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The moment of inertia of a uniform circular disc of radius $R$ and mass $M$ about an axis passing from the edge of the disc and normal to the disc is
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The correct answer is:
$\frac{3}{2} M R^2$
We should use parallel axis theorem.
Moment of inertia of disc passing through its centre of gravity and perpendicular to its plane is
$I_{A B}=\frac{1}{2} M R^2$
Using theorem of parallel axes, we have
$\begin{aligned} I_{C D} & =I_{A B}+M R^2 \\ & =\frac{1}{2} M R^2+M R^2 \\ & =\frac{3}{2} M R^2\end{aligned}$
NOTE : The role of moment of inertia in the study of rotational motion is analogous to that of mass in study of linear motion.
Moment of inertia of disc passing through its centre of gravity and perpendicular to its plane is
$I_{A B}=\frac{1}{2} M R^2$
Using theorem of parallel axes, we have
$\begin{aligned} I_{C D} & =I_{A B}+M R^2 \\ & =\frac{1}{2} M R^2+M R^2 \\ & =\frac{3}{2} M R^2\end{aligned}$
NOTE : The role of moment of inertia in the study of rotational motion is analogous to that of mass in study of linear motion.
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