Question

A solid sphere of mass m and radius r rolls on a horizontal surface without slipping. The ratio of rotational kinetic energy to total kinetic energy is -

Answer 1

The ratio of rotational kinetic energy to total kinetic energy is 2/7

explanation : rotation kinetic energy , K.E_(rotation) = 1/2 Iω²

here, sphere is rolling on a horizontal surface without slipping.

so, it follows v = ωr

then, K.E_(rotation) = 1/2 I(v/r)²

now moment of inertia of sphere , I = 2/5 mr²

so, K.E_(rotation) = 1/2 (2/5 mr²) (v/r)²

= 1/5 mr²

now translational kinetic energy, K.E_(trans) = 1/2 mv² .

so, total kinetic energy , K.E_(total) = K.E_(trans) + K.E_(rotation)

= 1/2 mv² + 1/5 mv²

= 7/10 mv²

now, ratio of rotational kinetic energy to total kinetic energy = K.E_(rotation)/K.E_(total) = (1/5 mv²)/(7/10 mv²)

= 2/7

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Answer 2

$\huge\bold\purple{Answer:-}$

rotation kinetic energy , K.E_(rotation) = 1/2 Iω²

here, sphere is rolling on a horizontal surface without slipping.

so, it follows v = ωr

then, K.E_(rotation) = 1/2 I(v/r)²

now moment of inertia of sphere , I = 2/5 mr²

so, K.E_(rotation) = 1/2 (2/5 mr²) (v/r)²

= 1/5 mr²

now translational kinetic energy, K.E_(trans) = 1/2 mv² .

so, total kinetic energy , K.E_(total) = K.E_(trans) + K.E_(rotation)

= 1/2 mv² + 1/5 mv²

= 7/10 mv²

now, ratio of rotational kinetic energy to total kinetic energy = K.E_(rotation)/K.E_(total) = (1/5 mv²)/(7/10 mv²)

= 2/7