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Cylinder rotating; coeff of friction?

  • Thread starter cugirl
  • Start date
  • #1
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Homework Statement


At an amusement park, some physics students are on a ride where they stand against the wall in a cylindrical apparatus that begins to rotate around its center. Once they are moving fast enough the floor drops and they “stick” to the wall due to friction. The radius of the cylinder is R = 3.5 m and the minimum speed the ride must move the people at to keep them from sliding down the wall is 10 m/s. What is the coefficient of static friction between the student and wall?


Homework Equations


Friction = mu * normal
a = V^2 / r


The Attempt at a Solution


I am not sure how to start and which equation I need to use. The answer is supposedly 0.34, but I don't know how to get there.
 

Answers and Replies

  • #2
2,149
555
Try drawing a FBD and write the equations of motion from there.
 
  • #3
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I did. I have Fn perpendicular to the person and pointing toward the center, down is mg, and up is Fk.
 
  • #4
2,149
555
What do you mean by Fk?

Where are your equations of motion?
 
  • #5
17
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Fk = friction

I have Fk - mg = ma
Fk - mg = m*v^2 / R
 
  • #6
14
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Ok. Let's imagine you're in this rotating thing. The force which is pressing you to the wall is centrifugal. So the Fn has the same amount but opposite direction of centrifugal force. Now you have realized which force is Fn, you can determine the Fk. If you want to be motionless in this rotating system, you aren't allowed neither to move up or down. In other words, the resultant force looking on the y-axis should be 0.

I drew you the picture of it. You should be able to understand it now.

http://img9.imageshack.us/img9/5894/assign1.jpg [Broken]
 
Last edited by a moderator:
  • #7
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555
Fk suggests that you think this is kinetic friction, but the whole point is that you do not want to slide, thus static friction is involved, not kinetic.

Think about the direction of your forces. In you equation of motion, you have set two vertical forces equal to a horizontal M*a term; this cannot be correct.
 
  • #8
minger
Science Advisor
1,495
1
Perhaps the reference frame is confusing you. Try drawing the problem rotated 90° such that the wall is horizontal, and the persons weight (ma) is acting sideways. You then have the frictional force acting in the opposite direction (opposite of motion), and the normal force from centripital acceleration acting downwards.
 
  • #9
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555
How will it help to have the wall horizontal, which puts gravity horizontal and the whole world on edge, so to speak? Why not work with things as they really are? The friction force always acts opposite to (impending) motion in any case, so standing the world on its head can't improve that.
 
  • #10
minger
Science Advisor
1,495
1
Because we typically think of gravity as acting downward. Therefore, the normal force that friction acts upon is normally downward.

In this example, gravity isn't necessarily the normal force that is resisting movement, it is the moving force; the normal force is generated by centripital force. By drawing it sideways, you get the classic "block on a plane" type question where there is one force acting on a block in plane. Friction acts opposite with the normal force (being centripital) acting downward.

It might not necessarily help for you, but the OP seemed to be having difficulty with the problem. Often times one just needs to look at it from a different perspective. More often than not, examining things in a different reference plane can help.
 
  • #11
17
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Thank you for the person (Lopina) who drew the FBD. Mine did not look like that, and I was able to solve it!
 

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