Question:

A solid sphere is rolling without slipping on a horizontal surface. The ratio of its rotational kinetic energy and translational kinetic energy is:

Updated On: Jul 28, 2022
  • $ \frac{2}{9} $
  • $ \frac{2}{5} $
  • $ \frac{2}{7} $
  • $ \frac{7}{2} $
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The Correct Option is B

Solution and Explanation

Moment of inertia of sphere $I=\frac{2}{5} M R^{2}$ ...(1) and for pure rolling $v=R \omega$...(2) $\frac{\text { Rotational KE }}{\text { Translational } KE }=\frac{\frac{1}{2} I \omega^{2}}{\frac{1}{2} m v^{2}}$ from (1) and (2) $=\frac{\frac{1}{2} \times \frac{2}{5} M R^{2} \omega^{2}}{\frac{1}{2} M R^{2} \omega^{2}}=\frac{2}{5}$
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Concepts Used:

System of Particles and Rotational Motion

  1. The system of particles refers to the extended body which is considered a rigid body most of the time for simple or easy understanding. A rigid body is a body with a perfectly definite and unchangeable shape.
  2. The distance between the pair of particles in such a body does not replace or alter. Rotational motion can be described as the motion of a rigid body originates in such a manner that all of its particles move in a circle about an axis with a common angular velocity.
  3. The few common examples of rotational motion are the motion of the blade of a windmill and periodic motion.