# What force acts on a small piece of a free, spinning ring?

Suppose i have a hollow ring and i take it in some place where there is no gravity and there is no air. Then i just start to rotate it and then because of it's inertia it will continue rotating and won't ever stop because there is no friction.

Question - Suppose i take a point on it of mass $dm$ and i see that it is also rotating around the COM of course. Which force is really providing it the centripetal acceleration? It can't be tension because the tension by the adjoining masses on will cancel out. Then what force is it?

• "the tension by the adjoining masses on will cancel out" That's quite an assumption. Can you justify it? – garyp Sep 21 '16 at 16:05
• Well just because the tension force by the adjoining particles on it will have an angle = 180 degrees. – Aaryan Dewan Sep 21 '16 at 16:12
• I don't think that the tension force would be exactly 180 as if it was then you would have a line instead of a ring – J. Shupperd Sep 21 '16 at 16:18
• Spin a rod instead of a hoop. There's no curvature to play a role. In both cases, though, the forces do not cancel. – garyp Sep 21 '16 at 16:29
• I hope you realize that the angle is not the whole story. If that's all there was to it, the unbalanced force would be directed toward the center of the hoop at all points on the hoop. But that is not the case. The unbalanced force points toward the axis of rotation. – garyp Sep 21 '16 at 16:34

It's the tension, or more generally the internal stress in the ring. The specific name is that this force results from the hoop stress.

Most likely, you drew a circle, considered a point on the edge of it, and drew two tension forces on that point. The tension forces point opposite directions and so exactly cancel. Therefore, the force from tension on a point is zero.

True, but not helpful. The mass of a point is also zero, so of course the net force on it is zero.

Instead, if your element has mass $dm$, you must also acknowledge that it is curved a bit and subtends an angle of $d\theta = 2\pi \frac{dm}{M}$, with $M$ the mass of the entire ring.

The two tension forces on a finite size piece of the ring do not cancel. They don't point at 180 degrees from each other, but a little bit less. Using the small angle approximation, you should be able to derive the hoop stress formula linked in the Wikipedia article

• Suppose you spin a rod instead of a hoop. Then your angle argument breaks down. For the rod and the hoop, the forces do not cancel. – garyp Sep 21 '16 at 16:27
• Thank You Mark. I got my mistake. And thanks to you too @garyp – Aaryan Dewan Sep 21 '16 at 16:31

I was also interested in the question how large the tension force on any point in a ring is due to its spinning. The logical answer is already given by Mark Eichenlaub; I just want to add some geometry and mathematical considerations.

We can try to find the correct value of the tension force in the ring at any given point by assuming that the ring of mass $$m_R$$ resembles a chain of $$N$$ finite point masses $$M_N$$ connected by tethers. The tethers then experience the tension force and have to produce the centripetal force which keeps the point masses on their circular path (i.e. in the ring).

Here's a picture of two such chains, once for $$N = 4$$ and once for $$N = 8$$: If $$N$$ goes to infinity we will find out the tension force in a real ring. So that's our goal.

First, have a look at the diagrams. There I drew accelerations instead of forces. This seemed appropriate because the point mass and thus the centripetal force change with $$N$$ while the centripetal acceleration stays constant. Since $$F = m \cdot a$$ and since $$M_N = \frac{m_R}{N}$$ is constant for each $$N$$, this is simply proportional and will give us the tension force eventually.

$$a_z$$ is this constant centripetal acceleration, $$a_1$$ and $$a_2$$ are the accelerations due to the pulling of the tethers which we know result in this centripetal acceleration. $$a_{t1}$$ and $$a_{t2}$$ are the accelerations (which cancel each other out unless the ring breaks) of the pure tension forces pulling on the point masses to both sides.

We can see that with growing $$N$$ the accelerations along the tension $$a_{t1}$$ and $$a_{t2}$$ will grow to infinity as well (notice the increment from the left to the right diagram). But since $$M_N$$ will shrink to 0 in the same process, their product (the force) might be finite, so let's go on.

Some basic geometry as formulas:

$$a_{t1} \cdot \tan\alpha = \frac{a_z}{2}$$

$$\implies a_{t1} = \frac{a_z}{2 \cdot \tan\alpha}$$

$$\alpha = \frac{2\pi}{2 \cdot N} = \frac{\pi}{N}$$ (It is half the angle between the point masses. And concerning your original question: Here you can see again, no matter how large a finite $$N$$ is, the angle $$\alpha$$ will never be zero.)

$$\implies a_{t1} = \frac{a_z}{2 \cdot \tan{\frac{\pi}{N}}}$$

For very small arguments to $$\tan$$ (which we have here if N goes to infinity), $$\tan x \approx x$$, so we can simplify this to:

$$a_{t1} = \frac{a_z}{2 \cdot \frac{\pi}{N}} = \frac{N}{2\pi} a_z$$

If we now multiply with the mass $$M_N$$ we get our forces $$F_T$$ (tension force) and $$F_z$$ (centripetal force):

$$F_T = a_{t1} M_N = \frac{N}{2\pi} a_z M_N = \frac{N}{2\pi} a_z \frac{m_R}{N} = \frac{1}{2\pi} a_z m_R = \frac{1}{2\pi} F_z$$

So $$F_T = \frac{F_z}{2\pi}$$.

Be aware that the centripetal force $$F_z = a_z m_R$$ is given in relation to the mass of the whole ring. What does that mean?

Well, consider "splitting" the ring into just one single point mass. So $$F_z$$ is the force we need to to apply on an object of the mass of the ring to rotate that object around the center in a distance equal to the radius of the ring.

Example:

We have a metal ring with a radius $$r = 1\mbox{m}$$ and a mass $$m_R = 1\mbox{kg}$$. We rotate the ring with a velocity $$v = 1\frac{\mbox{m}}{\mbox{s}}$$.

The formula for the centripetal force to rotate a point mass of 1kg in a radius of 1m with this speed is $$F_z = m\frac{v^2}{r} = 1\frac{1}{1} \mbox{kg}\frac{\mbox{m}}{\mbox{s}^2} = 1\mbox{N}$$. According to our computation above the tension force $$F_T$$ in any point of the ring is then $$\frac{1}{2\pi} \mbox{N}$$.

Centripetal force is provided by molecular bonds holding the ring together, If these molecular bonds aren't strong enough to hold for a specific angular velocity $\omega$ and you spun it up at that $\omega$ angular velocity it will break (See this video of disk actually breaking because of this). For visualization suppose that ring is made out of rubber (can stretch easily) and you spin it, it's radius will increase because it's molecular bonds aren't strong enough to provide centripetal acceleration which is essential to circular motion, by increasing the radius it provides more force through tension.

Mathematically centripetal force must equal Tension Force so:

$$F_c=m\frac{v^2}{r}$$

must equal to Force of Tension provided by string:

$$F=kx\Rightarrow$$

$$x=m\frac{v^2}{kr}$$

From equation we can see that more velocity and mass an object has more string has to stretch to keep it moving in a circular path.