The total angle turned through during this interval is

  1. 48 rad

  2. 24 rad

  3. 96 rad

  4. 72 rad

Answers (1)
S Sayak

Initial Angular Velocity, \omega _{0}=0

Final Angular Velocity, \omega=24rev/s=48\pi \ rad/s

Time, t = 8 s

\\\alpha =\frac{\omega -\omega _{0}}{t}\\ \alpha =\frac{48\pi }{8}=6\pi \ rad\ s^{-2}

\\\omega^{2} -\omega _{0}^{2}=2\alpha \theta \\ \theta =\frac{\omega^{2} -\omega _{0}^{2}}{2\alpha }\\ \theta =\frac{(48\pi )^{2}}{2\times 6\pi }\\ \theta =192\pi

i.e. 96 revolutions

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