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

Question

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 5, 2 5, 2 7, 7 10

MARKS App · Accepted Answer

Under pure rolling condition: v = ω R Translational K . E = 1 2 m v 2 Rotational K . E = 1 2 × 2 5 m R 2 v R 2 = 1 5 m v 2 Total K . E = 1 2 m v 2 + 1 5 m v 2 = 7 10 m v 2 So, Rotational K · E Total K · E = 1 5 m v 2 7 10 m v 2 = 2 7

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