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JEE Mains · Physics · STD 11 - System of particles and rotational motion
  • JEE Mains 2022
  • Medium
  • MCQ

A ball is spun with angular acceleration \(\alpha=6 t ^{2}-2 t\) where \(t\) is in second and \(\alpha\) is in \(rads\) \(^{-2}\). At \(t=0\), the ball has angular velocity of \(10\,rads\) \(^{-1}\) and angular position of \(4\,rad\). The most appropriate expression for the angular position of the ball is

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Answer & Solution

Correct Answer

(B) \(\frac{t^{4}}{2}-\frac{t^{3}}{3}+10 t+4\)

Step-by-step Solution

Detailed explanation

\(\frac{d \omega}{d t}=6 t^{2}-2 t\)

\(\int \limits_{10}^{m} d \omega=2 t^{5}-t^{2}\)

\(\omega=10+2 t^{5}-t^{2}\)

\(\frac{d \theta}{d t}=10+2 t^{5}-t^{2}\)

\(\int \limits_{4}^{\theta} d \theta=10+2 t^{3}-t^{2}\)

\(\int \limits_{4}^{\theta} d \theta=10 t+\frac{t^{4}}{2}-\frac{t^{3}}{3}\)

\(\theta=4+10 t+\frac{t^{4}}{2}-\frac{t^{5}}{3}\)

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