Question

The centre of a disc rolling without slipping on a plane surface moves with speed u. Find the speed of the particle at the lower half of the rim making an angle ${60}^0$.

Answer 1

velocity of centre, $V_{cm}=u\hat{i}$

tangential velocity at $P,V_t=R\overrightarrow{w}$

also in pure rolling , $V_{cm}=R_w,\Rightarrow u=R_w$

From, figure,

angle between $V_{t}K{V}_{cm}$ at $P={30}^0$

So,

$|{\overrightarrow{V}}_p|=\sqrt{V_{c{m}^2}+V_{t^2}+2V_{cm}.V_t\cos {30}^0}$

$=\sqrt{(RW{)}^2+(RW{)}^2+2(RW{)}^2\cos {30}^0}$

$\Rightarrow |{\overrightarrow{V}}_p|R_w\left(\sqrt{2+\sqrt{3}}\right)=u\left(\sqrt{2+\sqrt{3}}\right)$