A thin ring of radius `R` is made of a material of density `rho` and Young's modulus `Y`. If the ring is rotated about its centre in its own plane with angular velocity `omega` , find the small increase in its radius.

Answer

2

Explanation

Consider an element AB of length dl. Let T be the tension and a be the area of cross-section of the wire.

Mass of the element is dm =(Volume)(Density)`=adl)rho`
The component of T, i.e., `(Tsin(theta)/(2)+Tsin(theta)/(2))` provides required centripetal forces to the element considered.
`thereforeF_(C)=2Tsin(theta)/(2)`
`(dm)R_(omega^(2))=2T((theta)/(2))(becausetheta " is small,sin"(theta)/(2)=(theta)/(2))`
`T=((adl)rhoRomega^(2))/(theta),theta=(dl)/(R)`
`thereforeT=aR^(2)rhoomega^(2)`... (i)
Let `DeltaR` be the increase in radius of the ring
Longitudinal strain `(Deltal)/(l)=(Delta(2piR))/(2piR)=(DeltaR)/(R)`
`Y=(T//a)/(DeltaR//R)rArrDeltaR=(TR)/(aY)`
From Eq . (i) , `DeltaR=(R)/(aY)(aR^(2)rhoomega^(2))thereforeDeltaR=(rhoR^(3)omega^(2))/(Y)rArrn=2`