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Can Rotational Dynamics be derived from Translational Dynamics?

  1. Aug 25, 2013 #1
    Torque, in particular, is always a concept that has confused me.

    Can the rules related to torque be derived or are they natural tendencies? (For lack of a better way to phrase this).

    For example, if I were trying to solve the problem of a ladder leaning against a frictionless wall, where I know the mass of the ladder and am trying to find the coefficient of friction between the ladder and the ground, would there be a way to derive the solution to this without knowing anything about torque and rotational dynamics? I understand that using torque works, but I don't understand why, and every textbook I've looked at hasn't made it clear to me....

    This same type of question applies to all of rotational dynamics...

  2. jcsd
  3. Aug 26, 2013 #2
  4. Aug 27, 2013 #3

    Philip Wood

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    You are right: the laws of rotational dynamics (involving torque and angular momentum) can be deduced from Newton's Laws using a few lines of vector algebra.

    For example, start with Newton's second law, for a particle of mass m.
    [tex]\sum{\vec{F}} = \frac{\mathrm{d}\vec p}{\mathrm{d}t}[/tex]
    Now form the cross product of each term with the displacement vector [itex]\vec{r}[/itex] of the particle from a point O.
    [tex]\sum{\vec{r}\times\vec{F}} = \vec{r}\times\frac{\mathrm{d}\vec p}{\mathrm{d}t}[/tex]

    But [itex]\vec{r}\times \vec{F}[/itex] is by definition the torque, [itex]\vec{G}[/itex], due to force [itex]\vec{F}[/itex] about point O. Note also that:
    [tex]\vec{r} \times \frac {\mathrm d \vec p}{\mathrm{d}t} = \frac{\mathrm{d}(\vec r \times \vec p)}{\mathrm {d} t}\ - \ \frac{\mathrm d \vec r}{\mathrm{d}t} \times \vec p[/tex]
    Now [itex]\vec r \times \vec p[/itex] is, by definition, the angular momentum, [itex]\vec J[/itex], of the particle about point O, and [itex]\frac{\mathrm d \vec r}{\mathrm{d}t} \times \vec p = 0[/itex] because [itex]\frac{\mathrm{d} \vec r}{\mathrm{d}t}[/itex], the particle’s velocity, and [itex]\vec p[/itex], its momentum, are in the same direction!

    So we can conclude that
    [tex]\sum{\vec{G}} = \frac{\mathrm{d}\vec J}{\mathrm{d}t}.[/tex]

    [It's also easy to derive the principle of conservation of angular momentum, but to do this we also need to assume that forces between particles are central, that is attractions or repulsions acting along the line joining the particles.]

    It turns out that working with torques simplifies many problems. As the above derivation shows, this amounts to saying that the mathematical device of vector multiplication by [itex]\vec r[/itex] of each term in a vector equation from linear dynamics simplifies many problems.

    Despite this, it does seem that the concept of torque is indispensable. How, for example, can one derive the 'law of the see-saw' ([itex]W_1 d_1 = W_2 d_2[/itex]) without using torques? It can be done, as I show in the attachment, by admitting that the see-saw bar must have some internal structure. Otherwise, how could the part of the bar near the fulcrum (say) 'know' how far away the weight on its side of the fulcrum is placed? Something must be changed inside the bar. In the attachment I've given the bar the simplest structure I could think of: a pin-jointed lattice of struts and ties. The 'law of the see-saw' can then be deduced by force equilibrium at each pin!

    Attached Files:

  5. Aug 28, 2013 #4
    You meant to say that forces between particles satisfy Newton's third law. Conservation of linear momentum has the same requirement.
  6. Aug 28, 2013 #5

    D H

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    He meant what he said. There are two forms of Newton's third law. The weak form, that third law force pairs are equal but opposite, yields conservation of linear momentum but not angular momentum. The strong form, that third law force pairs are equal but opposite and are central forces, yields conservation of angular momentum as well as linear momentum.
  7. Aug 28, 2013 #6
    You mean that the weak form includes forces that are not directed along the line of interaction?

    In principle, I agree that mathematically the distinction is important. Physically, though, what interaction between particles has such forces? I have a mental block, please help me out.
  8. Aug 28, 2013 #7

    D H

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    Classical electromagnetism.

    Newton's third law assumes forces are instantaneous, even at at a distance. The electromagnetic field stores energy, linear momentum, and angular momentum. The conservation laws are more basic than Newton's third.
  9. Aug 28, 2013 #8
    This I understand. But these are field interactions. Is there a particle-particle interaction that is not central, without having to resort to a field view?
  10. Aug 28, 2013 #9


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    That's a nice answer to a common request, to explain torque for a static case, without using virtual work and conservation laws. It could be simplified to single triangle with 3-pins (fulcrum & 2 force point of attack), that are not on the same line (every rod must bend a little under load).
  11. Aug 29, 2013 #10

    Philip Wood

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    A.T. I've presented the single triangle approach on the new attachment - hope it's what you had in mind. It's certainly a lot simpler than my original 'lattice' treatment, though that treatment did show the progressive changes in compressive and tensile 'forces' as we move along the bar; these would have shown up even more clearly if I'd used a thinner 'bar' with more triangles, and even taken this to the limit of a continuous medium.

    The triangle approach is particularly good for addressing the original poster's concern. It's not essential to use the torque concept, but if you do use it, taking moments about the fulcrum, F (see attachment) [itex]T_3 [/itex] pulls to the right on pin E and to the left on pin G, giving no net torque. Similarly for [itex]T_1[/itex] and [itex]T_2[/itex]. [Note how the pin-jointed structure of struts and ties ensures 'central forces' between pins (see posts 3 and 5) without which the argument I've just used wouldn't work.] And, of course, the 'external force' upwards through F has no torque. So using torques is a neat trick for eliminating 'irrelevant' forces from consideration.

    Attached Files:

    Last edited: Aug 29, 2013
  12. Sep 1, 2013 #11
    An aside: if angular momentum can be derived from linear momentum, then for example in 2D statics why can we get three independent equations using force balance in 2 directions and moment balance?
  13. Sep 2, 2013 #12

    Philip Wood

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    The derivation I gave in post 3 was for a particle (point mass). Your three independent equations are for an extended object. This can be regarded as a system of particles, with certain constraints. Analyse this system using force equilibrium, and the Principle of (balancing) Moments will arise. For a simple example, see the attachment to post 10.
    Last edited: Sep 2, 2013
  14. Sep 2, 2013 #13
    As in NLM



    everything analogous,
    F produce acceleration
    ζ produce angular acceleration

    mass of body analogues to moment of inertia (both try in resisting motion)
    no need to say on α i suppose.
    hence a rigid body at rest can be solved using torque balance [rotational equilibrium] as it is done in newtons laws.....
  15. Sep 12, 2013 #14
    What about torque produced on a "free body," with no fulcrum or pivot around which it must rotate? A metal rod in a gravity-less vacuum perhaps? This situation is MUCH more difficult for me to grasp for some reason. Does the center of mass act as the fulcrum? There seems to be no intuitive reason that would be the case. I keep thinking in terms of the unequal accelerations of the parts of the object on either side of the applied force/torque, but can't work it out completely.
  16. Sep 13, 2013 #15
    this motion is called rotational + translation motion
    for solving these kind of motion you must refer a book as i myself would've told you but the problems are big so if you want start a new thread with some numerical and i or anyone may be helping you
  17. Sep 13, 2013 #16
    If a body is acted upon by "pure" torque, meaning that the net force is zero, then indeed the body will rotate about its center of mass. Why? Because when the net force is zero, total momentum is conserved; when total momentum is conserved, the center of mass continues its uniform straight-line motion.
  18. Sep 13, 2013 #17
    Fair enough. This topic has been very confusing for me, and has only recently been made lucid. Learning rotational dynamics has solidified the beauty of physics for me. Quite elegant.
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