A solid sphere of mass 1 kg rolls without slipping on a plane surface. Its kinetic energy is 7 × 1 0 − 3 J. The speed of the center of mass of the sphere in cm s − 1 is

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Total kinetic energy of a solid sphere rolling without slipping is the sum of its translational and rotational kinetic energies.

$K = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$

For a solid sphere, $I = \frac{2}{5} m r^{2}$ and for pure rolling, $ω = \frac{v}{r}$ .

$K = \frac{1}{2} m v^{2} + \frac{1}{2} (\frac{2}{5} m r^{2}) (\frac{v}{r})^{2}$

$K = \frac{1}{2} m v^{2} + \frac{1}{5} m v^{2} = \frac{7}{10} m v^{2}$

Given $K = 7 \times 1 0^{- 3}$ J and $m = 1$ kg.

$7 \times 1 0^{- 3} = \frac{7}{10} \times 1 \times v^{2}$

$v^{2} = 1 0^{- 2}$

$v = 0.1$ m s $^{- 1}$

$v = 10$ cm s $^{- 1}$

Answer: $10$

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