Is the concept of reactive centrifugal force valid?

[olivermsun]

Try this: Imagine the ball on the end of rope being divided into an inner and an outer half (the inner half being the half connected to the rope).

Great example.

One could also imagine a Ferris wheel or a roller coaster or something, where there's the car which goes in a circle, and then there's you in the car (being pushed inward by the seat or the floor of the the car). Or a merry-go-round where there's the horsey and then there's you hanging on to the horsey... (wait, that example has been done already in this thread, right?)

 

[Doc Al]

One could also imagine a Ferris wheel or a roller coaster or something, where there's the car which goes in a circle, and then there's you in the car (being pushed inward by the seat or the floor of the the car). Or a merry-go-round where there's the horsey and then there's you hanging on to the horsey... (wait, that example has been done already in this thread, right?)

Exactly. The merry-go-round and rotating spaceship examples have been discussed, but their analyses keep getting ignored (by some) in favor of less interesting examples.

 

[Dale]

Well, let me limit it to just 2 bodies for the moment to try and get this straight, since I don't really see why this is so different from the application of the 3rd law in the rope-and-cart example.

It is no different. The third law is the same regardless of the situation. Andrew Mason is incorrect in his assertions that there is something different in rotational motion which violates or modifies Newton's laws.

Suppose you have the regular old problem, two balls connected by a rope, twirling around a common center of mass. Pick any old inertial reference frame, e.g., the magic one in which the COM is not moving. Now it seems to me that the entire centripetal acceleration Fr on Ball 1 is exerted by Ball 2 through tension on the string. Conversely, Ball 2 must pull back on the string with exactly the same force, just like the cart pulling back on the rope in the earlier example. This pull is outward along the radius, hence this is the "reactive" centrifugal force, and it must be exactly equal to -Fr or else the COM would start to move.

So where is the constant offset?

Again, the "offset" is for the 2nd law on a single body, not the 3rd law between two bodies. If a single body undergoing uniform circular motion is experiencing a centripetal force Fi (inwards) and a centrifugal force Fo (outwards) then by Newton's 2nd law Fi-Fo = ma where a is directed inwards. So the offset between Fi and Fo is given by Fi = Fo+ma which follows directly from the 2nd law.

Are you clear on the fact that the forces involved in uniform circular motion obey Newton's laws?

 

[Andrew Mason]

Gravity pushes me down toward the ground. I infer that the ground pushes back, more or less, since I am not currently sinking into the ground. I don't happen to believe that it misleading and wrong to say the ground is exerting an upward "force" on me, even if the force only exists as long as I am here to push down on it (it won't suddenly fling other, lighter, objects into space, for example).

There is absolutely nothing wrong with calling the normal force that the Earth exerts on you an outward or force. You can call it centrifugal if you like but that usually refers to rotation.

So long as you understand that the "centrifugal" or normal force is NOT in any way shape or form a reaction to the centripetal acceleration that you experience due to the Earth's rotation, you can call it a centrifugal force. The centripetal acceleration is provided by gravity. The reaction to the centripetal force on you is the attractive gravitational force that you exert on the earth.AM

 

[Andrew Mason]

Ok, so you are saying A and B are held apart by a spring, but a rope is used to impose an additional force which compresses the spring and moves A and B closer together. So far, so good.


Of course there are outward forces in the rotating system. The rope or whatever is holding the balls together is pulling with the amount of inward, "centripetal" force (through tension) required to maintain the circular motion of each ball. An outward, "centrifugal" force (as viewed from the inertial, non rotating frame) is exerted by each ball on the line and keeps the line taut. Otherwise the balls would just collapse inward and the line would crumple. It also happens that the outward force of one ball also equals the inward force on the other, since they're connected by the same line, after all.

Think of the rope as a spring. The spring stretches to provide the necessary centripetal force (F = -kx). The only way that spring can exert an inward centripetal force is if it is also pulling inward on the other ball with the same force. So the spring is pulling inward.

[Note: We assume the rope is of negligible mass (otherwise you have to factor in the centripetal force on each part of the rope, which complicates things), the net force on each part of the rope is 0. So the net force is between the ends.]

There is nothing pushing outward. The spring stretches in order to supply the centripetal force that is required to maintain rotational motion. If it is not strong enough, the inertial motion of the balls will carry them away from the centre. They would not crumple to the centre.

By the way, you are talking about the centrifugal force that everyone agrees is fictitious, not the reactive centrifugal force.

AM

 

[Andrew Mason]

It is no different. The third law is the same regardless of the situation.

Agreed.

Andrew Mason is incorrect in his assertions that there is something different in rotational motion which violates or modifies Newton's laws.

I said that?! I certainly assert that there is something about rotational motion that is different than linear motion or linear acceleration: ie acceleration that is constantly changing direction. But it does not violate or require modification to the third law to explain.

Again, the "offset" is for the 2nd law on a single body, not the 3rd law between two bodies. If a single body undergoing uniform circular motion is experiencing a centripetal force Fi (inwards) and a centrifugal force Fo (outwards) then by Newton's 2nd law Fi-Fo = ma where a is directed inwards. So the offset between Fi and Fo is given by Fi = Fo+ma which follows directly from the 2nd law.

Are you clear on the fact that the forces involved in uniform circular motion obey Newton's laws?

You have to be careful here. You may confuse people into making the same mistake that you incorrectly accused me of making.

There is no centrifugal force that has to be subtracted here to determine its centripetal acceleration. That, in fact, is how you prove there is no centrifugal force on the body - by measuring the centripetal force and showing that it is equal to its centripetal acceleration times the mass of the body.

Rather the "centrifugal reaction force" is (said to be) on the matter that is in contact with the rotating mass. We all agree that there is a reaction force to the centripetal acceleration. I am just saying it is the force of one rotating mass on the other and both are inward. The others, except for DH it seems, says that there is an outward component of force that is a reaction to the force of one rotating body on the other.

AM

 

[A.T.]

There is certainly a centrifugal effect.

"Centrifugal effect" is still just vague gibberish.

But it is not a force.

Calling forces "effects" doesn't change physics. It is just pointless obfuscation.

 

[olivermsun]

Again, the "offset" is for the 2nd law on a single body, not the 3rd law between two bodies.

I guess it wasn't completely clear to me due to the title and the ongoing discussion about finding the 3rd law "reaction" to the centripetal force. But thanks.

Are you clear on the fact that the forces involved in uniform circular motion obey Newton's laws?

No need to condescend. It's the shifting language and incessant bickering about definitions that confuses me, not Newtonian mechanics properly expressed.

By the way, you are talking about the centrifugal force that everyone agrees is fictitious, not the reactive centrifugal force.

Actually, I was talking about the reactive centrifugal force. But thanks anyway.

 

[Dale]

No need to condescend. It's the shifting language and incessant bickering about definitions that confuses me, not Newtonian mechanics properly expressed.

I was not intending to be condescending, but only making sure that you were clear on that point since it is extremely important. It appears to be a point of confusion with others in this thread.

 

[Dale]

I said that?! I certainly assert that there is something about rotational motion that is different than linear motion or linear acceleration: ie acceleration that is constantly changing direction. But it does not violate or require modification to the third law to explain.

Yes, when you say that a centrifugal force is an effect and not a force you are saying that Newton's laws don't apply or that they need modifications.

There is no centrifugal force that has to be subtracted here to determine its centripetal acceleration. That, in fact, is how you prove there is no centrifugal force on the body - by measuring the centripetal force and showing that it is equal to its centripetal acceleration times the mass of the body.

This is a good suggestion. Let's take the following example: In the absence of gravity, two balls are attached to two ropes as shown. Each ball has a mass of 1 kg. Each rope is massless with a length of 1 m and a gauge attached to measure the tension in the rope. The whole assembly is swung about the left end of the rope in a uniform circular motion at a rate of 1 revolution per second. Using Newton's laws of motion find:
1) the acceleration of masses A and B
2) the net forces acting on masses A and B and their directions
3) the measured tensions in gauges 1 and 2
4) all forces acting on masses A and B and their directions

 

[Andrew Mason]

The whole assembly is swung about the left end of the rope in a uniform circular motion at a rate of 1 revolution per second.

I think that means that the masses are not rotating about their centre of mass. If so, you have to add in the acceleration and mass or moment of inertia of whatever it is that it is tied to. So we need more information. You can avoid this problem by having the two tethered balls rotate in space about their centre of mass.

To demonstrate the third law you need objects that are "not hindered by any other impediments".

AM

 

[Dale]

I think that means that the masses are not rotating about their centre of mass. If so, you have to add in the acceleration and mass or moment of inertia of whatever it is that it is tied to. So we need more information.

No further information is needed. The system is not isolated, but Newton's laws still apply and can be used without modification or additional information.

 

[Andrew Mason]

No further information is needed. The system is not isolated, but Newton's laws still apply and can be used without modification or additional information.

Of course Newton's laws still apply. But you need to know what other mass is involved. A mass cannot just rotate in space. It can only rotate about a centre of mass of a rotating system. The rope cannot rotate all by itself.

The common example of a centrifugal reactive force that engineers use is a car turning on a road. The centripetal force supplied by the road results in a reaction force away from the car, it is said. So they conclude that the reaction force is a centrifugal force. It appears that way but it isn't. This is a very subtle and understandable mistake.

The reason: it is not the road that is supplying the centripetal force - it is the earth. If the road was not connected to the earth, the car would not turn. So the reaction force to the centripetal force of the Earth on the car is a force of the car on the earth. To see whether it is centripetal or centrifugal you have to determine the centre of mass of the car/earth system and determine how the Earth centre of mass moves in response to the car's acceleration. It always moves toward the centre of mass of the car/earth system.

Of course it is impossible to see or measure because of the difference in mass between the Earth and car. But we don't have to do that if we accept Newton's third law. We just do the physics.

Anyone who thinks that classical mechanics is simple or easy has never tried to work out the physics of a precessing and nutating spinning top. It is difficult. It is counter-intuitive.

So what we have to do is put our intuition aside, quit arguing with words and apply the laws strictly and correctly.

AM

 

[rcgldr]

The common example of a centrifugal reactive force that engineers use is a car turning on a road. ... To see whether it is centripetal or centrifugal you have to determine the centre of mass of the car/earth system and determine how the Earth centre of mass moves in response to the car's acceleration.

Much of the reaction at the surface of the Earth would be angular (versus linear) acceleration where the surface of the Earth would accelerate "outwards" from the center of mass.

This is getting off-topic. Earlier in this thread, there have been examples where there is is a reactive centrifugal force and others were there is not. Just because there are cases where there isn't reactive centrifugal force doesn't invalidate the usage of that term to describe the cases where it does apply.

 

[A.T.]

To demonstrate the third law you need objects that are "not hindered by any other impediments".

Not true. You could use soft objects in a static case to demonstrate the deformation on both, as a result of the equal opposite forces. My space station can be covered with soft foam on the inside, that will deform and demonstrate the centrifugal force exerted on it by the astronaut.

Acceleration is just easier to visualize. That's why the special case "no other impediments" is used as an example. But you are stuck at this simplistic special case and fail to understand the general meaning of the law.

We just do the physics.

Well, you don't. Any high school student could "just do the physics" and solve the problem posed by DaleSpam without any further information. But you apparently can't. You demand more information to answer questions that were never asked.

So what we have to do is put our intuition aside, quit arguing with words and apply the laws strictly and correctly.

Yes that is what YOU have to do. Note that DaleSpam did not ask you about the names of all the forces. You can just name them F₁, F₂ ... F_n. But instead of "just do the physics" you continue to argue with words.

 

[A.T.]

To see whether it is centripetal or centrifugal you have to ... acceleration...

You have obviously your private definition of "centripetal" & "centrifugal" in respect to forces, and that's fine. But the more common definition, referred to in Wikipedia, doesn't depend on acceleration. It depends only on:

- center of rotation
- point of application of the individual force (not some net force!)
- direction of the individual force (not some net force!)

Centripetal : Force direction is the same as the direction from the point of force application to the center of rotation.
Centrifugal : Force direction is the same as the direction from the center of rotation to the point of force application, or center of rotation and point of force application are the same point.

There is no point in arguing which definition is correct. Definitions are neither false not correct. It is just a convention. You should just be aware that many people use the convention above, so you don't get confused when they use those terms.

 

[Dale]

Of course Newton's laws still apply. But you need to know what other mass is involved.

No, you don't. There obviously is another mass involved since the system is not isolated, but none of the details of that other mass affect the answers to the questions asked. The problem is fully and completely specified, no additional information is required.

So what we have to do is put our intuition aside, quit arguing with words and apply the laws strictly and correctly.

I agree completely. If you cannot apply the laws strictly and correctly to this freshman-level problem, then perhaps you should listen to and learn from those who can.

I encourage you to work the problem yourself before peeking at the answer:

Spoiler

1)For uniform circular motion the acceleration is r\omega^2 where r is the radius and w is the angular frequency. From the problem \omega=2\pi s^{-1} = 6.28 s^{-1}. So
a_A = (1 m)(6.28 s^{-1})^2 = 39.5 m/s^2 \text{inwards}
a_B = (2 m)(6.28 s^{-1})^2 = 79.0 m/s^2 \text{inwards}

2) From Newton's 2nd law F_{net} = ma. So
F_{A\,net}=(1 kg)(39.5 m/s^2)=39.5 N \text{inwards}
F_{B\,net}=(1 kg)(79.0 m/s^2)=79.0 N \text{inwards}

3) The force from rope 2 on mass B F_{2B} is the only force acting on B, so
F_{2B}=F_{B\,net}=79.0 N \text{inwards}
By Newton's 3rd law the force from mass B on rope 2 F_{B2}=-F_{2B} so the tension in rope 2 is
T_2=|F_{B2}|=79.0 N

Since the rope is massless, by Newton's 2nd law F_{A2}+F_{B2}=ma=0 so by Newton's 3rd law F_{2A}=-F_{A2}=F_{B2}=-F_{2B}=79.0 N \text{outwards}. The forces on mass A are the force from rope 1 and the force from rope 2 F_{1A}+F_{2A}=F_{A\,net}. So F_{1A}=F_{A\,net}-F_{2A}=(39.5 N \text{inwards}) - (79.0 N \text{outwards}) = 118.5 N \text{inwards}. By Newton's 3rd law the force from mass A on rope 1 F_{A1}=-F_{1A} so the tension in rope 2 is
T_1=|F_{A1}|=118.5 N

4) So the forces acting on B are:
F_{2B}=79.0 N \text{inwards}

And the forces acting on A are:
F_{2A}=79.0 N \text{outwards}
F_{1A}=118.5 N \text{inwards}

 

[Andrew Mason]

Much of the reaction at the surface of the Earth would be angular (versus linear) acceleration where the surface of the Earth would accelerate "outwards" from the center of mass.

Lets use centripetal and centrifugal because that is the real issue here. The question is not the direction that the surface moves. It is the direction that the Earth moves as required by Newton's third law.

That direction in relation to the centre of rotation determines whether the reaction to the centripetal force on the car is centripetal or centrifugal. That is the problem that we seem to be having here. It is not a problem in the simplest case where everyone agrees that the reaction to a centripetal force/acceleration is another centripetal force/acceleration eg: gravitational orbit.

Let's say that the road is a huge piece of steel plate mounted on a sheet of ice. As I drive on the plate in a circle, the centre of mass of the steel plate prescribes a rotation. The car and the centre of mass of the steel plate both rotate about the centre of mass of the car/plate system which remains fixed. As measured in the inertial frame of reference of the system centre of mass, the reaction force of the car on the plate is the centripetal force on the plate. That is all I am saying.

This is getting off-topic. Earlier in this thread, there have been examples where there is is a reactive centrifugal force and others were there is not. Just because there are cases where there isn't reactive centrifugal force doesn't invalidate the usage of that term to describe the cases where it does apply.

I say simply that the third law reaction to a centripetal force causing centripetal acceleration is always an equal and opposite centripetal force causing centripetal acceleration as measured in an inertial frame of reference. I would point out that when something is rotating on or around the earth, the centre of mass of the Earth is not an inertial frame of reference.

AM

 

[olivermsun]

It is not a problem in the simplest case where everyone agrees that the reaction to a centripetal force/acceleration is another centripetal force/acceleration eg: gravitational orbit.

Let's say that the road is a huge piece of steel plate mounted on a sheet of ice. As I drive on the plate in a circle, the centre of mass of the steel plate prescribes a rotation.

What's the essential difference between this and the other examples that have already posed, e.g., gravitational orbits, masses connected by strings, merry-go-rounds, etc.?

 

[Andrew Mason]

No, you don't. There obviously is another mass involved since the system is not isolated, but none of the details of that other mass affect the answers to the questions asked. The problem is fully and completely specified, no additional information is required.

The whole point of this exercise is to determine the magnitude and direction of the reaction force to the centripetal force. The reaction to the centripetal force is not on that rope. It is on what is connected to the "fixed" end which, by the way, cannot possibly be "fixed" to an inertial reference frame. If you refuse to tell us what it is fixed to so we can choose an inertial reference frame you cannot analyse the problem. End of story.

If you cannot apply the laws strictly and correctly to this freshman-level problem, then perhaps you should listen to and learn from those who can.

I would not put it in such a condescending fashion but I would suggest that you apply that to yourself. Some things that appear to be "freshman-level problems" raise profound issues. Feynman's freshman level lectures will be studied and discussed for a long time.

I encourage you to work the problem yourself before peeking at the answer:

You are measuring the forces in from a non-inertial frame and you are not including the other half of the rotating system (ie the Earth or what ever it is that you won't say the rope is connected to).

So let's do it right. We have two such ropes and masses and we let them rotate about the centre of mass. We measure all forces in the non-rotating inertial reference frame of the centre of mass/rotation. We will call the balls A1, B1 and A2, B2 and the respective ropes the same. Here is my analysis:

1)Accelerations:
a_{A1} = (1 m)(6.28 s^{-1})^2 = 39.5 m/s^2 \text{centripetal}
a_{B1} = (2 m)(6.28 s^{-1})^2 = 79.0 m/s^2 \text{centripetal}
a_{A2} = (1 m)(6.28 s^{-1})^2 = 39.5 m/s^2 \text{centripetal}
a_{B2} = (2 m)(6.28 s^{-1})^2 = 79.0 m/s^2 \text{centripetal}2) Forces:
F_{A1}=(1 kg)(39.5 m/s^2)=39.5 N \text{centripetal}
F_{B1}=(1 kg)(79.0 m/s^2)=79.0 N \text{centripetal}
F_{A2}=(1 kg)(39.5 m/s^2)=39.5 N \text{centripetal}
F_{B2}=(1 kg)(79.0 m/s^2)=79.0 N \text{centripetal}

3) The rope tensions supply the centripetal forces. So the tension in ropes A1 and in A2 is the sum of these centripetal forces on masses A1 and B1: 118.5 N. The tension in ropes B1 and B2 provide only the centripetal force on masses B1 and B2 respectively: 79N.

The force acting on A1 is equal and opposite to the force acting on A2 and is the difference in rope tensions: 39.5N. Both are centripetal.

Similarly the force acting on B1 is equal and opposite to the force on B2 and is 79 N. Both are centripetal.

Those are the action and reaction pairs and they are the only forces.

AM

 

[A.T.]

I say simply that the third law reaction to a centripetal force causing centripetal acceleration is always an equal and opposite centripetal force causing centripetal acceleration as measured in an inertial frame of reference.

Yes, we got that. It is still not true in general, based on the common definition of "centripetal" & "centrifugal" (see post #196). Changing definitions, calling forces "effects" and pointing out special cases proves nothing.

 

[Dale]

I say simply that the third law reaction to a centripetal force causing centripetal acceleration is always an equal and opposite centripetal force causing centripetal acceleration as measured in an inertial frame of reference.

Which has been demonstrated to be incorrect by many counter-examples where the 3rd law reaction was an equal and opposite centrifugal force opposing the centripetal acceleration.

 

[Dale]

The whole point of this exercise is to determine the magnitude and direction of the reaction force to the centripetal force. The reaction to the centripetal force is not on that rope. It is on what is connected to the "fixed" end which, by the way, cannot possibly be "fixed" to an inertial reference frame. If you refuse to tell us what it is fixed to so we can choose an inertial reference frame you cannot analyse the problem. End of story.

And yet, I analyzed it fine without any additional information. Study my analysis and show me any point which violates any of Newton's laws. For each object Newton's 2nd law is satisfied because the net force acting on that object is equal to ma, and for each interaction Newton's 3rd law is satisfied because the forces are equal and opposite. Show me any contradiction where Newton's laws are not satisfied. There is no requirement that Newton's laws only be applied to isolated systems where all of the forces are internal.

So let's do it right. We have two such ropes and masses and we let them rotate about the centre of mass.

I will be glad to analyze your new problem after you have analyzed mine.

Without reference to your new scenario, can you point to any of the 4 objects in my analysis for which Newton's 2nd law is violated or any pair of those objects for which Newton's 3rd law is violated? If not, then the analysis is correct.

 

[A.T.]

It is on what is connected to the "fixed" end which, by the way, cannot possibly be "fixed" to an inertial reference frame.

The point on the rope 1m left from mass A is at rest in an inertial frame. You can assume that as given. This is trivial to achieve but the details of it are irrelevant for the questions.

 

[A.T.]

So let's do it right. We have two such ropes and masses and we let them rotate about the centre of mass. We measure all forces in the non-rotating inertial reference frame of the centre of mass/rotation. We will call the balls A1, B1 and A2, B2 and the respective ropes the same. Here is my analysis:

1)Accelerations:
a_{A1} = (1 m)(6.28 s^{-1})^2 = 39.5 m/s^2 \text{centripetal}
a_{B1} = (2 m)(6.28 s^{-1})^2 = 79.0 m/s^2 \text{centripetal}
a_{A2} = (1 m)(6.28 s^{-1})^2 = 39.5 m/s^2 \text{centripetal}
a_{B2} = (2 m)(6.28 s^{-1})^2 = 79.0 m/s^2 \text{centripetal}


2) Forces:
F_{A1}=(1 kg)(39.5 m/s^2)=39.5 N \text{centripetal}
F_{B1}=(1 kg)(79.0 m/s^2)=79.0 N \text{centripetal}
F_{A2}=(1 kg)(39.5 m/s^2)=39.5 N \text{centripetal}
F_{B2}=(1 kg)(79.0 m/s^2)=79.0 N \text{centripetal}

3) The rope tensions supply the centripetal forces. So the tension in ropes A1 and in A2 is the sum of these centripetal forces on masses A1 and B1: 118.5 N. The tension in ropes B1 and B2 provide only the centripetal force on masses B1 and B2 respectively: 39.5N.

The force acting on A1 is equal and opposite to the force acting on A2 and is the difference in rope tensions: 79N. Both are centripetal.

Similarly the force acting on B1 is equal and opposite to the force on B2 and is 39.5 N. Both are centripetal.

Those are the action and reaction pairs and they are the only forces.

AM

Where is the answer to point 4) for your scenario, listing ALL the forces (not only net forces) acting on each mass?

 

[Andrew Mason]

There is no requirement that Newton's laws only be applied to isolated systems where all of the forces are internal.

There is if you want to analyse third law pairs of forces. If you disagree, try explaining the third law reaction force to the force on me when I fall from a tree accelerating at 9.8 m/sec^2 toward the ground without viewing the earth/me as an isolated system.

I will be glad to analyze your new problem after you have analyzed mine.

I will when you give me all the information needed to analyse it properly.

Without reference to your new scenario, can you point to any of the 4 objects in my analysis for which Newton's 2nd law is violated or any pair of those objects for which Newton's 3rd law is violated? If not, then the analysis is correct.

Certainly.

You say:

"Since the rope is massless, by Newton's 2nd law F_{A2}+F_{B2}=ma=0 so by Newton's 3rd law F_{2A}=-F_{A2}=F_{B2}=-F_{2B}=79.0 N \text{outwards}"

The problem is that you are measuring a pseudo force. You are applying Newton's laws in a rotating frame of reference. Of course, that is how it appears in the rotating reference frame. That is why the concept of centrifugal force is useful. B appears to be pulling on A but, in fact, it is not. This is because A is not pulling on B. Both are pulling on whatever it is the other end of the rope is connected to and the reaction force is on whatever it is connected to. You can prove this by cutting the rope that connects A and B. If you cut the rope between B and A, A's circular rotation motion does not change at all.

AM

 

[Andrew Mason]

Where is the answer to point 4) for your scenario, listing ALL the forces (not only net forces) acting on each mass?

I said that those are ALL the forces. Those are the ONLY forces.

AM

 

[A.T.]

I said that those are ALL the forces. Those are the ONLY forces.

No, you gave only the net forces. You computed them by taking the "difference in rope tensions" at A1 & A2. This is a cheap semantic trick to hide that fact that you are adding the individual force vectors to get the net force.

In point 4) you were supposed to list all the individual forces acting on A1. There are two ropes at A1. So there are two individual forces acting on A1. Please list those two individual interaction forces and their directions.

 

[Andrew Mason]

No, you gave only the net forces. You computed them by taking the "difference in rope tensions" at A1 & A2. This is a cheap semantic trick to hide that fact that you are adding the individual force vectors to get the net force.

This is why centrifugal forces create confusion!

If you add into this configuration centrifugal forces on A1 (and A2) you end up with the following tension on the rope between A1 and the centre of mass (and A2, its the same rope):

Centripetal force on A1: 39.5N plus
Centripetal force on B1: 79 N plus
Centrifugal force on A1: 79N

Total tension: 197.5 N

Since the rope provides the force required to accelerate BOTH A1 and B1, the tension includes the centripetal forces required for A1 and B1.If you add to that the centrifugal forces you get 197.5 N which is certainly not the tension that is measured.

In point 4) you were supposed to list all the individual forces acting on A1. There are two ropes at A1. So there are two individual forces acting on A1.

No there is not. There is a real centripetal force and a pseudo centrifugal force. I listed only the REAL forces.

AM

 

[A.T.]

There are two ropes at A1. So there are two individual forces acting on A1.

No there is not. There is a real centripetal force and a pseudo centrifugal force. I listed only the REAL forces.

Nonsense. Forces from attached ropes are electro-magnetic interaction forces, and act in every frame. Inertial forces in non-inertial frames are not acting trough attached ropes, but directly on every piece of mass.

Two ropes under tension attached to A1 means two individual interaction forces acting at A1. Please list those two forces and their directions.