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Question

A disc having mass 1 kg and radius 2 m is pivoted about its centre. Initially, it is at rest. A force F=(5t2+9) N starts acting tangentially on it. Find the angular velocity of the disc after 3 seconds in rad/s

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Solution

Given
Mass of the disc =1 kg
Radius of the disc =2 m
Force acting on the disc F=(5t2+9)
Moment of inertia of the disc about its centre,
I=MR22
I=1×222=2 kg m2
Initially, disc is at rest, hence initial angular momentum of the disc L1=Iω1=0
As we know, angular impulse J=t2t1τ.dt
Here,
τ=F.R=(5t2+9)×2=(10t2+18) N-m
(because force acts tangentially)
J=30(10t2+18)dt=[10t33+18t]30
=90+54=144
But J=t2t1dL=L2L1
144=I.ω20
Here, ω2 is the angular velocity of the disc after 3 seconds
144=2×ω2
ω2=72 rad/s

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