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An elastic band is stretched using a known force and then placed around a cylinder. How are the forces or tensions distributed? I assume there will be two components: firstly, a tangential or circumferential component, and secondly, a radial or centripetal component. How are these components calculated?

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Assuming no friction, the tension in the elastic band will be uniformly distributed. It will apply a pressure on the cylinder. Assuming an infinitely thin band, the units would be $force / length$ rather than $force / length^2$. – Brandon Enright May 2 '13 at 21:18

The upshot is that there will be a net radial component only.

To see this, pick a point on the circumference of the cylinder and compute the net force on that point due to the rest of the rubber band. I'm going to say that the radius of the cylinder is $R$, that the force applied to stretch the band is $T$, and I'm going to ignore the width of the band. Then the force per unit length $dl$ is $dF = T(dl/2\pi R)$.

Now let me take the point I'm interested as $\theta = 0$, and ask what the force on that point is due to a point at some $\theta$ around the circle. That point applies the force $dF$ at the angle $\theta$. Decompose that into radial and tangential components. But notice that there is a matching point at $-\theta$ that supplies the same radial force but an exactly opposing tangential force. The net force due to $dl$ is therefore $dF(\theta) = T(dl/2\pi R)|\sin\theta|$, all directed radially. I'm taking the absolute value of the $\sin$ because the points on the far side of the cylinder will have a negative value for $\sin\theta$, but will still supply a positive force.

To actually do the integral, we need to convert the differential: $dl = Rd\theta \Rightarrow dF = T|\sin\theta|d\theta/2\pi$. Finally, the total radial force applied at the point $\theta = 0$ is $F = \int_0^{2\pi} d\theta T|\sin\theta|/2\pi$. To simplify this, note that $\int_0^{\pi/2}|\sin\theta|d\theta = \int_{\pi/2}^\pi|\sin\theta|d\theta = \int_\pi^{3\pi/2}|\sin\theta|d\theta = \int_{3\pi/2}^{2pi}|\sin\theta|d\theta $. So $$F = \frac{4F}{2\pi} \int_0^{\pi/2} \sin\theta d\theta = \frac{4T}{2\pi}$$.

A few notes on that result. As I mentioned above, I've neglected the width of the band. The only real effect of adding the width should be do divide this result by the width. If you do that, then you get a pressure. In this form, I've calculated a force per unit length. This result does not depend on the radius of the cylinder; the force you supplied initially to stretch the band is spread evenly over the entire circumference. As a result, there will be a larger total force applied if you stretch the rubber band over a larger cylinder. This is what we expect for an elastic band. The force is spread evenly along the length of the band, and the total force you need to apply scales linearly with the stretching length.

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