# Slingshot Physics

1. Mar 4, 2013

### FredericChopin

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

and

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. Mar 4, 2013

### 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.

3. Mar 4, 2013

### e.chaniotakis

I didn't quite understand the question..

4. Mar 4, 2013

### rcgldr

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):

Elastic_hysteresis.htm

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):

rubberdata.htm

Last edited: Mar 4, 2013
5. Mar 5, 2013

### FredericChopin

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).