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Marble Shoot Loop-the-Loop

  1. Aug 9, 2010 #1
    1. The problem statement, all variables and given/known data
    A popular pastime used to be to construct helter-skelters: tracks along which to
    roll (without slipping) a ball (a marble say, or ball bearing). I made one that mimics a
    fairground loop-the-loop (see attached .jpg). If the drop from start to base of the loop is D, and the top of the loop is a height H above the bottom, then what is the largest theoretically
    possible value for H/D?

    2. Relevant equations
    ?? - I am very confused.

    Also, if my picture is bad, the height D is obviously supposed to be larger than the height H.
    Thanks in advanced

    Attached Files:

  2. jcsd
  3. Aug 9, 2010 #2


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    Staff: Mentor

    Welcome to the PF. The easiest way to solve this type of problem is to use energy considerations. When the ball is at the top of the D ramp, it has all potential energy (PE) and no kinetic energy (KE) because it is not rolling yet. Are you familiar with how to calculate the PE of an object of mass m raised by some height D?

    At the bottom of the ramp, the ball has all KE, although part of that KE is from its linear motion (what is the equation for KE as a function of an object's mass and velocity?), and part is from its rotation (are you familiar with a ball's Moment of Inertia and the energy associated with rolling?).

    At the top of the loop, the ball has to be moving fast enough not to fall off the track. Are you familiar with how to figure that out?
  4. Aug 9, 2010 #3
    Thank you for the help! I should have realized this was an energy problem.

    PE = mgD
    KE (at the beginning of the ramp) = .5mv2 + .5Iw2

    Now I don't know how fast the ball has to move at the top of the loop to not fall off....

    Thank you so far though!
  5. Aug 9, 2010 #4
    I guess the marble is supposed to complete one round in the loop? Or it just has to go and stop at the highest point of the loop?
    Anyway, think of contact. If it loses contact with the loop, then what happens is the normal force = ..... And when it comes to force, you can calculate it, since you know speed by the energy conservation law and weight, right? :wink:
  6. Aug 9, 2010 #5
    you can use kinematics to do it that would take you a long time and its tedious, so use gravitational and kinetic energy, basically conservation of energy to solve it. which would be very fast
  7. Aug 9, 2010 #6
    So at the top of the loop-the-loop,
    mgD = mgH + .5Iw2 + .5mvH2.

    The marble will not fall with the same force it will fall. So at the top of the loop, the marble would fall if both rotational and kinetic energy were not up to par???

    But I can't seem to figure out when the normal force of the track will be less than the force of gravity pulling the marble down. I can't seem to relate force to the energy equations.
  8. Aug 9, 2010 #7
    i dont think you can relate force to energy..
  9. Aug 9, 2010 #8
    hmmm....What should my next step be? At height H on the loop-the-loop what will cause the marble to fall/not fall?
  10. Aug 9, 2010 #9
    do you have to finish this by today.
    cuz im kinda busy got physic finals tomorrow morning.
  11. Aug 9, 2010 #10
    oh don't worry about it. Its not actually homework or anything. Its just to help me practice for the midterm which is on Thursday for me.

  12. Aug 10, 2010 #11
    Well, in this problem, dynamics is inevitable.

    @topgun08: Think of my hint :wink: The key is the normal force. The energy conservation law is just for calculating speed.
  13. Aug 10, 2010 #12
    Okay here is what I have After checking out the equations and realizing what your normal force hint was all about.

    KEat any point = .5mv2 + .5Iw2

    Using the known formulas for a sphere's moment of inertia and for angular acceleration...
    I = 2/5mr2
    w = v/r
    I substitute these formulas into my KE equation and compute:
    KE = .5mv2 + .5(2/5mr2)(v/r)2
    KE = .5mv2 + .2mv2
    KE = .7mv2

    Now once inside the loop-the-loop the marble will have centripetal acceleration. For this problem we want the normal force at the top of the loop-the-loop to be 0. So the only force acting on it will be the centripetal force whose acceleration is equal to gravity's acceleration. Thus the Fc = mac where ac = g

    Using the formula that ac = v2/R where R is the radius of the loop-the-loop, I can solve for v2
    ac = g = v2/R ==> v2 = Rg

    Plug this back into the KE equation:
    KE = .7mv2
    v2 = Rg
    KE = .7mRg

    Applying the conservation of Energy Law at the top of the loop-the-loop:
    PEtotal = KEtop + PEtop
    mgD = .7mRg + mgH
    Realize that H = 2R
    D = .7R + 2R
    D = 2.7R
    (1/2.7)D = R
    .74 = 2R/D
    .74 = H/D

    Is my logic and math right? Any mistakes or is this correct?
  14. Aug 10, 2010 #13
    It's correct :wink:
  15. Aug 10, 2010 #14
    thanks for the help. much appreciated!
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