What is the magnitude of the angular acceleration of the salad spinner as it slows down?

The magnitude of the angular acceleration is 8.39 1/(s^2)

Explanation:

The complete question is:

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?

The frequency (f) at the beginning is calculated as follows:

f = 20 rotations/5 seconds = 4 1/s

The angular frequency at the beginning (ωi) is calculated as follows:

ωi = 2*π*f = 2*π*4 = 8π 1/s

At the end angular frequency is zero (the spinner stops moving). ωf = 0

The angular displacement  in 6 times is:

θ = 6*2π = 12π

The angular acceleration (α) can be obtained from the following equation of rotational motion:

ωf^2 = ωi^2 + 2*α*θ

α = (ωf^2 - ωi^2)/(2*θ)

α = [0 - (8π)^2]/(2*12π)

α = 8.39 1/(s^2)

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