Question

In the figure, the velocities are in ground frame and the cylinder is performing pure rolling on the plank, velocity of point ' $A$ ' would be

• A. $2 v_{ C }$
• B. $2 v_{ C }+v_{P}$
• C. $2 v_{ C }-v_{P}$
• D. $2\left(v_{ C }-v_{P}\right)$

Answer 1

Answer: (C)

For pure rolling velocity of the point of contact has to be equal to the velocity of the surface.

Let us say cylinder rolls with angular velocity $\omega$.
At point $B$,
$v_{C}-\omega r=v_{P} $
$\Rightarrow \omega r=v_{C}-v_{P}$
At point $A$,
$v_{A}=v_{C}+\omega r=2 v_{C}-v_{P}$