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Speed of a roller coaster

  1. May 30, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider a roller coaster with length = 10m and mass 4000kg consisting of 5 identical carts. Ignore the length of the connector between the carts. Suppose the rolercoaster is on a track where it loses energy at a rate of 1300 J/m. The rollercoaster approaches a loop with a speed of 40m/s with radius 20m. Find the speed of the first and the last cart of the rollercoaster when each of them is at the top of the loop.


    2. Relevant equations
    Kinetic and potential energy formulas


    3. The attempt at a solution

    we can solve the speed of the center of the mass of the rollercoaster using the following:

    [tex] E_{K1} = E_{K2} + mg (2r) + 1300 (\pi r ) [/tex]

    However, how do we accoutn for the actual length of the rollercoaster? can we treat each cart independantly? If that is the case, then would simply change the value of the last term in the above expression to account for the length of the rollercoaster - for example, if we consider the lead cart, 1300 would be multiplied by pi* r - 4. Is this correct? Do we have to include the friction of each cart in this calculation?

    Thanks for your help
     
  2. jcsd
  3. May 30, 2013 #2

    collinsmark

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    I advise against bringing the entire rollercoaster's center of mass into things. See more advice below.

    That looks close to okay for the lead car, but not quite there yet.

    A couple of things:
    • You should be able to substitute something into [itex] E_{K1} [/itex] easily enough.
    • Treat the gravitational potential energy separately for each of the 5 cars (the cars are not all at the same height :wink:).

    Yes, I would treat the gravitational potential energy of each of the cars independently, since they are not all that the same height. A little bit of trigonometry is involved.

    On the other hand, there is no need to do this with the rollercoaster's kinetic energy since they are all traveling at the same speed. Similarly, you don't need to break up the 1300 J/m either.

    I'm not sure where your "- 4" part comes from. But anyway, if you treat the gravitational potential energy of each cart separately, things should fall into place.

    [Edit: But yes, as part of the process of finding the speed when the end car is at the top, you'll need to calculate the distance from the center of the lead car to the center of the end car.]

    No, I don't think so.
     
  4. May 30, 2013 #3

    haruspex

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    I second that. You could consider the KE of the mass centre, but then you'd have to add in the rotational KE around that centre. Much simpler to treat it as a number of point masses that all happen to have the same KE. (Of course, this ignores the rotational KE of each cart about its own centre, but that's a rather smaller error.)
    Really? Is that necessary for finding the PE of the system?
     
  5. May 30, 2013 #4

    collinsmark

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    It's necessary for the "it loses energy at a rate of 1300 J/m" aspect of the problem. There's some energy lost from when the first cart is on top of the loop to when the last cart is on top.
     
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