# Bead confined to a circular hoop

1. Dec 1, 2007

### issisoccer10

[SOLVED] Bead confined to a circular hoop

1. The problem statement, all variables and given/known data
A bead with mass m is confined to traveling along a cirlcular hoop with radius r. The bead can slide without friction along this hoop. The hoop, which is vertical, rotates with an angular velocity (omega). What is the angle (theta), measured from the bottom of the hoop and with counterclockwise as positive, in which the bead will not move for this angular velocity?

2. Relevant equations
angular momentum (L) = I * (omega)
torque = dL/d(theta)
Force = torque/R

3. The attempt at a solution
I have been working on this problem for a solid day and a half and really haven't made too much progress with it. The bead, at this certain value of theta, will basically be rotating in a circle. Then I found the angular momentum (at least what I thought it was) by taking m*(R sin (theta))^2 * (omega). Then I took the derivative of L to get torque and then divided to torque by R, thinking I would be obtaining the force. The only other force that I knew to be at work on bead was the force of gravity. So I set the net force to equal zero and then solved for theta. Yet upon reflection, I realize that the angular momentum has a vertical and horizontal component going out from the center, but i don't know how to find it. But that could be entirely wrong as well.

I have also thought that this senario seems as if it could be explained by a string attached to the bead and anchored at the origin. As the angular velocity increases, the angle would increase, and the particle would be moving along as if confined to circular hoop like in the original problem. However, I'm not able to figure out where this force in the vertical direction comes into play that cancels out the force of gravity...

I also considered the position of the bead could be described parametrically with x = sin(theta) and y = -cos (theta). Maybe I could differentiate this to get acceleration and then eventually force, but this course of action didn't seem all that logical, so I didn't really pursue it.

All in all i'm pretty lost, but I hope that I've got some bits and pieces of "correctness" in there...but in any case, I would certainly enjoy some guidance on this problem

thanks a lot

2. Dec 1, 2007

### muppet

Can I clear up a few points?
1.Does the hoop rotate in its own plane? Or around an axis perpendicular to the normal vector to that plane?
2.The question states that bead will not move... from your working I take it you mean relative to the initial position on the the hoop?
3.Does the angle describe the position of the bead on the hoop? Or are we changing the angle of the hoop from the vertical?

I think the hoop is rotating around the vertical, and theta describes an angular position on the circumference of the hoop at which position the bead will not move relative to that point?

3. Dec 1, 2007

### ozymandias

Hey issisoccer10,

You should analyze the problem by transforming it to a frame of reference which rotates with the hoop.
Let the center of your system be at the center of the hoop and assume your frame of reference rotates with the same angular velocity as the hoop.
What fictitious forces does this introduce? Where does it/do they point?
In the new frame of reference, your hoop is stationary and so is your bead.
What forces act on the bead? (don't forget to take the fictitious force(s) into account!)
Can you figure out the stationary point for the bead now?

--------
Assaf
http://www.physicallyincorrect.com/" [Broken]

Last edited by a moderator: May 3, 2017
4. Dec 1, 2007

### Dick

The bead is indeed rotating in a circle whose radius is determined by theta. At that position the forces acting on it are mg vertically and a normal force from the hoop with is directed towards the center of circle. The vertical component of that normal force has to cancel the mg and the horizontal has to match the centripetal acceleration at that angle of theta. Forget angular momentum, torque etc. Its' much simpler than that.

5. Dec 1, 2007

### issisoccer10

Thanks Dick,

Your explanation really seems to make sense to me. As for ozymandias, I honestly have no idea what what fictitious force you are refering to. But as for the normal force from the hoop, how would you go about calculating that?

6. Dec 1, 2007

### Dick

It's unknown but it's direction is known, it's perpendicular to the hoop. And it's vertical component has to cancel the mg downward force. That let's you find the horizontal component.

7. Dec 1, 2007

### ozymandias

I meant the centripetal force, pointing towards the middle of the hoop. :).
(when the bead is in motion there is another force, proportional to $$R \ddot{\theta}$$ present, but since we're analyzing a system in equilibrium $$\theta$$ does not vary with time and this term is zero).
Now, at equilibrium, what would be the condition for the bead not to move?

--------
Assaf
http://www.physicallyincorrect.com/" [Broken]

Last edited by a moderator: May 3, 2017
8. Dec 1, 2007

### issisoccer10

ok so it seems to me that sin^-1 (mg/normal force) = theta. This is amazing thanks...but is there a way to describe normal force in terms of the angular velocity and radius and mass

9. Dec 1, 2007

### issisoccer10

wait...centripetal acceleration equals -R*(omega)^2...so the horizontal force would be
-mR(omega)^2?

10. Dec 1, 2007

### Dick

If you mean R is the radius of the hoop, no. R should be the distance from the axis of rotation.

11. Dec 2, 2007

### ozymandias

Yep, sorry, my mistake. I stand corrected.