B String constant of a folded and a cut up rubber

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1. Oct 11, 2016

ChessEnthusiast

Imagine that we wanted to build a slingshot, using one piece of rubber attached to two hooks - each on a side.

We use a rubber, whose spring constant is k.

My questions are:

1. If we began to string that slingshot and the rubber would begin to fold, what would happen with the string constant? Would we need to consider this one rubber as two, separate rubbers with the string constant k?
$$\frac{1}{2}mv^2 = \frac{1}{2}kx^2 + \frac{1}{2}kx^2$$

or consider it as one rubber, with string constant 2k
$$\frac{1}{2}mv^2 = \frac{1}{2}(2k)(x^2)$$

2. If we were to cut the rubber in half and build two slings with the two rubbers we have just gotten, what would be the string constant of each of these rubbers?
k or k/2?

Could you attach some reference to your answers?

2. Oct 11, 2016

Staff: Mentor

What is your assessment?

3. Oct 11, 2016

ChessEnthusiast

Well, if we were to put an object into the sling, there would be two sources that exert a force F = kx, thus I would say that the first interpretation is more accurate.

4. Oct 11, 2016

Staff: Mentor

The spring constant of a section of rubber is determined by $k=EA/L$, where E is the elastic modulus of the rubber (a material constant), A is the cross sectional area (normal to the stretch direction), and L is the length of the piece of rubber. Does this help?

5. Oct 11, 2016

ChessEnthusiast

Thank you. Your answer has dispelled my doubts about the second part of my question.

As of the first part, would it be more appreciate to analyze this situation as two rubbers of length 0.5L, or one rubber of length L?
The math will be the same, but the concept is way different.

6. Oct 11, 2016

Staff: Mentor

$$K=k_1+k_2=\frac{EA}{0.5L}+\frac{EA}{0.5L}=4\frac{EA}{L}$$where K is the equivalent spring constant of the combination, and L is the total original length.

7. Oct 11, 2016

ChessEnthusiast

Thank you.
Last question:
If we were to increase the number of rubbers, would this equation hold:
$$K = \sum_{i=1}^{n} k_i$$

8. Oct 11, 2016

Yes