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 angular acceleration of the salad spinner as it slows down in d/s^2?

Answer 1

Answer:

-480.1°/s²

Explanation:

The angular acceleration (α) can be calculated using the following equation:          

$\alpha = \frac{\omega_{f}^{2} - \omega_{0}^{2}}{2*\Delta \theta}$                                

Where:

$\omega_{f}$: is the final angular speed = 0  

$\omega_{0}$: is the initial angular speed  

Δ$\theta_$: is the angular displacement  

The initial angular speed (ω₀) can be calculated as follows:

$\omega_{0} = \frac{\theta_{0}}{t}$

Where t is the time = 5.00 s

$\omega_{0} = \frac{\theta_{0}}{t} = \frac{20 rev*\frac{2\pi rad}{1 rev}}{5 s} = 25.13 rad/s$    

Now, we can find the angular acceleration:

$\alpha = \frac{\omega_{f}^{2} - \omega_{0}^{2}}{2*\Delta \theta} = - \frac{(25.13 rad/s)^{2}}{2*(6 rev*(\frac{2\pi rad}{1 rev}))} = - 8.38 rad/s^{2}$

$\alpha = -8.38 rad/s^{2}*\frac{360 ^\circ}{2\pi rad} = -480.1 ^\circ/s^{2}$  

Therefore, the angular acceleration is -480.1°/s².

I hope it helps you!                            

Answer 2

Answer:

$-\frac{8\pi}{3}rad/s^2$

Explanation:

To solve this problem we need to apply the concept related to Angular Acceleration. We can find it through the equation

$\omega_f^2-\omega_i^2=2\alpha\theta$

Where for definition,

$\omega_i = \frac{\theta}{t}$

The number of revolution $(\theta)$was given by 20 times, then

$\omega_i = \frac{20*2pi}{5}$

$\omega = 8\pi rad/s$

We know as well that the salad rotates 6 more times, therefore in angle measurements that is

$\theta = 6*2\pi rad = 12\pi rad$

The cook at the end stop to spin, then using our first equation,

$0-8\pi = 2\alpha (12\pi)$

re-arrange to solve$\alpha$ ,

$\alpha = \frac{-8\pi}{2*12\pi}$

$\alpha = -\frac{8\pi}{3}rad/s^2$

We can know find the required time,

$\omega_f-\omega_i = \alpha t$

Re-arrange to find t, and considering that $\omega_f=0$

$t= \frac{\omega_i}{\alpha}$

$t=\frac{-8\pi}{-8\pi/3}$

$t=3s$

Therefore take for the salad spinner to come to rest is 3 seconds with acceleration of $-\frac{8\pi}{3}rad/s^2$