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If the moment of inertia of a disc about an axis tangential and parallel to the surface is I.Then find the moment of inertia about the axis tangential but perpendicular to the surface in terms of the I?

\[\begin {array} {1 1} (a)\;\frac{6}{5}\;I & \quad (b)\;\frac{3}{4}\;I \\ (c)\;\frac{3}{2}\;I & \quad  (d)\;\frac{5}{4}\;I \end {array}\]

1 Answer

Parallel axis theorem:
$I= \large\frac{MR^2}{4}$$-MR^2$
$\quad= \large\frac{5}{4}$$MR^2$
$I= \large\frac{MR^2}{2}$$+MR^2$
$\quad= \large\frac{3}{2}$$MR^2$
$I'= \large\frac{5}{4} \times \frac{4}{5} \bigg(\frac{3}{2}\bigg)$$MR^2$
$\qquad= \large\frac{6}{5}$$I$
answered Nov 26, 2013 by meena.p

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