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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
 
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B
vQ<vC<vP
 
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C
vQ=vP, vC=12vP
 
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D
vQ>vC>vP
 
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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>vP

Physics

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