A solid sphere of mass \(\mathrm{m}\) and radius \(\mathrm{R}\) is gently placed on a conveyer belt moving…
A solid sphere of mass \(\mathrm{m}\) and radius \(\mathrm{R}\) is gently placed on a conveyer belt moving with constant velocity \(\mathrm{v}_{0}\). If coefficient of friction between belt and sphere is \(2 / 7\) the distance traveled by the centre of the sphere before it starts pure rolling is
\(\frac{\mathrm{v}_{0}^{2}}{7 \mathrm{~g}}\)
\(\frac{2 v_{0}^{2}}{49 g}\)
\(\frac{2 \mathrm{v}_{0}^{2}}{5 \mathrm{~g}}\)
\(\frac{2 \mathrm{v}_{0}^{2}}{7 \mathrm{~g}}\)
Solution
$\begin{aligned}
v_{p} &= v + R \omega = v_{0} \text{ or, at } + R(\alpha t) = v_{0} \\
\therefore \left(\frac{2}{7} gt\right) + \frac{5}{7} gt &= v_{0} \\
\therefore t &= \frac{v_{0}}{g} \\
S &= \frac{1}{2} at^{2} = \frac{1}{2}\left(\frac{2}{7} g\right)\left(\frac{v_{0}}{g}\right)^{2} = \frac{v_{0}^{2}}{7 g}
\end{aligned}$
$\begin{aligned}
a &= \frac{\mu mg}{m} = \mu g = \frac{2}{7} g \\
\alpha &= \frac{\mu mg / R}{\frac{2}{5} mR^{2}} = \frac{5 \mu g}{2 R}
\end{aligned}$
\(\frac{5}{7} \frac{g}{R}\) Pure rolling will start when,
