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

Dario, a prep cook at an Italian restaurant, spins a salad spinner 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?

Express your answer numerically in degrees per second per second.


For convenience, convert the given speed to rad/s :

20.0rev / 5.00s = 4.00rev / s(2πrad / rev) = 25.1rad / s

So when Dario stops spinning the spinner, it has this angular velocity (ω). Its final angular velocity (ω₀) is zero because it comes to rest. From the time it stops spinning to when it stops it goes through an angle (Δθ) of 6.00rev(2πrad / rev) = 37.7rad. From :

ω² = ω₀² + 2αΔθ

α = (ω² – ω₀²) / 2Δθ

= [(25.1rad/s)² – 0] / (2 x 37.7rad)

= 8.36rad/s²

Hope this makes sense.

Algol’s answer is correct except for one small thing. Instead of going through the whole thing again, just notice the answer needs to be in degrees, not radians. To convert from radians to degrees, multiply by 180 and divide by Pi

8.36rad/s² = 479 deg/s²

(note that you should give the answer to 3 significant figures because all values in the question are to 3.s.f as well. That is why I havent included decimal places in the answer)

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