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A disc of mass m and radius R has a concentric hole of radius r. Find the M.I about a normal axis to its plane and passing through centre of disc. |
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Answer» Solution :Consider a ring of radius .X. and of thickness dx of the disc The M.I. of the ring about a normal AXIS through its centre is `dI=(m)/(pi(R^(2)-r^(2)))2pix.dx.x^(2)` The total M.I of the disc is obtained by integrating the above expression. `intdI=int_(r)^(R)(m)/(pi(R^(2)-r^(2)))2pix.dx.x^(2)` on solving `I=(m)/(2)(R^(2)+r^(2))` |
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