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Question

A wheel is rolling without slipping on ground along a straight line. If the axis of the wheel has a linear speed v, the instantaneous velocity of point P at an angle θ from vertical on the rim, relative to the ground will be



A
vcos(θ2)
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B
2vcos(θ2)
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C
v(1+sinθ)
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D
v(1+cosθ)
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Solution

The correct option is B 2vcos(θ2)
The velocity at point P on the wheel will be a vector sum of translational velocity of COM of wheel and tangential velocity of point P due to rotational motion.


In case of pure rolling, v=rω ....(i)
The angle between translational velocity (v) and tangential velocity (rω) of point P is θ as shown in figure.
|vP|=v2+(rω)2+2(v)(rω)cosθ ....(ii)
From Eq. (i) and (ii):
|vP|=r2ω2+(r2ω2)+2(r2ω2)cosθ
|vP|=rω2(1+cosθ)
1+cosθ=2cos2θ2
|vP|=rω2×2cos2θ2
|vP|=2rωcosθ2=2vcos(θ2)

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