Question

Operating speed of an automatic washing machine is 5.5 rad s-1. After loading dirty clothes and pressing a start button, the tub of the washer can reach its operating speed with an average angular acceleration of 4.15 rad s-2. Later at the spin-dry mode, it is starting from rest and reaching an angular speed of 6.0 revolution per second in 7.0 s.
a) How long does it take for the clothes to come up to the speed during the washing mode?
b) Now the tub of the washer goes into it spin-dry cycle. 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 13.0 s. How many revolutions does the tub turn during this 20.0 s interval?

Answer 1

The equations for rotary motion is based on motion in a circular path

a) Time taken by clothes to come up to speed during washing is approximately 1.325 s

b) The tub turns approximately 24.016 revolutions in the 20.0 s interval

Reasons:

Known parameter are;

Operating speed of an automatic washing machine = 5.5 rad·s⁻¹

Average angular acceleration of the washing machine, α = 4.15 rad·s⁻²

Angular speed in the spin-dry mode, ω = 6.0 revolution per second

Time it takes to reach angular speed in spin-dry mode = 7.0 s

a) The relationship between angular acceleration and time are;

$\alpha = \dfrac{\Delta \omega}{\Delta t}$, $\Delta t = \dfrac{\Delta \omega}{\alpha}$

Where;

Δω = ω₂ - ω₁ = 5.5 rad·s⁻¹

Δt = The time it takes to accelerate

Therefore;

$\Delta t = \dfrac{5.5 \ rad\cdot s^{-1}}{4.15 \ rad \cdot s^{-2}} \approx 1.325 \ s$

The time it takes for the clothes to come up to speed during the washing mode, Δt ≈ 1.325 s

b) The angular acceleration of the spin-dry mode is given as follows;

6 revolutions per second = 2·π ×6  rad per second = 12·π rad/s

$\alpha = \dfrac{\Delta \omega}{\Delta t}$

The acceleration during spin up

$\therefore \alpha = \dfrac{12 \cdot \pi }{7.0} \ rad \cdot s^{-2} \approx 5.39 \ rad \cdot s^{-2}$

The acceleration during slowing down;

$\therefore \alpha = \dfrac{12 \cdot \pi }{13.0} \ rad \cdot s^{-2} \approx 2.9 \ rad \cdot s^{-2}$

The angle turned in the 20.0 second interval is therefore;

$\dfrac{1}{2} \times 5.39 \times 7^2 + 12 \cdot \pi \times 7-\dfrac{1}{2} \times 2.9 \times 13^2 \approx 150.9$

The angle turned in the 20.0 s is approximately 150.9 radians

$Number \ of \ revolutions \ =\dfrac{\theta}{2 \cdot \pi}$

Therefore;

$Number \ of \ revolutions \ in \ 20.0 \ s =\dfrac{150.9 }{2 \cdot \pi} \approx 24.016$

The number of revolutions of the tub in the 20.0 s interval is approximately 24.016 revolutions

Learn more about constant angular acceleration here:

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