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Question

A thin hoop initially at rest with a mass of 2.5kg and a radius of 2.5m is subject to a net torque τ(t)=500t+50, where τ is measured in Nm.
Which of the following best represents the angular velocity of the cylinder after 3 secods?

A
465rad/s
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B
144rad/s
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C
154.6rad/s
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D
2250rad/s
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E
2400rad/s
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Solution

The correct option is C 154.6rad/s
Given : m=2.5 kg r=2.5 m
Moment of inertia of the hoop I=mr2=2.5×(2.5)2=15.625 kgm2
Using τ=Iα
α=τI=500t+5015.625
Integrating w.r.t time we get angular velocity w(t)=αdt
w(t)=115.625×(500t+50)dt=250t2+50t15.625 rad /s

wt=3s=250×32+50×315.625=154.6 rad/s

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