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The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 6.0 rev/s in 9.0 s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 9.0 s. Through how many revolutions does the tub turn during this 18 s interval? Assume constant angular acceleration while it is starting and stopping.
we use the equation
angular displacement = w0 t + 1/2 at^2
where w0=initial ang vel
we first find ang accel from
ang accel = (change in ang vel)/time = 6rev/s/9s
6 rev/s = 6x2pi rad/s , a = 4.19rad/s/s
the initial speed starting up is 0, so for the first 9 secs we have
ang displacment =1/2(4.19rad/s/s)(9s)^2=169.7 rad
in the second 9 s, the tub has a speed of 6 rev/s (=37.7rad/s, and we use this as the initial vel of the interval) we have
ang displacement =
37.7rad/sx9s - 1/2(4.19rad/s/s)(9s)^2 =
339.1rad - 169.7 rad
so the total displacement is 339.1 rad
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