I Bead on a vertical frictionless ring

[TahirMaqbool]

TL;DR Summary

How can the problem be solved from an innertial frame!

A bead is kept at the bottom of a vertical frictionless ring , if the ring is given an acceleration a horizontally, what is the maximum angular displacement of the particle?
The problem is easily solved from the frame of reference of the moving ring, I am however Intrested in a solution without use of pseudo force.

 

[Vanadium 50]

Without solving the problem, the fact that you want to know the angle of the bead with respect to the center (or the bottom) of the ring suggests that a non-ring-centered frame is not the right setup, Why convert twice?

 

[TahirMaqbool]

Without solving the problem, the fact that you want to know the angle of the bead with respect to the center (or the bottom) of the ring suggests that a non-ring-centered frame is not the right setup, Why convert twice?

Just curious, I got the answer from a ring centerd frame . I want to solve the problem from ist principle without involving pseudo forces .
Infact I did get a matching answer from a non ring centerd static frame ,but I am not comfortable with some of the things I did there

 

[kuruman]

I assume that, since the ring is said to be vertical, the bead is subject to the force of gravity. If so, what is the direction of the given acceleration relative to the acceleration of gravity?

 

[TahirMaqbool]

I assume that, since the ring is said to be vertical, the bead is subject to the force of gravity. If so, what is the direction of the given acceleration relative to the acceleration of gravity?

I apologise, I should have mentioned the ring is accelerated horizontally

 

[kuruman]

I apologise, I should have mentioned the ring is accelerated horizontally

Have you ever solved the problem of the mass hanging by a string from the ceiling of an accelerating car? The only difference between that problem and this one is that the tension in the string is replaced by the normal force exerted by the ring on the bead. Just draw a free body diagram and apply Newton's second law.

 

[TahirMaqbool]

That would be true if one was to ask for the equilibrium location, however for maximum displacement this one isn't becoming easy(equilibrium doesn't mean maximum angular displacement) rather velocity should become equal to velocity of frame .thanks for the reply though

 

[kuruman]

Are you saying that the accelerating bead is not at a fixed angle relative to the vertical? You say you got the right answer in the non-inertial frame. What is that answer?

 

[TSny]

I don't see much difference between working in the inertial frame and working in the frame of the ring.

In the inertial frame, the 2nd law is $\sum \vec F = m \vec a_{\small B}\,$, where the forces are the "true" forces acting on the bead and $\vec a_{\small B}$ is the acceleration of the bead relative to the inertial frame.

In the inertial frame, you can write the acceleration of the bead as $\vec a_{\small B} = \vec a_{\small C} + \vec a_{\small B/C}$ where $\vec a_{\small C}$ is the acceleration of the center of the ring relative to the inertial frame and $\vec a_{\small B/C}$ is the acceleration of the bead relative to the center of the ring.

So, the 2nd law in the inertial frame may be written as $\sum \vec F = m \vec a_{\small C} + m\vec a_{\small B/C}$.

But this is equivalent to working in the accelerated frame of the ring where the 2nd law is written as $\sum \vec F - m \vec a_{\small C} = m\vec a_{\small B/C}$. The second term on the left is the fictitious force in the accelerated frame. So, you get essentially the same equations to solve in either frame.

$\vec a_{\small B/C}$ is the same vector in either frame and can be decomposed into a centripetal component, $r \dot \theta^2$, and a tangential component, $r \ddot \theta$. Here, $r$ is the radius of the ring and $\theta$ is the angular displacement of the bead on the ring.

 

[TahirMaqbool]

Are you saying that the accelerating bead is not at a fixed angle relative to the vertical? You say you got the right answer in the non-inertial frame. What is that answer?

I have attached the solution w.r.t a non inertial frame you may take a look

 

[TahirMaqbool]

I don't see much difference between working in the inertial frame and working in the frame of the ring.

In the inertial frame, the 2nd law is $\sum \vec F = m \vec a_{\small B}\,$, where the forces are the "true" forces acting on the bead and $\vec a_{\small B}$ is the acceleration of the bead relative to the inertial frame.

In the inertial frame, you can write the acceleration of the bead as $\vec a_{\small B} = \vec a_{\small C} + \vec a_{\small B/C}$ where $\vec a_{\small C}$ is the acceleration of the center of the ring relative to the inertial frame and $\vec a_{\small B/C}$ is the acceleration of the bead relative to the center of the ring.

So, the 2nd law in the inertial frame may be written as $\sum \vec F = m \vec a_{\small C} + m\vec a_{\small B/C}$.

But this is equivalent to working in the accelerated frame of the ring where the 2nd law is written as $\sum \vec F - m \vec a_{\small C} = m\vec a_{\small B/C}$. The second term on the left is the fictitious force in the accelerated frame. So, you get essentially the same equations to solve in either frame.

$\vec a_{\small B/C}$ is the same vector in either frame and can be decomposed into a centripetal component, $r \dot \theta^2$, and a tangential component, $r \ddot \theta$. Here, $r$ is the radius of the ring and $\theta$ is the angular displacement of the bead on the ring.

Basically the problem involves using workenergy theorem, a non inertial frame eats away work done by normal reaction. since the bead motion in a non inertial frame is always perpendicular to the the Normal reaction . Same , however,doesn't hold when you allow the ring to move and thus a labouries variable normal reaction force awakens . You can see my solution attached as an image in the original question .thanks for the response though

 

[kuruman]

I have attached the solution w.r.t a non inertial frame you may take a look

Thank you for posting your solution. I took a look and it does not make sense because the bead cannot be above the horizontal diameter of the ring without violating Newton's second law. Here is why.

I assume that the contact between the bead and the ring is without friction. There are two forces acting on the bead, gravity in a direction straight down and the normal force. The normal force is radial and can be directed either towards the center or away from it. Below are free body diagrams in the inertial frame showing the two possibilities. The acceleration is also shown directed horizontally to the left.

Look at the diagram on the left. The vector sum of the normal force and the weight, which is the net force, cannot possibly be in the horizontal direction.

Look at the diagram on the right. The The vector sum of the normal force and the weight, which is the net force, cannot possibly a zero vertical component which is required if the bead is to accelerate horizontally.

 

[TahirMaqbool]

Ist of all thank you so much for taking time to think over it, but you are upon error here .bead will have attained some velocity before the horizontal diameter and then it will deccelerate because of the forces which you have already shown.
And even for the sake of argument it was to stop somewhere below the horizontal diameter (which it can if acceleration is small). The work energy theorem still applies.
Ah!! This is giving me sleepless

 

[Dale]

I would just use a Lagrangian approach. That way you would have no pseudo forces because you aren’t using forces at all. And you still can have as simple a solution as possible.

 

[kuruman]

The work energy theorem still applies.

Sure. The work theorem always applies. Which force does the necessary work on the bead to lift it above the horizontal diameter? In your solution you invoke the pseudo-force, but that points in the horizontal direction. It can provide a component tangent to the ring to lift the bead against the weight but only up to the horizontal diameter and it needs to be infinite to reach that point.

 

[TahirMaqbool]

I would just use a Lagrangian approach. That way you would have no pseudo forces because you aren’t using forces at all. And you still can have as simple a solution as possible.

Thanks for that answer , I am interested in solving from ist principles ,I think I do have a solution already but why can't it be done from the ground frame is my real question here.
Work done by gravity bis same in both the frames just work done by normal reaction is bothering me.

 

[TahirMaqbool]

Sure. The work theorem always applies. Which force does the necessary work on the bead to lift it above the horizontal diameter? In your solution you invoke the pseudo-force, but that points in the horizontal direction. It can provide a component tangent to the ring to lift the bead against the weight but only up to the horizontal diameter and it needs to be infinite to reach that point.

By the time particle reachs the horizontal diameter it would have some velocity,which would lift it further up, just like we throw a ball upwards ,just because gravity is pulling it down doesn't mean the it can't rise to some height.

 

[A.T.]

By the time particle reachs the horizontal diameter it would have some velocity,which would lift it further up, just like we throw a ball upwards ,just because gravity is pulling it down doesn't mean the it can't rise to some height.

I think the solution depends on how the horizontal acceleration changes from zero to a fixed value.

If the change in acceleration is instantaneous, then I think the largest angle will be twice the equilibrium angle. Reasoning: Without friction the particle will oscillate forever around the equilibrium angle at a constant amplitude, which is the initial deflection form the equilibrium. The equilibrium angle is easy to find as atan(a/g).

 

[TahirMaqbool]

I think the solution depends on how the horizontal acceleration changes from zero to a fixed value.

If the change in acceleration is instantaneous, then I think the largest angle will be twice the equilibrium angle. Reasoning: Without friction the particle will oscillate forever around the equilibrium angle at a constant amplitude, which is the initial deflection form the equilibrium. The equilibrium angle is easy to find as atan(a/g).

Thats correct and if you see my solution above (attached as an image)I got exactly same answer using pseudoforces. I just wanted to see is it possible to solve it by simply workenergy theorem applied from ground frame.

 

[Dale]

I am interested in solving from ist principles

Ok, so Hamiltonian then. That would work too.

 

[TahirMaqbool]

I finally got what I wanted, thank you all who showed interest