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^(@)`?

The figure shows a thin ring of mass `M=1kg` and radius `R=0.4m` spin

1.	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^(@)`?
Answer» `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`

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