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 12.0 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 11.0 s. Through how many revolutions does the tub turn during this 23 s interval? Assume constant angular acceleration while it is starting and stopping in revolutions.
The average velocity during slowdown is 3rev/s
during spinup, it goes: 3rev/sec*12 sec
during slowdown, it goes:3rev/sec*11sec
To find the number of revolutions the tub turns during the 23 s interval, we need to calculate the angular displacement during the different phases of motion.
1. Finding angular displacement during acceleration phase:
Initial angular speed (ω₁) = 0 rev/s
Final angular speed (ω₂) = 6.0 rev/s
Time (t₁) = 12.0 s
Using the equation for angular acceleration:
ω₂ = ω₁ + αt₁
Rearranging the equation:
α = (ω₂ - ω₁) / t₁
α = (6.0 - 0) / 12.0
α = 0.5 rev/s²
Now, using the equation for angular displacement during constant acceleration:
θ₁ = ω₁t₁ + 0.5αt₁²
θ₁ = 0(12.0) + 0.5(0.5)(12.0)²
θ₁ = 36.0 rev
The tub turns 36.0 revolutions during the acceleration phase.
2. Finding angular displacement during deceleration phase:
Initial angular speed (ω₃) = 6.0 rev/s
Final angular speed (ω₄) = 0 rev/s
Time (t₂) = 11.0 s
Using the same equation for angular displacement during constant acceleration:
θ₂ = ω₃t₂ - 0.5αt₂²
θ₂ = 6.0(11.0) - 0.5(0.5)(11.0)²
θ₂ = 66.0 - 33.0
θ₂ = 33.0 rev
The tub turns 33.0 revolutions during the deceleration phase.
3. Finding total angular displacement:
Total angular displacement (θ) = θ₁ + θ₂
θ = 36.0 + 33.0
θ = 69.0 rev
Therefore, the tub of the washer turns 69.0 revolutions during the 23 s interval.
To find the number of revolutions the tub turns during the 23 s interval, we need to calculate the angular displacement during each phase (accelerating and decelerating) and then sum them up.
1. Accelerating Phase:
Given:
Initial angular speed, ω1 = 0 rev/s
Final angular speed, ω2 = 6.0 rev/s
Time, t1 = 12.0 s
We can use the formula for angular acceleration:
ω2 = ω1 + α * t1
Solving for α:
6.0 rev/s = 0 rev/s + α * 12.0 s
α = (6.0 rev/s) / (12.0 s)
α = 0.5 rev/s^2
Now, we can calculate the angular displacement using the formula:
θ1 = ω1 * t1 + (1/2) * α * t1^2
θ1 = 0 rev/s * 12.0 s + (1/2) * 0.5 rev/s^2 * (12.0 s)^2
θ1 = 0 + 0.5 rev/s^2 * 144.0 s^2
θ1 = 72.0 rev
2. Decelerating Phase:
Given:
Initial angular speed, ω3 = 6.0 rev/s
Final angular speed, ω4 = 0 rev/s
Time, t2 = 11.0 s
Using the same formula for angular acceleration:
ω4 = ω3 + α * t2
0 rev/s = 6.0 rev/s + α * 11.0 s
α = (-6.0 rev/s) / (11.0 s)
α = -0.5455 rev/s^2
Calculating the angular displacement:
θ2 = ω3 * t2 + (1/2) * α * t2^2
θ2 = 6.0 rev/s * 11.0 s + (1/2) * (-0.5455 rev/s^2) * (11.0 s)^2
θ2 = 66.0 rev + (-0.2728 rev/s^2) * 121.0 s^2
θ2 = 66.0 rev - 33.0132 rev
θ2 = 32.9868 rev
Adding the angular displacements for both phases:
Total angular displacement = θ1 + θ2
Total angular displacement = 72.0 rev + 32.9868 rev
Total angular displacement = 104.9868 rev
Therefore, the tub turns approximately 104.9868 revolutions during the 23 s interval.