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Rotational Dynamics — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsRotational Dynamics3 Marks
Question
State the expression for the Ml of a thin spherical shell (i.e., a thin-walled hollow sphere) about its diameter. Hence obtain the expression for its $\mathrm{Ml}$ about a tangent.
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Answer

Consider a uniform, thin-walled hollow sphere radius $R$ and mass M. An axis along its diameter is an axis of spherical symmetry through its centre of mass. The $\mathrm{Ml}$ of the thin spherical shell about its diameter is
$
\mathrm{I}_{\mathrm{CM}}=\frac{2}{3} M R^2
$
Let I be its $\mathrm{Ml}$ about a tangent parallel to the diameter. Here, $\mathrm{h}=\mathrm{R}=$ distance between the two axes. Then, according to the theorem of parallel axis,
$
\begin{aligned}
I & =I_{\mathrm{CM}}+M h^2 \\
& =\frac{2}{3} M R^2+M R^2=\frac{5}{3} M R^2
\end{aligned}
$
[Note : The corresponding radii of gyration are
$
k_{\mathrm{CM}}=\sqrt{\frac{I_{\mathrm{CM}}}{M}}=\sqrt{\frac{2}{3}} R \simeq 0.8165 R \text { and }
$
$
\left.k=\sqrt{\frac{1}{M}}=\sqrt{\frac{5}{3}} R \simeq 1.291 R\right]
$

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