# Physics help?

A solid sphere (mass 9.14kg], radius 88 cm) is rolling without slipping on a horizontal table. What fraction of its total kinetic energy is translational kinetic energy?

### 2 Answers

- WhomeLv 72 months agoFavorite Answer
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- FiremanLv 72 months ago
Let the velocity of the solid sphere is v m/s

=>Angular velocity (ω) = v/r

Now KE(translational) = 1/2mv^2 = 0.5mv^2------------(i)

& KE (rotational) = 1/2Iω^2

{As I = 2/5mr^2 for solid sphere}

=>KE(rotational) = 1/2 x 2/5 x mr^2 x (v/r)^2

=>KE(rotational) = 1/5mv^2 = 0.20mv^2--------------(ii)

Thus KE (Total) = KE(translational) + KE (rotational)

=>KE(Total) = 0.5mv^2 + 0.2mv^2

=>KE(translational)= 0.7mv^2 -----------------------(iii)

Thus fraction of its total kinetic energy is translational kinetic energy by =>KE(translational)/KE(Total) = 0.5mv^2)/(0.7mv^2)

=>KE(translational)/KE(Total) = 0.71

=>KE(translational) = 0.71 x KE(Total)

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