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Homework Help: Cons. of energy with rotational motion

  1. Nov 18, 2008 #1
    In Figure 11-32, a solid brass ball of mass m and radius r will roll without slipping along the loop-the-loop track when released from rest along the straight section. For the following answers use g for the acceleration due to gravity, and m, r, and R, as appropriate, where all quantities are in SI units.


    (a) From what minimum height h above the bottom of the track must the marble be released to ensure that it does not leave the track at the top of the loop? (The radius of the loop-the-loop is R. Assume R > r.)

    (b) If the marble is released from height 6R above the bottom of the track, what is the magnitude of the horizontal component of the force acting on it at point Q?

    For part a I came up with the equation

    mgh = (1/2)mv2 + (1/2)mv2 + 2mgR
    gh = (7/10)v2 + 2gR

    I think that's right, but I can't figure out how to solve for v. I'm guessing that It has something to do with centripetal force.

    Part b probably needs an answer from part a. The horizontal component of the force is the normal force (= centripetal force?).

    Help please.
  2. jcsd
  3. Nov 18, 2008 #2
    your logic is good (except I don't get your second 1/2mv2).

    You are right about centripital force. At the top the ball needs enough velocity so that its centripital inertia= its weight. The force of gravity must equal its centripital force. This will translate directly into an equation for velocity from centripital acceleration or froce.
  4. Nov 18, 2008 #3

    Doc Al

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    Staff: Mentor

    OK. (That second term in your first equation should be rotational KE.)

    Right. Apply Newton's 2nd law to solve for v.

    Not really.
  5. Nov 18, 2008 #4
    For that I meant to say 1/5mv2

    For part a I set N = 0 at the top of the loop

    N + mg = mv2/(R-r)

    g = v2/(R-r)

    v2 = g(R-r)

    gh = (7/10)g(R-r) + 2gR

    h = (7/10)g(R-r) + 2R

    Part b

    Using the equation from earlier...

    g(6R) = (7/10)v2 + gR
    v2 = (50/7)gR

    N = mv2 / (R-r)
    N = m(50gR)/(7(R-r))

    But it won't take my answer. =(

    What did I do wrong now?
  6. Nov 18, 2008 #5

    Doc Al

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    Staff: Mentor

    The height of the ball's center at the top of the motion should be 2R-r, not 2R.

    This looks OK to me.
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