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If the angular momentum of a particle is conserved and it is also 0, then is it true that the particle moves along a line? If so, how can we derive the equation for the trajectory from both the above fact and knowledge of initial position?
(This question was inspired by the fact that if angular momentum is conserved and not 0 then a particle always moves on the same plane, but nothing is said if it is 0).
As always, any hint or comment is highly appreciated!
Yes, if the angular momentum of a particle is conserved and it is zero, the particle must move along a straight line.
Indeed, by using the triple product identity, we have $$ {\bf r} \times ( {\bf r} \times {\bf v} ) = ( {\bf r} \cdot {\bf v} ) {\bf r} - r^2 {\bf v}. $$ Therefore, if the angular momentum is zero, we must have at every time $t$ $$ ( {\bf r} \cdot {\bf v} ) {\bf r} = r^2 {\bf v} $$ which means that velocity always has the direction of the initial position vector, i.e., the motion is always along a line.
The angular momentum is defined relative to a chosen point. For it to be zero, the velocity vector of the center of mass of a non-rotating rigid object must be directed toward that point.
First of all, the angular momentum is defined only with respect to a point. If we treat that point as the origin, and the position of the particle is $P(t)$, then angular momentum is $mP(t)\times P'(t)$. If this is zero, then at least one of these quantities is zero, or the position and velocity are parallel. Them being parallel corresponds to the object moving along a straight line that goes through the origin. If the velocity is zero, then the object is stationary, which is a subset of moving along a line. Presumably, the mass is non-zero. That leaves the possibility that the position is zero. If the position is constantly zero, then the object is stationary. But the object could travel through the origin, in between travelling along lines through the origin. It could travel along different lines, going through the origin each time it changes lines.
