Question
Download Solution PDFIf the equation for the displacement of a particle moving on a circular path is given by \(\theta = 2t^3 + 0.5\), where \(\theta\) is in radians and \(t\) in seconds, then the angular velocity of the particle after \(2s\) from its start is
Answer (Detailed Solution Below)
Option 3 : \(24 \, \text{rad/s}\)
India's Super Teachers for all govt. exams Under One Roof
Detailed Solution
Download Solution PDFAngular displacement of the particle: \(\theta(t) = 2t^3 + 0.5\) radians.
Therefore, angular velocity: \(\omega = \dfrac{d[\theta(t)]}{dt}\)
=\(6t^2 \, \text{rad/s}\)
Angular velocity at \(t = 2 \, \text{s}\): \(\omega|_{t=2} = 6 \times 2^2 = 24 \, \text{rad/s}\)
India’s #1 Learning Platform
Start Complete Exam Preparation
Daily Live MasterClasses
Practice Question Bank
Mock Tests & Quizzes
Trusted by 7.3 Crore+ Students