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JEE Main 2021 (25 Jul, Shift 2) · Rotational Kinematics & Dynamics

Rotational Kinematics & Dynamics question from JEE Main 2021 (25 Jul, Shift 2)

Q41JEE Main 2021 · 25 JUL · SHIFT 2Easy

Rotational Kinematics & Dynamics

A solid disc of radius 20 cm and mass 10 kg is rotating with an angular velocity of 600 rpm, about an axis normal to its circular plane and passing through its centre of mass. The retarding torque required to bring the disc at rest in 10 s is ____________ π×\pi \times 10−-1 Nm.

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

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Answer: 4

  1. I=12MR2=12×10×0.22=0.2I = \frac{1}{2}MR^2 = \frac{1}{2} \times 10 \times 0.2^2 = 0.2 kg m2^2
  2. ω0=600\omega_0 = 600 rpm =20π= 20\pi rad/s, so α=20π10=2π\alpha = \frac{20\pi}{10} = 2\pi rad/s2^2
  3. τ=Iα=0.2×2π=0.4π=4π×10−1\tau = I\alpha = 0.2 \times 2\pi = 0.4\pi = 4\pi \times 10^{-1} N m
  4. So the answer is 4.
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