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A tangential force of F = 1.5 N acts on ...

A tangential force of F = 1.5 N acts on a particle of mass m = 2 kg revolving in a circular path of radius r = 3 m. What is the work done by the torque for a complete revolution of the particle ?

Text Solution

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Work done by the torque `tau` is given by
`W = int tau d theta ` where `tau = r F`
`W = tau F int_(0)^(theta) d 0`
Putting `theta = 2 pi` for a complete revolution we obtain
`W = 2 pi r F`
`rArr" "W = 2 pi (3)(1.5) = 9 pi J`
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