A bead on the wire loop

Question

In my physics book there is a question which stated is as follows:

A thin circular loop of radius $R$ rotates about its vertical diameter with an angular frequency $\omega$. Show that a small bead on the wire loop remains at its lowermost point for $\omega \leq \sqrt{g/R}\,$. What is the angle made by the radius vector joining the centre to the bead with the vertically downward direction for $\omega=\sqrt {2g/R}$? Neglect friction.

When I started doing this question, I ran into the problem that I cannot gain any intuition that why should the bead move upward the loop if $\omega \geq \sqrt {g/R}$? What other forces might be acting on it (other than normal force by the loop and the gravitational force)?

Most important of these two is the former, as when I imagined this I cannot think why it should move upward because it looks like it should remain at the lowest point whatsoever be the angular velocity (because if it is like a particle and is at the lowermost point then it would lie on the axis of rotation and hence should have a zero acceleration.)

Please correct me if I'm going wrong anywhere.

Answer 1

What other forces might be acting on it (other than normal force by the loop and the gravitational force)?

Actually, those are the only forces acting on the bead. However, keep in mind that there will be "inertial effects" coming into play here. For another example of this, think of the example of a bead on a rotating, horizontal, straight wire. In that case the only force acting in the horizontal plane is the normal force that acts perpendicular to the wire, yet the bead will slide outwards with an increasing acceleration.

In your case and in the bead on the horizontal wire, it is much easier to gain intuition looking at the system in a rotating reference frame that rotates with the system. For the horizontal wire case, we then have a centrifugal force that pulls the bead outwards. The same is true in your case. The hoop needs to be rotating fast enough so that the centrifugal force is strong enough to overcome the weight of the bead.

However, as I stated before, in an inertial reference frame the only forces acting on the bead are its weight and the normal force supplied by the hoop. The bead still follows $\mathbf F=m\mathbf a$, but to get an actual expression for the normal force is pretty involved. Nevertheless, this technically is all you would need to describe the behavior of the bead.

The most important of these two is the former, as when I imagined this in my head I cannot think why it should move upward because it looks like it should remain at the lowest point whatsoever be the angular velocity (because if it is like a particle and is at the lowermost point then it would lie on the axis of rotation and hence should have a zero acceleration.)

Your reasoning here is correct if the bead started at the lowest point. Then it will ideally remain there forever (the same is true for the horizontal wire example with the bead starting on the axis of rotation). However, if the hoop is spinning fast enough this point becomes an unstable equilibrium. A slight disturbance will cause it to move away from this point on the hoop (this will happen on its own if you did this in real life). I suppose the problem could have been more precise by stating the additional assumption that the bead is not starting at the lowest point on the hoop.

Answer 2

Absolutely. For your problem, if the bead is at rest at the bottom initially, it will always stay there.

The possible forces on the bead in the rotating reference frame are:

1. Normal reaction from the wire

2. Weight

3. Centrifugal force

4. Coriolis force (irrelevant for motion on the wire, as it is perpendicular to the wire and taken care of by the azimuthal component of the normal reaction)

To get the equilibrium positions, you just need to find the angles $\theta$ for which the tangential force on the stationary bead (i.e. along the wire) is zero.

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More Perspective

This is a mechanical example of what is known as Landau's theory of phase transitions in condensed matter.

When $\Omega$ is small, there is only one equilibrium position of the bead in the rotating frame of reference, and that's the lowermost position, at the bottom.

As one increases $\Omega$, at some point, the centrifugal force becomes strong enough to balance out the tangential component of the bead at a non-trivial angle (on either side), and two new equilibrium positions occur. If you now do a stability analysis of the problem, it will turn out that as long there is one equilibrium position, it is stable. However, as soon as the new equilibrium positions occur, the lowermost position becomes unstable (while the new ones are stable).

You may now ask what happens if the bead is at rest at the bottom initially. The answer is that the bead should stay there forever. In practice, however, this doesn't happen because there is no perfectly isolated & frictionless/dissipationless system. The system might pick up some stray disturbance and it will spontaneously sway to one side. Once that happens, the bead will move towards the more stable position and in the presence of infinitesimal dissipation eventually settle in that position. This phenomenon in more exotic condensed matter/particle physics setups is called spontaneous symmetry breaking and is responsible for solids, ferromagnetism, superconductivity, superfluidity, Higgs mechanism etc.

Answer 3

On further consideration, I feel that my previous answer was wrong. If the loop is rotating at a constant angular velocity, w, with the bead at an angle, A, from the vertical. Then the the vertical component of the normal force from the loop is N cos(A) = mg and the the horizontal component N sin(A) provides the centripetal force. mRw^2. Dividing these: sin(A)/cos(A) = (R w^2)/g and w^2 = [g tan(A)]/R. Since the tangent approaches zero as, A, approaches zero, I see no justification for the limitation given in the problem. (There would be a limitation imposed by the distance from the center of the axle to the center of the bead.