Q. A solid sphere is rolling on a frictionless surface, shown in figure with a transnational velocity $v\ m/s$. If sphere climbs up to height $h$ then value of $v$ should be
A) $\geq \sqrt{\frac{10}{7}gh}$
B) $\geq \sqrt{2gh}$
C) $2gh$
D) $\frac{10}{7}gh$
Explanation
(b) As body is moving on a frictionless surface. Its mechanical energy is conserved. When body climbes up the inclined plane it keeps on rotating with same angular speed, as no friction force is present to provide retarding torque so
$\frac{1}{2}I\omega^{2} + \frac{1}{2}mv^{2} \geq \frac{1}{2}I\omega^{2} + mgh \Rightarrow v \geq \sqrt{2\ gh}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.