A solid sphere of mass $M$ and radius $R$ is attached to a spring of negligible mass kept on a horizontal plane such that it can roll without slipping. The sphere is made to execute SHM by stretching through a distance and released, then the time period of such oscillation is $(\mathrm{K}=$ spring…

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A solid sphere of mass $M$ and radius $R$ is attached to a spring of negligible mass kept on a horizontal plane such that it can roll without slipping. The sphere is made to execute SHM by stretching through a distance and released, then the time period of such oscillation is $(\mathrm{K}=$ spring constant $)$

Aicharya · Accepted Answer

Total energy $=\mathrm{E}=\frac{1}{2} \mathrm{mV} V^2+\frac{1}{2} \mathrm{I} \omega^2+\frac{1}{2} \mathrm{Kx}^2$ $\begin{aligned} & =\frac{1}{2} \mathrm{mV}^2+\frac{1}{2}\left(\frac{2}{5} \mathrm{mR}^2ight) \frac{\mathrm{V}^2}{\mathrm{R}^2}+\frac{1}{2} \mathrm{Kx}^2 \ & =\frac{7}{10} \mathrm{mV}^2+\frac{1}{2} \mathrm{Kx}^2\end{aligned}$ As $\mathrm{E}=$ constant $\begin{aligned} & \Rightarrow \frac{\mathrm{dE}}{\mathrm{dt}}=0 \ & \Rightarrow \frac{7}{10} \mathrm{~m}(2 \mathrm{~V}) \frac{\mathrm{dV}}{\mathrm{dt}}+\frac{1}{2} \mathrm{~K}(2 \mathrm{x}) \frac{\mathrm{dx}}{\mathrm{dt}}=0 \ & \Rightarrow \frac{7}{5} \mathrm{mVa}+\mathrm{KVx}=0 \ & \Rightarrow \frac{7}{5} \mathrm{ma}+\mathrm{Kx}=0 \ & \Rightarrow \mathrm{a}=\frac{-5}{7} \frac{\mathrm{K}}{\mathrm{m}} \mathrm{x}\end{aligned}$ In SHM, $a=-\omega^2 x$ So, $\omega=\sqrt{\frac{5 \mathrm{~K}}{7 \mathrm{~m}}}$ Hence, $\mathrm{T}=2 \pi \sqrt{\frac{7 \mathrm{~m}}{5 \mathrm{~K}}}$

A solid sphere of mass $M$ and radius $R$ is attached to a spring of negligible mass kept on a horizontal…

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