Q118 / 180 · JIPMER 2015 · Physics

A solid sphere is set into motion on a rough horizontal surface with a linear speed $v$ in the forward direction and an angular speed $\frac{V}{R}$ in the anticlockwise direction as shown in figure. Find the linear speed of the sphere when it stops rotating and $\omega=\frac{V}{R}$. Figure

  • Correct answer
    A
    $\frac{3 v}{5}$
  • B
    $\frac{2 v}{5}$
  • C
    $\frac{4 v}{3}$
  • D
    $\frac{7 v}{3}$

Solution

If we take moment at $A$, then extemal torque will be zero. Therefore, the initial angular momentum = the angular momentum after rotation stop (i.e. only linear velocity exist) $\begin{aligned} \Rightarrow & M v \times R-\left(\frac{2}{5}\right) I \omega & =M v_0 \times R \\ \Rightarrow & M v R-\frac{2}{5} M R^2 \times \frac{V}{R} & =M v_0 R \\ \Rightarrow & v_0 & =\frac{3 v}{5}\end{aligned}$

Hence, the correct answer is option: 'A'.

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