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JEE Main 2026 (2 Apr, Shift 1) · Non-Uniform Circular Motion

Non-Uniform Circular Motion question from JEE Main 2026 (2 Apr, Shift 1)

Q3JEE Main 2026 · 2 APR · SHIFT 1Easy

Non-Uniform Circular Motion

A particle is rotating in a circular path and at any instant its motion can be described as θ=5t440−t33\theta = \frac{5t^4}{40} - \frac{t^3}{3}.

The angular acceleration of the particle after 10 seconds is _________ rad/s2.

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Answer: (c)

  1. θ=t48−t33\theta = \frac{t^4}{8} - \frac{t^3}{3}
  2. ω=dθdt=t32−t2\omega = \frac{d\theta}{dt} = \frac{t^3}{2} - t^2
  3. α=dωdt=3t22−2t\alpha = \frac{d\omega}{dt} = \frac{3t^2}{2} - 2t
  4. At t=10t = 10 s: α=150−20=130\alpha = 150 - 20 = 130 rad/s2^2
  5. So the answer is (c) 130.
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