A solid sphere of mass 2 kg is making pure rolling on a horizontal surface with kinetic energy 2240 J . The velocity of centre of mass of the sphere will be ______ m s - 1 .

Question

TestHub · Accepted Answer

Let velocity of centre of mass of sphere be $v$ .

Total kinetic energy of solid sphere is $K E = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$ , where, $I$ is moment of inertia of the solid sphere about the axis passing through it's centre of mass and $ω = \frac{v}{R} (pure rolling condition)$ .

Therefore,

$2240 = \frac{1}{2} 2 (v)^{2} + \frac{1}{2} [\frac{2}{5} (2) R^{2}] {(\frac{v}{R})}^{2}$

$\Rightarrow 2240 = v^{2} + \frac{2}{5} v^{2}$

$\Rightarrow v = 40 m s^{- 1}$ .

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