A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The…

A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is
  1. 15
  2. 25
  3. 27
  4. 710

Solution

Under pure rolling condition: v=ωR

Translational K.E=12mv2

Rotational K.E=12×25mR2vR2=15mv2

Total K.E=12mv2+15mv2=710mv2

So, Rotational K·ETotal K·E=15mv2710mv2=27

Asked in: JEE Main 2022 (26 Jun Shift 2)

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