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MHT CET 2025 (21 Apr, Shift 1) · Rotational Kinematics & Dynamics

Rotational Kinematics & Dynamics question from MHT CET 2025 (21 Apr, Shift 1)

Q16MHT CET 2025 · 21 APR · SHIFT 1Medium

Rotational Kinematics & Dynamics

A disc of mass 25 kg and radius 0.2 m is rotating at 240 r.p.m. A retarding torque brings it to rest in 20 second. If the torque is due to a force applied tangentially on the rim of the disc then the magnitude of the force is

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Answer: (c)

  1. I=12MR2=12(25)(0.2)2=0.5 kg m2I = \frac{1}{2}MR^2 = \frac{1}{2}(25)(0.2)^2 = 0.5\text{ kg m}^2
  2. ω0=240 rpm=8π rad/s\omega_0 = 240\text{ rpm} = 8\pi\text{ rad/s}, α=8π20=0.4π rad/s2\alpha = \frac{8\pi}{20} = 0.4\pi\text{ rad/s}^2
  3. τ=Iα=0.2π N m\tau = I\alpha = 0.2\pi\text{ N m}, F=τR=0.2π0.2=π NF = \frac{\tau}{R} = \frac{0.2\pi}{0.2} = \pi\text{ N}
  4. So the answer is (c) π\pi N.
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