Question

The center of a disc rolling without slipping on a plane surface moves with speed u. A particle, on the lower half of the rim making an angle ${60}^{\circ }$ with vertical, will be moving at speed:

• A. Zero
• B. u
• C. $\sqrt{2u}$
• D. 2u

Answer 1

The correct option is B u

Disc is rolling without slipping so $\omega r=u$

Velocity diagram is shown in figure , angle made by speed is 120 because $\theta ={60}^o$ is given,

Resultant speed $u_R=\sqrt{u^2+(\omega r{)}^2+2\times u\times \omega rcos({120}^0)}$,

$\Rightarrow u_R=\sqrt{u^2+u^2-u^2}$, ( $\because \omega r=u$ )

$\Rightarrow u_R=u$