Consider a uniform wire of mass M and length | vedclass

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(C) The length of the wire is $L$. When bent into a semicircle of radius $r$,the arc length is $\pi r = L$. Thus,the radius is $r = \frac{L}{\pi}$. Si

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$\frac{ML^2}{\pi^2}$

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(C) The length of the wire is $L$. When bent into a semicircle of radius $r$,the arc length is $\pi r = L$. Thus,the radius is $r = \frac{L}{\pi}$. Since all points of the wire are at the same distance $r$ from the center of the semicircle,the moment of inertia $I$ about an axis perpendicular to the plane passing through the center is given by $I = \int r^2 dm = r^2 \int dm$. Since $\int dm = M$,we have $I = Mr^2$. Substituting $r = \frac{L}{\pi}$,we get $I = M \left( \frac{L}{\pi} \right)^2 = \frac{ML^2}{\pi^2}$.

Consider a uniform wire of mass $M$ and length $L$. It is bent into a semicircle. Its moment of inertia about a line perpendicular to the plane of the wire passing through the centre is:

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