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Homework Help: Fairground 'Slingshot' physics

  1. Apr 10, 2013 #1
    1. The problem statement, all variables and given/known data
    For an assignment I am to find the physics behind one of those fairground slingshots. Specifically the maximum height and speed. We're given no data but we have to present our method, so I assume we just make it as realistic as possible and go from the most basic concept up until we have those answers. I have requested help from my lecturer but I haven't yet had a reply yet and I'm working on my own whereas everyone else has a group, thus, I'm reaching out to this forum.


    2. Relevant equations
    [itex]F=kx[/itex] (Due to the elastic ropes)
    I derived that the maximum height = [itex]\frac{V_{0}^{2}}{2g}[/itex] but I would need the max speed for that.
    I also know that when [itex]KE=0[/itex] the max height would be reached.

    I just have no idea how to start all of this.

    3. The attempt at a solution
    So far I've simplified the situation so that there's no air resistance, the cage is a particle of constant mass, instead of 2 strings there's just 1 in the middle, that the force of gravity is constant and that the particle shoots up at a perfect 90°s to the horizontal.

    I'm just not sure where to tackle this from. I've struggled to get confidence to make friends on my course and now I see that its possibly more damaging on my results than my psyche as I can't bounce ideas around.
  2. jcsd
  3. Apr 10, 2013 #2


    Staff: Mentor

    You need to draw a force diagram and include gravity. You just cant equate the F=ma and F=kx equations together.

    So basically you have two cords which behave according to hooke's law (assumption?) and gravity.

    When at rest with a person in the harness, the forces are at equilibrium.
    Last edited: Apr 10, 2013
  4. Apr 10, 2013 #3
    So, when they're pulled all the way down and clipped to the floor this would occur?

    EDIT: Actually, no?

    If they were in equilibrium, [itex]F = T[/itex]

    meaning [itex]mg = T[/itex]?
  5. Apr 10, 2013 #4


    Staff: Mentor

    In your force diagram, the downward arrow is simply -mg and the sum total of forces F = T - mg.

    Eventually though you need to draw a force diagram for the whole system including each cord because you're assuming the sum total of cord tension behaves according to hooke's law and I don't think it does even if individually the cords do.

    At equilibrium, the cage is at rest (= zero acceleration) hence sum total of forces is zero 0 = T - mg so T = mg in that case and from there you can compute k given the elastic rope displacement (the x0 in the T=kx0 equation).

    Another thing you need to consider is that as the person rises the tension drops to zero and he continues to rise a bit more and then the cord tension will start to pull him back down.

    Anyway one thing that may help you here is to actually build a jig using rubber bands and some small weight that you can play with to understand the forces and how and when they are applied.
    Last edited: Apr 10, 2013
  6. Apr 11, 2013 #5
    Okay, so I can calculate k with [itex]k=\frac{T}{x_{0}}[/itex] where [itex]{x_{0}}[/itex] is some change in length that I decide. I'm not sure how this would help me find the maximum height and speed though as I've already technically worked out T from [itex]T=mg[/itex]
  7. Apr 11, 2013 #6


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    Staff: Mentor

    Or the minimal height.
    Did you take gravitational potential energy into account?
    The two points of zero kinetic energy (together with general data about the setup) are sufficient to find the maximal height.

    Those are good assumptions. Hooke's law for the string is another one you used already.
  8. Apr 11, 2013 #7

    Maybe something along the lines of,
    [itex]E_{MAX}=\frac{1}{2}mv^{2}+mgh+\frac{1}{2}kx^{2}[/itex] ([itex]E_{MAX}=KE+GPE+EPE[/itex])
    I can then rearrange that to find h.

    The point at the bottom where [itex]KE=0[/itex] is easier because [itex]v=0[/itex], but then that means the height is just 0 as you could guess anyway?

    But then again, I would need to know v. Hmm.
  9. Apr 11, 2013 #8


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    Staff: Mentor

    Setting the height to 0 for the lowest point is a good idea, indeed.
    You have to be careful with x then, however: x=0 does not correspond to the ground (otherwise, the mass would not move at all).
    0 for both points, as you already pointed out.
  10. Apr 11, 2013 #9
    Oh yes of course. Because it's at it's peak, [itex]v=0[/itex].
    Okay so x is the extension in length of the rope, I guess I just make a realistic idea up for that? Let's say 10m?
  11. Apr 11, 2013 #10


    Staff: Mentor

    Don't forget if your object passes a certain height then the cords will work against you and try to pull you back down in addition to the gravity acting on the objects. There are three states here:

    -- the cords pulling you up while gravity pulls you down,
    -- the cords not doing anything but gravity is pulling you down and then when
    -- the cords start pulling you down as gravity is pulling you down.
  12. Apr 12, 2013 #11
    Just realised that rearranging the [itex]E_{MAX}[/itex] formula, I believe I get [itex]h=\frac{E_{MAX}-\frac{1}{2}kx^{2}-\frac{1}{2}mv^{2}}{mg}[/itex] which means I'll need to know [itex]E_{MAX}[/itex] first. :S Grr.

    Yeah good point. I think I'll state in my assumptions that from looking at various clips of this ride in action, the particle will never rise high enough for the cords to start pulling downwards again.
  13. Apr 13, 2013 #12


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    Staff: Mentor

    I think it is all in the thread now, you just have to combine it.
  14. Apr 13, 2013 #13
    Hmm, still a bit stuck, but I'll give it another hour or so.
  15. Apr 13, 2013 #14
    EDIT: I worked out the Pythagorean theorem issue before this.
    Last edited: Apr 13, 2013
  16. Apr 13, 2013 #15


    Staff: Mentor

    your cords are at an angle to the object so pythagorean theorem is used to compute the length for hooke's law
  17. Apr 13, 2013 #16
    Yeah sorry, I just worked that out.

    Now all there is, is...

    It says how [itex]E=\frac{1}{2}mv^2+k(\sqrt{H^{2}+l^{2}}-l_{0})^{2}-mgH[/itex] which I understand, EXCEPT, they've put [itex]k(\sqrt{H^{2}+l^{2}}-l_{0})^{2}[/itex] instead of [itex]\frac{1}{2}k(\sqrt{H^{2}+l^{2}}-l_{0})^{2}[/itex]. What's the reason for this? Is it because there's 2 cords so the half cancels out, maybe? :S
  18. Apr 13, 2013 #17


    Staff: Mentor

    Yes that's right there's two cords and the usage is symmetric.
  19. Apr 13, 2013 #18
    Hmm very interesting. Never encountered that before.

    EDIT: How come the kinetic energy has a [itex]\frac{1}{2}[/itex]? :eek:
  20. Apr 13, 2013 #19


    Staff: Mentor

    That comes from integrating hooke's law over distance aka work.

    f=kx work = 1/2 k x^2
  21. Apr 13, 2013 #20
    Oh gosh now I'm confused. So, the [itex]\frac{1}{2}kx^{2}[/itex] doesn't need the [itex]\frac{1}{2}[/itex] because it's symmetrical, but why does the [itex]\frac{1}{2}mv^{2}[/itex] keep the [itex]\frac{1}{2}[/itex]?
  22. Apr 14, 2013 #21
    Bump for the previous question above this post? Should I start a new thread as I've now pushed past the original issue?

    This help then goes on to say that,
    [itex]E=k(\sqrt{H^{2}+l^{2}}-l_{0})^{2}-mgH[/itex], [itex]H[/itex] being the distance the particle is lowered from the tops of the poles.
    [itex]E=k(\sqrt{h^{2}+l^{2}}-l_{0})^{2}+mgh[/itex], [itex]h[/itex] being the maximumg height above the tops of the poles.
    I understand those formulae, but the next bit is confusing.

    It says that by equating those 2 equations, we get;
    which then,
    so that the max height is then,

    I've got everything I need to complete the task, but I really would appreciate it if one of you could explain my previous question about why the [itex]\frac{1}{2}kx^{2}[/itex] no longer needs the [itex]\frac{1}{2}[/itex] but the [itex]\frac{1}{2}mv^{2}[/itex] keeps the [itex]\frac{1}{2}[/itex] and also if someone could run me through this part about equating those 2 formulae. Do you believe this is the simplest way?

    I really would appreciate it, if there's anything I can do in return, please say.
    Last edited: Apr 14, 2013
  23. Apr 14, 2013 #22
    You asked about the factor of 1/2 that disappears in one equation but not the other.

    There are two elastic bands, each of which provides half the force. Thus you have total energy stored in the bands of 2*(1/2 k sqrt(...

    You only have one mass m, with one velocity. The kinetic energy of such a mass is 1/2 m v^2

    Does that clarify it for you?
  24. Apr 14, 2013 #23
    Thank you, thank you, thank you! That makes a lot of sense, thanks. :)

    Now just to wrap my head around the equating and I'm done! :D
  25. Apr 14, 2013 #24
    The other half of your question was "why do we equate these two"?

    Remember these are the _extreme_ points of the trajectory - the very bottom, and the very top. What do these two have in common? The kinetic energy is zero in both cases (velocity = 0). Thus the only thing that is different is the elastic energy stored in the cords in the two positions, and the potential energy. Equating the two is like saying "energy is conserved". By the way I think there is a typo in your summary of the two E= equations: the second = sign should be a + sign (the energy is the sum of kinetic, elastic and gravitational energy).

    Then I get confused by how you solve for h - somehow, the angle of the bungee (if you like, the spacing between the support points) does not feature in your solution, and you just know that can't be right. If I have the bungees almost vertical, then a certain initial displacement will result in more stored energy than if they are almost horizontal (support points space far apart). This means that for the same initial displacement I should get catapulted higher if the supports were close together. That's not what your solution is saying - so I suspect something is going wrong in your "set these two equal" step. I think you should take a closer look at it.

    And while you're there, consider what happens if the bungee is slack at the top of the trajectory. In that case the elastic term should be zero (it can't go negative).

    I think you're not quite there yet..
  26. Apr 14, 2013 #25
    (You're correct about the typo, my bad)

    Ahh okay, so the first [itex]E=[/itex] equation is the total energy using the lowest point variable [itex]H[/itex] and the second [itex]E=[/itex] equation is the total energy using the highest point variable [itex]h[/itex]. So then it says we equate them by saying [itex]k(\sqrt{H^{2}+l^{2}}-l_{0})^{2}-mgH = k(\sqrt{h^{2}+l^{2}}-l_{0})^{2}+mgh[/itex]
    which then rearranges to [itex]h^{2}+\frac{mgh}{k}+\frac{mgH}{k}-H^{2}=0[/itex] which then in turn simplifies to [itex](h+H)

    Is that correct?
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