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Homework Help: Loop the Loop (circular motion)

  1. Feb 8, 2009 #1
    1. The problem statement, all variables and given/known data
    The two problems below are related to a cart of mass M = 500 kg going around a circular loop-the-loop of radius R = 10 m, as shown in the figures. All surfaces are frictionless. In order for the cart to negotiate the loop safely, the normal force exerted by the track on the cart at the top of the loop must be at least equal to 0.8 times the weight of the cart. You may neglect the size of the cart. (Note: This is different from the conditions needed to "just negotiate" the loop.)
    a) For this part, the cart slides down a frictionless track before encountering the loop. What is the minimum height h above the top of the loop that the cart can be released from rest in order that it safely negotiate the loop?

    2. Relevant equations
    a_c = (V^2)/R
    F = ma
    V = sqrt(2gh)

    3. The attempt at a solution
    Top of circle
    a_c equals the sum of the other forces, so
    V = 10.295 m/s, a_c = 10.6 m/s/s
    height to have given velocity - 5.41m
    This is where I don't know. How does having centripital acceleration help in finding the height at all? Does it?
  2. jcsd
  3. Feb 8, 2009 #2


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    Think about this:

    The potential energy due to the cart's initial position goes entirely into kinetic energy when it is at the bottom of the loop (I have assumed the reference level for potential energy to be PE = 0 at the bottom). From there that energy becomes part kinetic and part potential energy. At the top of the loop you have then mgh = K.E. + P.E. Knowing the centripetal acceleration gives you the V^2 that you need for the kinetic energy.
    (The h in my equation above is the initial height where the cart is released.)
  4. Feb 8, 2009 #3


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    Homework Helper

    First of all at the top of the loop, you must have an additional contact force of .8*m*g.

    That means your Centripetal acceleration needs to be 1.8*m*g

    To determine the height ... that's the potential energy that you will draw from to give you the necessary kinetic energy.
  5. Feb 8, 2009 #4
    So it's .8 the weight, not the mass. And because energy is never lost, then the sum of the potential and kinetic will always be the same, no matter where on the loop it is. That makes much more sense than the textbook. Thanks.
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