Question

# A disc of radius R is rolling (without slipping) on a frictionless surface as shown in figure. C is it's centre and Q and P are two points equidistant from C. Let vP,vQ and vC be the magnitudes of velocities of points P, Q and C respectively, then

A
vQ>vP>vC

B
vQ<vC<vP

C
vQ=vP, vC=12vP

D
vQ>vC>vP

Solution

## The correct option is C vQ>vC>vP  Points P and Q are equidistant from C, then CQ=CP=r Let ω be the angular speed of disc. For pure rolling, v=Rω, where v is the translational velocity of COM of the disc. ⇒Net velocity at any point on the disc will be  due to vector sum of translational velocity and tangential velocity due to rotational motion. ⇒→vQ will be greater in magnitude compared to →vP, since taking resultant of velocities, θ<90∘,  cosθ is +ve hence component vectors will add up together at point Q ⇒ At point P, angle between velocity vectors is θ>90∘,  cosθ is −ve so component vectors will get substracted at point P to give resultant →vP.  and at point C,  |→vC|=v. Order of magnitude of velocity, ∴vQ>vC>vPPhysics

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