Q 11-06-161JEE MainJEE Main 2021 (22 Jul, Shift 1)Easy
The centre of a wheel rolling on a plane surface moves with a speed $v_0$. A particle on the rim of the wheel at the same level as the centre will be moving at a speed $\sqrt x\,v_0$. Then the value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 2
The particle has the translational velocity $v_0$ (horizontal) plus the rotational velocity $\omega R = v_0$ (vertical, perpendicular to the radius):
$$v = \sqrt{v_0^2 + v_0^2} = \sqrt2\,v_0 \Rightarrow x = 2$$
Solution by Sreeraj P, M.Sc Physics