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How to Calibrate a Catapult?

  1. Apr 22, 2012 #1
    My current physics project is to design a catapult that will launch a marble to a given distance (between 2-3m). I have successfully made one out of wood, using a bungee cord to snap the arm back after being pulled on.

    My question is how do I calibrate this properly? Our teacher said we need to use projectile motion equations and energy equations. I am not entirely sure how to do this so can someone help me please?
     
  2. jcsd
  3. Apr 22, 2012 #2
    You use the energy equations to determine the velocity of the projectile as it leaves the sling. Once it leaves you use the projectile motion equations to determine how far it will travel.
     
  4. Apr 22, 2012 #3
    I thought of that, but the question arises as to how to fire the marble at that specific velocity?
     
  5. Apr 22, 2012 #4
    How about measuring your spring constant(by a newtonmeter) and you can calc. the spring energy: 1/2kx^2
    Then you can calulate energies set the zero-point for potential energy when the catapult is loaded and then use energy conservation:
    [itex] E_{initial} = 1/2kx^2\\
    E_{final} = mgy + 1/2mv^2 = 1/2kx^2\\
    1/2mv^2 = 1/2kx^2 -mgy \\
    v = \sqrt{\dfrac{k}{m}x^2-2gy}[/itex]
     
  6. Apr 22, 2012 #5
    Actually nevermind my previous post. Would I use the following equation:
    Elastic energy = gravitational potential + kinetic energy
    Ee = Eg + Ek

    If so how would I calculate the elastic constant for my specific bungee cord?
     
  7. Apr 22, 2012 #6
    dikmikkel, I do not have a newtonmeter. Are there any alternative methods to calculating the coefficient ?
     
  8. Apr 22, 2012 #7
    You take an object which you know the weight, let it hang free and see how long it stretches the bungee and the gravity is the only force:
    [itex] mg - k\Delta x = 0 \\
    k = mg/\Delta x[/itex]
    So you would need to measure the initial length of the bungee.
     
  9. Apr 22, 2012 #8
    Ok so once I have the coefficient, I will use your derived equation to solve for velocity. Then how do I find the specific range? would I use this equation:

    ΔDx = (V1sin2θ)/g
     
  10. Apr 22, 2012 #9
    You would use your eq.s for projectile motion:
    [itex]
    x = v_0 \cos(\theta)t \\
    y = y_0 + \sin(\theta) t -1/2gt^2
    [/itex]
    Where y0 is the height of the catapult just before shooting. Now the task is to eliminate the time from the above i.e. isolate the time in x and insert into y. then you would solve your quadratic equation for x with y = 0. Can you see why?
     
  11. Apr 22, 2012 #10
    Also in your derived equation, the velocity you solved for is initial right? Just making sure
     
  12. Apr 22, 2012 #11
    Sorry I don't follow, could you explain why I need the y as well as the x?
     
  13. Apr 22, 2012 #12
    oops i made a booboo:
    [itex] y = y_0 + \sin(\theta) v_0 t -1/2gt^2[/itex]
    But yeah that is right the v0 is the velocity from #4.
    Edit: You need the time to be eliminated, because you don't know how many seconds your marble will fly.
    So the only way is to define the ground as y=0 and therefore the y is needed aswell as the x.
     
  14. Apr 22, 2012 #13
    You dont need the y it is only interesting when set to zero.
     
  15. Apr 22, 2012 #14
    I understand what you're saying but shouldnt the ground be -0.48m? The distance from the ground to where the marble is launched on my catapult is 48cm above ground level. Or is this distance insignificant enough to rule out?
     
  16. Apr 22, 2012 #15
    No it is not. But i choose ground to be y=0 hence the y0 term in the expression for y.
    Edit: With your suggestion the y_0 should be zero, and we would get the same result.
     
  17. Apr 22, 2012 #16
    So setting y0 = -0.48 I derived the following equation:
    y = -0.48 + xsinθ/vcosθ - 1/2g(x/vcosθ)^2
    Is this correct? Now all I have to do is set the equation to zero, and plug in the x which is the given distance, and solve for v. Then from there I work backwards and use the energy eq'ns to solve for how much to pull it back?

    Also is θ the angle at which the arm is to the ground?
     
  18. Apr 22, 2012 #17
    You probably want the y0 to be positive 0.48 else you have to make some odd changes.
    Make a sketch with your definition of the coordinate system and see.
    the theta you got right.
    Hmm you could do that, or you could derive a equation which only need to take the desired firing range x? It could be done from our discussion and eq.s.
    But i get the same eq. except from the sign on y0
     
  19. Apr 22, 2012 #18
    When doing that you would insert the v from #4 as v0 and arrive at the following result:
    [itex] 0 = y_0 + \dfrac{\sin(\theta)}{\cos(\theta)} x -1/2 g x^2\dfrac{1}{\cos(\theta)^2 (\dfrac{k}{m}\Delta x^2-2gy)}[/itex]
    Which you would need to solve for delta x :)
    Edit: That is good physics: Obtain a finished equation and plug in your numbers at last then you could also optimize it by varying the catapult height and so on.
    EditEdit: The y is now the catapult height. sorry.
     
    Last edited: Apr 22, 2012
  20. Apr 22, 2012 #19
    So as a final result i get (by sin/cos=tan):
    [itex] \Delta x = \sqrt{\dfrac{gx^2m}{2k(y_0+\tan(\theta ) x)\cos(\theta )^2}+\dfrac{2gym}{k}}[/itex]
     
  21. Apr 22, 2012 #20
    Going back to the energy equations, I think you missed an 'm'. It's supposed to say -2mgy right? And for the y, would that be the distance above ground at the end? So wouldn't that be 0, making Eg negligible?
     
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