MHT CET · Physics · Rotational Motion
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
- A \(M R^{2}\)
- B \(\frac{2}{5} M R^{2}\)
- C \(\frac{3}{2} M R^{2}\)
- D \(\frac{1}{2} M R^{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{3}{2} M R^{2}\)
Step-by-step Solution
Detailed explanation
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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