What Is the Minimum Coefficient of Friction Needed in a Rotating Cylinder Ride?

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Homework Help Overview

The problem involves a rotating cylinder ride where riders are suspended against the wall after the floor drops away. The objective is to determine the minimum coefficient of friction required to prevent the riders from slipping. The subject area includes concepts of circular motion, forces, and friction.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the riders, including gravity, normal force, and friction. There are attempts to clarify the directions of these forces and their roles in maintaining equilibrium.

Discussion Status

The discussion is ongoing, with participants exploring the relationships between the forces involved. Some guidance has been offered regarding the nature of the normal force and friction, but there is no explicit consensus on the complete understanding of the forces at play.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or reference. There is a focus on accurately identifying and interpreting the forces in the context of the problem.

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Homework Statement



a rotating cylinder of radius 3 m is set in rotation at an angular speed of 5 radians/second. the floor then drops away, leaving the riders suspended against the wall in a vertical position. what minimum coefficient of friction between a rider's clothing and the wall is needed to keep the rier from slipping?

Homework Equations





The Attempt at a Solution



i've attempted to draw a free body diagram, cause I'm pretty sure that's the first thing i have to do. but i have to say, this FBD, is VERY VERY tricky.
 
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What forces are acting on the people?(list them please :)) What does the fact that the people do not accelerate upwards or downwards tell you about the vertical forces acting on the people?
 
blader324 said:

Homework Statement



a rotating cylinder of radius 3 m is set in rotation at an angular speed of 5 radians/second. the floor then drops away, leaving the riders suspended against the wall in a vertical position. what minimum coefficient of friction between a rider's clothing and the wall is needed to keep the rier from slipping?

Homework Equations





The Attempt at a Solution



i've attempted to draw a free body diagram, cause I'm pretty sure that's the first thing i have to do. but i have to say, this FBD, is VERY VERY tricky.


What forces did you come up with and in what direction do you think they are pointing? It's actually not that hard once you knwo what forces are involved and you get their directions right. Hint: there are three forces.
 
okay...so i chose the position when the person is standing to the far most right hand side so that the centripetal acceleration is towards the left. i also have a normal force from the floor, her mass and gravity acting downwards, and then friction acting in the downwards direction ( i think)
 
blader324 said:
okay...so i chose the position when the person is standing to the far most right hand side so that the centripetal acceleration is towards the left. i also have a normal force from the floor, her mass and gravity acting downwards, and then friction acting in the downwards direction ( i think)

Watch out.

You are ok with the centripetal acceleration. Just be careful to remember that it's not a force (so, to be strictly speaking, it is not suppoed to be in a free body diagram).

The floor is no longer there so there is no normal force exerted by the floor!

Gravity is acting downward, yes.

But the person has no vertical acceleration, right? So what does it tell you about the net vertical force?
 
okay...so i do have friction acting on the person right (towards the right?) but nrged said that there were three forces...i still have no idea what force I'm missing.
 
and also...does that mean that the net vertical force is equivalent to the mass and gravity?
 
blader324 said:
and also...does that mean that the net vertical force is equivalent to the mass and gravity?


well, the vertical acceleration is zero so the net vertical force is zero, right?
This means that there must be an other force upward which cancels gravity (well, there could be in general several other forces but it turns out here that there is only one other vertical force, acting straight up).

Now, picture the situation and think: what could be the force acting straight up??
 
my only guess is that friction is preventing the person from flying out of the cylinder or falling out from under it.
 
  • #10
blader324 said:
my only guess is that friction is preventing the person from flying out of the cylinder or falling out from under it.

Let's tackle these two questions one at a time. It's clear that friction plays a role here but the question is: in what direction is the friction force acting?
 
  • #11
okay...if friction isn't acting upwards, that means that friction is acting towards the right right?
 
  • #12
blader324 said:
okay...if friction isn't acting upwards, that means that friction is acting towards the right right?

Well, if friction is not acting upward, what force could be acting upward to cancel the force of gravity?
 
  • #13
okay. no normal force, gravity we already have. not friction in that direction. what other forces are there? acceleration is not a part of the diagram. is there tension? no...that can't be it.
 
  • #14
blader324 said:
okay. no normal force, gravity we already have. not friction in that direction. what other forces are there? acceleration is not a part of the diagram. is there tension? no...that can't be it.

I never said there was no normal force:wink:

There is a normal force. And there is friction. The trick is to figure out in what directions they are acting. Can you tell?
 
  • #15
really...alright...so if the normal force isn't acting up...does that mean that it's acting down? ...trying to push the person down? but i don't understand how that would make sense.
 
  • #16
OR..since the normal force is a contact force, does that mean that the normal force is between the person and the cylinder?
 
  • #17
blader324 said:
really...alright...so if the normal force isn't acting up...does that mean that it's acting down? ...trying to push the person down? but i don't understand how that would make sense.

You must recall a basic property of a normal force. First, why is it called a normal force in the first place? why the adjective "normal"?
 
  • #18
normal means a force caused by contact perpendicular to the contact surface. so that does mean that the normal force is between the person and the cylinder right?
 
  • #19
blader324 said:
normal means a force caused by contact perpendicular to the contact surface. so that does mean that the normal force is between the person and the cylinder right?

Yes, a normal force is always perpendicular to the surface. And it points away from the surface. In your drawing, the wall i sto the right of the person so the wall pushes left.

What about friction now? In what direction is the friction force between an object and a surface?
 
  • #20
friction acts along the contact surface, but it tries to prevent acceleration, which means that it's acting in the same direction as the normal force?
 
  • #21
no, that's not right. you still need a force to cancel out gravity. so does that mean that friction is going to be upwards?
 
  • #22
blader324 said:
friction acts along the contact surface, but it tries to prevent acceleration, which means that it's acting in the same direction as the normal force?

The normal force is perpendicular to the surface and the friction force is parallel to the surface so they can't be in the same direction. (even if you think about a simple case like an object at rest on an inclined plane, the static friction force and the normal are not in the same direction)


It's a bit misleading to say that the friction force tries to prevent acceleration. It's better to say that a static friction force tries to prevent the object from slipping on the surface. In other words, imagine that the surface was perfectly frictionless. In what way would the object (the person) slide? The static friction force will be opposite to that.

So in what direction does it point?

EDIT Yes, the friction force is up and it's the force that cancels gravity!
 
  • #23
Whoever said there were 3 forces, I disagree. I count 4.

Ask yourself these questions:

What is the ride doing? Which way is it accelerating me?
What surfaces am I in contact with?
If there's a normal force perpindicular to ANY SURFACE I PUSH ON, what way is that normal force pushing? (remember perpinducular)
What way is gravity pushing/pulling me?
If there's some frictional force, what does friction do? Friction IMPEDES MOTION. So which way should the friction be forcing to impede my motion?
 
  • #24
exactly...so...without friction the person would fall down, but the fact that friction is present is preventing the person from slipping down, which means that friction is going upwards. THANK YOU SOOOOOOO MUCH! now all i have to do is the net forces in the x and y direction and then i'll definitely get it. THANKS SO MUCH...IT ALL MAKES SENSE NOW!
 
  • #25
K.J.Healey said:
Whoever said there were 3 forces, I disagree. I count 4.

Ask yourself these questions:

What is the ride doing? Which way is it accelerating me?
What surfaces am I in contact with?
If there's a normal force perpindicular to ANY SURFACE I PUSH ON, what way is that normal force pushing? (remember perpinducular)
What way is gravity pushing/pulling me?
If there's some frictional force, what does friction do? Friction IMPEDES MOTION. So which way should the friction be forcing to impede my motion?


There is a normal force (left), a static friction force (up) and gravity (down).
What is the fourth force?
 
  • #26
blader324 said:
exactly...so...without friction the person would fall down, but the fact that friction is present is preventing the person from slipping down, which means that friction is going upwards. THANK YOU SOOOOOOO MUCH! now all i have to do is the net forces in the x and y direction and then i'll definitely get it. THANKS SO MUCH...IT ALL MAKES SENSE NOW!

Great!:-p You are very welcome.

As you now see, the friction force is indeed the force that keeps the person from sliding down, it's the force that cancels gravity. And the normal force provides the centripetal force.

The equation along x will give you the value of the normal force (since you know the rate of rotation). Now, using the fact that the net force along y is zero will give you the value of the coefficient of static friction.

best luck!
 

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