The moment of inertia of a uniform annular di | vedclass

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(B) The moment of inertia of a uniform solid disc of mass $M'$ and radius $R$ about an axis through its center and perpendicular to its plane is $I =

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$\frac{1}{2}M(R^2 + r^2)$

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(B) The moment of inertia of a uniform solid disc of mass $M'$ and radius $R$ about an axis through its center and perpendicular to its plane is $I = \frac{1}{2}M'R^2$.An annular disc can be considered as a large solid disc of radius $R$ minus a smaller solid disc of radius $r$.Let $\sigma$ be the surface mass density,$\sigma = \frac{M}{\pi(R^2 - r^2)}$.The mass of the large disc is $M_R = \sigma \pi R^2 = M \frac{R^2}{R^2 - r^2}$.The mass of the small disc is $M_r = \sigma \pi r^2 = M \frac{r^2}{R^2 - r^2}$.The moment of inertia of the annular disc is $I = I_R - I_r = \frac{1}{2}M_R R^2 - \frac{1}{2}M_r r^2$.Substituting the masses: $I = \frac{1}{2} \left( M \frac{R^2}{R^2 - r^2} \right) R^2 - \frac{1}{2} \left( M \frac{r^2}{R^2 - r^2} \right) r^2$.$I = \frac{M}{2(R^2 - r^2)} (R^4 - r^4)$.Using the identity $R^4 - r^4 = (R^2 - r^2)(R^2 + r^2)$,we get:$I = \frac{M}{2(R^2 - r^2)} (R^2 - r^2)(R^2 + r^2) = \frac{1}{2}M(R^2 + r^2)$.

The moment of inertia of a uniform annular disc of internal radius $r$,external radius $R$,and mass $M$ about an axis passing through its center and perpendicular to its plane is:

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