JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Rotational Mechanics

A uniform sphere of mass 500 g rolls without slipping on a plane horizontal surface with its centre moving at a speed of

5.00 cm / s

. Its kinetic energy is:

Answer: (A)

Solution

A uniform solid sphere rolling without slipping stores both translational and rotational kinetic energy. With moment of inertia

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

and the rolling condition

ω = v / R

, the rotational part is

\frac{1}{2} I ω^{2} = \frac{1}{5} m v^{2}

, so the total is

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

. With

m = 0.5 kg

and

v = 5.00 cm s^{- 1} = 0.05 m s^{- 1}

, this gives

\frac{7}{10} \times 0.5 \times (0.05)^{2} = 8.75 \times 1 0^{- 4} J

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Single correct option