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Position vector in torque (and angular momentum)

  1. May 1, 2015 #1
    Torque is defined as the cross product of position vector and force, i.e. [itex]\vec \tau = \vec r \times \vec F[/itex] .

    However the force vector [itex]\vec F[/itex] is fixed, but the choice of origin is arbitrary, making [itex]\vec r[/itex] also arbitrary. Does it make the torque vector also arbitrary, which apparently shouldn't be?

    So let's say in a very general case, a force [itex]\vec F(t)[/itex] acts on a particle between times t1 and t2 with position vector of the particle [itex]\vec r(t)[/itex]. Is the torque at time t simply [tex]\vec r(t) \times \vec F(t)[/tex] or [tex](\vec r(t) - \vec r(t_1)) \times \vec F(t)[/tex]? And if I want to find the angular momentum, is the linear momentum [itex]\vec p(t)[/itex] or [itex]\vec p(t_1) - \vec p(t)[/itex]? How do you justify the choice of origin as the centre of rotation of most standard cases? (as in https://commons.wikimedia.org/wiki/File:Angular_momentum_circle.svg)

    Please correct me if I have any conceptual problems, but I am really confused.
     
  2. jcsd
  3. May 1, 2015 #2
    Yes, the origin is arbitrary thus the torque. Whatever origin you define, you will find the angular momentum with respect to that point, so angular momentum is also not an absolute concept.
     
  4. May 1, 2015 #3
    Thank you so much. So in my case the torque is simply [itex]\vec r(t) \times \vec F(t)[/itex], right?
     
  5. May 1, 2015 #4
    Yes.
     
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