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 2 years ago
The tub of a washer goes into its spin cycle, starting from rest and gainiing angular speed steadily for 8.0 s when it is turning at 5.0 rev/s. At this time the person doing the laundry opens the lid and a safety switch turns off the washer. The tub smoothly slows to rest in 12.0 s. Through how many revolutions does the tub turn whil it is in motion.
 2 years ago
The tub of a washer goes into its spin cycle, starting from rest and gainiing angular speed steadily for 8.0 s when it is turning at 5.0 rev/s. At this time the person doing the laundry opens the lid and a safety switch turns off the washer. The tub smoothly slows to rest in 12.0 s. Through how many revolutions does the tub turn whil it is in motion.

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Inspired
 2 years ago
Best ResponseYou've already chosen the best response.1Look at the problem in two separate parts. Before the lid is opened, the initial angular speed is 0. Your final w is 5 rev/s. Time is 8 seconds. Angular acceleration is change of angular velocity over time. Use the common form: wf^2wi^2=2(alpha)(theta) and isolate theta to get the number of revolutions of the first part. You should get 20rev.. Afterwards, when the system slows down, you now have your initial speed as 5 rev/s. It comes to rest so wf is 0. Solve for alpha. Plug it in to common form and solve for theta. You get 30 rev. Add the two parts, you should get 50 rev.

Shane_B
 2 years ago
Best ResponseYou've already chosen the best response.0@Inspired's answer is correct. Doing it another way you get: \[\large \Delta \theta = (\bar{\omega_1}t_1)+(\bar{\omega_2}t_2)\]\[\large \Delta \theta = \frac{(0+5(2\pi \space rads/s))(8s)}{2} + \frac{(5(2\pi \space rads/s)+0 \space rads/s)(12s)}{2}\]\[\large \Delta \theta = 20(2\pi \space rad)+30(2\pi \space rad)=50(2\pi\space rad)=100\pi \space rad\]\[\large \frac{100\pi \space rad}{2\pi \space rad}=50 \space revolutions\] For more info see: http://hyperphysics.phyastr.gsu.edu/hbase/rotq.html
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