Q 11-06-052JEE MainJEE Main 2026 (2 Apr, Shift 1)Easy
A particle is rotating in a circular path and at any instant its motion can be described as
$\theta=\dfrac{5t^4}{40}-\dfrac{t^3}{3}$.
The angular acceleration of the particle after $10$ seconds is ______ $\text{rad/s}^2$.
Answer: (C) $130$
$\theta=\dfrac{t^4}{8}-\dfrac{t^3}{3}$
$\omega=\dfrac{d\theta}{dt}=\dfrac{t^3}{2}-t^2,\qquad\alpha=\dfrac{d\omega}{dt}=\dfrac{3t^2}{2}-2t$
At $t=10$ s: $\alpha=150-20=130\ \text{rad/s}^2$
Solution by Sreeraj P, M.Sc Physics