# Homework Help: Bead on spinning hoop

1. Nov 10, 2012

### thisisbenbtw

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

τ = r F cos θ = I α
α = a / r

3. 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.

2. Nov 10, 2012

### 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.

Last edited: Nov 10, 2012
3. Nov 10, 2012

### 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?

Last edited: Nov 10, 2012
4. Nov 10, 2012

### 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).

5. Nov 10, 2012

### thisisbenbtw

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

6. Nov 10, 2012

### 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...

7. Nov 10, 2012

### thisisbenbtw

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

8. Nov 10, 2012

### lewando

Yes.

9. Nov 10, 2012

### thisisbenbtw

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

And solving Fn 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?

10. Nov 10, 2012

### lewando

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

11. Nov 10, 2012

### 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))

12. Nov 10, 2012

### lewando

That looks right--good job by you!

13. Nov 10, 2012