Question

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.

Answer 1

For no slipping $v_0=\omega R$

Now, $v_A=v_B=\sqrt{v_0^2+(\omega R{)}^2}$

$=\sqrt{2}{v}_0$

Given, $v_A=v_B=\sqrt{x}{v}_0$

$\Rightarrow x=2$