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The moment of inertia of a uniform circular disc of radius $R$ and mass $M$ about an axis touching the disc at its diameter and normal to the disc is
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Verified Answer
The correct answer is:
$\frac{3}{2} M R^{2}$
The moment of inertia about an axis passing through centre of mass of disc and perpendicular to its plane is
$$
I_{\mathrm{CM}}=\frac{1}{2} M R^{2}
$$
Where $M$ is the mass of disc and $R$ its radius According to theorem of parallel axis, moment of inertia of circular disc about an axis
touching the disc at its diameter and normal to the disc is
$$
\begin{aligned}
I &=I_{\mathrm{CM}}+M R^{2} \\
&=\frac{1}{2} M R^{2}+M R^{2} \\
&=\frac{3}{2} M R^{2}
\end{aligned}
$$
$$
I_{\mathrm{CM}}=\frac{1}{2} M R^{2}
$$
Where $M$ is the mass of disc and $R$ its radius According to theorem of parallel axis, moment of inertia of circular disc about an axis
touching the disc at its diameter and normal to the disc is
$$
\begin{aligned}
I &=I_{\mathrm{CM}}+M R^{2} \\
&=\frac{1}{2} M R^{2}+M R^{2} \\
&=\frac{3}{2} M R^{2}
\end{aligned}
$$
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