ExamBro
ExamBro
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\(^2\).

  1. A \(150\)
  2. B \(120\)
  3. C \(130\)
  4. D \(170\)
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…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app