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A solid sphere of radius $\mathrm{R}$ is set into motion on a rough horizontal surface with a linear speed $v_0$ in forward direction and an angular velocity $\omega_0=v_0 / 2 R$ in counter coefficient of friction is $\mu$, then find the time after which sphere starts pure rolling.
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The correct answer is:
$t_o=\frac{3 v_o}{7 \mu g}$
As the sphere is slipping, kinetic friction will act,
$\begin{aligned}& f_k=\mu m g, a_{c m}=\frac{f_k}{m}=\mu g, \\& a=\frac{f_k R}{I_{c m}}=\frac{\mu m g R}{\frac{2}{5} m R^2}=\frac{5 \mu g}{2 R}\end{aligned}$
when the sphere starts pure rolling, velocity $v$ of com will be $\omega R$.
$\begin{aligned}& v=v_0-\mu g \mathrm{t} \\& \omega=-\omega_0+a t\end{aligned}$
for pure rolling,
$\begin{aligned}& v=\omega R \\& \Rightarrow v_0-\mu g t=-\left(-\omega_0+a t\right) R \\& \Rightarrow v_0-\mu g t=\left(\frac{-v_0}{2 R}+\frac{5 \mu g}{2 R}\right) R \\& \Rightarrow \frac{3 v_0}{2}=\frac{7}{2} \mu g t \Rightarrow t_0=\frac{3 v_0}{7 \mu g}\end{aligned}$
$\begin{aligned}& f_k=\mu m g, a_{c m}=\frac{f_k}{m}=\mu g, \\& a=\frac{f_k R}{I_{c m}}=\frac{\mu m g R}{\frac{2}{5} m R^2}=\frac{5 \mu g}{2 R}\end{aligned}$
when the sphere starts pure rolling, velocity $v$ of com will be $\omega R$.
$\begin{aligned}& v=v_0-\mu g \mathrm{t} \\& \omega=-\omega_0+a t\end{aligned}$
for pure rolling,
$\begin{aligned}& v=\omega R \\& \Rightarrow v_0-\mu g t=-\left(-\omega_0+a t\right) R \\& \Rightarrow v_0-\mu g t=\left(\frac{-v_0}{2 R}+\frac{5 \mu g}{2 R}\right) R \\& \Rightarrow \frac{3 v_0}{2}=\frac{7}{2} \mu g t \Rightarrow t_0=\frac{3 v_0}{7 \mu g}\end{aligned}$
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