Bead on a rotating vertical ring

[issacnewton]

Hi

I have some conceptual questions about one situation. I have posted the picture. We have a bead on a ring which is rotating about the vertical axis passing through its center. The bead is not tied to the center of the ring , though it appears like that in the figure. Let m be the mass of the bead and R be the radius of the ring. Then at the angle \theta, the force on the bead which is directed towards the axis of rotation is N\sin \theta, where N is the normal reaction by the ring on the bead. Now since the ring is rotating about its axis, there has to be a centripetal force on the bead. It is \frac{mv^2}{R\sin \theta}. If T is the period of the rotation then , we can write

v=\frac{2\pi R\sin \theta}{T}

so the required centripetal force on the bead, at a given angle is give by

F_{cetr}=\frac{mv^2}{R\sin \theta} = \frac{4\pi^{2}R\sin \theta}{T^2}

and some force must play the role of centripetal force and as we have seen above, that is
F=N\sin \theta

The second equation for the bead , in the vertical direction is

N\cos \theta=mg so

N=\frac{mg}{\cos \theta}

plugging this into the expression of F, we get

F=\frac{mg}{\cos \theta}\, \sin \theta=mg\tan \theta

so F is the force which will play the role of the centripetal force and the centripetal force is
F_cetr.

Now if we plot both F and F_cetr versus angle on the same graph, we see that
the graphs intersect only at two points, one at \theta=0 and another at
\theta=\theta_{o}.

So its only at these two points that F = F_cetr. So that means, if we leave the bead at some angle between these two angles, bead will slide upwards so that F is equal to the required centripetal force.

Now my question is as follows. When the bead is between these two angles and bead starts
sliding upwards, what force is acting on the bead ? Can we express this force as a function
of m, g, and \theta_{o} ?

I couldn't come up with any thought . Any opinions ?

 

[dgOnPhys]

Hi,

interesting problem. Here it's quite late so take this with a grain of salt; also I did not check your equations...

If the constraint is frictionless, in the inertial frame along the tangential direction you have no force but (a component of) gravity: the ring will always slide downwards.

If the constraint has friction than you have a tangential component to the constraint reaction than can provide a static solution for \theta \neq 0 or even an upward (along the circle) acceleration.

So if you know the bead climbs up, the force acting on it is friction and its value is easy to find as equal to the tangential component of the centripetal force.

It might be easier to reason in the rotating frame.

Hope this helps

 

[issacnewton]

Hi. didn't follow what you are saying. so it its frictionless , and bead is between these two critical angles, will it slide downwards ? makes sense...

now in practice, there is always a small amount of friction, so you are saying there will be upward friction force along the tangent. but friction gets 'activated' because of some applied force. so what is this force which causes friction force to get activated.

 

[dgOnPhys]

In absence of friction is the component of gravity tangent to the ring that moves the bead, friction will be activated in reaction to this force.

 

[issacnewton]

Ok, but why would bead slide upwards then since the friction force will be equal to the tangential component of the weight.

 

[dgOnPhys]

Ok, but why would bead slide upwards then since the friction force will be equal to the tangential component of the weight.

This is a tricky one but notice that friction can be larger than the force that triggers it... otherwise your car would not accelerate, right?

EDIT: ignore this one, I guess I need to wake up completely before answering...

 

[issacnewton]

Do you think Lagrangian or Hamiltonian way is better approach to analyze this ?

 

[dgOnPhys]

A lagrangian approach when you are interested in the constraint forces at play is... the long way there...

I finally spent some time on it and found that there is no need for friction to get the behavior you described in the initial post; it is hard to see things in the inertial frame though; it is a lot easier to start in a rotating frame [see http://en.wikipedia.org/wiki/Rotati...tion_between_accelerations_in_the_two_frames"] (say origin at center of circle, xz plane on the circle plane,z in the vertical direction opposite to gravity): there Newton equations are easier to write and interpret
m \vec a' = \vec N + m \vec g -m \vec{\omega} \times (\vec{\omega} \times \vec r') -2m \vec{\omega} \times \vec{v'}
- the last term (Coriolis force) is normal to the circle plane so it is canceled by a component of the normal constraint force so we are left with
m \vec a' = \vec N_{//} + m \vec g -m \vec{\omega} \times (\vec{\omega} \times \vec r')
- all these forces lie in the circle plane
- the last term on the force side is the centrifugal force (due to the non-inertial frame): it is directed along x
- part of the acceleration \vec a' is centripetal to allow for the circular trajectory and is caused by a balance of normal reaction, weight and centrifugal force
- the residual tangential component gives
m a'_{\phi} = -m g sin \phi +m \omega^2 R sin \phi cos \phi

There are 4 equilibrium points, 2 stable (+/- acos (g/\omega^2 R)), two unstable (0 and pi) and you can see that the forces in the equation describe your motion: leaving the bead free at a small phi angle it will accelerate in the direction of growing phi towards the static equilibrium point, etc.

If we move back to the inertial frame, the equation of motion in the tangential direction can be read
m a'_{\phi} -m \omega^2 R sin \phi cos \phi = -m g sin \phi
so there is no force other than gravity along the circle tangent but the bead still manages to move upwards because of the trajectory enforced by the normal constraint reaction.

Even in absence of gravity (g=0) the bead will move upwards towards the stable position (+/-pi/2 in that case) so it is clear that there are no forces along the tangent necessary to create the acceleration. This happens because the direction along which we are projecting the equations of motion is itself rotating... interesting!

 

[issacnewton]

Wow, that makes sense. I thought that since this is rotational motion, some advanced thinking is necessary.

And for +/- you can use \pm like this \pm

 

[dgOnPhys]

Hi Issac,

I am still a little puzzled by the result; getting no acceleration (in the inertial frame) on the tangential direction is a bit surprising; I think you can basically think that the in the ring is instantaneously moving of uniform motion along that direction... definitely not intuitive to me!

Thanks for the tip on \pm