A thin wire of length ell and mass m is bent | vedclass

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(C) The length of the wire is $\ell$,which forms a semicircle of radius $R$. Therefore,$\ell = \pi R$,which gives $R = \frac{\ell}{\pi}$.Every point o

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$\frac{m \ell^2}{\pi^2}$

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(C) The length of the wire is $\ell$,which forms a semicircle of radius $R$. Therefore,$\ell = \pi R$,which gives $R = \frac{\ell}{\pi}$.Every point on the semicircular wire is at a distance $R$ from the axis passing through its free ends (the diameter of the semicircle).The moment of inertia $I$ of a particle of mass $m$ at a distance $R$ from the axis is $I = m R^2$. Since the entire mass $m$ of the wire is at a distance $R$ from the axis,the moment of inertia of the semicircular wire about this axis is $I = m R^2$.Substituting $R = \frac{\ell}{\pi}$ into the formula,we get:$I = m \left( \frac{\ell}{\pi} \right)^2 = \frac{m \ell^2}{\pi^2}$.

$A$ thin wire of length $\ell$ and mass $m$ is bent in the form of a semicircle as shown. Its moment of inertia about an axis joining its free ends will be ...........

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