A chain of mass m forming a circle of radius R is slipped on a smooth round cone with half- angle ` theta`. Find the tension in the chain if it rotates with a constant angular velocity `omega` about a vertical axis coinciding with the symmetry axis of the cone .

Explanation

The sought tensile stress acts on each element of the chain. Hence divide the chain into small, similar elements so that each element may be assumed as a particle. We consider one such element of mass `dm`, which subtends angle `dalpha` at the centre. The chain moves along a circle of known radius R with a known angular speed `omega` and certain forces act on it. We have to find one of these forces.
From Newton's second law in projection form, `F_x=mw_x` we get
`2Tsin (dalpha//2)-dN cos alpha=dm omega^2R`
and from `F_z=mw_z` we get
`dN sin theta=gdm`
The putting `dm=mdalpha//2pi` and `sin (dalpha//2)=dalpha//2` and solving, we get,
`T=(m(omega^2R+gcottheta))/(2pi)`