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With the aid of a string, a gyroscope is accelerated from rest to 32 rad/s in 0.40 s under a constant angular acceleration. (a) What is its angular acceleration in rad/s²? (b) How many revolutions does it go through in the process?

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a)\textbf{a)} From  the kinematics of the rotational motion\textbf{ the kinematics of the rotational motion} we know that:

ωf=ωi+αt\omega_{f} = \omega_{i} + \alpha t

where:

  • ωf\omega_{f} is the final angular velocity of the body.
  • ωi\omega_{i} is the initial angular velocity of the body.
  • α\alpha is the angular acceleration of the body.
  • Δθ\Delta \theta is the angular displacement .
  • tt is the time .

Form givens\textbf{givens} we know that : ωi=0\omega_{i} = 0 rad/s because it starts from rest , ωf=32 rad/s\omega_{f} = 32\ \mathrm{rad/s} and t=0.4t = 0.4 s.

Plugging\textbf{Plugging} known information into :

ωf=ωi+αtα=ωfωit=320.4=80\begin{align*} \omega_{f}& = \omega_{i} + \alpha t \\ \alpha &=\dfrac{\omega_{f} - \omega_{i} }{t}\\ &= \dfrac{32}{0.4}\\ &=80 \end{align*}

α=80   rad/s2\boxed{\alpha = 80\ \; \mathrm{rad/s^2}}

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