# Question #d8ba2

Apr 1, 2017

I got $6.4 r e v s$

#### Explanation:

I would use the general relationship for rotational motion (analogous to the linear one):
${\omega}^{2} = {\omega}_{0}^{2} + 2 \alpha \theta$
where:
$\omega =$ angular velocity (initial and final, with $\omega = 0$ the final one);
$\alpha$ is angular acceleration;
$\theta$ is the angle in radians.
We get:
$0 = {20}^{2} - 2 \cdot 5 \cdot \theta$ as it is slowing down $\alpha < 0$;
rearranging:
$\theta = \frac{400}{10} = 40 r a d$
Giving:
$\frac{40}{2 \pi} = 6.36 \approx 6.4 r e v s$