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Slingshot Physics

  1. Mar 4, 2013 #1
    I am investigating how the displacement of an elastic band affects the distance an attached object will travel (the relationship between how far back you pull a slingshot and the distance an object will travel). What I am trying to do is find a formula where distance travelled is a function of elastic band displacement (in other words, a formula which says "x = ... d..."), but I ended up very confused. I tried equation manipulation with the following equations:

    W = F*d

    W = ΔEKinetic

    EKinetic = (1/2)*m*v2


    EElastic Potential = (1/2)*k*x2

    I got close to the final equation, but I couldn't make it. Here are a few things to keep in mind:

    * In the final equation, where "x = ... d...", I don't want the force, F, to be in the equation.

    * We are assuming that initial kinetic energy was 0 (meaning that the initial velocity was 0).

    * We are assuming that the effects of friction and air resistance are negligible.

    Thank you.
  2. jcsd
  3. Mar 4, 2013 #2


    Staff: Mentor

    I don't think the F of the elastic band is constant with distance.

    You could try hooke's law on it and measure the F values for various distances of pullback.

    Then you can use that to compute the work involved.
  4. Mar 4, 2013 #3
    I didn't quite understand the question..
  5. Mar 4, 2013 #4


    User Avatar
    Homework Helper

    You won't need F, but you will need the "sprint constant" k for the sling shot. Note this is an approximation, since the graph of force versus stretch of rubber bands is not a straight line, but a curve. Example graph from a wiki article, which is also somewhat idealized. Usually the slope for the initial stretch is steeper, then decreases to a near straight line, then increases again at the limit of stretch (permanent deformation can occur if stretched close to the limit):


    Archived web page showing a stretch versus tension graph for latex rubber used to launch radio control gliders (tension at 300% is about 175 lbs per square inch cross sectional area):

    Last edited: Mar 4, 2013
  6. Mar 5, 2013 #5
    Thank you very much, but I have found a solution:

    First, due to conservation of energy, the elastic potential energy of the elastic band is equal to the kinetic energy of release. So:

    EElastic Potential = (1/2)*k*x2

    EKinetic = (1/2)*m*v2

    (1/2)*k*x2 = (1/2)*m*v2

    (1/2)*x2 = (m*v2)/(2k)

    x2 = (2m*v2)/(2k)

    x = √((m*v2)/(k))

    By finding v2 using the equations of motion, the mass, m, will be the function of spring displacement, x, while v2 and k remain constant (v2 will change depending on the desired displacement for the object to travel in the x or y axis).
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