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System of Particles and Rotational Motion
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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?
\[\begin {array} {1 1} (a)\;\frac{6}{5}\;I & \quad (b)\;\frac{3}{4}\;I \\ (c)\;\frac{3}{2}\;I & \quad (d)\;\frac{5}{4}\;I \end {array}\]
jeemain
physics
class11
unit5
rotational-motion
moment-of-inertia
easy
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asked
Nov 25, 2013
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meena.p
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Jul 10, 2014
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1 Answer
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Parallel axis theorem:
http://clay6.com/mpaimg/jee%205%20n%2018.JPG
$I= \large\frac{MR^2}{4}$$-MR^2$
$\quad= \large\frac{5}{4}$$MR^2$
http://clay6.com/mpaimg/1_jee%205%20n%2018-1.JPG
$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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