What is the reaction force between the ring and the hoop?

[Olivia Lam]

Homework Statement

A circular hoop rotates with uniform angular velocity w about a vertical diameter AOB, O being the center. A smooth ring P of mass m can slide on the hoop. If θ be the angle of inclination of the radius OP to the vertical when the ring is in equilibrium with respect to the hoop, (a) find cos θ in terms of the radius of the hoop r, angular velocity w and g. (b) Find the reaction force between the ring and the hoop.

Relevant Equations

N cos θ=mg
N sin θ=mrw^2 sin θ
cos θ=g/rw^2

a.)N cos θ=mg
N sin θ=mrw^2 sin θ
cos θ=g/rw^2

b.) My question is reaction force =N ? or =F=mg tanθ ?
If it is N then N=mg cosθ =mg^2/r w^2 or N=mg/cosθ =mrw^2 ?
Thank you

 

[TSny]

I would interpret "the reaction force between the ring and the hoop" to be the normal force N.

Your answer to (a) looks good.

 

[Olivia Lam]

I would interpret "the reaction force between the ring and the hoop" to be the normal force N.

Your answer to (a) looks good.

Thank you! Then N=mg cosθ =mg^2/r w^2 ?

 

[TSny]

Thank you! Then N=mg cosθ =mg^2/r w^2 ?

Check your work here. Note that you had the correct equation Ncosθ = mg in your first post.

Also, there is something interesting about the result for part (a): cosθ = g/(ω²r).
cosθ cannot be larger than 1. But the right hand side g/(ω²r) can be larger than 1 if ω is small enough. What do you conclude from this?