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

Answer 1

Given

20 rev in 5 seconds

that is the angular velocity w = 20 /5 = 4 rev/s = 4*2pi rad/s = 8pirad/s

the displacemnt is theta = 20*2pi rad = 40 pi rad

the angular velocity is w1 = theta1/t = 40 pi /5 = 25.133 rad/s

and making 6 more rotations before come to rest

the displacement covered in 6 rotations is theta2 = 6*2pi = 12*pi rad = 37.7 rad

from equations of motion  

w2^2 -w1^2 = 2*alpha(theta)

here W2 = 0 rad/s comes to rest

alpha = -W1^2/(2(theta2))

alpha = -(25.133)^2/(2(37.7) rad/s2

alpha = -8.378 rad/s2