Q 11-06-102JEE MainJEE Main 2023 (29 Jan, Shift 1)Easy
A solid sphere of mass $2\ \text{kg}$ is making pure rolling on a horizontal surface with kinetic energy $2240\ \text{J}$. The velocity of centre of mass of the sphere will be ______ $\text{m s}^{-1}$.
Numerical value type. Enter your answer.
Answer: 40
For rolling, $K=\tfrac12mv^2\left(1+\dfrac{k^2}{R^2}\right)=\tfrac12mv^2\left(1+\tfrac25\right)=\tfrac{7}{10}mv^2$.
$$2240=\frac{7}{10}\times2\times v^2\ \Rightarrow\ v^2=1600\ \Rightarrow\ v=40\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics