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Work and Torque

  1. Jul 23, 2004 #1
    After I read that Torque is a force that is both proportional to moment arm and force applied T = f *r , I tried to explain this fact by using the concept of Work which i learned earlier. When the force F is applied perpendiculary to some point at a distance r it does work equal to W = F*(angle measure*r) or W = F*(length of the arc).Thus, if the same work needs to be done at r/2 the force twice the magnitude should be exerted on the point. However the movement in an arc is not exactly in the direction of the force so i am kinda confused. Any advice?
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  3. Jul 23, 2004 #2


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    The torque is NOT a force!!!!!

    Secondly, while it's great that you've noticed that the dimensions of work and torque are in one sense the same (both are Newton*meter), a crucial difference (that you haven't covered yet), is that work is a "scalar quantity" (a number), whereas the torque is a "vector quantity" (which in addition to a number serving as the magnitude, has a specified direction).

    There are other differences as well; I would strongly advise you not to ponder on the weak analogy work/torque.

    Try instead to regard the torque as a fundamentally new concept to learn and master.
  4. Jul 23, 2004 #3
    I am not trying to show that Torque is Work, what im trying to understand is how Torque became to be defined this way and why. And definately work done by the lever arm at various radii offers some clues.
  5. Jul 23, 2004 #4


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    I remember having conceptual problems when I was learning this stuff too. My professor helped me greatly by explaining that all of rotational mechanics, is nothing more than a fancy, dressed up version of linear mechanics(the stuff you know and love.) In other words, torque is just a contrived quantity that is used to make solving problems easier. The physics haven't changed. The math surrounding it has.
  6. Jul 23, 2004 #5


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    Work done on turning something is torque times angle turned. Angles of course have no dimensions, though conventionally called radians (or degrees). So if it helps, think of torque as energy per radian (e.g. joule per radian). In the same way, you can think of force as energy per linear distance e.g. a newton is a joule per meter.
  7. Jul 23, 2004 #6


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    I think I might disagree with your definition of work being equivalent to energy. Work causes changes in the energy of a system, but is not energy in itself; in the same way force causes changes in acceleration, but force is not acceleration in itself.
  8. Jul 23, 2004 #7


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    That's really quibbling. Work is mechanical energy transferred. So force is energy transferred per linear distance. This isn't meant to be a rigorous definition. It's to help someone gain insight.
  9. Jul 23, 2004 #8
    If Torque = F*r
    then F must be the force perpendicular to the line from centre of rotation to the point on which you apply the force.

    The nice thing of defining Torque like this is that you then have:
    Torque*angle = work
    This is the case because:
    F*r*angle = F*(r*angle) = F*(length_of_arc)

    (the force may need to change direction, or it may pull in the same direction on different parts of the object as with a chainwheel and chain)
  10. Jul 24, 2004 #9
    Originally Posted by gerben:
    I shall definetely agree with that. For i assume torque was first brought up to find a way of balancing objects in rotation.However, when a pair of oppositely directed forces was applied to balance an object like a seesaw application at different radii yielded different results.And, indeed the only concept that accurately described this phenomenon was work.Thus, if force F is applied to the point at a certain angle on radius r, the work done by that point through the length of arc is the vector product of force and radius times the angle in radians or W = F x r *(theta).However work is independent of direction in a sense that one can't one work negative and another positive and cancel them out.Therefore, if one can think of something like potential work, that if applied through an angle will yield actual work, it can be assigned direction for it hasn't been done yet.This "potential work" can be than defined as F x r or Torque. Am i getting it?
  11. Jul 25, 2004 #10


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    Thanks for the clarification krab, I never really thought of it that way before.
  12. Jul 25, 2004 #11
    I do not really understand what you are saying here. I do agree that torques can be balanced, while work cannot, if torques are balanced there will be no work. However, the reason that torque is defined this way is more like this:

    There are two different sorts of movement, linear movement and rotational movement. There is a relation between linear movement and force. It would be nice to have something, which we will call torque, that has a similar relation to rotational movement as force has to linear movement. Therefore torque is defined like this. Torque has a similar relation to rotational movement as force has to linear movement.

    A nice way to define force is to say how much work it does when it acts over a certain distance, now we also want something that is defined by how much work it does when it acts over a certain angle and we call that torque.
    This means that torque has to be force times radius.
  13. Aug 1, 2004 #12
    scalar and vector definitions

    What I was surprised to see was that no one at the beginning of this thread pointed out to the person asking the question that Torque and Work are defined using the cross and dot product respectfully. This might have ended the confustion:

    T = F x r (x means cross-product from linear algebra - cross product produces another vector that is perpendicular to the first two).

    W = F * d (* means dot product, also from linear algebra which produces a scalar, ie a number, from two vectors).
  14. Aug 1, 2004 #13

    well actually, [itex] \vec{\tau} = \vec{r} \times \vec{F} [/itex]

    I think the best way to learn this rather confusing subject matter is to start out with a good book, i.e. Introduction to Mechanics, by Kleppner and Koleknow and see how all of these values are derived. That's how I've learned and it's stuck, and I have an intuitive feel for this stuff, and it's not magical and crazy anymore. Just repeated applications of newton's 2nd and 3rd...

    Hope this helps...

  15. Aug 2, 2004 #14
    All fancy complex technical derivations aside, torque is defined the way it is because when you apply a force with the same magnitude on a rotating arm at two different points: R and at R/2, such that you apply the force until the arm has rotated through an angle theta, you have simply done more work on the object when the force is applied at R (as the arc length is proportional to the Radius), which increases both the object's momentum and kinetic energy. It's quite simple: you apply more work on an object, you increase it's kinetic energy.

    You are completely right on this...in reality, in order to keep the force to stay perpendicular to the surface of the rotating arm, you must do an extra tidbit of work in keeping the object that applies the force pointing in the direction perpendicular to the surface of the rotating object. There are two equivalent ways of defining torque, both mathematically correct. I use the latter in physics simulations.

    "Torque is the perpendicular component of force, multiplied by the distance from the pivot point"

    it is equivalent to say:

    "Torque is the force multiplied by the *perpendicular distance from the pivot point*"

    The perpendicular distance is the distance between the pivot point, and the projection of the pivot point onto the vector direction represented by the force.
  16. Aug 2, 2004 #15


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    Consider very tiny angular motion (quasilinear, like how a regular n-gon starts to look like a circle for large n).
  17. Aug 2, 2004 #16
    Lets think of motion on a plane for now, so [itex]\tau[/itex] loses its vector properties.

    Think of a force applied on a particle moving it a an infitesimal distance [itex]ds[/itex].
    Then we can say [itex]dW=Fds[/itex]. If you think of the torque instead, about some point (doesn't matter here for such a small distance), [itex]F=\tau / r[/itex] where [itex]r[/itex] is the radius about which the torque is being calculated. then [itex]dW=\tau ds / r[/itex] but [itex]ds/r[/itex] is only the very small angular displacement, [itex]d\theta[/itex]. (remember, for small angles, we can make the assumption that [itex]ds = rd\theta[/itex]). so in other words, [itex]dW = \tau d\theta[/itex]. angles are dimensionless... explaining why they have the same unit.

    This is exactly what turin said in the previous post.
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