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Question

# A pulley is rotated about its axis by a force F=(20t−5t2) N (where t is measured in seconds) applied tangentially. Find the magnitude of angular impulse [in Nm-s] on the pulley in the initial 2 seconds. Given radius of pulley =0.5 m.

A
10/3
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B
20/3
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C
40/3
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D
zero
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Solution

## The correct option is C 40/3As we know, Angular Impulse (J)=t2∫t1τ.dt where →τ=→d×→F where d is the perpendicular distance of force vector from the axis of rotation. i.e τ=F dsinθ=Fd (since force acts tangentially, θ=90∘) d=R=0.5 m Hence, J=2∫0(20t−5t2)×0.5dt J=[10t22−2.5t33]20 J=[5×4−2.5×83] Or, J=[20−203] Or, J=403 Nm-s

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