Question

Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates 20.0 times in 5.00 seconds and then stops spinning it. The salad spinner rotates 6.00 more times before it comes to rest. Assume that the spinner slows down with constant angular acceleration. What is the magnitude of the angular acceleration of the salad spinner as it slows down?

Answer 1

Answer:

8.37758 rad/s²

Explanation:

$\omega_i$ = Initial angular velocity = $20\times \frac{2\pi}{5}$

$\omega_f$ = Final angular velocity = 0

$\theta$ = Angle of rotation = $6\times 2\pi\ rad$

Equation of rotational motion

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-(20\times \frac{2\pi}{5})^2}{2\times 6\times 2\pi}\\\Rightarrow \alpha=-8.37758\ rad/s^2$

The the magnitude of the angular acceleration of the salad spinner as it slows down is 8.37758 rad/s²