- #1
chuchung712
- 2
- 0
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.
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.