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Understanding torque

  1. Dec 14, 2005 #1
    We all know [tex]\tau=\bold{r} \times \bold{F}[/tex]

    Also we know about moments. My question is, can we prove/derive the equations for these quantities from Newton's laws of motion without a priory setting the relation to r as proportional from experiment?

    Meaning that if we define [tex]\tau=f(\bold{r}) \times \bold{F}[/tex], can we find [tex]f(\bold{r})[/tex] from first principles, i.e Newtons laws of motion, conservation laws, e.t.c.?

    I've been trying for many hours now but can't seem to get there...
     
    Last edited: Dec 14, 2005
  2. jcsd
  3. Dec 14, 2005 #2
    so far I have written:

    [tex]m \frac{d \vec{\omega} }{dt} \times\vec{r}=\vec{F}[/tex]

    by a considering the force acting on a stationary body fixed at a distance from a point,

    and:

    [tex]\vec{\tau}=\frac{m}{2}dvdv\hat{\tau}=\frac{m}{2}r^2d\omega d\omega}\hat{\tau}[/tex]

    by defining torque as the effort required to increase the angular velocity of a stationary body fixed about a point.
     
    Last edited: Dec 14, 2005
  4. Dec 14, 2005 #3

    Doc Al

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    Staff: Mentor

    Pick up any classical mechanics textbook!

    In the meantime, try this. First define angular momentum of a set of particles as being the sum of [itex]\vec{L}_n = \vec{r}_n \times \vec{p}_n[/itex], then take the derivative with respect to time to get Newton's 2nd law for rotation, etc. (Is this what you are looking for?)
     
  5. Dec 14, 2005 #4
    No, I'm afraid it's not. Doing what you suggest would give us an quantity with dimensions of energy, but not equivalent to the expression for torque I defined in my previous post. I am 100% happy and sure of my definition of torque, meaning that it perfectly expresses the effort I have to make to angularly accelerate a body fixed about a point. Using your suggestion we arrive at the conventional expression of torque... why are the conventional and mine different? This is why I don't understand torque I guess. Is it not the effort required to increase the angular velocity of a stationary body fixed about a point? If not, what is it?
     
    Last edited: Dec 14, 2005
  6. Dec 14, 2005 #5

    Doc Al

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    Sorry, I should have read your earlier post more carefully. Maybe someone will catch what you are trying to do; I don't understand it. I don't understand your definition of torque as "effort required to increase the angular velocity of a stationary body fixed about a point". (I don't even know what you mean by "stationary body".)
     
  7. Dec 14, 2005 #6
    I guess what I was trying to do is to arrive at the expression for torque from a physical definition of torque in terms of work done, a definition of angular velocity, and by considering Newton's laws for linear motion. What you showed me defines torque as the rate of change of angular momentum. In fact my definition is different from the conventional definition, which is why I couldn't get anywhere.

    However thinking about it I am now happy with the conventional defintion: the expression tells me that the larger the radius over which torque is applied the larger the rate of change of angular momentum. (This is something I hadn't previously realised about torque; I didn't know how it was defined) That's great, thank you very much. =)

    And to get to my expression for "effort required to increase the angular velocity of a stationary body fixed about a point" you simply have to integrate the expression for [itex]\tau[/itex] from 0 to [itex]d\theta[/itex], just as to get to the energy done by applying a linear force you have to multiply the force applied by the distance moved. =)
     
    Last edited: Dec 14, 2005
  8. Dec 14, 2005 #7

    Doc Al

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    When you apply a torque over a given time dt, you have supplied an angular impulse that gives rise to a change in angular momentum. (Exactly analogous to the impulse-momentum theorem for linear motion.)

    The energy supplied depends on the angular displacement (angle) that the torque was applied over. (Again, just like the work done by a linear force equals the force times distance.) You may find this helpful: http://hyperphysics.phy-astr.gsu.edu/hbase/rke.html#rke
     
  9. Dec 14, 2005 #8
    yup already got there =) thanks doc, i really should stop posting before properly considering. heh
     
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