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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?​

Operating Speed Of An Automatic Washing Machine Is 55 Rad S1 After Loading Dirty Clothes And Pressing A Start Button The Tub Of The Washer Can Reach Its Operati class=

Sagot :

whitneytr12

a) The time at which the tub of the washing machine reaches its maximum velocity (5.5 rad/s) is:

t = 1.33 s.

b) The total revolutions that the tub turn during the interval of 20 s, it is 7 second to reach 6 rev/s and 13 s to get rest, is:

[tex]\theta_{tot}=60.20\: rev[/tex]

a)

We know that the final angular velocity of the tub is:

[tex]\omega_{f}=5.5\: rad/s[/tex]

We have the angular acceleration in the washing mode:

[tex]\alpha_{1}=4.15\: rad/s^{2}[/tex]

Now, we need to find the time at which the tub reaches this final angular velocity. We can use this equation:

[tex]\omega_{f}=\omega_{i}+\alpha_{1} t[/tex]

The initial angular velocity is 0, so we have:

[tex]t=\frac{\omega_{f}}{\alpha_{1}}[/tex]

[tex]t=\frac{5.5}{4.15}[/tex]

Therefore, the tub reaches this velocity at t = 1.33 s

b)

First of all, let's find the revolutions in the interval of 7 s.

The final angular velocity here is 6 rev/s, so the angular acceleration in this stage is:

[tex]\alpha_{2}=\frac{\omega_{f}-\omega_{i}}{t}[/tex]  

[tex]\alpha_{2}=\frac{6}{7}[/tex]

[tex]\alpha_{2}=0.86\: rev/s^{2}[/tex]

Now, we can use the following equation to find the revolution in this interval. We do the initial angular velocity equal to zero because it starts from the rest.

[tex]\theta_{1}=\omega_{i}t+0.5\alpha t^{2}[/tex]

[tex]\theta_{1}=0.5*0.86*7^{2}[/tex]

[tex]\theta_{1}=21.07\: rev[/tex]

In the second interval of 13 s, the tub slows down to rest, we need to find the new angular acceleration.

[tex]\alpha_{3}=\frac{\omega_{f}-\omega_{i}}{t}[/tex]  

[tex]\alpha_{3}=\frac{0-6}{13}[/tex]

[tex]\alpha_{3}=-0.46\: rev/s^{2}[/tex]

Using the same equation above, the revolutions in this interval will be:

[tex]\theta_{2}=\omega_{i}t+0.5\alpha t^{2}[/tex]

[tex]\theta_{2}=6*13-0.5*0.46*13^{2}[/tex]

[tex]\theta_{2}=39.13\: rev[/tex]

Therefore, the total revolutions that the tub turn during 20 s are:

[tex]\theta_{tot}=\theta_{1}+\theta_{2}[/tex]

[tex]\theta_{tot}=21.07+39.13[/tex]    

[tex]\theta_{tot}=60.20\: rev[/tex]

You can learn more about angular motion here:

https://brainly.com/question/15522059

I hope it helps you!

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