Question

A solid sphere is rolling on a frictionless surface, shown in figure with a translational velocity $v m / s$. If it is to climb the inclined surface then $v$ should be:

• A. $\geq \sqrt{\frac{10}{7} g h}$
• B. $\geq \sqrt{2 g h}$
• C. $2 g h$
• D. $\frac{10}{7} g h$

Answer 1

Answer: (A)

Kinetic energy is converted to potential energy.
From law of conservation of energy, energy can neither be created nor destroyed but it remains conserved. In the given case the sum of kinetic energy of rotation and translation is converted to potential energy.

$\Rightarrow \frac{1}{2} m v^2+\frac{1}{2}\left(\frac{2}{5} M R^2\right) \frac{v^2}{R^2}=m g h$
where $\nu=R \omega, \omega=$ angular velocity
$ \Rightarrow \frac{7}{10} m v^2=m g h$
$ \Rightarrow v=\sqrt{\frac{10}{7}} g h$
Hence, to climb the inclined surface velocity should be greater than $\sqrt{\frac{10}{7} g h}$.