How do we provide centripetal force in this situation?

In summary, the conversation discusses the concept of centripetal force and how it applies in a situation where a bead is sliding through a wire that is being swung in a vertical plane. The participants explore the different forces acting on the bead, including gravity and centripetal force, and discuss how the wire provides the necessary centripetal force to keep the bead in circular motion. They also clarify the role of centrifugal force and normal force in this scenario. The conversation ends with a suggestion to physically demonstrate the concept to gain a better understanding.
  • #1
Adesh
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Situation: Let’s say we have a wire bent into a circular shape, there lies a bead through the wire and it can slide through it. The wire is kept in vertical plane and is swung along the axis AB.
image.jpg

My question : How the centripetal force is provided to the bead?

The bead will go into a circular motion as well (the wire is being swung in the vertical plane) and force acting on the bead are : gravity, and centripetal force. By help of Perok and haruspex, I know that centripetal force is just a component of forces (like tension, normal) it’s not something which exists on its own.

When we tie a stone to the rope and swing it round the circle in a horizontal plane, it is the tension force which is causing the centripetal acceleration (acceleration towards the center) and this tension exists because we are pulling the stone towards us, hence it’s we who are actually doing the real work.

But in the situation that I have described at first, I cannot seem to imagine how the centripetal force will ever come into play. How swinging around AB would provide the the centripetal force to the bead? Will there be some stretches or starins? Some people are saying that there are two forces acting on the bead when it swung: gravity and normal force from the wire, and it’s the horizontal component of the normal force which is causing the centripetal acceleration. What I don’t understand is why the wire would push the bead normally at all? (Well, I know why it cannot push tangentially, because we have assumed the wire to be friction-less hence all the forces from the wire will be perpendicular to it) but why wire will push the bead normally? What causes it to push on the bead? A mere contact cannot produce a normal force, we have to have something more because two pens touching each other on my desk aren’t pushing on each other apart until I push them together.

I hope I made myself clear, I tried my best to explain myself.
 
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  • #2
If you sit on a merry-go-round (I won't ask why...), and it starts to spin up, you'll "feel" yourself getting pushed back into your seat away from the axis of rotation. This "force" pushing you back is a consequence of analysing the motion in a rotating frame of reference; i.e. a pseudo-force which exactly balances the horizontal normal force. In the lab frame, the only horizontal force present is a normal contact force between yourself and the seat, which exists to provide the necessary centripetal force.

The rotating wire frame and bead really isn't much different conceptually. In the rotating frame of the bead, the bead gets thrown away from the axis by a centrifugal force causing the bead and wire frame to come into contact, producing that necessary reaction force. Switch back to the lab frame, and this normal force is still necessary to keep it going around in a circle.

It would be easiest to see if you try and build some sort of physical model and spin it around. The bead is "thrown" outward, and the inner face of the bead suddenly comes into contact with the wire frame. I think I drew a pretty picture of the interface in the previous thread. This one's not so pretty, and the bead/wire frame should be in contact (i.e. there's a gap drawn here), but I think it illustrates what you might be confused about:

1588695730430.png
 
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  • #3
etotheipi said:
In the lab frame, the only horizontal force present is a normal contact force between yourself and the seat, which exists to provide the necessary centripetal force.
If you’re talking about this type of merry-go-round then the normal contact force between me and the seat will be vertical. And one more point, in the lab frame centripetal force acts on the horse and so it will act on me too and that’s the reason I will go round in a circle with the horse. The normal contact force will cancel the gravity exactly.

etotheipi said:
In the rotating frame of the bead, the bead gets thrown away from the axis by a centrifugal force causing the bead and wire frame to come into contact, producing that necessary reaction force. Switch back to the lab frame, and this normal force is still necessary to keep it going around in a circle.
But centrifugal force will act perpendicularly away from the axis AB while the normal force will act towards the center of circle, not perpendicular to AB.
 
  • #4
Adesh said:
If you’re talking about this type of merry-go-round then the normal contact force between me and the seat will be vertical.
That link is to a food blogger called Mary's website. Is that what you meant?
Adesh said:
And one more point, in the lab frame centripetal force acts on the horse and so it will act on me too and that’s the reason I will go round in a circle with the horse. The normal contact force will cancel the gravity exactly.
Yep, it's a slightly different setup (must admit I didn't think of horses...), but the same concepts apply, you still get a vertical and horizontal component of normal force applied to you to a) balance weight and b) provide the centripetal force. Perhaps it's easier to forget the horse and to just think of a chair facing toward the centre of a disk, being rotated in a circle. Or a horizontal James Bond-style centrifuge.
Adesh said:
But centrifugal force will act perpendicularly away from the axis AB while the normal force will act towards the center of circle, not perpendicular to AB.
That's right; though we need to be careful since there are two circles in question here. The circle traced out by the bead and the circle of the wire frame. The normal force points toward the centre of the wire frame, so there is a horizontal and vertical component. In the rotating frame, the vertical component balances weight and the horizontal component balances the centrifugal force.
 
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  • #5
Look at a spherical pendulum swinging in a horizontal circle at constant ##\omega##.
Look at the bead in the wire going around in a horizontal circle at constant ##\omega##.
What's the difference? The force producing the acceleration is the same; only its name and origin differ. In the pendulum case it's called "tension" and is provided by the string, in the bead case it's called "normal force" and is provided by the wire. The free body diagrams are identical and so are the accelerations; if you understand one, you understand the other.
 
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  • #6
kuruman said:
in the bead case it's called "normal force"
If we forget about gravity for a moment, would then also wire going to push normally on the bead?
 
  • #7
Adesh said:
What causes it to push on the bead?
This is a like all normal forces: Two bodies get too close and start repulsing each other electromagnetically. But these microscopic details are irrelevant to solving the macroscopic problem, so I don't understand why you care.
 
  • #8
Adesh said:
If we forget about gravity for a moment, would then also wire going to push normally on the bead?
Yes. "Normal" in this context means "perpendicular to the surface" not "opposite to gravity." If the contact of the bead and the wire is frictionless, the only force the wire can exert on the bead is perpendicular to its surface and away from (not towards) the surface. So whether gravity is on or off does not matter to the kind of force that the wire exerts on the bead; it wll be normal regardless.
 
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  • #9
I want to know why there will be a normal force on the bead at all? When we keep a book on the table, the book feels a normal force but this normal is just a reactionary force, reaction caused by the weight of the book. In our wire and bead situation there is nothing pushing on the wire then why would the wire will push on the bead ?
 
  • #10
Adesh said:
In our wire and bead situation there is nothing pushing on the wire then why would the wire will push on the bead ?

If they are in contact, there will be a normal force between them

Adesh said:
When we keep a book on the table, the book feels a normal force but this normal is just a reactionary force, reaction caused by the weight of the book

Careful; the weight and the normal force are not Newton's third law pairs; NIII pairs are of the same type and act on different bodies. The normal forces between the book and table are pairs, as are the gravitational forces between the book and the Earth.
 
  • #11
Adesh said:
When we keep a book on the table, the book feels a normal force but this normal is just a reactionary force, reaction caused by the weight of the book.
All that reaction-talk doesn't actually mean much in terms of physics. A force is a force. And it's actually wrong if you are thinking about Newtons 3rd Law here: The normal force is not the 3rd Law reaction to weight.

My advice: Abandon your misguided notions about forces causing forces. The book and the table repulse each other, because of their proximity. Same for the wire and the bead.
 
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  • #12
Adesh said:
I want to know why there will be a normal force on the bead at all? When we keep a book on the table, the book feels a normal force but this normal is just a reactionary force, reaction caused by the weight of the book. In our wire and bead situation there is nothing pushing on the wire then why would the wire will push on the bead ?
The segment of wire that is in contact with the bead accelerates, because its velocity is continuously changing direction. What do you think keeps the wire loop going around at constant ##\omega## if not some kind of force external to it?

Suppose you're right and there is no normal force acting on the bead.
Question 1: What other force is there acting on the bead?
Answer: Gravity, nothing else.
Question 2: What if gravity alone acted on the bead?
Answer: The bead will be in free fall and at best it will describe a parabola in a vertical plane.
Question 3: Then how come the bead goes around in a horizontal circle?
Answer: There must be an additional force that prevents it from undergoing parabolic motion and constrains it to undergo uniform circular motion in a horizontal plane.
Question 4: What is the name of that force and what entity provides it?
Answer: ##~~\dots##
 
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  • #13
A.T. said:
My advice: Abandon your misguided notions about forces causing forces. The book and the table repulse each other, because of their proximity. Same for the wire and the bead.
I think your advice is really good, thinking about “force causing force” might lead me to some unnecessary sophisticated analysis.

You say any two object in contact with each other will exert force on each other (due to microscopic phenomena). If I keep a book on the table, the gravity tries to increase the contact between the book and table by attracting the book and the closer the book going to come to the table the more normal force it will feel (as microscopic forces are short range) and hence the normal force does indeed depends on the exertenal force’s magnitude. But it is quite unproductive to think in terms of “forces causing forces”.
 
  • #14
kuruman said:
The segment of wire that is in contact with the bead accelerates, because its velocity is continuously changing direction. What do you think keeps the wire loop going around at constant ωω\omega if not some kind of force external to it?
That’s exactly what my question is? And how that external force is causing the wire to push on the bead?

Question 4: What is the name of that force and what entity provides it?
Answer:
The name of that force is Centrieptal force and it’s the horizontal component of normal force provided by the wire, therefore, the entity is wire.
 
  • #15
@A.T. But how the bead is coming closer to the wire for normal force to act on the bead? Since, the normal force is caused when two objects come close to each other. If I keep two pens touching each other on my table they won’t exert obersvable normal forces on each other because microscopically they are too far apart. How the bead is coming closer to the wire, just passing the bead through the wire won’t make normal force to come into play as they are not close microscopically?
 
  • #16
Adesh said:
That’s exactly what my question is? And how that external force is causing the wire to push on the bead?
The name of that force is Centrieptal force and it’s the horizontal component of normal force provided by the wire, therefore, the entity is wire.
"Centripetal" denotes direction and means "directed towards the center". It is not a special kind of force. Although the horizontal component comes from the normal force, it is better to think of this in terms of the net force, (the sum of all the forces) that is centripetal; the acceleration is also centripetal and Newton's 2nd law says that the net force and the acceleration mst be in the same direction. Stated differently, at all times the resultant of gravity and the normal force is in the horizontal direction and provides the observed acceleration. Had you drawn a proper free body diagram, you would have seen this. The normal force act in the radial direction as shown in the drawing by @etotheipi in #2 and gravity acts vertically down. The fact remains, if the wire did not push on the bead, the bead would be in free fall.
 
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  • #17
Adesh said:
But how the bead is coming closer to the wire for normal force to act on the bead?
The wire moves in a circle, while the bead has the tendency to stay at rest of move uniformly along a line, due to inertia. These two motions are not compatible, so there must be a collision at some point.
 

Related to How do we provide centripetal force in this situation?

1. What is centripetal force?

Centripetal force is a force that acts on an object moving in a circular path, directed towards the center of the circle. It keeps the object moving along the circular path instead of moving in a straight line.

2. How is centripetal force different from centrifugal force?

Centripetal force is the force that keeps an object moving in a circular path, while centrifugal force is the apparent outward force that seems to act on an object moving in a circular path. Centrifugal force is not a real force, but rather an apparent force due to the inertia of the object.

3. How do we calculate centripetal force?

Centripetal force can be calculated using the formula F = (mv²)/r, where m is the mass of the object, v is its velocity, and r is the radius of the circular path.

4. What are some examples of providing centripetal force?

Some examples of providing centripetal force include swinging a ball on a string, rotating a bucket of water in a vertical circle, and driving a car around a curve.

5. How do we provide centripetal force in a situation where gravity is the only force acting?

In a situation where gravity is the only force acting, we can provide centripetal force by changing the direction of the object's velocity, causing it to move in a circular path. This can be achieved by using a curved track or by applying a force perpendicular to the object's velocity, such as with a string or a rod.

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