The moment of inertia of a uniform annular di | vedclass

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Question

(B) Let $M$ be the mass of the annular disc with outer radius $R$ and inner radius $r$.The surface mass density $\sigma$ is given by $\sigma = \frac{M

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $M$ be the mass of the annular disc with outer radius $R$ and inner radius $r$.The surface mass density $\sigma$ is given by $\sigma = \frac{M}{\pi(R^2 - r^2)}$.Consider an elementary ring of radius $x$ and thickness $dx$. The mass of this elementary ring is $dm = \sigma 	imes (2\pi x dx) = \frac{M}{\pi(R^2 - r^2)} 	imes 2\pi x dx = \frac{2Mx dx}{R^2 - r^2}$.The moment of inertia $dI$ of this elementary ring about the central axis is $dI = (dm)x^2 = \frac{2Mx^3 dx}{R^2 - r^2}$.To find the total moment of inertia $I$,we integrate $dI$ from $x = r$ to $x = R$:$I = \int_{r}^{R} \frac{2Mx^3 dx}{R^2 - r^2} = \frac{2M}{R^2 - r^2} \int_{r}^{R} x^3 dx$.Evaluating the integral: $I = \frac{2M}{R^2 - r^2} \left[ \frac{x^4}{4} ight]_{r}^{R} = \frac{2M}{R^2 - r^2} \left( \frac{R^4 - r^4}{4} ight)$.Since $R^4 - r^4 = (R^2 - r^2)(R^2 + r^2)$,we get $I = \frac{2M}{R^2 - r^2} 	imes \frac{(R^2 - r^2)(R^2 + r^2)}{4} = \frac{1}{2}M(R^2 + r^2)$.

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 centre and perpendicular to its plane is:

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