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Homework Help: Circular Motion Loop question

  1. Jan 20, 2015 #1
    1. The problem statement, all variables and given/known data

    Bob starts at rest from the top of a frictionless ramp. At the bottom of the ramp, he enters a frictionless circular loop. The total mass of the child and the cart he sits in his m. What must the height of the ramp be in order for the cart to successfully traverse the loop.

    r = radius of loop
    h = height of ramp
    theta = angle of the ramp (irrelevant though)

    2. Relevant equations

    3. The attempt at a solution

    I solved for the minimum speed at the top of the loop.

    Fy = F + mg = mv^2/r

    v= sqrt(rg)

    I then used conservation of energy.

    Initial : mgh
    Final : mg2r + (m(sqrt(rg))^2)/2

    mgh = mg2r + mrg/2

    mgh = 5mgr/2

    Cancel stuff out h = 5r/2 (WRONG)

    Instead the solution calls for using kinematics not energy conservation.

    v= sqrt(rg) stills hold.

    vf^2 = vi^2 + 2ax

    rg = 0 + 2gsin(theta)*(h/sin(theta)

    rg = 2gh

    r = 2h

    h = r/2 (CORRECT answer)

    I understand the mathematical process of the correct solution.
    However, I don't understand why I can't use conservation of energy(gives me wrong answer) instead of kinematics.


    Attached Files:

  2. jcsd
  3. Jan 20, 2015 #2


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    The second answer (r/2) is clearly wrong since it would not provide enough energy to reach the top of the loop even with no remaining KE.
    The calculation goes wrong because it equates the speed at the bottom of the ramp to that required at the top of the loop.
    The first answer is correct.
  4. Jan 20, 2015 #3
    The solutions I have showed the second answer as the correct answer.
    Really? That's the solution provided to me.
    Glad to know I wasn't paranoid or something.

    Is there any additional information that would make solving this question using kinematic possible, then?
  5. Jan 20, 2015 #4


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    The v2-u2=2as equation is effectively KE+PE constant. All that's different is factoring out the mass.
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