Question bank/JEE Main/Physics

A solid sphere of mass m and radius r is gently placed on a
conveyer belt moving with constant velocityV. If the coefficient of
friction between the belt and sphere is$\frac{2}{7}$, the distance
travelled by the centre of the sphere before it starts pure rolling is.

A$\frac{v^{2}}{7g}$B$\frac{2V^{2}}{49g}$C$\frac{2V^{2}}{5g}$D$\frac{2V^{2}}{7g}$

Show answer and solution▾

Initial situation w.r.t. the conveyer belt:
$\mu mgR = \frac{2}{5}{mR}^{2} \cdot \alpha$ $\Rightarrow$
$\alpha = \frac{5}{7} \cdot \frac{g}{R}$
At pure rolling:
$v^{'} = v - (\mu g)t$and $\omega^{'} = 0 + \left( \frac{5}{7} \cdot \frac{g}{R} \right)t$
(Since at pure rolling)
$\Rightarrow$ $v - \mu gt = \frac{5}{7} \cdot g \cdot t$
$\Rightarrow$ $t = \frac{v}{g}\left( \because\mu = \frac{2}{7} \right)$
For centre of the sphere:$S = v\left( \frac{v}{g} \right) - \frac{1}{2} \cdot \left( \frac{2}{7} \cdot g \right)\left( \frac{v^{2}}{g^{2}} \right)$
$S = \frac{v^{2}}{g} - \frac{v^{2}}{7g}$ (w.r.t. belt)
In this time, the belt with speed will move a distance
of$s_{b} = vt = v\left( \frac{v}{g} \right) = \frac{v^{2}}{g}$
[IMAGE] Distance travelled w.r.t. ground will be:$S_{g} = \left| S - S_{b} \right| = \left| \left( \frac{v^{2}}{g} - \frac{v^{2}}{7g} \right) - \left( \frac{v^{2}}{g} \right) \right| = \frac{v^{2}}{7g}$

Source: Grass-G / P-23 Physcis Paper 31 Dec FST.docx