The angular velocity of a rotating rigid body increases from 500 to 1500 rev/min in 120 s. (a) What is the angular acceleration of the body? (b) Through what angle does it turn in this 120 s?

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

\dfrac{1 \mathrm{rev}}{\mathrm{min}} \cdot \dfrac{2 \pi\; \mathrm{rad}}{\mathrm{rev}} \cdot \dfrac{1 \mathrm{min}}{60\ \mathrm{ s}}= 0.10479\ \;\mathrm{ rad/s}

$\textbf{For}$ $\omega_{i} = 500$ rev/min :

\dfrac{500 \mathrm{rev}}{\mathrm{min}} \cdot \dfrac{2 \pi\; \mathrm{rad}}{\mathrm{rev}} \cdot \dfrac{1 \mathrm{min}}{60\ \mathrm{ s}}= 52.3598\ \;\mathrm{ rad/s}

$\textbf{For}$ $\omega_{f} = 1500$ rev/min

\dfrac{1500 \mathrm{rev}}{\mathrm{min}} \cdot \dfrac{2 \pi\; \mathrm{rad}}{\mathrm{rev}} \cdot \dfrac{1 \mathrm{min}}{60\ \mathrm{ s}}= 157.079\ \;\mathrm{ rad/s}

From the equation of the angular velocity we know that :

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

where:

From $\textbf{givens}$ we know that : $t = 120$ s , $\omega_{i} = 52.3598$ rad/s and $\omega_{f} = 157.079$ rad/s .

$\textbf{Entering }$ known information into :

\begin{align*} \omega_{f}& = \omega_{i} + \alpha t \\ \alpha t&= \omega_{f} - \omega_{i}\\ \alpha &= \dfrac{\omega_{f} - \omega_{i} }{t}\\ &= \dfrac{157.079 - 52.3598}{120}\\ &= 0.87266 \end{align*}

\boxed{\alpha = 0.87266\ \; \mathrm{rad/s^2}}

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