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How high to complete the loop?

  1. Apr 23, 2014 #1
    Problem:

    A little roller coaster (see attached image) is at a certain height h and starts moving downwards, Neglecting the friction and drag. From what height does the coaster have to start in order to complete a full loop with radius a without coming of the track? h should be expressed in a.

    Thoughts:

    I feel like somehow the potential energy formulas should be involved here. But how do I calculate how much energy/momentum is required to have such a centripetal force so that the coaster stays on its track. Obviously the slope doesen't seem to affect anything so it's just the height that should be related to the radius of the loop.
     

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  2. jcsd
  3. Apr 23, 2014 #2

    mfb

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    Right
    Start with the centripetal force - what is its formula?
    That is a good observation. What does the height give to the roller coaster?
     
  4. Apr 23, 2014 #3
    Well, for the centripetal force we have that

    [tex]F_c = mv^2/a[/tex]

    and for the kinetic energy given from the height is

    [tex]E_k = mv^2/2[/tex]

    But how do I relate h? How do I convert the to force from kinetic energy?
     
  5. Apr 23, 2014 #4

    mfb

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    Which effect delivers this force? What is its strength?

    Where does the kinetic energy (in the loop) come from and how is that related to the initial height?
     
  6. Apr 23, 2014 #5

    jbriggs444

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    Hold up. Don't go there yet.

    We're aiming for a condition in which the roller coaster is just barely touching the track without falling off. In that condition, what force is acting to provide the centripetal acceleration?
     
  7. Apr 23, 2014 #6
    I'm not sure about your first question. But for the second one, It can be related like this: First it's potential/stored energy in the form of

    [tex] E_p = mgh [/tex]

    that is upon motion converted to kinetic of the form

    [tex] E_k = \frac{mv^2}{2} [/tex]

    So

    [tex]E_k = E_p \Leftrightarrow mgh = \frac{mv^2}{2} \Leftrightarrow h = \frac{v^2}{2g}.[/tex]

    Si how many a's does [tex]v^2/2g[/tex] correspond to?
     
  8. Apr 23, 2014 #7
    Gravity, initially?
     
  9. Apr 23, 2014 #8

    jbriggs444

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    Right. That answers the first part of mfb's question 1.

    Now, what about the "how strong" part?
     
  10. Apr 23, 2014 #9
    I understand that it's the acceleration, and the only thing that gives rise to it is gravity and the magnitude of it depends only on the height. The strength of this should equal to the strength of the centripetal force. If we have the acceleration a we get that

    [tex] F = mr = mv^2/a [/tex]

    not sure where this leads me?
     
  11. Apr 23, 2014 #10

    jbriggs444

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    What's r and what's a? There are two definitions for "a" so far in this thread.
     
  12. Apr 23, 2014 #11
    r is the acceleration and a is the radius. I don't see any contradiction to this anywhere in this thread?
     
  13. Apr 23, 2014 #12

    mfb

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    A very unconventional notation... and you mixed them in post #9 ("If we have the acceleration a [...]").

    And what is the strength of the gravitational force at this point?
     
  14. Apr 24, 2014 #13
    Yes I see it now. I know it's more intuitive to have a=acceleration and r=radius but the picture already had a=radius. MY apologies

    Well the stregth of the gravitational force is always constant? 9.82*m = 9.82m N where m is the mass of the body. But how come? Mass shouldn't be an issue here.
     
  15. Apr 24, 2014 #14

    mfb

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    Right.
    The units are wrong.
    Don't worry, it will cancel later. Mass is an issue for the required force to keep the rollercoaster on the circular track, but required force is not your final result.
     
  16. Apr 24, 2014 #15
    I'm so sorry man but I really don't follow here? I need the acumulated momentum from the initial slide to be at least equal to the required centripetal force within the loop. However the centripetal force needed has to be constant but the momentum decreases with time, I'm probabaly over-complicating things right? Thanks for having patience!
     
  17. Apr 24, 2014 #16

    mfb

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    Momentum cannot be equal to a force.

    At the top of the loop, gravity acts downwards. What is the gravitational force of an object with mass m?
    This force is exactly enough to keep the car in the loop, so the gravitational force is equal to mv^2/a with the radius a.

    Afterwards you just have to make the connection between the initial height and the speed at the top of the loop. Hint: energy conservation.
     
  18. Apr 29, 2014 #17
    So, the gravitational force of an object with mass [itex] m [/itex] is

    [tex] F_g = mg [/tex]

    and this should be equal to

    [tex] F_c = \frac{mv^2}{a}. [/tex]

    Setting [itex] F_g = F_c [/itex] gives [itex] a = \frac{v^2}{g}.[/itex] For the speed we can set the potential energy of the object before it's released from height [itex] h [/itex] equal to the kinetic energy of it throughout the loop and solve for the speed:

    [tex] E_p = E_k \Leftrightarrow mgh = \frac{mv^2}{2} \Leftrightarrow v^2 = 2gh.[/tex]

    Substituting this into the expression for [itex]a [/itex] above we get that

    [tex] a = \frac{2gh}{g} = 2h \Leftrightarrow h = \frac{a}{2}. [/tex]

    Answer should be [itex]h\geq\frac{5a}{2}.[/itex] Why is my answer 5 times smaller than it's supposed to be?
     
  19. Apr 29, 2014 #18

    mfb

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    Your answer is the height above the top of the loop, the other answer is above the bottom of the loop.
     
  20. Apr 29, 2014 #19
    Oh, I see. I suppose I should just add

    [tex] h = 2a + \frac{a}{2} = \frac{5}{2}a.[/tex]

    But how did you know that my answer is from the top of the loop and not from the bottom?
     
  21. Apr 29, 2014 #20

    mfb

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    From your approach to calculate it - you used the height difference between starting point and the upper point of the loop.
     
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