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Homework Help: Loop a loop

  1. Nov 24, 2011 #1
    1. The problem statement, all variables and given/known data
    A marble rolls down a track and around a loop-the-loop of radius R. The marble has mass m and radius r. What minimum height h must the track have for the marble to make it around the loop-the-loop without falling off?

    2. Relevant equations

    What minimum height h must the track have for the marble to make it around the loop-the-loop without falling off?

    3. The attempt at a solution
    Find initial energy: mgh
    Final energy tabulation: mg(2R) + 0.5mrg + 0.5(2/5)mr^(2)(v/r)^2
    Set equal to initial energy, I obtain 2R+7/10 =h, which isn't the right answer, what did I do wrong?

    I found the velocity at top by using newton's 2nd law, i get v = squareroot(rg). I find w using constraint of wr=v. I get 2/5(mr^2) using moment of inertia for a solid sphere.

    Any help would be appreciated. I'm so confused why this doesn't work out.
     
  2. jcsd
  3. Nov 24, 2011 #2
    I'm still new to this, but it looks like you've got everything right except for the last part of your final energy calculation (the 0.5(2/5)mr^(2)(v/r)^2 part). I'm not quite sure why it's there, but I'm pretty sure that's the part that's throwing you off.
     
  4. Nov 24, 2011 #3
    The last part is the kinetic rotational energy using I = (2/5)(M)(R^2)
    Using kinetic rotational energy = 0.5Iw^2, I have
    0.5(2/5)(MR^2)w.
    using constraint wr=v, knowing that v=(rg)^0.5, I get
    0.5(2/5)mr^(2)(v/r)^2

    The answer isn't right.
     
  5. Nov 24, 2011 #4
    The energy at the top of the loop is the potential (which you have) plus it's kinetic, since it is moving in a circle there must be an acceleration at the top equal to [itex] \frac{V^2}{R} [/itex], for this to be a minimum (IE only just making it around the loop) this acceleration must be set equal to [itex]g[/itex] so we have
    [tex] V^2 = gR \Rightarrow KE = \frac{1}{2}mgR [/tex]
    [tex]\displaystyle{\dot{. .}} mgh = mg(2R) + \frac{1}{2}mgR [/tex]
    Which you can then solve for the minimum height :)

    EDIT: Just reread your question, the "effective" radius of the loop is changed because of the radius of the marble (the center of mass of the marble moves around a slightly smaller loops) and there is indeed rotational energy equal to [itex]\frac{1}{2}\frac{2}{5}\frac{gR}{r^2}mr^2 [/itex] (again this has to be changed because the [itex]R[/itex] isn't really the radius of the loop that the center of mass moves over, but I'll leave that to you)

    :)
     
    Last edited: Nov 24, 2011
  6. Nov 24, 2011 #5
    Never mind. Thanks for all your suggestions. I've just solved them. I left out the little tiny details which messed up the problems. I didn't take into account of the center of mass, which I paid dearly.
    Here is how I solved it.
    Initial energy: mg(h+r)
    At the loop, the radius is R-r because we're taking into account of the center of mass. Hence using Newton's 2nd law.
    m(v^2)/(R-r) = mg, solve for v = (g(R-r))^0.5 (1)

    Final energy at top of the loop is:
    mg(2R-r) + 0.5(m)(g(R-r)) + 0.5(2/5)(mr^2)(w^2).
    Using velocity constraint, w=v/r=(1/r)(g(R-r))^0.5
    Plug that in (1), and set (1) equal to initial energy, then solve for h.
     
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