Question 456: Centre of Mass and Rotational Motion

Question

The centre of a wheel rolling on a plane surface moves with a speed v0. 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 _____________.

Accepted Answer

$$\left| \omega  \right| = {{{v_0}} \over R}$$
$${\overrightarrow v _p} = {v_0}\widehat i + \omega R( - \widehat j) = {v_0}\widehat i - {v_0}\widehat j$$
$$\left| {{{\overrightarrow v }_p}} \right| = \sqrt 2 {v_0}$$
$$x = 02$$

Centre of Mass and Rotational Motion

The centre of a wheel rolling on a plane surface moves with a speed v0. 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 _____________.

Choose the correct answer: