A ring of radius `r` is suspended from a point on its circumference. Determine its angular frequency of small oscillations.

Explanation

It is physical pendulum , the time period of which is ,
`T=2pisqrt((I)/(m gl))`
Here ,I=moment of ineria of the ring about point of suspension
`mr^(2)+mr^(2)=2mr^(2)`
and l=distance of point of suspension from centre of gravity =r
`therefore T=2pisqrt((2mr^(2))/(m gr))=2pisqrt((2 r)/(g))`
`therefore` Angular frequency `omega=(2pi)/(T)` or `omega=sqrt((g)/(2 r))`