JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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
- A \(\frac{3}{2} t^{4}-t^{2}+10 t\)
- B \(\frac{t^{4}}{2}-\frac{t^{3}}{3}+10 t+4\)
- C \(\frac{2 t ^{4}}{3}-\frac{ t ^{3}}{6}+10 t +12\)
- D \(2 t^{4}-\frac{t^{3}}{2}+5 t+4\)
Answer & Solution
Correct Answer
(B) \(\frac{t^{4}}{2}-\frac{t^{3}}{3}+10 t+4\)
Step-by-step Solution
Detailed explanation
\(\frac{d \omega}{d t}=6 t^{2}-2 t\) \(\int \limits_{10}^{m} d \omega=2 t^{5}-t^{2}\) \(\omega=10+2 t^{5}-t^{2}\) \(\frac{d \theta}{d t}=10+2 t^{5}-t^{2}\) \(\int \limits_{4}^{\theta} d \theta=10+2 t^{3}-t^{2}\)…
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