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Circular Motion Problem with a Motorcycle Driving in a Sphere

  1. Apr 1, 2012 #1
    I have tried this problem multiple times without success. Please provide a full solution. Thanks!

    A local circus act features a motorcycle rider inside a hollow sphere. He accelerates his bike to a maximum velocity in order to allow himself to be able to drive parallel to the ground inside the sphere. The sphere has an inner diameter of 8.00m, the motorcycle and rider have a mass of 250kg and his tires have a coefficient of static friction of 0.80 with the inside surface of the sphere. What is the minimum speed he must attain in order to keep from falling inside the sphere?
     
  2. jcsd
  3. Apr 1, 2012 #2

    cepheid

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    We don't do your homework for you on this site. That would be of no help to you whatsoever anyway. You might want to take a look at the PF rules. You need to show your attempt at a solution, and if you do, then people can begin to guide you in the right direction. What have you done so far on this problem?

    Also, this should have been posted in the homework help section (but it will be moved there).
     
  4. Apr 1, 2012 #3

    DaveC426913

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    :bugeye:
    Boy have you ever come to the wrong place...
     
  5. Apr 1, 2012 #4
    Okay..I was trying to save time. Here is what I have thus far:

    EF = Fs - Fg

    Therefore: Fs - Fg = mv^2 / r

    mu_s x mg - mg = mv^2 / r

    mg (mu_s - 1) = mv^2 / r

    r mg (mu_s - 1) = mv^2 / r

    [r mg (mu_s - 1)] / m = v^2

    rg (mu_s -1 ) = v^2

    This, however will not give me the minimum velocity because I will be squaring a negative. The final formula should be like this, but I can't find out how the mu_s is the denominator.

    v^2 = rg / mu_s

    I am not trying to have my homework done for me. I have made legitimate attempts at this problem for the past half hour and would like some help. Thanks.
     
  6. Apr 1, 2012 #5
    First you have to draw a diagram- Free-body diagram
    http://en.wikipedia.org/wiki/Free_body_diagram

    Then run the video in your head what happen when the bike is at the top of hollow sphere and parallel to ground BUT not falling.
     
  7. Apr 1, 2012 #6
    1. A local circus act features a motorcycle rider inside a hollow sphere. He accelerates his bike to a maximum velocity in order to allow himself to be able to drive parallel to the ground inside the sphere. The sphere has an inner diameter of 8.00m, the motorcycle and rider have a mass of 250kg and his tires have a coefficient of static friction of 0.80 with the inside surface of the sphere. What is the minimum speed he must attain in order to keep from falling inside the sphere?

    2. Ʃf = mv^2 / r

    Fs = μs Fn


    3. EF = Fs - Fg

    Therefore: Fs - Fg = mv^2 / r

    mu_s x mg - mg = mv^2 / r

    mg (mu_s - 1) = mv^2 / r

    r mg (mu_s - 1) = mv^2 / r

    [r mg (mu_s - 1)] / m = v^2

    rg (mu_s -1 ) = v^2

    This, however will not give me the minimum velocity because I will be squaring a negative. The final formula should be like this, but I can't find out how the mu_s is the denominator.

    v^2 = rg / mu_s
     
  8. Apr 1, 2012 #7

    K^2

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    Picture the whole thing rotated so that the bike seems to ride on a horizontal surface. Remember critical angle? You have "horizontal" gravity fixed at mg. You have vertical "gravity" due to centrifugal effect at mv²/r. Now you need to find the ratio of the two that allows the motorcycle to stay put without slipping.
     
  9. Apr 1, 2012 #8
    I would really like some real help, since I have been working on this problem for over two hours without an answer. I am not getting the right answer and an algebraic solution for what I attempted above would be helpful. I have a FBD and know where the forces are directed.
     
  10. Apr 1, 2012 #9

    K^2

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    Then why do you have up there in your equations the centripetal force, gravity, and friction all running in the same direction?
     
  11. Apr 1, 2012 #10

    cepheid

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    Just to clarify (and maybe help the OP out a little bit here), you're considering the instant at which the motorbike first goes vertical (i.e. halfway up the loop), because this is the first instant at which the bike gets no help from the normal force for vertical force balance, and has to rely entirely on friction, right?
     
  12. Apr 1, 2012 #11

    K^2

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    The motorcycle in this problem travels in horizontal circle, not vertical. Quick check against the solution verifies it.

    Besides, for a vertical loop, friction coefficient is not a factor at the top point, and it's the only one that matters.
     
  13. Apr 1, 2012 #12
    I know that the motorcycle is going horizontal. Therefore, Fs acts upward, Fn acts toward the center of the circle, and Fg acts downward. This isn't where I am having trouble. I have a correct FBD and have labelled all of the forces correctly.
     
  14. Apr 1, 2012 #13

    cepheid

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    Sorry, my bad. I misread the problem.
     
  15. Apr 1, 2012 #14

    cepheid

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    Right, exactly. Therefore, equating Fs - Fg to mv^2/r (like you did in your algebra above) is just wrong. You're equating the sum of the vertical forces to the horizontal force, which makes no sense.

    If the bike is not falling, then what, then, should the sum of vertical forces be equal to?
     
  16. Apr 1, 2012 #15
    It will be 0
     
  17. Apr 1, 2012 #16

    cepheid

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    Having another look at your equations, you also seem to be under the impression that the normal force (the thing that you multiplied by mu) on the bike is equal to mg (in spite of the fact that this force acts horizontally here). I would rethink that. The normal force is just the contact force between any two surfaces that are touching. It doesn't always have to be equal to mg.

    There is another criterion that determines what the normal force must be at all points on the circle in this problem :wink:
     
  18. Apr 1, 2012 #17

    cepheid

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    Great, so can you re-work your equations with that in mind?
     
  19. Apr 1, 2012 #18
    Ef = Fs - Fg
    0 = Fs - Fg
    Fs = Fg
     
  20. Apr 1, 2012 #19
    This is where I am lost; I know that Fs = mu_s Fn and since Fn is acting inward, it is equal to Fac (centripetal acc.) and therefore Fs = mu_s Fac. However, what can Fac be? I know that Fac = V^2 / r , but that is of no help.
     
  21. Apr 1, 2012 #20

    cepheid

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    You're absolutely right with your reasoning. Since the thing is moving in a circle, there is a requirement that there be a centripetal force. But since the circle is horizontal, the centripetal force is also always horizontal (i.e parallel to the ground). Of the physical forces present in the problem neither friction nor gravity is horizontal, therefore they cannot be providing the centripetal force. The only thing that remains to provide a centripetal force is the normal force of the contact between the bike and the wall. Conclusion: the normal force is equal to the centripetal force. This is what I was hinting at with my post #17 and you got it.

    You're very close with the math. Centripetal acceleration is equal to v2/r. Centripetal force is that multiplied by m.

    I don't understand why you think this insight is of no help? It's actually very useful to you. Now that you know what the normal force is (in terms of v and r), you can multiply it by mu to get the friction and then plug that into your force balance equation. A little bit of algebra later, and you'll be done!
     
    Last edited: Apr 1, 2012
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