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Q.
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_______.
JEE MainJEE Main 2021System of Particles and Rotational Motion
Solution:
For no slipping $v_{0}=\omega R$
Now $v_{A}=v_{B}=\sqrt{v_{0}^{2}+(\omega R)^{2}}$
$=\sqrt{2} v_{0}$
$\Rightarrow x=2$
