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Moment of inertia of a portion of a disc about an axis perpendicular to its plane; Mass of the object is M and radius R is

\[\begin {array} {1 1} (a)\;\frac{MR^2}{2} & \quad (b)\;\frac{1}{6}\frac{MR^2}{2} \\ (c)\;6.\frac{MR^2}{2} & \quad  (d)\;none\;of\;the \;above \end {array}\]

1 Answer

Let I be moment of inertia of point.
This is a symmetric part of larger dise.
$\therefore 6I= \large\frac{(6M)R^2}{2}$
$\therefore I= \large\frac{MR^2}{2}$
If an object is a symmetric part of a larger object, then its moment of inertia will have the same form as the large object.
answered Nov 26, 2013 by meena.p

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