# Elastic band around a cylinder

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

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 $$F = \frac{4F}{2\pi} \int_0^{\pi/2} \sin\theta d\theta = \frac{4T}{2\pi}$$.