InterviewSolution
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InterviewSolution
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From a circular disc of radius R and mass 9 M , a small disc of radius R/3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is A. `4 MR^(2)`B. `(40)/(9)MR^(2)`C. `10 MR^(2)`D. `(37)/(9)MR^(2)` |
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Answer» Correct Answer - A Mass per unit area of disc `= (9M)/(piR^(2))` Mass of removed portion of disc `= (9M)/(piR^(2)) xx pi ((R )/(3))^(2) = M` Moment of inertia of removed portion about an axis passing through centre of disc and perpendicular to the plane of disc, using theorem of parallel axis is `I_(1) = (M)/(2) ((R )/(3))^(2) + M ((2R)/(3))^(2) = (1)/(2)MR^(2)` when portion of disc would not have been removed, then the moment of inertia of complete disc about the given axis is `I_(2) = (1)/(2)MR^(2)` So moment of inertia of the disc with removed portion, about the given axis is `I = I_(2) - I_(1) = (9)/(2)MR^(2) - (1)/(2)MR^(2) = 4MR^(2)` |
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