# Rubber Band Spinning in midair

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1. Jan 14, 2017

### Chozen Juan

1. The problem statement, all variables and given/known data
A uniform thin circular rubber band of mass M and spring constant k has an original radius R. Now it is tossed into the air. Assume it remains circular when stabilized in air and rotates at angular speed ω about its center uniformly. Derive an expression for the new radius of the rubber band. Express your answer in terms of ω, k, M, R, and any fundamental constants.

2. Relevant equations
Fspring = kΔx
Fcp = mω2R
F = ma
C/(2r) = π
3. The attempt at a solution

I first considered a mass element Δm in the rubber band. This Δm has acceleration ω2R', where R' is the new radius. Of course, this means that the mass element has centripetal force of magnitude ΔF = (Δm)ω2R'.

Furthermore, a rubber band stretched a distance Δx behaves just like a spring; so if a rubber band with initial length 2πR (original radius of the rubber band) is stretched to a final length of 2πR', the magnitude of the force on the rubber band is T = 2πk(R' - R).

But I don't know how to relate the force T to ΔF. Since the center of mass of the rubber band only moves in the up-down direction, I don't think it's true that the sum of all the small ΔF's is T; I think the sum of all the small ΔF's is actually 0 since there's 0 net force on the rubber band in the horizontal plane. Because of this, I don't know how to even use these two equations together.

The correct answer is R' = (4π2kR)/(4π2k - Mω2).

2. Jan 14, 2017

### haruspex

Consider a short element of the band subtending an angle dθ at its centre. The tensions at each end are not quite in line. Find an approximate expression for their resultant.

3. Jan 14, 2017

### Staff: Mentor

Another way to do this is to imagine the rubber band hypothetically cut in half, and obtain the net resultant acceleration of half the rubber band. Then, from a free body diagram on half the rubber band, write the force balance, including the tensions on the two cut ends.

4. Jan 14, 2017

### Chozen Juan

Ok I tried haruspex's method, and it seemed to work. However, I'm still a bit confused as to why it works.

I figured that the small mass element would have tensions on both sides equal to T. The two tensions must be equal since the mass element has no tangential acceleration. So now the free body diagram of the mass element is an isosceles triangle with two equal sides T. The resultant of these two tensions is 2Tsin(Δθ/2), where Δθ is the angle subtended at the center. This resultant vector is the centripetal force and has magnitude Δmω2R'. Since Δθ/2 is so small, sin(Δθ/2) = Δθ/2. The equation is then

TΔθ = Δmω2R'

Integrating both sides, we obtain:

T∫dθ =ω2R' ∫dm ⇒ 2πT = Mω2R' ⇒ T = Mω2R' / (2π)

Another equation we have is F = 2πk(R'-R), where F is the force on the rubber band when the rubber band is stretched linearly a distance of 2π(R'-R). If we equate T to F, then we can solve for R' and obtain the final answer.

But why is F = T? Stretching the rubber band linearly is different from having it stretch in midair. It's difficult for me to intuitively understand this.

5. Jan 14, 2017

### Staff: Mentor

Actually, the rubber band should be treated as an elastic solid, rather than a spring. For an elastic solid, the relationship between the tensile force and the change in length is given by: $$T=EA\frac{\Delta l}{l}$$ where E is the Young's modulus, A is the cross sectional area, l is the original length of a portion of the rubber band, and $\Delta l$ is the change in length of the same portion of the rubber band. For this problem, you can use $$\frac{\Delta l}{l}=\frac{2\pi (r'-r)}{2\pi r}=\frac{(r'-r)}{r}$$. So $$T=EA\frac{(r'-r)}{r}$$

6. Jan 14, 2017

### haruspex

Tension is not exactly a force. It is more like a pair of equal and opposite forces. When you maintain a steady pull on a light spring with force F, there is a tension F at each point of the spring. That is, in each small segment of the spring there is a force F pulling at each end. Thus, F=T.