Centripetal force problem involving a washing machine

[Caw]

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!

 

[kuruman]

Hi Caw and welcome to PF.

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.

 

[Caw]

Hi Caw and welcome to PF.
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.

 

[kuruman]

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.

 

[Caw]

F vertical is equal to mg and F horizontal is equal to (umv^2)/r.
So, the final equation is mg=(umv^2)/r?

 

[kuruman]

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.