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Moment of force and Newtons 2nd law

  1. Dec 1, 2011 #1
    By the end of the day, I must admit that the rotational analog of Newtons 2nd law:
    Τ = I*α
    contradicts my intuition. Suppose that you take a part of a body in the radius r from the rotational axis and exert a force on that.
    It will then get a linear acceleration. We can then find the angular acceleration using:
    a = r*α
    Which means that the angular acceleration is bigger the closer we are to the axis of rotation. I mean that if we take 2 identical rotating disk, and measure the linear acceleration of a point on the first disk to be a at r=1m, and then find for the second disc, that the linear acceleration is a at r=2m, then obviously the first disk must have a higher angular acceleration... But then this suggests that a force exerted closer to the rotational centre means a higher angular acceleration than one exerted further away, and that is not implied by the rotational analog of Newtons second law. What am I doing wrong in my assumptions?
     
  2. jcsd
  3. Dec 1, 2011 #2

    Doc Al

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    OK. The linear acceleration is given by a = F/m.
    No, that's not true. (Are you thinking of something rolling without slipping, perhaps?)

    The angular acceleration about the center of mass is given by T/I = Fr/I.
     
  4. Dec 1, 2011 #3
    I am sorry. Could you tell me what is the 'T' and what is the 'I'?
     
  5. Dec 1, 2011 #4
    Sorry yeh. I should have been more discriptive.
    a is the linear acceleration
    α is the angular acceleration
    T is the net torque
    I is the moment of intertia with respect to the rotational axis.
    I don't see how you wouldnt be able to find the angular acceleration using:
    a = α*r
    Its moving in a circle with radius r.
     
  6. Dec 1, 2011 #5
    OK , I think I have got your mean.
    In my opinion, the following is wrong.
    1)When you exert a force on the disc, you forget the force exerted by the rotational axis.
    2)F=ma is right for the particle, and your disc aren't treated as particle. So if you want use F=ma, you must treat the disc as the system of particles(or rigid body) and consider the internal force.
     
  7. Dec 1, 2011 #6

    Doc Al

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    OK. Let's use an example of a disk that is free to rotate about some axis.

    You apply a torque, which gives the disk a certain angular acceleration α. You can then calculate the tangential acceleration of a point on the disk a distance r from the axis using a = r*α. Sounds good to me.

    Going back to your original post:
    That's not a helpful way to look at it. When you exert a force at some distance r from the axis, you exert a torque with gives the entire object an angular acceleration α.

    Actually, you'll find the angular acceleration using α = T/I. Then you can find the tangential acceleration of any point on the disk using a = r*α.
    Only if you assume that the tangential acceleration is the same everywhere, which it isn't.
    Definitely. So that means that you're exerting a greater torque on the first disk.
    I think you are assuming that a force applied at some distance r will create the same tangential acceleration at that point no matter what r is. That's not true.

    The way to figure it out is this. First compute the angular acceleration: α = T/I = Fr/I. Now compute the tangential acceleration at point r: a = r*α = Fr2/I. So the closer you apply the force to the axis the smaller the resulting linear acceleration of that point is. Which should make sense.

    Let me know if that clarifies things a bit.
     
  8. Dec 1, 2011 #7
    It did clear some things up, or at least hinted me towards the right understanding.
    Though I still can't quite get the right idea. The key to understanding it correctly is I think in the fact that a force exerted a distance r away will according to you NOT give an object the same tangential acceleration.
    I don't see why this is however as force exerted on the i'th particle in the object does do that, so I think my problem lies in the fact that I don't consider the object as a collection of particles, but rather that exerting a force on the i'th particle is the same as exerting a net force on the object as a whole. When my book derives the analog of Newtons 2nd law in rotational coordinates, it starts by taking the i'th particle with a mass mi and saying that if we exert a net, tangential force on it, that gives it a linear acceleration:
    Ftan = mi*a
    =>
    Ftan = mi*α*r
    And multiplying that by r we get:
    T = mi*r^2*α = Ii * α , where Ii denotes the moment of intertia for the i'th particle. And they then sum all the torque for the particles in object and thereby derive:
    T = I*α
    So I think my problem is maybe that I don't consider the moment of intertia, since the radii that I think of as important just go into that term. But rotation a body in any distance requires the same moment of inertia, so I don't see how it could be that..
     
  9. Dec 1, 2011 #8

    Doc Al

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    That's correct.
    Careful. Since the object--the disk--is rotating about an axis, there are other forces acting besides the one you exert on the i'th particle. The axis exerts a counterforce so the disk ends up with no net translational acceleration. (A point on the disk will have a tangential acceleration.)
    That's cool. (There's a bit of handwaving there, as you must show that the net torques on all the particles add up to the net torque due to the external force.) Realize that they are talking about the net tangential force on the i'th particle. If you apply an external force F to the i'th particle of the disk, for instance, that does not mean that the net force on that particle is F.

    Also realize that due to the fact the particles are connected together as an object, they all have the same angular acceleration (but different tangential accelerations). That's what allows you to get to the last step.
    I don't quite understand that last sentence, so maybe you can restate it.
     
  10. Dec 1, 2011 #9
    Hmm... the last sentence didn't really make any sense, when I think about it..

    But I now see that my problem maybe lies in the fact that I took the external force as the net force, which as you say is wrong, since there is also a radial component which makes it stay in a circular motion - is that what you mean by a counterforce?
    But doesn't this force only act to change the direction of the velocity, so can't we neglect that? It is neglected in my books derivation. What they actually do is put up the equation:
    Fi(tangential) = mi * a(tangential)
    So what i meant was not a as in the net acceleration but a is in the tangential acceleration - my mistake.
    Still not sure about it all though - I'm obviously thinking about it the wrong way, and it's driving me insane. Let me go back to the first example:
    Suppose you take a disc and exert the same force on two different particles in the disc, one in a radius of r1=1m and one i a radius of r2=2m from the rotational centre. Then the particle will no matter what get a linear acceleration of a.
    BUT this is wrong according to you. Is it because you can't exert the same force in different radii. Why is that? All you're doing is exerting a force...
     
  11. Dec 1, 2011 #10
    This is really the key to it all I think.. Can you please elaborate on this not using a torque-argument, since it's torque as a whole that I don't quite get. Or well I know how torque is defined and all, but don't see it as an intuitive quantitive.
     
  12. Dec 1, 2011 #11

    Doc Al

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    Not just a radial component. Their proof uses the net tangential force, but the net tangential force does not equal the external force.

    Here's an example. Push that disk anywhere you want. Do you agree that all points on the disk have some kind of tangential acceleration? Even points that you aren't pushing against, right? So the material of the disk transmits internal forces between the particles.

    See my comments above. The key, which is easy to show, is that [STRIKE]external[/STRIKE] internal forces exert no net torque on the object.

    It's wrong because the force you exert is not the net force on the particle (not even the net tangential force). F = ma only applies when F is the net force.

    I would certainly agree that if the net tangential force on each particle were the same, then they would have the same tangential acceleration. But that's not the situation. All you can say is that you exerted the same external force on the two particles. And exerting the same external force at different points produces different results.
     
    Last edited: Dec 1, 2011
  13. Dec 1, 2011 #12
    Don't you mean the internal forces?
    Okay, so I think I got closer to understanding there. The net tangentialforce is not some particular force that we know is being exerted on the object. Because as it is a rigid body the external force that creates this tangential force for every particle is transmitted between all the particles to create a uniform angular velocity. But I think I'll have to think a little more about torque as a whole
     
  14. Dec 1, 2011 #13

    Doc Al

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    Oops... you're right. I'll correct that.
     
  15. Dec 1, 2011 #14
    hmm thought about it about more, but wasn't really illuminated. Perhaps you could do this: Tell me the physical explanation (don't use any equations) why a force applied a greater distance from the rotational centre, i.e. a larger torque, causes a faster rotation.
     
  16. Dec 1, 2011 #15

    Doc Al

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    I'm not sure I have a good answer for that one off the top of my head (or even deep inside it), but I'll think about it. This stuff is quite subtle!
     
  17. Dec 1, 2011 #16

    Doc Al

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    This isn't exactly what you want, but it might help:

    Imagine you were to rotate the object through an angle θ by applying a tangential force F at a distance r from the axis. The distance through which the force acts is the arc length, which is proportional to r; distance = r*θ (the angle is in radians). Thus if you apply the force at a greater distance from the center, the force must act through a greater distance and thus do more work. And more work means more rotational kinetic energy and a faster rotation. Thus we can conclude that the force applied further from the center makes the object rotate faster.
     
  18. Dec 1, 2011 #17
    hmm yes. I have considered that, but came to the conclusion that an energy observation isn't directly applicable to the problem, that troubles me :(
     
  19. Dec 1, 2011 #18

    Doc Al

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    How did you conclude that? Why isn't it applicable?
     
  20. Dec 1, 2011 #19
    Well I thought that this is about applying a force on a certain point of a circle, which would be something that happens instantly and not over a distance but not sure. Anyways I don't think it shows in the equations leading to T = I*alpha
    But maybe you are right, and I have to think more of it.
     
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