Tension in a chain around a cone A chain of some mass, forming a circle, is slipped on a smooth cone.
If we consider an infinitesimally small section of the chain, a component of gravity will try to accelerate it along the surface of the cone. So there must be some tension in the string that prevents the acceleration. What will be the direction of the tension on this section of the chain?
Will it be towards the height of the cone, parallel to the ground?
 A: Suppose the angle of the cone is $\theta$, the radius of the ring is $r$ and the length density of the ring is $\rho$. Take a small segment of the ring that corresponding to an angle $\phi$, then the mass of this segment would be 
$$m=r\phi \rho$$
the normal force from the cone would be
$$N=\frac{mg}{\mathrm{sin}(\theta/2)}$$
If the tension on the ring is $T$, then we have 
$$2T\mathrm{sin}(\phi /2)=N\mathrm{cos}(\theta/2)$$
and consequently we have upon substituting in the expression for $N$
$$2T\mathrm{sin}(\phi /2)=r\phi \rho g \mathrm{cot}(\theta/2)$$
Now we take $\phi$ to be very small, which means $\mathrm{sin}(\phi /2) \approx \phi /2$, then we have 
$$T=r\rho   g \mathrm{cot}(\theta/2)$$
which only depends on the size and density of the ring and the angle of the cone, just as expected.
A: Let the cone angle at the apex be $2\alpha$.
Method 1 (Force considerations)
Consider a small arc of the ring subtended by angle $d\phi$. The tension $T$ acts tangential to both ends of this arc. This results in a radially-inward force of $2T \sin{(d\phi /2)}$.
Now consider the entire ring. The forces acting on it are:


*

*$mg$ vertically downards

*normal force $N$ all around the ring acting perpendicular to cone surface

*radially inward force due to tension


Resolving along vertical direction yields $N = mg / \sin{\alpha}$. The normal force acting along a small arc of the ring subtended by angle $d\phi$ is  $\frac{d\phi}{2\pi} N$. Now resolving along the horizontal direction yields
$$\frac{d\phi}{2\pi} N \cos {\alpha} = 2T \sin{(d\phi /2)}$$
$$\frac{d\phi}{2\pi} mg \cot {\alpha} = 2T \sin{(d\phi /2)}$$
In the limit $d\phi \to 0$, $\sin{(d\phi /2)} \to d\phi /2$, so
$$ T = \frac{mg}{2\pi} \cot \alpha $$
Method 2 (Work-energy)
At equilibrium, the drop in gravitation potential energy by losing a height of $dh$ equals the work done by the tension $T$ in stretching the ring. (The normal force acts perpendicular to the direction of ring movement if the ring slides, so does no work).
A change of height $dh$ changes the ring radius by $dh \tan \alpha$, and thus changes the ring circumference by $2\pi (dh) \tan \alpha$. Therefore
$$T 2\pi (dh) \tan \alpha = mg(dh) $$
$$ T = \frac{mg}{2\pi} \cot \alpha $$
