	heta]_{0}^{\pi} = \frac{MR^2}{\pi} (\pi - 0) = MR^2$.Thus,the moment of inertia is $MR^2$.

The moment of inertia of a semicircular ring of mass $M$ and radius $R$ about its centre is:

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