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

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 +vP (C) 2 v C -vP (D) 2(v C -vP)

Tardigrade Admin · Accepted Answer

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

ω

.
At point

B

,

v_{C} - ω r = v_{P}

\Rightarrow ω r = v_{C} - v_{P}

At point

A

,

v_{A} = v_{C} + ω r = 2 v_{C} - v_{P}

Q. 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

$2 v_{C}$

$2 v_{C} + v_{P}$

$2 v_{C} - v_{P}$

$2 (v_{C} - v_{P})$

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 $ω$ .
At point $B$ ,
$v_{C} - ω r = v_{P}$
$\Rightarrow ω r = v_{C} - v_{P}$
At point $A$ ,
$v_{A} = v_{C} + ω r = 2 v_{C} - v_{P}$

Q. 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

2vC​

2vC​+vP​

2vC​−vP​

2(vC​−vP​)