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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.

    Slingshot:
    MrzKRXL.jpg

    2. Relevant equations
    [itex]F=kx[/itex] (Due to the elastic ropes)
    [itex]F=ma[/itex]
    [itex]∴kx=ma[/itex]
    [itex]EPE=\frac{1}{2}kx^{2}[/itex]
    [itex]KE=\frac{1}{2}mv^{2}[/itex]
    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

    jedishrfu

    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?
    ubwOFUe.jpg

    EDIT: Actually, no?

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

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

    jedishrfu

    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

    mfb

    User Avatar
    2016 Award

    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
    Hmm,

    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

    mfb

    User Avatar
    2016 Award

    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

    jedishrfu

    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

    mfb

    User Avatar
    2016 Award

    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

    jedishrfu

    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

    jedishrfu

    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

    jedishrfu

    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]?
     
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