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NEET 2019 · Rotational Kinematics & Dynamics

Rotational Kinematics & Dynamics question from NEET 2019

Q53NEET 2019Medium

Rotational Kinematics & Dynamics

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\pi revolutions is:

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Answer: (b)

  1. I=12MR2=12(2)(0.04)2=1.6×10−3 kg m2I = \frac{1}{2}MR^2 = \frac{1}{2}(2)(0.04)^2 = 1.6\times10^{-3}\text{ kg m}^2
  2. ω0=3 rpm=3×2π60=0.1π rad/s\omega_0 = 3\text{ rpm} = \frac{3\times 2\pi}{60} = 0.1\pi\text{ rad/s}; θ=2π rev=4π2 rad\theta = 2\pi\text{ rev} = 4\pi^2\text{ rad}
  3. α=ω022θ=0.01π28π2=1.25×10−3 rad/s2\alpha = \frac{\omega_0^2}{2\theta} = \frac{0.01\pi^2}{8\pi^2} = 1.25\times10^{-3}\text{ rad/s}^2
  4. τ=Iα=1.6×10−3×1.25×10−3=2×10−6 N m\tau = I\alpha = 1.6\times10^{-3}\times1.25\times10^{-3} = 2\times10^{-6}\text{ N m}
  5. So the answer is (b) 2×10−62\times10^{-6} N m.
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