Question

The velocities are in ground frame and the cylinder is performing pure rolling on the plank, velocity of point 'A' would be
a) 2Vc
b) 2Vc- Vp
C) 2Vc+ Vp
D) 2(Vc-Vp)​

Answer 1

Answer:

2Vc-Vp

Explanation:

Velocity of the center is due to translation with velocity Vp and rotation about the instantaneous axis through the point in common(to the ground and cylinder).

Let the angular velocity about this instantaneous axis of rotation be omega (w)

Hence,

$V_c=V_p+ \omega R\\</p><p>\text{Where R is the radius of cylinder}\\</p><p>V_a=V_p+ \omega (2R)\\</p><p>V_a= V_p+ 2(V_c-2V_p)\\</p><p>\boxed{V_a=2V_c-V_p}$