I_{CM} is the moment of inertia of a circular | vedclass

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Question

(D) The moment of inertia of a circular disc about an axis passing through its center and perpendicular to its plane is given by $I_{CM} = \frac{1}{2}

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(D) The moment of inertia of a circular disc about an axis passing through its center and perpendicular to its plane is given by $I_{CM} = \frac{1}{2}MR^2$.According to the parallel axis theorem,the moment of inertia about an axis parallel to the center of mass axis is $I = I_{CM} + Md^2$,where $d$ is the distance between the axes.Here,$d = \frac{2}{3}R$.Therefore,$I_{AB} = \frac{1}{2}MR^2 + M\left(\frac{2}{3}Right)^2$.$I_{AB} = \frac{1}{2}MR^2 + M\left(\frac{4}{9}R^2ight) = \left(\frac{1}{2} + \frac{4}{9}ight)MR^2 = \left(\frac{9+8}{18}ight)MR^2 = \frac{17}{18}MR^2$.Now,the ratio $\frac{I_{AB}}{I_{CM}} = \frac{\frac{17}{18}MR^2}{\frac{1}{2}MR^2} = \frac{17}{18} 	imes 2 = \frac{17}{9}$.Given the ratio is $x:9$,we have $x = 17$.

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