Q104 / 160 · JIPMER 2006 · Physics

If a spherical ball rolls on a table without slipping the friction of its total energy associated with rotational energy is :

  • A
    $3 / 5$
  • Correct answer
    B
    $2 / 7$
  • C
    $2 / 5$
  • D
    $3 / 7$

Solution

Rotational energy of sphere $E_R=\frac{1}{2} I \omega^2$ For sphere, moment of inertia $\begin{aligned} I & =\frac{2}{5} m R^2 \\ \therefore \quad E_R & =\frac{1}{2}\left(\frac{2}{5} m R^2\right)\left(\frac{v}{R}\right)^2 \\ & =\frac{1}{5} m v^2\end{aligned}$ Translational kinetic energy $E_r=\frac{1}{2} m v^2$ $\begin{aligned} \therefore \text { Total energy } & =\frac{1}{5} m v^2+\frac{1}{2} m v^2 \\ & =\frac{7}{10} m v^2\end{aligned}$ $\therefore$ Required fraction $=\frac{\frac{1}{5} m v^2}{\frac{7}{10} m v^2}$ $=\frac{2}{7}$

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

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