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 increases in its radius.

Explanation

Consider an element PQ of length dl. Let T be the tension and A the area of cross-section of the wire .
Mass of element `dm = volume xx density`
`=A(dl)_(rho)`
The component of T, towards the centre provides the necessary centripetal force to portion PQ.
`:.F = 2T sin((d theta)/(2))= (dm)Romega^(2)`
For small angles `sin (d theta) /(2) =(d theta)/2= ( dl//R)/2`
or d theta `= (dl)/(R)`
Substituting in Eq. (i), we have
`T.(dl)/(R) =A(dl)_(rho)Romega^(2)`
or `T = A_(rho)omega^(2)R^(2)`
Let `Delta`R be the increase in radius.
Longitudinal strain `= (Deltal)/(l) = (Delta(2piR))/(2piR)=(DeltaR)/(R)`
Now, `Y=(T//A)/(DeltaR//R)`
`:. DeltaR= (TR)/(AY)`
`= ((A_(rho)omega ^(2)R^(2))R)/(AY)`
or `DeltaR=(rhoomega^(2)R^(3))/(Y)`