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JEE Main 2026 · 2 Apr · Shift 1, Physics question 31

A particle is rotating in a circular path and at any instant its motion can be described as θ=5t440−t33\theta=\dfrac{5t^4}{40}-\dfrac{t^3}3. The angular acceleration of the particle after 10 seconds is ________ rad/s².

Choose one option
Answer and worked solution

Official answer: C

Differentiate twice. ω=t32−t2\omega=\dfrac{t^3}2-t^2 and α=3t22−2t\alpha=\dfrac{3t^2}2-2t.

At t = 10 s. α=150−20=130\alpha=150-20=130 rad/s².

(C)

Topics:Torque and rotational dynamicsUniform circular motion
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