A prep cook spins a salad spinner 25.0 times in 10.0 seconds (at a constant rate) and then stops. The salad spinner rotates 8.00 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 comes to rest?

Divide 2 pi x 25 radians by 10 seconds for the initial angular velocity. I get 15.71 rad/s

Divide that angular rate by the time it takes to stop, to calculate the angular deceleration rate.

The time that it takes to stop is 16 pi radians divided by the average angular velocity. I get 6.40 s for that

To find the angular acceleration of the salad spinner as it comes to rest, we can use the formula that relates the number of revolutions, the initial angular velocity, the final angular velocity, and the angular acceleration:

θ = ω_i * t + (1/2) * α * t^2

Where:
θ is the total number of revolutions
ω_i is the initial angular velocity
t is the time
α is the angular acceleration

From the information given, we know that the prep cook spins the salad spinner 25.0 times in 10.0 seconds, and then it rotates an additional 8.00 times before it comes to rest.

Let's break it down step by step:

1. Calculate the initial angular velocity (ω_i):
We know that the prep cook spins the salad spinner 25.0 times in 10.0 seconds, so we can use the formula:

ω_i = (θ - θ_initial) / t

Where:
θ_initial is the initial number of revolutions

θ_initial = 0 (since it starts from rest)

ω_i = (25.0 - 0) / 10.0

ω_i = 2.5 revolutions per second

2. Calculate the final angular velocity (ω_f):
The spinner rotates an additional 8.00 times before it comes to rest, so the final number of revolutions is:

θ_final = 25.0 + 8.00

θ_final = 33.0 revolutions

We also know that the final angular velocity is zero since it comes to rest:

ω_f = 0 revolutions per second

3. Calculate the angular acceleration (α):
Using the formula mentioned earlier, we rearrange it to solve for α:

α = (θ - θ_initial - ω_i * t) / (0.5 * t^2)

α = (33.0 - 0 - 2.5 * 10.0) / (0.5 * 10.0^2)

α = 0.8 revolutions per second squared

Therefore, the angular acceleration of the salad spinner as it comes to rest is 0.8 revolutions per second squared.