Roller coaster: kinetic energy

  1. A roller coaster car may be approximated by a block of mass m. The car, which starts from rest, is released at a height h above the ground and slides along a frictionless track. The car encounters a loop of radius R, as shown. Assume that the initial height h is great enough so that the car never loses contact with the track.

    http://session.masteringphysics.com/problemAsset/1011023/13/MPE_ug_2.jpg

    so, i know kinetic energy = .5m(v^2) and centripetal force in circle is m(v^2)/R so that if i play around with the equations i can get KE = 0.5m(gR) ... but how do i take into account the height if i have to give my answer in terms of m, g, h, and R?

    Anyone out there who can solve this thing?
     
  2. jcsd
  3. Find an expression for the kinetic energy of the car at the top of the loop.
    Express the kinetic energy in terms of m, g, h, and R.

    forgot to add that what's above is the problem!
     
  4. learningphysics

    learningphysics 4,123
    Homework Helper

    Just use conservation of energy... I don't think you need to deal with centripetal motion or anything.
     
  5. i can't work just on that. what do you mean by 'just use conservation of energy'. i need to express my answer in terms of those variables i listed above.
     
  6. learningphysics

    learningphysics 4,123
    Homework Helper

    What is the energy of the coaster at the beginning when it's at rest? Take the gravitational potential energy on the ground to be 0.
     
  7. The energy of the coaster at the beginning is potential energy which is just mgh.
     
  8. learningphysics

    learningphysics 4,123
    Homework Helper

    Yes, now at the top of the loop, it has potential energy mg(2R), and kinetic energy. can you use conservation of energy to solve for kinetic energy?
     
  9. thanks! i got it ... now what if i want to find the minimum initial height h at which the car can be released that still allows the car to stay in contact with the track at the top of the loop? how would i approach this one?
     
  10. learningphysics

    learningphysics 4,123
    Homework Helper

    Now you'd use the centripetal acceleration... what does the velocity need to be at the top of the loop for the car to maintain contact?
     
  11. well acceleration is V^2/R . and so min velocity would be v = gR ...
     
  12. sorry, v = (gr)^(1/2)
     
  13. learningphysics

    learningphysics 4,123
    Homework Helper

    yup. so what is the total energy of the coaster when it is at the top of the loop? It has that same energy when it is released from rest.
     
  14. mgh... sorry i'm not getting it
     
  15. i set mgh-mg2R = 0.5(mv^2) and plugged in what i got for v ... so i ended up with h=3R but i think i'm off by some multiplicative factor ???

    the ans should be in terms of R by the way...
     
  16. learningphysics

    learningphysics 4,123
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    can you show how you get h = 3R, I'm getting h = 2.5R
     
  17. ah, stupid calc. error. thanks lots learningphysics. you made my day haha
     
  18. learningphysics

    learningphysics 4,123
    Homework Helper

    no prob.
     
  19. Thats very simple and straight forward. use the conservation of energy law, and solve for the unkwon K=kinetic energy at the top of the loop. NB: th energy at initial position is equal to the energy at final position = top of the loop. initial kinetic energy = 0, So u will left only with final unknown kinetic energy K. Min Velocity, v=gR^0.5
     
  20. any probs rply
     
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