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Centripetal Acceleration Question

  1. Nov 9, 2006 #1
    :rofl:

    An amusement park ride consists of a round room. Once you are placed against the outer wall, the room begins to spin. After the room attains a critical rotational speed, the floor drops out, but you remain against the wall. The radius of the room is 30m, and the coefficient of friction between you and the wall is u = .02. What is the minimum period of rotation that the room must spin in order for you not to fall? What restriction shold the park enforce regarding the maximum weight of the rider?

    Okay, well I know that
    a_c = (v^2)/r
    v = 2(pi)(r)/(T)
    a_c = 4(pi^2)r/(T^2)
    F_c = m(v^2)/r
    F_c = (u_s)(F_n) = (u_s)(m)(g) = (m)(v^2)/r
    v = sqrt ((u_s)(g)(r))

    I'm assuming "period of rotation" means centripetal acceleration. In that case,

    v = sqrt ((u_s)(g)(r)) = sqrt ((.02)(9.8m/s^2)(30 m)) = sqrt (5.88 m^2/s^2) = 2.425 m/s
    a_c = ((2.425 m/s)^2)/(30m) = .196 m/s^2

    From this equation, F_c = (u_s)(F_n) = (u_s)(m)(g) = (m)(v^2)/r, we see that the m's cancel out, so in regards to the weight of the rider, it shouldn't matter when he/she rides the ride.

    I was just wondering if someone could check my work and see if I got the right answer. If I did not, could someone please give me a lead of what to do next, but not tell me the answer until I get it myself? Thanks :biggrin:
     
  2. jcsd
  3. Nov 9, 2006 #2

    OlderDan

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    Looks to me like you lost it on the second line that has F_c. What are you trying to say there? Think again about the relationship between F_c and F_n and the relationship between F_n and weight.
     
  4. Nov 9, 2006 #3
    F_c = (u_s)(F_n) = (u_s)(m)(g) = (m)(v^2)/r
    I actually got this straight from my textbook. The centripetal force is equal to the coefficient of static friction times the normal force. In this case the normal force is mg cos 0, which is just mg. So F_c = (u_s)(m)(g), which equals m times a_c because F=ma. And a_c is (v^2)/r.

    What did I do wrong?
     
  5. Nov 9, 2006 #4

    OlderDan

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    If you got that equation from your textbook, it was probably in connection with something in circular motion on a horizontal surface, perhaps a car on a flat circular track. That is not what you have here. Look at the surface you do have (vertical) in relation to the centripetal force, and look at the direction of the friction force in relation to the centripetal force and gravity.
     
    Last edited: Nov 9, 2006
  6. Nov 9, 2006 #5
    Oh okay thanks... lets see... yeah. I think it was for a flat surface. Then let me rethink.. the fritcional force should be pulling the rider towards the middle of the circle... and I would use sine for the vertical component. fritcional force is to the left, and F_n is positive.. so maybe

    F_c = F_n sin (90) - uFf = m(v^2)/r
    F_c = mg sin (90) - (.02)mg cos(0) = m(v^2)/r
    I think m's still cancel out.. so
    F_c = g sin 90 - .02g cos 0 = (v^2)/r
    v^2 = r(g sin 90 - .02g cos 0)
    v^2 = 30(9.8sin90 - .02(9.8cos0)
    v^2 = 30(9.8) - 0.2(9.8)
    v^2 = 293.8
    v = 17.14 m/s
    a_c = 17.14^2 / 30
    a_c = 9.8 m/s^2

    Am I closer?
     
  7. Nov 9, 2006 #6
    Oh yeah and I've noticed you helped me out a few times before ^^. I appreciate all of the time and effort you put into helping me and everyone else. I really do admire your deeds, and hopefully when I go to college I'll have learned enough physics to help out people at least on the introductory level ^^. You really encourage me to learn the material =)
     
  8. Nov 9, 2006 #7

    OlderDan

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    You are still not getting the directions right. A normal force is always perpendicular to a surface and a frictional force is always parallel to the surface. In this problem, the surface is vertical. Therefore the normal force must be horizontal and the frictional force must be vertical.
     
  9. Nov 9, 2006 #8
    If the normal force is horizontal and the frictional force is vertical.. then

    F_c = F_n - uF_f = m(v^2)/r
    F_c = mg cos (0) - (.02)mg sin(90) = m(v^2)/r
    The m's cancel out.. so
    F_c = g cos(0) - .02g sin(90) = (v^2)/r
    v^2 = r(g cos (0) - .02g sin (90)
    v^2 = 30(9.8cos0 - .02(9.8 sin90)
    v^2 = 30(9.8) - 0.2(9.8)
    v^2 = 293.8
    a_c = (v^2)/r
    a_c = 293.8 / 30
    a_c = 9.8 m/s^2

    Hm... I seemed to get the same answer. I switched the directions (cos/horizontal for the normal force and sin/vertical for the frictional force.. right?) and still got the same answer. Was I supposed to get the same answer, and my error was just the directions? Or am I still wrong x_x;. I'm sorry.. I'm trying my best.
     
  10. Nov 10, 2006 #9

    OlderDan

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    Friction in this problem is vertical. It contributes nothing to the centripetal force. All of the centripetal force comes from the normal force. The faster the ride spins, the greater the centripetal force required, so the normal force increases with speed. Friction is what keeps the people in the ride from slipping down the wall. It acts upward, opposing the gravitational force (weight). If the ride is going fast enough, the normal force will be large enough so that the frictional force equals the weight of the people. The coefficient of static friction does not tell us the actual frictional force in most cases; it tells us the maximum possible frictional force associated with a normal force. The ride has to be going at least fast enough so that the maximum possible static friction equals the weight. If it goes faster, the frictional force will be less than the maximum possible, but that is OK. Static friction is whatever it has to be to keep something from sliding.
     
  11. Nov 10, 2006 #10
    Okay... let me take in what you said.
    The frictional force acts upward and the gravitation force acts downward. That means they cancel each other out. The normal force, then, must account for all of the centripetal force, since there are no other forces acting on the person. And in this case, the coefficient of static friction tells us minimum force required to keep the rider from falling. Therefore...

    F_c = uF_n = m(a_c)
    F_c = umgcos(0) = m(a_c)
    F_c = ug = (v^2)/r
    (v^2) = ugr
    v^2 = (.02)(9.8)(30) = 5.88 m/s
    a_c = (v^2)/r
    a_c = (5.88^2)/30 = .196 m/s^2

    Hm.. this is what I got before. I'll have to think again.. I'll modify this again when I come up with something. However, I think my mind may be clearer tomorrow because I am mentally exhausted because I got up early and have been up for almost 20 hours. So I will try this again tomorrow. I appreciate your patience and help very much... I'll do my best to find a correct method
     
    Last edited: Nov 10, 2006
  12. Nov 10, 2006 #11

    OlderDan

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    OK.. sleep on it. When you wake up, turn these words into equations

    centripetal force equals normal force

    coefficient of friction times normal force equals maximum frictional force

    maximum frictional force equals weight (condition at the minimum speed)
     
  13. Nov 10, 2006 #12
    F_c = F_n
    uF_n = F_f
    F_f = mg

    F_c = mgcos0
    .2mgcos0 = mgsin90
    mgsin90 = mg
    F_c = .2
    F_c = m(a_c)
    m(v^2/r) = .2
    v^2/r = .2/m
    v^2 = (.2r)/m
    v^2 = .2(30m - radius)/m - mass
    V = 6m/mass
    Uh.. I tried to follow what you said but the mass isn't given and I'm not sure what to do now :(.
    Oh.. and I realized that the coefficient of status friction is .2, not .02. Sorry.
     
  14. Nov 10, 2006 #13
    Maybe...
    F_c = mg/u
    m(a_c) = mg/u
    a_c = g/u
    (v^2)/r = g/u
    (v^2)= gr/u
    v = sqrt gr/u
    v = sqrt (9.8)(30)/(.2)
    v = sqrt 58.8
    v = 7.668
    a_c = (v^2)/r
    a_c = 58.8 / 30
    a_c = 1.98 m/s^2

    This makes a little more sense than my last post..
     
  15. Nov 10, 2006 #14

    OlderDan

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    I can't seem to get you to give up on trying to make a direct connection between F_c and mg. :smile:

    Look what happens if you just make substitutions based on the first 3 equations, and then equate F_c to mv^2/r

    F_c = F_n = F_f/μ = mg/μ = mv^2/r

    g/μ = v^2/r

    v^2 = r*g/μ

    v = sqrt(r*g/μ)

    T = 2πr/v (π is pi)

    The question is actually worded incorrectly. This is the maximum period you can have and not have people falling out the bottom. Shorter periods mean higher velocity and greater friction to hold everyone in place.

    You are correct that mass divides out, so nobody should be excluded on the basis of them falling out the bottom. What is a consideration is the maximum amount of mass the vertical wall can support in the horizontal direction. The more mass (and the greater the speed) the greater the force being exerted by the people on the walls. You would probably also be somewhat concerned about distributing the mass uniformly around the perimeter.
     
  16. Nov 10, 2006 #15
    Hm.. actually on my last post before yours I did get up to the v = sqrt(r*g/μ) part :)! So I got v = 7.668.

    So what exactly does "period of rotation" mean? Time? Centripetal acceleration? If it's time, then T=2pi(30)/(7.668) which is 24.58 s. If it's centripetal acceleration, then its 1.96 m/s^2.

    So.. period of rotation is in "seconds" for the unit?
     
  17. Nov 10, 2006 #16
    Yeah. I think I got my final answer now. T = 24.58 s, because period is expressed in seconds. And seconds are a measure of time. And the mass of the rider doesn't matter because the m's cancel out.

    Okay.. thanks so much ^^! You're the best.
     
  18. Nov 11, 2006 #17

    OlderDan

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    So you did. I guess I was too focused on your first response and did not look carerully enough at the second one. Good going.
     
  19. Nov 11, 2006 #18
    ^_^ It's all thanks to you in the end though :D
     
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