Physics, asked by sruthik16, 11 months ago

A disc of mass 'm' and radius 'R' has a concentric hole of radius 'r'.Its M.I. about an axis through centre and normal to its plane is​

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Answered by Anonymous
11

Answer:

Explanation: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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Answered by priti964
2

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A disc of mass 'm' and radius 'R' has a concentric hole of radius ... Its M.I. about an axis through centre and normal to its ...

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