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From a disc of mass m and radius r, a circular portion of radius (r )/(3) is removed from the edges. Find the M.O.I of the remaining system about a normal through center of original disc |
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Answer» Solution :Let `I_(2)` be the M.I. of remaining poriton. M.I of the removed partion about a normal axis through centre of the original disc is `I_(1)=((m)/(9)((r )/(3))^(2))/(2)+(m)/(9)((2r)/(3))^(2)`or `I_(1)=(MR^(2))/(18)` Hence, M.I. of the remaining portion is `I_(2)=I-I_(1)=(mr^(2))/(2)-(mr^(2))/(18)` or `I_(2)-(4mr^(2))/(9)` |
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