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

$\omega_{f} = \omega_{i} + \alpha t$

where:

• $\omega_{f}$ is the final angular velocity of the body.
• $\omega_{i}$ is the initial angular velocity of the body.
• $\alpha$ is the angular acceleration of the body.
• $\Delta \theta$ is the angular displacement .
• $t$ is the time .

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

$\textbf{Plugging}$ known information into :

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

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

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