Centripetal force problem involving a washing machine

In summary: The sum of all the forces in the vertical direction is mg + Ffr and the acceleration in the vertical direction at the moment the sock is just about ready to slip but has not started moving yet is ac. The sum of all the forces in the horizontal direction is mg + N and the acceleration in the horizontal direction at the moment the sock is just about ready to slip but has not started moving yet is ac.
  • #1
Caw
3
0
Summary:: Just want to know if I'm on the right track with this question.

Hi, so this is what I have for my assignment:
A washing machines drum is rotating rapidly about a vertical axis (a so-called toploader). A wet sock is stuck on the inside, halfway up the drum, and the drum begins to slow its rotation. How many rotations per second is the drum making when the sock falls to the bottom of the drum? The coefficient of static friction between the sock and the drum is 0.21, and the drum radius is 0.32 m.

And after deriving some formula, I got these statements:
- Sock falls when Fc<Fg
=> Fg=mac+Ffr
where Fc is centripetal force
m is mass
ac is centripetal acceleration
Ffr is force friction

- From Fg=mac+Ffr, I got: mg=mac+umg
then: mg=(mv^2)/r + umg
finally: v=squrt[(g-ug)r]

- Then I used the velocity to find rotation/sec

I just want to know if I'm on the right track. And if I'm not where should I start because I have no other clue. Thank you in advance!
 
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  • #2
Hi Caw and welcome to PF.
:welcome:

Caw said:
mg=(mv^2)/r + umg
There is a problem with this equation. The weight mg and the force of friction μmg are vertical but mv2/r is horizontal because it points towards the axis of rotation. You cannot add vertical and horizontal forces like this. Hint: The maximum force of static friction is not always ##f_s^{max}=\mu_s mg## but it is always ##f_s^{max}=\mu_s N##. What is the normal force ##N## in this case? Draw a free body diagram of the sock.
 
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  • #3
kuruman said:
Hi Caw and welcome to PF.
:welcome:There is a problem with this equation. The weight mg and the force of friction μmg are vertical but mv2/r is horizontal because it points towards the axis of rotation. You cannot add vertical and horizontal forces like this. Hint: The maximum force of static friction is not always ##f_s^{max}=\mu_s mg## but it is always ##f_s^{max}=\mu_s N##. What is the normal force ##N## in this case? Draw a free body diagram of the sock.
Oh, so N in this case would be equal to the centrifugal force since that's the only one available horizontally, right? Also should the mg on the other side be changed too (I don't really know because isn't it the only force that can pull the sock down? Thank you so much for pointing out that mistake and sorry for the late respond.
 
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  • #4
You need to write two separate F = ma equations, one for the horizontal direction and one for the vertical direction. You are correct in saying that the centripetal force and the normal force are horizontal. Note that the weight and force of friction are vertical. Please post the equations.
 
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  • #5
F vertical is equal to mg and F horizontal is equal to (umv^2)/r.
So, the final equation is mg=(umv^2)/r?
 
  • #6
Caw said:
F vertical is equal to mg and F horizontal is equal to (umv^2)/r.
So, the final equation is mg=(umv^2)/r?
You cannot say that a vertical force is equal to a horizontall force. To clarify in your mind how this works, please answer the following questions
1. What is the sum of all the forces in the vertical direction?
2. What is the acceleration in the vertical direction at the moment the sock is just about ready to slip but has not started moving yet?
3. What is the sum of all the forces in the horizontal direction?
4. What is the acceleration in the horizontal direction at the moment the sock is just about ready to slip but has not started moving yet?

You get the two equations I mentioned by setting your answer to 1 equal to mass times your answer in 2 and your answer in 3 equal to mass times your answer in 4. That's how it's supposed to work and don't worry that the mass of the sock is not given.
 

Related to Centripetal force problem involving a washing machine

1. What is centripetal force and how does it relate to a washing machine?

Centripetal force is a force that acts towards the center of a circular motion. In the context of a washing machine, it is the force that keeps the clothes and water moving in a circular motion during the spin cycle.

2. How is the centripetal force calculated for a washing machine?

The centripetal force in a washing machine can be calculated using the formula F = mv^2/r, where F is the centripetal force, m is the mass of the clothes and water, v is the velocity of the spin, and r is the radius of the drum.

3. Can the centripetal force in a washing machine be too strong?

Yes, if the centripetal force is too strong, it can cause the clothes and water to stick to the sides of the drum, resulting in an unbalanced load and potential damage to the machine. This is why washing machines have a maximum spin speed and load capacity.

4. How does the centripetal force affect the efficiency of a washing machine?

The centripetal force helps to remove excess water from the clothes during the spin cycle, making the drying process more efficient. However, if the force is too strong, it can cause the clothes to become tangled and reduce the efficiency of the machine.

5. Can the centripetal force in a washing machine be adjusted?

Yes, the centripetal force in a washing machine can be adjusted by changing the spin speed or the load capacity. Some newer models also have sensors that can detect an unbalanced load and adjust the spin speed accordingly to maintain a safe and efficient level of centripetal force.

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