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 ___.

For a wheel of radius $$R$$ rolling without slipping on a flat surface, the centre of the wheel moves with speed $$v_0$$. The rolling condition gives the angular velocity as $$\omega = v_0/R$$, directed (for rightward motion) clockwise, i.e. $$\vec{\omega} = -\dfrac{v_0}{R}\hat{k}$$.

A particle on the rim at the same height as the centre is positioned either to the left or right of the centre. Consider the particle at position $$\vec{r} = R\hat{i}$$ relative to the centre (i.e., at the rightmost point of the wheel, at the same height as the centre).

The velocity of this particle is the sum of the centre-of-mass velocity and the rotational velocity:

$$\vec{v} = v_0\hat{i} + \vec{\omega} \times \vec{r} = v_0\hat{i} + \left(-\frac{v_0}{R}\hat{k}\right) \times (R\hat{i}) = v_0\hat{i} - v_0(\hat{k}\times\hat{i}) = v_0\hat{i} - v_0\hat{j}$$

The speed is:

$$|\vec{v}| = \sqrt{v_0^2 + v_0^2} = v_0\sqrt{2} = \sqrt{2}\,v_0$$

Comparing with the given form $$\sqrt{x}\,v_0$$, the value of $$x$$ is $$2$$.