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Velocity w/ kinetic energy

  1. Nov 7, 2005 #1
    Riding a Loop-the-loop. A car in an amusement park ride rolls without friction around the track shown in the figure. It starts from rest at point A at a height h above the bottom of the loop. Treat the car as a particle.
    If the car starts at height h= 95.0 m and the radius is r= 19.0 m, compute the speed of the passengers when the car is at point c, which is at the end of a horizontal diameter.
    Take the free fall acceleration to be g= 9.80 m/s^2.
    This is a poorly worded problem in my opinion. For one thing, I don't know where C is by their description.
  2. jcsd
  3. Nov 7, 2005 #2
    Any help would be appreciated
  4. Nov 7, 2005 #3
    If you treat the loop as a circle, C is at x = pi. You want to use energy methods to solve this problem.
  5. Nov 7, 2005 #4
    How would I set that up if I don't know the mass or weight of the coaster. The only equation I know is (1/2)*m*v^2 + m*g*y_1 = (1/2)*mv^2 + mg*y_2
  6. Nov 7, 2005 #5
    Notice that any change in kinetic energy would only be caused by a change in potential energy. You know the PE at point A so you know the KE at the bottom of the circle. Now to displace 'r' in the upwards direction you'll need to gain a PE of 'mg(r)' by losing it from your KE at the bottom. Get the idea?
  7. Nov 7, 2005 #6
    Not Really. The potential energy at point a= mgy = m(9.8)(95) does it not? I still dont know the mass.
  8. Nov 7, 2005 #7
    Okay, the total energy E = PE + KE. Te total energy is constant.
    At the top KE = 0 so the total energy is PE.
    At the bottom the PE = 0 So the KE = (the old PE).

    Get that so far? You don't need to know the mass, it will cancel.
  9. Nov 7, 2005 #8
    so pe=g*95 ?
  10. Nov 7, 2005 #9
    No the PE = mg(95) like you said. Since you can't get over this hump I"ll give you a push:

    The change in kinetic energy will be due to the change in potential energy:

    [tex] \Delta KE = \Delta PE [/itex]

    [tex] KE_f - KE_i = PE_f - PE_i [/tex]

    [tex] KE_f - 0 = 0 - PE_i [/tex]

    [tex] \frac{1}{2}mv^2 = -mgh [/tex]
  11. Nov 7, 2005 #10
    Sorry, I just don't understand this at all, but I already said from the beginning that .5mv^2=-mgh, but I still don't know how im suppose to solve for v without m.
  12. Nov 7, 2005 #11
    [tex] \frac{1}{2}mv^2 = -mgh [/tex]

    [tex] \frac{\frac{1}{2}mv^2}{m} = \frac{-mgh}{m} [/tex]

    [tex] \frac{1}{2}v^2 = -gh [/tex]

    Since we are just looking for a change in potential, we don't need a magnitude and can take away the negative.

    [tex] \frac{1}{2}v^2 = gh [/tex].

    Can you find v now?
  13. Nov 7, 2005 #12
    NO I CANT!!!!

    v=43.15 = INCORRECT ANSWER!!!!
  14. Nov 7, 2005 #13
    Thank you whozom for completely wasting the past two hours of my life.
  15. Nov 8, 2005 #14

    I showed you how to do your problem. Your incompetance is not my fault.
  16. Nov 8, 2005 #15
    They say "the end of a horizontal diameter", which probably means "at half the circle's diameter above the ground" (i.e. a position on a horizontal line through the middle of the circle). In that case the difference in height between the starting point and point C would be 95-19=76.

    Does using this height difference (in the equation that whozum provided) give you the right answer?
  17. Nov 8, 2005 #16


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    Try that g*h multiplication again before you get pissy next time...
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