1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    using constraint wr=v, knowing that v=(rg)^0.5, I get

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook