Question

A wheel performs pure rolling on the ground. Find a point on the periphery of the body which has a velocity equal to the velocity of the centre of mass of the body.

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $\frac{2}{3}\pi$

Answer 1

The correct option is D $\frac{2}{3}\pi$

Given, pure rolling, thus $v_{\text{cm}}=\omega R$
Consider a point $P$ on the periphery of the body as shown. Assume velocity of point $P$ is $v_P$.


Then,
$v_P^2=v_{\text{cm}}^2+(\omega R{)}^2+2v_{\text{cm}}(\omega R)\cos \theta$]
[$\because v_{\text{cm}}=\omega R$ and $v_P=v_{\text{cm}}$ given]
$\Rightarrow v_{\text{cm}}^2=v_{\text{cm}}^2+v_{\text{cm}}^2+2v_{\text{cm}}^2\cos \theta$
$\Rightarrow \cos \theta =-\frac{1}{2}$
$\Rightarrow \theta ={120}^{\circ }=\frac{2\pi }{3}$