A solid sphere of mass 1 kg rolls without slipping on a plane surface. Its kinetic energy is $7 \times 10^{-3} \mathrm{~J}$. The speed of the centre of mass of the sphere is $\_\_\_\_$ $\mathrm{cm} \mathrm{s}^{-1}$.

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Rotational Mechanics

A solid sphere of mass 1 kg rolls without slipping on a plane surface. Its kinetic energy is

7 \times 1 0^{- 3} J

. The speed of the centre of mass of the sphere is

____

cm s^{- 1}

Answer: 10

Solution

For a solid sphere rolling without slipping the kinetic energy is shared between translation and rotation,

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

with

I = \frac{2}{5} M R^{2}

and

ω = v / R

, giving

K = \frac{1}{2} M v^{2} (1 + \frac{2}{5}) = \frac{7}{10} M v^{2}

. With

M = 1

kg,

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

, so

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

and

v = 0.1

m s

^{- 1} = 10

cm s

^{- 1}

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Integer answer

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