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Q.

A hoop rolls on a horizontal ground without slipping with linear speed v. Speed of a particle P on the circumference of the hoop at angle $θ$ is

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a

$vcos θ$

b

$2 vsin \frac{θ}{2}$

c

$2 vcos \frac{θ}{2}$

d

$vsin θ$

answer is A.

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Detailed Solution

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$v_{P} = \sqrt{v^{2} + v^{2} + 2 v^{2} \cos (180 - θ)} = v \sqrt{2 (1 - \cos θ)} = 2 vsin \frac{θ}{2}$

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