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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\(^2\).

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Verified Solution

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: \(130\)
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