a popular amusement park ride


by blader324
Tags: amusement, park, popular, ride
blader324
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#1
Jul31-07, 11:53 AM
P: 51
1. The problem statement, all variables and given/known data

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?

2. Relevant equations



3. 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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dontdisturbmycircles
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#2
Jul31-07, 11:57 AM
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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?
nrqed
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#3
Jul31-07, 11:59 AM
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Quote Quote by blader324 View Post
1. The problem statement, all variables and given/known data

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?

2. Relevant equations



3. 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.

blader324
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#4
Jul31-07, 12:02 PM
P: 51

a popular amusement park ride


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)
nrqed
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#5
Jul31-07, 12:06 PM
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Quote Quote by blader324 View Post
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?
blader324
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#6
Jul31-07, 12:19 PM
P: 51
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.
blader324
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#7
Jul31-07, 12:20 PM
P: 51
and also...does that mean that the net vertical force is equivalent to the mass and gravity?
nrqed
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#8
Jul31-07, 12:23 PM
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Quote Quote by blader324 View Post
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??
blader324
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#9
Jul31-07, 12:25 PM
P: 51
my only guess is that friction is preventing the person from flying out of the cylinder or falling out from under it.
nrqed
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#10
Jul31-07, 12:26 PM
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Quote Quote by blader324 View Post
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?
blader324
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#11
Jul31-07, 12:28 PM
P: 51
okay...if friction isn't acting upwards, that means that friction is acting towards the right right?
nrqed
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#12
Jul31-07, 12:43 PM
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Quote Quote by blader324 View Post
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?
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#13
Jul31-07, 12:49 PM
P: 51
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.
nrqed
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#14
Jul31-07, 12:52 PM
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Quote Quote by blader324 View Post
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

There is a normal force. And there is friction. The trick is to figure out in what directions they are acting. Can you tell?
blader324
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#15
Jul31-07, 12:55 PM
P: 51
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.
blader324
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#16
Jul31-07, 12:56 PM
P: 51
OR..since the normal force is a contact force, does that mean that the normal force is between the person and the cylinder?
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#17
Jul31-07, 12:56 PM
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Quote Quote by blader324 View Post
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"?
blader324
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#18
Jul31-07, 12:58 PM
P: 51
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?


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