Ans in revs.
The tub of a washer goes into its spin-
dry cycle, starting from rest and reaching an
angular speed of 5 rev/s in 10.5 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 5.6 s .
Through how many revolutions does the
tub turn? Assume constant angular accelera-
tion while it is starting and stopping. Answer
in units of rev.
While accelerating, the washer has an average speed of 2.5 rev/s and goes through 2.5x10.5= 26.25 revs.
While decelerating, it has the same average speed for 5.6 s, and turns 14 revs.
Add the two.
how did you get the 2.5 rev/s
To find the number of revolutions the tub turns, we need to determine the angular displacement while it is accelerating and decelerating.
First, we'll find the angular displacement during the acceleration phase. We know that the initial angular speed (ω_i) is 0 rev/s, the final angular speed (ω_f) is 5 rev/s, and the time (t) is 10.5 s.
We can use the formula:
ω_f = ω_i + αt
Since ω_i = 0, the formula becomes:
5 rev/s = α * 10.5 s
Next, we'll solve for α (the angular acceleration):
α = 5 rev/s / 10.5 s
≈ 0.476 rev/s^2
Now, we can use the formula for angular displacement during constant acceleration:
θ = ω_i t + 0.5 α t^2
Here, ω_i = 0, α = 0.476 rev/s^2, and t = 10.5 s:
θ = 0.5 * 0.476 rev/s^2 * (10.5 s)^2
≈ 27.825 rev
So, during the acceleration phase, the tub turns approximately 27.825 revolutions.
Next, we need to find the angular displacement during the deceleration phase. We know that the initial angular speed is 5 rev/s, the final angular speed is 0 rev/s, and the time is 5.6 s.
Using the same formula for angular displacement during constant acceleration:
θ = ω_i t + 0.5 α t^2
Here, ω_i = 5 rev/s, α = -0.476 rev/s^2 (negative because it's decelerating), and t = 5.6 s:
θ = 5 rev/s * 5.6 s + 0.5 * -0.476 rev/s^2 * (5.6 s)^2
≈ 14 rev
Thus, during the deceleration phase, the tub turns approximately 14 revolutions.
Now, to find the total number of revolutions, we add the angular displacements during acceleration and deceleration:
Total revolutions = 27.825 rev + 14 rev
≈ 41.825 rev
Therefore, the tub of the washer turns approximately 41.825 revolutions.