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A disc of mass m and radius R has a concentric hole of radius r. Moment of inertia about an axis through its centre and perpendicular to its plane
Consider a ring of radius x and thickness DX of the disc The MI of the ring about a normal axis through which its centre is dI =(2πx.dx.x^2)/π[R^2_r^2]The total MI of the disc is obtained by integrating the above expressionIntegrations I=integration {m/π(R^2-r^2)}.2πx.dx.x^2On solvingI={m(R^2 + r^2)}/2
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