Conceptual question about a rotating ring

In summary: As we have seen here, its the force exerted by the two springs on a bead acts as the centripetal force. So the net force acting on a bead is directed towards the center of mass of the whole system. So this is how I resolved the question. I want to know if the reasoning is correct.In summary, the speakers discuss the concept of a thin ring rotating about an axis with no gravity, and the role of centripetal force in this scenario. One speaker proposes a model using beads and springs to demonstrate how the springs act as the centripetal force, while another speaker questions the existence of centripetal force and discusses the role of tension forces instead. They ultimately agree that simple models can be helpful in understanding
  • #1
issacnewton
998
29
Hi

I was thinking about this conceptual problem. Consider a thin ring of radius R, which is rotating about the axis passing through its center of mass. Now let's says there is no gravity and the ring is rotating at some constant angular velocity [tex]\omega[/tex] , so the angular momentum is conserved. Now if we look at some infinitesimal element of mass dm, on the circumference
then it is rotating about the axis too. So there must be a centripetal force acting on this mass element. First I couldn't think of any force which can play the role of centripetal force here.
Then I thought about a model. Let's say that some 10 beads are connected in circle with springs. So we have a polygon with 10 vertices (which are beads)and 10 sides (which are springs). Now let's set this in circular motion with some constant angular velocity. Now if we look at one of the beads, then we can imagine that two springs which connect it to the neighboring beads will get stretched. If we increase the angular velocity, then the springs will stretch even more. So we can model the thin ring as the beads connected by the springs and then its easy to see which forces act as the centripetal force. As we have seen here, its the force exerted by the two springs on a bead acts as the centripetal force. So the net force acting on a bead is directed towards the center of mass of the whole system. So this is how I resolved the question. I want to know if the reasoning is correct.

Thanks
 
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  • #2
hi IssacNewton! :wink:
IssacNewton said:
… if we look at some infinitesimal element of mass dm, on the circumference then it is rotating about the axis too. So there must be a centripetal force acting on this mass element.

As we have seen here, its the force exerted by the two springs on a bead acts as the centripetal force. So the net force acting on a bead is directed towards the center of mass of the whole system.

yes, that looks fine :smile:

in a solid spinning ring, each tiny element rdθ has tension forces at an angle of (i think) dθ/2 to the element, so there's a small resultant force radially inward, much like your springs :wink:
 
  • #3
thanks tiny-tim. My reasoning was right...
 
  • #4
it's bound to happen sometimes! :biggrin:

keep eating the apples! :smile:
 
  • #5
tiny-tim said:
hi IssacNewton! :wink:


yes, that looks fine :smile:

in a solid spinning ring, each tiny element rdθ has tension forces at an angle of (i think) dθ/2 to the element, so there's a small resultant force radially inward, much like your springs :wink:

Are you sure about this?
Not saying your wrong, but I do not see how a rotating ring can generate inward force vectors. I would think the force vectors would be tangent and out; not tangent, out and inwards.
 
  • #6
pallidin said:
… I would think the force vectors would be tangent and out; not tangent, out and inwards.

sorry, pallidin, I'm confused :confused:

radial can be either in or out, but tangential has to be out …

poor ol' tangents never get to come in! :redface:

but the two (slightly angled) tangent forces combine to have an inward radial component :wink:
 
  • #7
tiny-tim said:
...but the two (slightly angled) tangent forces combine to have an inward radial component :wink:

OK, but what is the source, in this scenario, for a second tangential force which results in a small, downward vector? I do not see it.
A rotating mass ring expands outwardly.

This would be like saying that the ring expands AND compresses at the same time.
Fine, that's a balloon. No problem there.
But, I thought we are talking just about a rotating solid mass ring, not a ring made of, say, expansive balloon material.
 
  • #8
pallidin, is my model correct? I don't see how tangential vectors come into the picture.
 
  • #9
IssacNewton said:
pallidin, is my model correct? I don't see how tangential vectors come into the picture.

tiny-tim model is like yours when you have an infinite number of beads and springs: each bead will be pulled in two (almost) opposite tangential directions by its springs; this generates a (small) centripetal force
 
  • #10
Small? Is there ANY evidence of it's existence at all?
Centripetal is a fictitious force. It does not exist.

Perhaps I am wrong, but please show me any evidence otherwise.
 
  • #11
pallidin said:
Small? Is there ANY evidence of it's existence at all?
Centripetal is a fictitious force. It does not exist.

Perhaps I am wrong, but please show me any evidence otherwise.

well let's take an element ds of the rotating ring circumference; in the inertial frame it rotates with (say) constant angular velocity in a circular trajectory; for this to happen, kinematics requires a constant centripetal acceleration. Newton equation requires then a force causing this acceleration: a constraint reaction (or tension), not a fictitious force at all.

Fictitious forces would appear in the non-inertial frame where the element ds is at rest. The fictitious force will be the centrifugal force (the opposite of the product of mass and centripetal acceleration) which will balance the constraint force.

In the inertial frame, its free-body diagram will show two tension forces on its extremities tangent to the arc ds (if ds subtends an angle [itex]d \theta[/itex] the angle between the two forces will be [itex]\pi-d \theta[/itex]). Since the angular velocity of the ring is constant, the two tensions will have the same modulus F so the resulting force will be directed along the radial direction towards the circle center with modulus [itex]2 F sin(d \theta / 2) = F d \theta [/itex], which is a small force accelerating the small mass dm of the arc element ds to a finite centripetal acceleration [itex]\omega^2 R[/itex].
 
  • #12
This problem signifies why simple models are important in physics. It helps understanding.
 
  • #13
pallidin said:
Centripetal is a fictitious force. It does not exist.

no, https://www.physicsforums.com/library.php?do=view_item&itemid=84" is a fictitious force that does not exist

centripetal force is simply a name for whatever existing force (such as tension or friction) is providing the https://www.physicsforums.com/library.php?do=view_item&itemid=27" that keeps something in a circle

in this case, the centripetal force is provided by the radial components of the nearby tangential tensions

as IssacNewton :smile: said …
IssacNewton said:
… if we look at some infinitesimal element of mass dm, on the circumference then it is rotating about the axis too. So there must be a centripetal force acting on this mass element.

As we have seen here, its the force exerted by the two springs on a bead acts as the centripetal force. So the net force acting on a bead is directed towards the center of mass of the whole system.
 
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  • #14
The original post is a very clear and helpful example. I believe that, once this principle is accepted, we can generalize it to say the following:

Consider any rigid body, arbitrary shape, initially at rest. Any one atom or molecule of the body, which is not accelerating (neglecting thermal jiggling), is being pulled in many directions simultaneously by atomic or molecular bonds, and these bond forces add up to a net force of zero. But now have the object rotate because of its conserved angular momentum -- now all of the bonds pulling on anyone atom or molecule will add up to a net force directed toward the center of rotation.
 

1. What is the concept of a rotating ring?

The concept of a rotating ring refers to a physical system in which a ring or circular object is spinning or rotating around a central axis. This can be seen in various natural and man-made objects such as planets, wheels, and turbines.

2. What causes a rotating ring to spin?

The spinning or rotation of a ring is caused by the application of a force or torque on the object. This force can come from various sources such as gravity, magnetic fields, or mechanical energy. In the absence of any external forces, a rotating ring would continue to spin at a constant rate due to its inertia.

3. How does the rotation of a ring affect its properties?

The rotation of a ring can affect its properties in several ways. For example, the centrifugal force created by the rotation can cause the ring to expand or contract, depending on the materials it is made of. Additionally, the rotational speed can affect the stability, angular momentum, and other physical characteristics of the ring.

4. What are some applications of a rotating ring?

A rotating ring has various applications in different fields of science and technology. For instance, it is used in engines and turbines to convert rotational energy into mechanical work. In astronomy, the rotation of planets and other celestial bodies is studied to understand their formation and evolution. It is also used in gyroscopes for navigation and in many other industrial and scientific processes.

5. Can a rotating ring ever stop spinning?

In a perfect vacuum with no external forces, a rotating ring would continue spinning indefinitely due to its inertia. However, in the presence of external forces such as friction or drag, the ring will eventually slow down and come to a stop. This process is known as rotational deceleration.

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