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. part a what is the magnitude of the angular acceleration of the salad spinner as it slows down?

Answer 1

`\alpha = -((25.132 rad/s)^2)/(2* 37.7 rad)= -8.377 rad/s^2`

And we can convert this into degrees like this:

`\alpha= -8.377 rad/s^2 * ((180)/(\pi rad)) =-479.967 rad/s^2`

Step-by-step explanation:

For this case we assume that the angular acceleration is constant and the spinner slows down and come to rest at the end

We can calculate the distance traveled each revolution with this formula:

`\theta= 20 rev * (2\pi rad)/(1 rev)= 40 \pi rad`

And since we know that the time to reach the velocity 0 is 5 s we can find the angular velocity like this:

`w_o= (\theta)/(t)= (40 \pi rad)/(5 s)= 25.132 rad/s`

We know that the spinner rotates 6 more times before come rest, so the total distance traveled is:

`\theta= 6* 2\pi = 37.699 rad`

`w_f = 0 rad/s`

And we have the following formula :

`w^2_f = w^2_i + 2\alpha \theta`

Since we know that the final angular velocity is 0 we can solve for
`\alpha` the angular acceleration and we got:

`\alpha = -(w^2_o)/(2 \theta)`

And replacing the values that we found before we have this:

`\alpha = -((25.132 rad/s)^2)/(2* 37.7 rad)= -8.377 rad/s^2`

And we can convert this into degrees like this:

`\alpha= -8.377 rad/s^2 * ((180)/(\pi rad)) =-479.967 rad/s^2`

Answer 2

For an uniformly accelerated motion, the following relationship is used:

`2 a S=v_f^2 -v_i ^2` (1)

where a is the acceleration, S the distance covered, and vf and vi the final and initial speeds of the motion.

In our problem we are dealing with a rotational motion. Initially, the salad spinner has constant angular speed, which is given by

`\omega _i = 2 \pi f`

where f is the rotational frequency, which is the number of revolutions per second:

`f=  (20 rev)/(5 s)=4 Hz`

so the initial angular speed is

`\omega _i = 2 \pi (4 Hz)=25.2 rad/s`

Then, the salad spinner starts to decelerate with constant deceleration
`\alpha`, and during its deceleration it spins for other 6 revolutions, so covering a total angle of

`\theta = 2 \pi (6 rev)=37.7 rad`

until it stops, so until it reaches a final speed of
`\omega _f=0`.

To find the angular acceleration, we can use the equivalent of equation (1) for angular motions:

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

and so, since the final speed is
`\omega _f=0`:

`\alpha = -  (\omega _i^2)/(2 \theta)=- ((25.2 rad/s)^2)/(2\cdot 37.7 rad)=-8.4 rad/s^2`

where the negative sign means the salad spinner is decelerating.