JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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\(^2\).
- A \(150\)
- B \(120\)
- C \(130\)
- D \(170\)
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…
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