JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2026
- Medium
- MCQ
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 _______ rad/s\(^2\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(130\)
Step-by-step Solution
Detailed explanation
Given \(\theta = \dfrac{5t^4}{40} - \dfrac{t^3}{3} = \dfrac{t^4}{8} - \dfrac{t^3}{3}\)
Angular velocity \(\omega = \dfrac{d\theta}{dt} = \dfrac{4t^3}{8} - \dfrac{3t^2}{3} = \dfrac{t^3}{2} - t^2\)
Angular acceleration \(\alpha = \dfrac{d\omega}{dt} = \dfrac{3t^2}{2} - 2t\)
At \(t = 10\) s, \(\alpha = \dfrac{3(10)^2}{2} - 2(10)\)
\(\alpha = 150 - 20 = 130\) rad/s\(^2\)
Answer: \(130\)
Angular velocity \(\omega = \dfrac{d\theta}{dt} = \dfrac{4t^3}{8} - \dfrac{3t^2}{3} = \dfrac{t^3}{2} - t^2\)
Angular acceleration \(\alpha = \dfrac{d\omega}{dt} = \dfrac{3t^2}{2} - 2t\)
At \(t = 10\) s, \(\alpha = \dfrac{3(10)^2}{2} - 2(10)\)
\(\alpha = 150 - 20 = 130\) rad/s\(^2\)
Answer: \(130\)
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