The correct option is
A vQ>vC>vPLet the linear velocity of C.O.M be
v and angular velocity of the disc about an axis passing through C.O.M be
ω.
In pure rolling, velocity of any point is the resultant of translational velocity and the tangential velocity due to rotation about C.O.M,
→v=→vc.o.m+→ωc.o.m×→r
i.e
→v=→v1+→v2 ...(1)
[→vc.o.m=→v1 and →ωc.o.m×→r=→v2]
where,
v1 & v2 are representing the translational velocity and tangential velocity( due to rotation) for points.
⇒As the points
P and
Q are equidistant from centre.
(v2=ωr) for both points
P and
Q
For point
Q,
θQ<90∘
∴cosθ is +ve ...(2)
For point
P,
QP>90∘
∴cos θ is −ve ...(3)
For point
C, tangential velocity due to rotation is zero about
C.
⇒v2=0, v1=v
Hence, net velocity of point
C,
vC=v1=v
From Eq.
(1) for vectorial addition, we can see that:
(a). The component of tangential velocity
(ωr) will be opposite to translational velocity
(v) in case of point
P.
and
(b). The component of tangential velocity
(ωr) will be in the direction of translational velocity
(v) in case of point
Q.
Hence the resultant velocity will be in the order,
vQ>vC>vP