A tangential force $F$ acts at the top of a thin spherical shell of mass $m$ and radius $R$. The acceleration of the shell if it rolls without slipping is

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Torque due to the force $F$ on a thin spherical shell, $ \begin{aligned} \tau & =r \times F \\ & =2 R F \sin 90^{\circ}=2 R F\left[\because \sin 90^{\circ}=1\right] \end{aligned} $

Where, $I$ is moment of inertia of a thin spherical shell. From parallel axes's theorem, moment of inertia of spherical shell, $ I=I_{\mathrm{cm}}+M r^2 $ or $ I=\frac{2}{3} M R^2+M R^2 \quad(\because r=R) $ From Eqs. (i), we get $ \begin{gathered} \alpha=\frac{2 R F}{\frac{2}{3} M R^2+M R^2} \\ =\left(\frac{6}{5}\right) \frac{F}{R M} \end{gathered} $ Hence, the tangential acceleration, $a_T=R \alpha$ or $ a_T=\left(\frac{6}{5}\right) \frac{F}{R M} \times R=\frac{6}{5} \frac{F}{M} $

A tangential force $F$ acts at the top of a thin spherical shell of mass $m$ and radius $R$. The…

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