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Questions  

A solid cylinder of mass 2 kg and radius 4 cm is rotating about its axis at the rate of 3 rpm. The torque required to stop after 2π revolutions is [NEET 2019]

a
2×10-3 N-m
b
12×10-4 N-m
c
2×106 N-m
d
2×10-6 N-m

detailed solution

Correct option is D

Given, mass of cylinder,  m=2 kgRadius of cylinder,  r=4 cm=4×10-2 mRotational velocity,  3rpm=3×2π60=π10rads-1 and θ=2π× revolution =2π×2π=4π2radThe work done in rotating an object by an angle  θ from rest is given by  W=τθAs the cylinder is brought to rest, so the work done will be negative.According to work-energy theorem,Work done = Change in rotational kinetic energy-τθ=12Iωf2-12Iωi2=12Iωf2-ωi2⇒  τ=Iωi22θ   ∵ωf-0=1212mr2ωi2θ  ∵I=12mr2( for cylinder )=14mr2ω2θ  ∵ωi=ω=14×2×4×10-22×π102×14π2=14×2×16×10-4×π2100×14π2=2100×10-4=2×10-6 N-m

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