Rotation around a non fixed axis + linear motion of a system

In summary, the conversation discusses a thought experiment involving rotating bodies and a larger system, and the issue of combining the physics of rotating bodies with the motion of a larger system. The speaker is seeking help in finding a solution or hints towards resolving this question, which involves calculating the force acting on the system when a body is pushed to rotate around an axis. They also mention their understanding of what should happen in this situation and their difficulty in describing it with formulas. The conversation then moves on to discussing the concept of drawing free body diagrams and using equations to solve the problem.
  • #1
Trigu
7
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I have had a thought experiment in my head for a while now and I am unable to find clear enough examples/info that deal with similar issues, to solve it on my own. This is why I hope that someone in this forum can at least point me towards a solution or provide hints as to where should I be looking to resolve this question.

The main issue I am having is how to combine the physics of rotating bodies and the motion of a larger system that the rotating body is a part of. Most of the info I am able to find about rotating bodies is in the context of rotation around a fixed axis.
Unfortunately what I am interested in requires info about bodies that rotate around an axis that is not rigidly fixed. At least not fixed as in tied to an external system etc. The axis would still be fixed to an internal system but that system has the ability to move around.

The issue:
I am interested to know, how much force a system is subjected to when a body that is in that system is pushed to rotate around an axis.

Ideally, I would like to have a formula that describes how the forces interact in this system. For example, if I know the weights and dimensions of all of the bodies in the system and the force and time it is applied to rotate the body, I would like to be able to calculate the force that is acting on the system.
I think I understand what should happen to the system in a situation like that, but I am unable to describe it with formulas. My current understanding would be that the size of the force acting on the system depends on the masses of the rotating body and the rest of the system. So when the rotating body has low mass compared to the system, the body will rotate more and the system would move less. But if the rotating body has more mass than the rest of the system, then due to the large moment of inertia of the rotating body the system acts closer to a rigid body and the body rotates less and more force is exerted on to the system. If I am misunderstanding something please let me know.

To make this easier I created a pdf that I am attaching to this post. The pdf shows the question as a thought experiment, with I hope, enough descriptions and clarifications so that someone could provide some insight.
If it is not clear enough I am more than happy to try to explain with more diagrams etc.
 

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  • Uniform Disc System on surface-v1.PDF
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  • #2
A tiny comment on the statement "A force F is applied for a time t ". If force works only in a moment, no momentum is supplied to the system. With a duration of ##\triangle t##, impulse ##F \triangle t## would be better to be considered.

Momentum of ##F\triangle t## and angular momentum around axis ##F\triangle t \ R## are given to each disk-base set.

[tex]F\triangle t = (m_D+m_B)v[/tex]
[tex]F\triangle t R=I \omega[/tex]
 
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  • #3
Trigu said:
Summary:: I am interested to know, how much force a system is subjected to when a body that is in that system is pushed to rotate around an axis.

I would like to be able to calculate the force that is acting on the system.
I think I understand what should happen to the system in a situation like that, but I am unable to describe it with formulas
So the key concept in situations like this is to draw free body diagrams for each object with each relevant force labeled and then start writing equations. As you write equations keep track of the number of equations and the number of unknowns.

So here you would have two free body diagrams: one for the disk and the other for the base. We are only interested in horizontal forces. The disk would have two forces, the external force and the axle force. The base would have one force, the axle force. Both the disk and base accelerate and have mass.

So in total we have three forces, two masses, and two accelerations which is 7 variables. Of those the two masses are known and one of the forces is known. So we have four unknowns, which means we need four equations.

We can write Newton’s 2nd law for the disk and the base, so that is two equations. We can write Newton’s 3rd law for the pair of forces at the axle. So that is three equations.

At this point we have exhausted Newton’s laws, and we have 3 equations in 4 unknowns which is unsolvable. So we need to find additional equations called constraint equations that come not from Newton’s general laws but from our specific setup. Here, the constraint equation is that the acceleration of the disk equals the acceleration of the base. This gives us a total of 4 equations in 4 unknowns, which we can solve.
 
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  • #4
Thank you for the answers.
First,
anuttarasammyak said:
With a duration of △t, impulse F△t would be better to be considered.
I understand the concern and I will use the provided equation when I get to the point where I need to calculate the final results.
Second,
Dale said:
So here you would have two free body diagrams: one for the disk and the other for the base. We are only interested in horizontal forces. The disk would have two forces, the external force and the axle force. The base would have one force, the axle force. Both the disk and base accelerate and have mass.
I am a bit confused about free body diagrams and I hope that you can clear them for me. You say that the disc diagram should have 2 forces. The external force F and the axle force. I understand that there should be an axel force but I am unable to calculate its size or direction. How would one do that? How is the external force F affecting the center of mass of a disc? Every example I have found only looks at problems where the disc's center is fixed and no one is looking at how the point O moves or what forces are applied to it due to the external force F. Or am I misunderstanding and the body will never rotate and move as a rigid body to the right? That doesn't seem right.
I have attached the free body diagram of the disc as I see it. Please tell me if I have made a mistake.
Note that I moved the external force to the bottom of the disc, so the problem resembles the classic Pulled Spool problem. Just, in this case, the spool would be in outer space without any frictions etc.
Free body diagram of the disc
 
  • #5
Trigu said:
I understand the concern and I will use the provided equation when I get to the point where I need to calculate the final results.
In form of acceleration
[tex]\ddot x = \frac{F}{m_D+m_B}[/tex]
[tex]\ddot \theta = \frac{FR}{I}[/tex]
where
x: position of the set
##m_D,m_B##: mass of disk, base
##\theta##: rotation angle of disk
R: radius of disk
I: disk's momentum of inertia
 
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  • #6
Trigu said:
I understand that there should be an axel force but I am unable to calculate its size or direction. How would one do that?
At the time of drawing the free body diagrams you do not worry about that. It is very important to do the free body diagram as a purely physical analysis without considering how you will calculate them. If a force exists you simply draw it and label it on the diagram. You don’t even necessarily need to know the direction, in which case you just guess and then do everything consistent with that guess.

Your free body diagram looks good. I generally only draw forces on the object and place accelerations and velocities as notations outside the object, but your approach of using different colors as you did also works. As long as you can visually see the difference.

Only after you have completed your free body diagrams for all objects do you start worrying about how you calculate things. So finish the free body diagram for the base before worrying about calculating the force at the axle.
 
  • #7
Thanks for the replies once again!
Dale said:
So finish the free body diagram for the base before worrying about calculating the force at the axle.
Ok so here is the diagram for the base and the disc so they are side by side.
Free-body diagram - Base - v1.png
Free-body diagram - Disc- v1.png
I am probably overthinking this but I am confused as to what is the force that acts on the base. My gut feeling is that it should be the same force that acts on the disc, due to the fact that they are connected at point O. But that means they can't have the same acceleration since they have a different mass. And because they both do move as one object in the system a1 = a2 = asystem , so F2 ≠F1.
Dale said:
The disk would have two forces, the external force and the axle force. The base would have one force, the axle force. Both the disk and base accelerate and have mass.
So this now leaves me with the equations. Firstly, Newton’s 2nd law for the disk and the base. I presume that even though the disc can rotate you meant the linear version for both so you can use the constraint that the disc and the base have the same acceleration. Therefore:
F1=m1*a
F2=m2*a​
If not then here is the disc's 2nd law:
##\alpha =\frac {rF} {I} ##
##I =\frac {mr^2} {2} ##​
This version seems intuitively more important because this equation connects the moment of inertia of the rotating disc to the applied force. This alludes to one of the main questions I proposed in the first post. The question of how the system moves depending on the masses of the disc and the base. Without the moment of inertia, it seems the disc's properties are not fully accounted for when looking at the system as a whole. So once again I am stuck.

For the third equation, you said to use Newton’s 3rd law for the pair of forces at the axle. Unfortunately, I am failing to see what force pair exists at the axle. How is the base affecting the disc if both are not moving, the base has no momentum since there is no movement? The only thing is the mass of the base but without friction, I don't see how it will affect horizontal movements. Though on second thought once the force is applied the base will move and it will have momentum. But it seems to me that since the disc and the base move in the same direction with the same acceleration they are not really interacting with each other in that way either. So another point I could use a hint.

Finally the fourth equation.
So if I am not mistaken since you mentioned that the constraint equation is the fact that the base and the disc have the same acceleration the fourth equation looks like this:
##\frac {F _{1} } {m _{1} } =\frac {F _{2} } {m _{2} } ##​
Unless there is a better way to use/represent that system constraint.

I really appreciate all the help so far.
 
  • #8
Trigu said:
Ok so here is the diagram for the base and the disc so they are side by side.
I am probably overthinking this but I am confused as to what is the force that acts on the base. My gut feeling is that it should be the same force that acts on the disc, due to the fact that they are connected at point O. But that means they can't have the same acceleration since they have a different mass. And because they both do move as one object in the system a1 = a2 = asystem , so F2 ≠F1.
1) By Newton's 3rd Law F1 and F2 are equal but opposite.
2) In Newton's 2nd Law the "F" is the net force, which for the disc is the vector sum of F and F1,
 
  • #9
Trigu said:
My gut feeling is that it should be the same force that acts on the disc, due to the fact that they are connected at point O.
What does Newton’s 3rd law say?

Trigu said:
But that means they can't have the same acceleration since they have a different mass.
You are jumping to conclusions here. Remember that there is more than one force acting on the disk.

Trigu said:
And because they both do move as one object in the system a1 = a2 = asystem , so F2 ≠F1.
The constraint equation is correct, but you are jumping to conclusions here. You should first start by writing the equations and only solve them later. For now, you know the accelerations are the same, but you do not know the relationship between the forces.

Trigu said:
Therefore:
F1=m1*a
F2=m2*a​
Yes, I meant the linear ones. (you can do the rotational ones but they introduce both new equations and new unknowns so they neither help nor hurt).

Your linear equation is correct for the base, but not the disk. Can you see why? You left out something important for the disk.

Trigu said:
you said to use Newton’s 3rd law for the pair of forces at the axle. Unfortunately, I am failing to see what force pair exists at the axle
The disk and the base are connected at the axle. If the axle pushes forward on the base then the base pushes backwards on the axle by the same amount. That is Newton’s 3rd law. Here it means that F2 is equal and opposite F1.

Trigu said:
How is the base affecting the disc if both are not moving, the base has no momentum since there is no movement?
That is not relevant. Even when objects are at rest Newton’s laws still hold.

Trigu said:
So if I am not mistaken since you mentioned that the constraint equation is the fact that the base and the disc have the same acceleration the fourth equation looks like this:
F1m1=F2m2​
Unless there is a better way to use/represent that system constraint.
I would just write it as you did above: a1=a2. You will solve for it eventually, but first you need to fix your 2nd law equations.
 
  • #10
Thanks once again for all of the replies.

I have updated the base diagram with the correct direction of the force F2 to show that it is equal in size but opposite to the force F1 that was showed on the disc diagram. And acceleration is still pointed in the same direction as is the disc's acceleration.
Free-body diagram - Base - v2.png
So now on to fixing the 2nd law equations.
Dale said:
You left out something important for the disk.
I now see that I had completely omitted the external force F that acts on the disc. But I think I have another issue now. It seems to me that I cannot just add the external force F into the 2nd law equation for the disc, because not all of the force F is used to move the disc in a linear direction, some of it goes into rotating the disc. Therefore I think I need to break force F into two parts: FL (linear part) and FR (rotational part) as shown on the second disc diagram.
F = FL + FR
Free-body diagram - Disc- v1.png
Free-body diagram - Disc-Separated- v1.png
This idea came to me thanks to
A.T. said:
2) In Newton's 2nd Law the "F" is the net force, which for the disc is the vector sum of F and F1,
So with this new force, I could write the 2nd law for the disc like this:
##m _{1} a _{1}=F _{1} +F _{L} ##​
All of the equations that I have so far:
2nd law for the disc:
##m _{1} a _{1}=F _{1} +F _{L} ##​
2nd law for the base:
##m _{2} a _{2}=F _{2} ##​
Constraint equation:
## a _{1}=a _{2} ##​
Newtons 3rd law for the force pair:
## F _{1}=-F _{2} ##​
So it seems I have 4 equations now, but because I separated the external force F into FL and FR, I still have more missing unknowns than equations.
Writing out the equations for the disc's angular acceleration ##\alpha =\frac {\tau} {I}## has not helped me so far.
## \tau=rF _{R} ## and ##I =\frac {mr^2} {2}##
give me
##\alpha =\frac {2F _{R}} {m _{1} r}## from here ##F _{R} =\frac {\alpha m _{1} r} {2}##
and then FL from the 2nd laws of base and disc:
##F _{L} =a _{1}(m _{1}+m _{2})##
therefore
##F =F _{L}+F _{R}=a _{1}(m _{1}+m _{2})+\frac {\alpha m _{1} r} {2}##

Unfortunately, this doesn't help me because I don't know of any relationship between disc's angular acceleration α and the acceleration a.
 
  • #11
Trigu said:
It seems to me that I cannot just add the external force F into the 2nd law equation for the disc, because not all of the force F is used to move the disc in a linear direction, some of it goes into rotating the disc. Therefore I think I need to break force F into two parts: FL (linear part) and FR (rotational part) as shown on the second disc diagram.
Where in Newton’s laws does it say to do anything like that? Does any part of Newton’s laws say that a given force causes less linear acceleration if it also causes angular acceleration?

Trigu said:
2nd law for the base:
##m _{2} a _{2}=F _{2} ##​
Note that ##a_2## points in the opposite direction of ##F_2## so there should be a minus sign here.
 
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  • #12
Trigu said:
It seems to me that I cannot just add the external force F into the 2nd law equation for the disc...
Yes, you can.
 
  • #13
Dale said:
Does any part of Newton’s laws say that a given force causes less linear acceleration if it also causes angular acceleration?
No, the laws do not state that. But this means that I am now even more confused.
Just to check that I have not made any algebra mistakes again, I will write out the equations where I add the external force F to the disc's 2nd law equation.
Disc:
##m _{1} a _{1}=F _{1} +F ##
Base:
##-m _{2} a _{2}=F _{2}##

Since F1=-F2,
##-m _{2} a _{2}=F _{2} =-F _{1} ##
##m _{2} a _{2}=F _{1} ##
Substituting this into the Disc's equation:
##m _{1} a _{1}=m _{2} a _{2} +F ##
Since ##a _{1}=a _{2}## I can write:
##m _{1} a _{1}=m _{2} a _{1} +F ##
And from here
##m _{1} a _{1}-m _{2} a _{1}= F ##
##a _{1}(m _{1} -m _{2} )= F ##
##a _{1} =\frac {F} {m _{1} -m _{2}}##
So if all of this is correct then to me it seems that there are very disturbing conclusions, atleast when I compare this result to my initial predictions.
This equation says that it does not matter if the disc has the ability to rotate or not, which seems very odd. If this is truly the case I need a lot of help.
 
  • #14
Trigu said:
##a _{1} =\frac {F} {m _{1} -m _{2}}##
F is the only external force to the whole system of m1 and m2. So it should be:
a = F / (m1 + m2)

Trigu said:
This equation says that it does not matter if the disc has the ability to rotate or not, which seems very odd.
You assumed that you can apply a force F, regardless if the disc has the ability to rotate or not. Under that assumption the conclusion is correct.

The practicality of applying a constant force to the disc might depend on its ability to rotate. So your odd feeling might be based on these practical experiences.
 
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  • #15
I got to a feeling that I have not properly described the setup that I am trying to solve here. The force F is not moving with the disc when the disc is rotating. The force F is essentially a rope that is tied around the disc and the rope is being pulled with a force that is equal to force F. And it seems to me that in that situation the rotation of the disc is a fundamental part of how this system behaves, but if it really isn't, then could someone point me to some material where I can understand why?
 
  • #16
Trigu said:
Just to check that I have not made any algebra mistakes again, I will write out the equations where I add the external force F to the disc's 2nd law equation.
Disc:
##m _{1} a _{1}=F _{1} +F ##
Base:
##-m _{2} a _{2}=F _{2}##

Since F1=-F2,
##-m _{2} a _{2}=F _{2} =-F _{1} ##
##m _{2} a _{2}=F _{1} ##
Substituting this into the Disc's equation:
##m _{1} a _{1}=m _{2} a _{2} +F ##
Since ##a _{1}=a _{2}## I can write:
##m _{1} a _{1}=m _{2} a _{1} +F ##
And from here
##m _{1} a _{1}-m _{2} a _{1}= F ##
##a _{1}(m _{1} -m _{2} )= F ##
##a _{1} =\frac {F} {m _{1} -m _{2}}##
Excellent! Good job.

Trigu said:
So if all of this is correct then to me it seems that there are very disturbing conclusions, atleast when I compare this result to my initial predictions.
This equation says that it does not matter if the disc has the ability to rotate or not, which seems very odd. If this is truly the case I need a lot of help.
This is truly the case. What help would you like?

By the way, one of the reasons for the systematic approach is precisely to avoid the situation where we make a mistaken intuitive leap. We systematically draw the free body diagrams, determine our known and unknown values, and use Newton’s laws and constraints to write equations until we have as many equations as unknowns. This systematic approach allows us to come to the right conclusion even when our intuition fails us or betrays us.
 
  • #17
Trigu said:
I got to a feeling that I have not properly described the setup that I am trying to solve here. The force F is not moving with the disc when the disc is rotating. The force F is essentially a rope that is tied around the disc and the rope is being pulled with a force that is equal to force F. And it seems to me that in that situation the rotation of the disc is a fundamental part of how this system behaves, but if it really isn't, then could someone point me to some material where I can understand why?
The rotation will affect how much power you need to apply that force F. But if you assume you can apply the same F, then acceleration will also be the same.
 
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  • #18
Trigu said:
The force F is essentially a rope that is tied around the disc and the rope is being pulled with a force that is equal to force F.
So you are pulling with a force F on the rope. Let’s consider two situations, one where the axle is normal and free to rotate and another where the axle is welded and is fixed.

Now, consider the length of rope that you have to pull in the two cases to move the base a bit. In the welded case you just have to pull the rope a bit. If you pull the rope a distance, ##d##, then the base moves a distance ##d##.

On the other hand for the free axle, if you pull the base a distance d then you have to pull the rope a much longer distance because as you pull the rope the axle spins and feeds rope to you. So the distance you pull the free axle is some ##D>d##.

Now, if the force is equal in both cases that means that you did work of ##Fd## in the fixed case and work of ##FD## in the free case where ##FD>Fd##.

So the difference is not in how much a given force accelerates, but rather in how much energy is required to produce that given force. In the free case it requires more energy to produce the same ##F##. That additional energy is exactly the energy that goes into spinning the disk.

This is what your intuition is pointing you towards. But the systematic approach indicated that your intuition was leading you to the wrong understanding of what was happening. The force is not split up into a linear and a rotational part, but the energy is. Some of the energy goes into rotating the free axle, so it takes more energy to produce that same force.
 
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  • #19
Thanks once again for all the replies. I am starting to grasp the ideas now, but I still have one nagging issue that I would like to clarify. And to do that I will show a situation that tries to bring all of the ideas in this thread into a single problem.
Pulley-system-v1.png
So this is basically the classic pulley problem, where a block with a mass m3 is hanging on a thread that runs around a pulley that will only change the direction of the string. Then the string is wound around the disc. The base is on a frictionless surface.
Question: If the block starts falling down due to the force of gravity and travels the distance h, what is the distance L that the disc and base system travels? Also, we have two options: first, the disc is welded to the base and the second the disc can freely rotate.

I now think that when I first posted my question, I was not able to express what was the real question I was interested in, and that created a lot of confusion for me.
So I hope that with this setup, I am able to remedy that.
My understanding when examining the two options is this.
First when the disc is welded. Everything plays out exactly like one expects when two blocks are connected and one falls to the ground. Block falls distance h, the system moves to the right a distance L, which in this case is comparable to distance h.
The second case seems to be different. As the block falls the disc rotates and feeds thread until the block travels the full distance h. The system moves a distance L, but it is small compared to the distance h and much smaller than distance L when the disc is welded.
Now the part that is problematic. If bodies are connected via a thread, what happens is that the force that is applied to one is applied in some direction to the other, but it always has the same magnitude. It seems that the force that is applied by the thread to the disc is exactly the same in both cases (rotating and fixed disc). But as I discovered, if a force F is applied to that system, it will always move the same way regardless of the fact if the disc can or cannot rotate, which does not seem to match with the assumption of how the disc and base will move in this setup when rotation is possible.

So I seem to be missing a crucial step on how to calculate the distance L traveled by the system in this setup when the disc can rotate. Because the two cases behave differently even though both seem to have the same force acting on the same point in the system.
 
  • #20
Trigu said:
It seems that the force that is applied by the thread to the disc is exactly the same in both cases (rotating and fixed disc).
Think again about that part. Consider extreme cases of mass-ratios and moments of inertia.
 
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  • #21
Trigu said:
It seems that the force that is applied by the thread to the disc is exactly the same in both cases (rotating and fixed disc).
Again, you are jumping to an incorrect conclusion. You need to follow the systematic approach. (In this case you will need to use the rotational equations for the disk also)

Trigu said:
So I seem to be missing a crucial step on how to calculate the distance L traveled by the system in this setup when the disc can rotate.
Well, yes. You are missing all of the steps for the block.

You personally need to take a much more systematic approach. As you do dozens or hundreds of problems you will gain intuition and be able to skip steps. You are not there yet. Your intuition is unreliable at this point so rely on the method until you can build correct intuition.
 
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  • #22
Dale said:
You personally need to take a much more systematic approach. As you do dozens or hundreds of problems you will gain intuition and be able to skip steps. You are not there yet. Your intuition is unreliable at this point so rely on the method until you can build correct intuition.
@Trigu : I agree with @Dale that you should do the math systematically: just apply the laws, don't assume anything beyond them.

But if you also want to do the intuitive reasoning thing, you have have to check each of your generalizing claims (your "It seems like..." sentences). One heuristic is testing them with extreme values for the parameters of your scenario. What if the disc has negligible moment of inertia (so it takes almost no force to unspool it) vs. extremely high moment of inertia (so it's almost like fixed at the axle). You still think the string tension would be the same?
 
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Related to Rotation around a non fixed axis + linear motion of a system

1. How does rotation around a non-fixed axis affect the linear motion of a system?

Rotation around a non-fixed axis can cause changes in the linear motion of a system, depending on the location and direction of the axis. This rotation can result in changes in velocity, acceleration, and momentum of the system.

2. What is the difference between rotation around a fixed axis and a non-fixed axis?

Rotation around a fixed axis occurs when the axis of rotation remains constant, while rotation around a non-fixed axis means that the axis of rotation is constantly changing. This can have significant impacts on the motion of a system.

3. How can the center of mass of a system be affected by rotation around a non-fixed axis?

The center of mass of a system can be shifted or changed due to rotation around a non-fixed axis. This can result in changes in the overall motion and stability of the system.

4. What is the relationship between angular velocity and linear velocity in a system with rotation around a non-fixed axis?

In a system with rotation around a non-fixed axis, the angular velocity and linear velocity are directly related. This means that changes in one will result in corresponding changes in the other.

5. Can a system have both rotation around a non-fixed axis and linear motion simultaneously?

Yes, it is possible for a system to have both rotation around a non-fixed axis and linear motion simultaneously. This can occur in situations where the system is moving in a curved path while also rotating around a non-fixed axis.

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