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Homework Help: Rollercoaster problem

  1. Nov 10, 2007 #1
    1. The problem statement, all variables and given/known data
    A roller coaster is just barely moving as it goes over the top of one hill. it rolls nearly frictionless down the hill and then up over a lower hill that has a radius of curvature 15m. How much higher must the first hill be than the second if the passengers are to exert no force on the seat as they pass over the top of the lower hill?

    2. Relevant equations
    angular kinematic equations.

    3. The attempt at a solution
    No idea where to start.

    Can someone give me a boost on how I should take this problem?

    Are there going to be real values or this is formula manipulation?

    I think if i use KE=PE I can get the velocity for a certain height no?
  2. jcsd
  3. Nov 10, 2007 #2


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    Yes, now to get that velocity at the top of the lower hill, you must first look at the centripetal force at the top of the lower hill.
  4. Nov 10, 2007 #3

    Shooting Star

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    The normal reaction on top of the 2nd hill is zero. That means that the only force acting on the car is mg. So, centripetal force = mg.

    You know v^2 at this point. Try now. Take some h as the height difference.
  5. Nov 10, 2007 #4
    ok now.

    what about this:

    I use v=sqrt(2gh) -- simplified from mgh=0.5mv^2

    h=15m, so that will be the velocity at the top of the lower hill

    then i use Total Energy at top of higher hill = PE + KE = PE + KE (of the lower hill)

    KE of the coaster at the top of the hill will be 0 since v is almost 0 as it describes...

    am i heading the right direction?
  6. Nov 10, 2007 #5

    Shooting Star

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    No. Take the difference of heights betwwen top of 1st and 2nd hill as h. You have to find h. Initial energy = mgh = final energy = mv^2/2, where we are taking the pe on top of 2nd hill as zero.

    So now you have found v at the top of the 2nd hill interms of h. Now, centriptal force=mg. So, mv^2/r = mg, where r =15m.

    See what you get by solving the eqns I've given.
    Last edited: Nov 10, 2007
  7. Nov 10, 2007 #6
    ok let us start over I am getting confused.

    KE of hill one is 0 because of the low velocity. it only has PE of mgh. (initial energy).

    can you explain this to me?
    Since the passengers will leave their seats, that means the only force acting downwards(centripetal) is mg so Fc=mg or else it would have been Fc=mg+FN right?

    We are taking PE at the top of hill two as zero because we are taking the height of hill two as 0 because h is the difference of the heights.

    so mgh=mv^2/2


    h=v^2/2g ***

    now to find v at the top of the second hill


    v=sqrt(gr) where r=15m

    v= 12.13m/s

    going back to h=v^2/2g


    so the height of hill one is h+15=22.5m

    *phew...* so complicated for an introductory physics course.
  8. Nov 10, 2007 #7
    bumping this for the person who was helping me or whoever can provide help.
  9. Nov 10, 2007 #8


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    You more or less got it right. The difference in height between the top hill and lower hill is 7.5m. It is not necessarily 22.5 m high, however, since all that is known is the radius of curvature of the lower hill, not the height of the lower hill.
  10. Nov 10, 2007 #9

    Shooting Star

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    As I said, you only need h. But good work!
  11. Nov 11, 2007 #10
    sorry yes. thank you. but i want to know some theory on this.

    the step where i did mgh=mv^2/2

    arent the masses different? mgh has the mass of the passengers included. and KE of the second hill only has the mass of the rollercoaster. so why would they cancel?
  12. Nov 11, 2007 #11

    Shooting Star

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    Looking at the problem again, nowhere it says so. How are the passengers going to get out in the middle of a trip anyway? No, the two masses include the same passengers, and are equal.
  13. Nov 12, 2007 #12

    Shooting Star

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    Oh, I understand your confusion. You think that since the passengers exert no force on their seats, why we should include their masses on top of the second hill. Well, it’s not just the passengers who are “floating”, the roller coaster too doesn’t exert any force on the track at that point. So, the whole mass is “floating”, or in free fall for that instant only.
  14. Nov 12, 2007 #13
    thanks alot shooting star. i have some few questions.

    there is a question that a stone in a pail is being swung. what is the minimum speed so that the stone stays in contact with the pail.

    how do i know when the centripetal force is mg and when it is Fc=mass*centripetal acceleration? or are they always the same?
    Last edited: Nov 12, 2007
  15. Nov 13, 2007 #14

    Shooting Star

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    How is the pail being swung – in a horizontal plane or a vertical plane or in some other way? Without that info, the problem is not formulated. Also, is the pail being swung, however it may be, with constant angular speed? And what is the length of the rope? You must supply all these info.

    Centripetal force is always equal to mv^2/r. The thing to find out is what is applying this force.

    I’m anticipating that it’s being swung in a vertical plane. When do you think the stone is most likely to lose contact with the pail? At that point, the normal reaction again vanishes, and only its weight is equal to the centripetal force. Then you can find v.

    Tell me the answer.
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