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Moment of inertia of a semicircular ring about an axis perpendicular to its plane and passing through its centre of mass. (mass M,Radius R)

 

\[\begin {array} {1 1} (a)\;MR^2+M \bigg(\frac{2R}{\pi}\bigg)^2 & \quad (b)\;MR^2 \bigg[1- \frac{4}{\pi}\bigg] \\ (c)\;MR^2 & \quad  (d)\;\frac{MR^2}{2} \end {array}\]

1 Answer

$I_0=I_{CM}+Md^2$
$d= \large\frac{2R}{\pi}$
$I_{CM}=I_0 - M \bigg(\large\frac{2R}{\pi}\bigg)$
$\qquad=MR^2 \bigg[1- \frac{4}{\pi}\bigg]$
answered Nov 26, 2013 by meena.p
 

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