The figure shows a thin ring of mass `M=1kg` and radius `R=0.4m` spinning about a vertical diameter (take `I=(1)/(2)MR^(2))` A small beam of mass `m=0.2kg` can slide without friction along the ring When the bead is at the top of the ring the angular velocity is `5rad//s` What is the angular velocity when the bead slips halfwat to `theta=45^(@)`?

Explanation

`I_(1)omega_(1)=I_(2)omega_(2)`
`thereforeomega_(2)=((I_(1))/(I_(2)))omega_(1)`
`=(((1)/(2)MR^(2)))/([(1)/(2)MR^(2)+m((R)/(sqrt(2)))^(2)])omega_(1)`
`=((M)/(M+m))omega_(1)=((1)/(1+0.2))^(5)`
`=(25)/(6)rad//s`