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A solid cylinder of mass \(M\) and radius \(R\) rolls on a flat surface. Its moment of inertia about the line of contact is
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The correct answer is:
\(\left(\frac{3}{2}\right) M R^2\)
Moment of inertia of solid cylinder of radius \(R\) and mass \(M\) about its axis is given as
\(I_{\mathrm{CM}}=\frac{1}{2} M R^2\) ...(i)
Moment of inertia about contact line \(A B\) is given
According to parallel axes theorem, i.e. \(\quad I_{A B}=I_{\mathrm{CM}}+M R^2=\frac{1}{2} M R^2+M R^2=\frac{3}{2} M R^2\)
\(I_{\mathrm{CM}}=\frac{1}{2} M R^2\) ...(i)
Moment of inertia about contact line \(A B\) is given
According to parallel axes theorem, i.e. \(\quad I_{A B}=I_{\mathrm{CM}}+M R^2=\frac{1}{2} M R^2+M R^2=\frac{3}{2} M R^2\)
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