Q92 / 180 · AIIMS 2005 · Physics
A solid sphere is rolling on a frictionless surface, shown
in figure with a translational velocity $v \mathrm{~m} / \mathrm{s}$. If it is to climb the inclined surface then $v$ should be
- Correct answerA$\geq \sqrt{10 / 7 g h}$
- B$\geq \sqrt{2 g h}$
- C$2 g h$
- D$10 / 7 g h$
Solution
From conservation of energy
Potential energy $=$ translational kinetic energy + rotational kinetic energy
$$
\begin{array}{r}
m g h=\frac{1}{2} m v^2+\frac{1}{2}\left(\frac{2}{5}\right) m R^2 \cdot \frac{v^2}{R^2} \\
\Rightarrow \quad \frac{7}{10} m v^2=m g h \Rightarrow v \geq \sqrt{\frac{10}{7} g h}
\end{array}
$$
Hence, the correct answer is option: 'A'.