Question

The centre of a wheel rolling on a plane surface moves with a speed ${\nu }_0$. A particle on the rim of the wheel at the same level as the centre will be moving at speed
(a) zero

(b) ${\nu }_0$

(c) $\sqrt{2}{\nu }_0$

(d) $2{\nu }_0$.

Answer 1

(c) $\sqrt{2}{\nu }_0$

For pure rolling, $\omega r=v_0$



As shown in the figure, the velocity of the particle will be the resultant of v₀ and ωr.
Therefore, we have:
$$\begin{gathered} v_{net}=\sqrt{{v_0}^2+{\left(\omega r\right)}^2} \\ {v}_{net}=\sqrt{{v_0}^2+{v_0}^2} \\ {v}_{net}=\sqrt{2}{v}_0 \end{gathered}$$