Kinetics of a released elastic band

[mrelet]

My problem is to calculate the final velocity of a small mass, m, tied to the end of an elastic band that is pulled and released. I know how to do this in the simple case: potential energy of the elastic (spring)equals kinetic energy of the mass. The challenge is that the mass of the elastic band may be significant in that it too will have significant kinetic energy so that the final velocity of the small mass will be less if the band soaks up kinetic also. [The elastic of course stretches and the elemental kinetic energy varies along its length]. Any ideas?

 

[butelja]

The effective mass of the rubber band is 1/3 of its total mass. This is assuming 1 end fixed, the other end moving.

If a mode shape is defined by a function Phi, then the effective mass is equal to the integral of (mass per unit length) times Phi^2, integrated over the length. In your case, Phi(x) = x, m = mass/length, and effective mass is m*L/3.

 

[butelja]

OOPS, I hit post a little too quickly. My previous post gives the effective mass of the elastic band only. The mass of your object must be added to this to get the total effective mass. If you use the combined effective mass, then you can use conservation of energy to solve for the particle velocity.

 

[boo1]

The force in the band could be found by:

F=100(DeltaL/OrigL)/(Elonation%*TensleStress*Area)