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NEET 2016 (Phase 1) · Rotational Kinematics & Dynamics

Rotational Kinematics & Dynamics question from NEET 2016 (Phase 1)

Q65NEET 2016 · PHASE 1Medium

Rotational Kinematics & Dynamics

A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad s−2{}^{-2}. Its net acceleration in m s−2{}^{-2} at the end of 2.0 s is approximately

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Answer: (c)

  1. At t=2t = 2 s: ω=αt=4 rad/s\omega = \alpha t = 4\text{ rad/s} (rim point, r=0.5r = 0.5 m).
  2. at=αr=2×0.5=1 m s−2a_t = \alpha r = 2 \times 0.5 = 1\text{ m s}^{-2}, ac=ω2r=16×0.5=8 m s−2a_c = \omega^2 r = 16 \times 0.5 = 8\text{ m s}^{-2}
  3. a=at2+ac2=65≈8 m s−2a = \sqrt{a_t^2 + a_c^2} = \sqrt{65} \approx 8\text{ m s}^{-2}
  4. So the answer is (c) 8.0.
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