The moment of inertia of a uniform semicircul | vedclass

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(C) The moment of inertia of a semicircular wire about an axis passing through its centre and perpendicular to its plane is $I = m r^2$.Let $I_{cm}$ b

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$m r^2\left(1-\frac{4}{\pi^2}\right)$

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(C) The moment of inertia of a semicircular wire about an axis passing through its centre and perpendicular to its plane is $I = m r^2$.Let $I_{cm}$ be the moment of inertia about an axis passing through its centre of mass and perpendicular to its plane.The distance of the centre of mass from the centre of the semicircle is $d = \frac{2r}{\pi}$.Using the parallel axis theorem,$I = I_{cm} + md^2$.Substituting the values,$m r^2 = I_{cm} + m\left(\frac{2r}{\pi}\right)^2$.Therefore,$I_{cm} = m r^2 - m\left(\frac{4r^2}{\pi^2}\right) = m r^2\left(1 - \frac{4}{\pi^2}\right)$.

The moment of inertia of a uniform semicircular wire of mass $m$ and radius $r$,about an axis passing through its centre of mass and perpendicular to its plane is ..........

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