What is the angle of equilibrium for a spinning bead constrained to a hoop?

[thisisbenbtw]

Homework Statement

A bead of mass m is constrained to move on a hoop of radius R which is spinning with angular velocity ω. There is no friction.

Determine the angle θ at which the bead does not move for angular velocity ω. Do not consider the solution θ = 0.

Homework Equations

τ = r F cos θ = I α
α = a / r

The Attempt at a Solution

Tried using formulas for torque - ended up with F = m R tan θ * α and α=α/ cos θ. Can't figure out what to do next. Also not sure about my free body diagram.

Free body diagram:
Force upwards due to angular velocity
Force downwards due to gravity.

 

[lewando]

Angular velocity is constant, so no torque involved. There is a normal force from the hoop acting on the bead. There is also a centripetal force.

 

[thisisbenbtw]

So there would be a downward force for gravity, a force towards the center of the hoop for centripetal force, and another force in the same direction for the normal force?

 

[lewando]

The normal force will be pointing to the center of the hoop (normal to a tangent line where the bead and hoop intersect). Downwards for mg--correct, centripetal force is an outward force (think of the bead in orbit about the vertical axis of the hoop).

 

[thisisbenbtw]

So the centripetal force would be perpendicular against the vertical axis, and point torwards it?

 

[lewando]

I always struggle on how to word this properly so here goes: The force that holds the bead in orbit is the centripetal force. You are correct that it is an inward force, not outward as I had incorrectly stated*. This force is provided by the horizontal component of the normal force. You can say that the centripetal force is equal to the horizontal component of the normal force.

*I tend to look at these types of problems from the perspective of rotating along with the hoop...

 

[thisisbenbtw]

So there would be -mg and F_n cos θ in the vertical direction, and F_n sin θ = m v² / r in the horizontal direction?

 

[lewando]

Yes.

 

[thisisbenbtw]

So F_n = m v² / (r sin θ) = m ω² r / sin θ

And solving F_n cos θ - m g = 0 :
ω² r cos θ/ sin θ - g = 0
ω² r cos θ/ sin θ = g
cot θ = g / (ω² r)
and θ = arccot(g / (ω² r))

Is that right?

 

[lewando]

I think the little r is not the same as the big R.

 

[thisisbenbtw]

Little r is the distance from the bead to the axis, correct?

So that would be r = R sin θ

ω² r cos θ/ sin θ = g
ω² (R sin θ) cos θ/ sin θ = g
cos θ = g / (ω² R)
and θ = arccos(g / (ω² R))

 

[lewando]

That looks right--good job by you!

 

[thisisbenbtw]

Thanks for your help

 

[lewando]

You're welcome. This problem is almost exactly similar to a tetherball-type problem. Other than checking your math for errors (to validate your answer), and if you have any extra non-existant free time, grab a string, length R, and tie it to a mass of your choosing and see how fast it needs to rotate to achieve an eyeball-45 degree angle. See if this reality matches your result.