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JEE Main 2022 (30 Jul, Shift 2) · Rotational Kinematics & Dynamics

Rotational Kinematics & Dynamics question from JEE Main 2022 (30 Jul, Shift 2)

Q32JEE Main 2022 · 30 JUL · SHIFT 2Medium

Rotational Kinematics & Dynamics

A fly wheel of mass 10 kg10\text{ kg} and radius 50 cm50\text{ cm} is rotating at a rate of 360360 rpm. Assuming mass to be concentrated at the rim, the constant retarding torque required to stop the rotation of wheel in 66 rotations will be xπ N mx\pi\text{ N m}. The value of xx is ________

Numerical answer type. Work it out, then open the answer.

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Answer: 15

  1. Mass at rim: I=MR2=10(0.5)2=2.5 kg m2I = MR^2 = 10(0.5)^2 = 2.5\text{ kg m}^2.
  2. ω0=360 rpm=12π rad/s\omega_0 = 360\text{ rpm} = 12\pi\text{ rad/s}, θ=6×2π=12π rad\theta = 6\times2\pi = 12\pi\text{ rad}
  3. α=ω022θ=144π224π=6π rad/s2\alpha = \frac{\omega_0^2}{2\theta} = \frac{144\pi^2}{24\pi} = 6\pi\text{ rad/s}^2
  4. τ=Iα=2.5×6π=15π N m\tau = I\alpha = 2.5\times6\pi = 15\pi\text{ N m}, so x=15x = 15.
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