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Simple argument that a force applied farther from the rotation axis

  1. Aug 20, 2012 #1
    Torque is defined as:

    τ = rxF

    this means that the farther from the rotation axis of a body a force is applied, the more it will tend to rotate the body.

    My question is:

    Can anybody give me a simple argument that a force applied farther from the rotation axis should be better at rotating a body than one closer. You can use energy considerations to say that the work done by a force applied at the farthest distance does a greater work, but I am not looking for that since it is not general enough. Because imagine that two tangential forces of equal length but opposite direction are being applied to our body. Then it will still rotate, and energy considerations cant really explain that.

    So yeah, please give me an intuitive, yet rigorous reason that it MUST be so greater torque = no equilibrium.
     
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  3. Aug 20, 2012 #2

    Filip Larsen

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    Re: Torque

    Perhaps you can hone your intuition by thinking about what happen when you apply the force directly towards the axis (or the center of mass) of a rigid body. Why don't you get any rotational acceleration in this case and how does that relate to your question?
     
  4. Aug 20, 2012 #3
    Well in that case you are obviously not speeding up the particles in going around the centre of rotation. They do need a centripetal force but that will be provided by the forces keeping the rigid body together. But how does this relate to my question?
     
  5. Aug 20, 2012 #4

    russ_watters

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    Your question is unclear to me: your two tangential forces of equal length but opposite direction: where are they being applied? Opposite sides of the shaft...? Could you draw us a diagram?
     
  6. Aug 20, 2012 #5
    What I wanted to say with that is that you can't always use an energy consideration to show that force applied at greater distance from rotation centre = bigger acceleration. Because that wont do to explain why our rigid body on the picture STARTS to move in the drawn situation.
     

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  7. Aug 20, 2012 #6

    russ_watters

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  8. Aug 20, 2012 #7
    Yes yes! I just dont see WHY a couple will tend to rotate our body according to the net torque. What is it that makes a force applied farther from the rotation centre "BETTER" at rotating our object than one applied closer? And do not use an energy consideration, because that wont do. I want a simple argument in terms of the internal forces inside the body.
     
  9. Aug 20, 2012 #8

    russ_watters

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    I don't understand: internal forces will not cause a body to move.

    I'm not sure if what you are asking can be proven mathematically without using energy conservation and I'm not sure why you object to using energy conservation to analyze a situation that you recognize involves motion. But back to the original question:
    A simple argument? Sure: it is observed to be true.
     
  10. Aug 20, 2012 #9
    Okay when you push with a tangential force the force is transmitted through our body to make it rotate. That's what I meant by internal forces - thos transmitting the tangential forces to an overall rotation. I want an argument with forces that shows why a force applied farther away is better at rotating. Because you must be able to prove it. As for energy considerations those will only be valid once the body is actually rotating and the forces are doing work. When it stands still like one the picture you can't use an energy consideration to show that it must start to rotate due to the couple. Understand my problem now? :)
     
  11. Aug 20, 2012 #10

    russ_watters

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    If by that you mean you must be able to prove it mathematically, no, I'm sorry but that is not correct. Physics is a mathematical study of reality. The only "proof" required is experimental evidence.
    No, that isn't correct. Looking at a previous thread you started on a similar subject, you're making the mistake of not counting the effect of inertia in opposing the force, so you're not summing all of the forces and you're not recognizing that the instant you apply the forces, there is an acceleration and a power. Ie, an object that is not moving, but has an unbalanced force (couple) is still accelerating.
    Yes, I think so: your problem is you misunderstand what "standing still" means. "Standing still" means not moving and not accelerating.

    [edit] Put another way: you're asking 'why is there an acceleration?' when you could be asking 'is there an acceleration?', and calculating that it is non-zero.
     
    Last edited: Aug 20, 2012
  12. Aug 20, 2012 #11
    Okay, I think you are the one misunderstanding me once again.

    My problem is NOT that I don't know what standing still means. My problem is that I geometrically can't see why a force applied at greater distance = bigger acceleration = greater tendency to rotate.

    Let me try to be more clear: In translational dynamics you learn that if the resultant force on an object is zero then it will not accelerate. This is actually where I have a problem. Because the resultant force on our rigid body is zero. Yet the body tends to rotate. I want to understand why the force applied at greater distance has the property of being able to rotate the body. I have said that in terms of energy it is clear once the body is actually rotating. Then you can see that the force applied at greater distance does greater work etc. etc. But we remain to show that the body will actually start rotating when there is a couple on it. That cannot be shown by energy considerations since in that instant no work will be done. Do you understand me better now? :)
     
  13. Aug 20, 2012 #12

    russ_watters

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    I was mistaken about my use of power since you are right that at the moment of the beginning of the acceleration, power is zero. It is change in momentum that is observed at the moment of application of the torque: http://en.wikipedia.org/wiki/Angular_momentum#Conservation_of_angular_momentum

    Ok....
    Ok....but if it is non-zero, it does accelerate, right? So what equation describes that acceleration?
    The linear force is zero, so the linear acceleration is zero. The 'angular force' (torque/couple) is non-zero, so the angular acceleration is nonzero. I don't see why you see a contradiction between the two examples you gave. They look to be the same, with the same logic and the same outcome if you are consistent in the construction of the examples.
     
  14. Aug 20, 2012 #13
    I know all about angular momentum, but for me that is all just math and does not necessarily provide an intuition or an understanding of what happens on microscopic level.
    And just to show you, I do know that (in vector notation):

    dL/dt = Ʃτ

    Which again can be regarded as an analogue of F = dp/dt for linear motion.
    But that equation does not say much for me.
    I want an argument purely in terms of forces and interactions of the particles that make up our rigid body.
    Even though linear motion and rotations are different the basic ingredient for any of the two motions is this: A force. Therefore you should be able to show purely through the use of force analysis that a force applied at a greater distance should mean that the moment of our body changes. Shouldn't you?
     
  15. Aug 20, 2012 #14

    russ_watters

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    How about f=ma? I find that to be more of a "why" than f=dp/dt, which is more of a definition or description. At least f=ma speaks to me more (not that the universe cares how it speaks to me...)
    You're looking too deep. This problem deals with theoretically perfect bodies that don't have particles. There are no inter-object forces to deal with. You would not ask for such a proof regarding linear motion: f=ma deals with completely solid, monolithic (or point) objects only.
    I don't know -- you're declaring what seem to me to be relevant points to be unacceptable to you. Do you have similar issues with f=ma? Do you need a proof of it as well?

    I guess the point here is that ultimately you're going to have to accept that some parts of what we understand of physics cannot be derived. They are simply observed to be true. May I ask why this is a problem for you? What would you gain by seeing a derivation that speaks to you that you are losing by not seeing it? Do you accept that the physics we currently have works?
     
  16. Aug 20, 2012 #15
    hmm it is not a problem to me when things are observed rather than derived. I just want to be clear of what can be derived and what will always be an observation. Newtons laws, e.g. F = dp/dt, are for instance no problem to me since they are clearly empirical.
    But I am still not sure that you can't show the above stated from Newtons laws, and I don't see why you are so sure either.
    Also I don't agree that you can't divide a rigid body into a lot of small mass elements. That is exactly what you do for moment of inertia calculations etc. etc. The important property of a rigid body as I understand it is not that the small mass elements can't interact through forces, but rather that it is undeformable.
     
  17. Aug 20, 2012 #16

    russ_watters

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    We've actually had this exact question asked a number of times before and I've never seen an answer for the static case. Indeed, asking for a static answer to a non-static situation appears to me to be a contradiction in terms.

    People don't tend to realize that the term "unbalanced force" is at best sloppy and at worst is simply nonexistent: Forces are never really unbalanced, it is just that externally applied forces that don't sum to zero are opposed by the inertia of the object, which does then result in a zero sum of forces.
    I'm not saying that you can't divide up an object, just that you don't. Doing so for the purpose of finding a moment of inertia isn't a correct analogy: the different pieces aren't being treated as interacting, they are just being summed to a single number. Ultimately, the arrangement of those specific pieces makes no difference; only the sum matters.
     
  18. Aug 20, 2012 #17
    Can you elaborate on the point with the inertia of the object. I didn't quite understand that part - probably because my english skills are not always too good.
    Anyways, I'm going to bed and will think more about it tomorrow, though I should like to think that you wrapped it up with the points in your last post.
     
  19. Aug 20, 2012 #18

    russ_watters

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    Physicists don't tend to like the way I used the term inertia, but for engineers, inertia is the ability of mass to resist an acceleration. So a=f/m quantifies how much acceleration an object can resist under a given force. Or put another way, when you push on an object and it accelerates, it is the inertia that is pushing back at you.

    For example, consider the following forces, acting in opposite directions against the same point:

    F1 = 5N
    F2 = 3N

    You might say the force is imbalanced at F1-F2=2N or you might say the force is balanced at F1-F2-ma=0, where ma=-2N.
     
  20. Aug 20, 2012 #19
    Let me try and justify this for you.

    Let f_2 be the force acting at greater distance from axis, and f_1 acting from the smaller distance.
    K=∫(ƩF dot dr)
    =∫((F_2-F_1) dot dr)
    =∫(F_2 dot dr)-∫(F_2 dot dr)

    Consider both of these forces acting tangentially and for a quarter revolution (arbitrary). Since F_2 acts over an arclength of a_2 and F_1 acts over an arclength of a_1, where a_1 is less than a_2, the first integral is evaluated from 0 to a2 and the second is evaluated from 0 to a1, which results in the net force doing nonzero work over a time interval, even though the forces are equal. Trippy.
     
  21. Aug 21, 2012 #20

    A.T.

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    Try to ask the opposite question:

    What is it that makes a force applied closer to the rotation centre "WORSE" at rotating the object?

    And to answer it, consider the extreme case: A force acting exactly at the rotation center.
     
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