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System of Particles and Rotational Motion
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
by
meena.p
retagged
Jul 10, 2014
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1 Answer
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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