# Can angular momentum not be conserved in a straight line motion?

Consider a particle moving an a straight line, with constant velocity $v$. The angular momentum (pivot point $O$) can be calculated as $$L=mr v_{\theta}$$Where $v_{\theta}$ is the velocity perpendicular to the vector $r$ at each istant.

Now if I calculate the angular momentum in $A$ it get $L_A=mr_A v_x$, while in $B$ I get $L_B=mr_B v_y$.

In general $L_A\neq L_B$ but how can that be? How can angular momentum not be conserved? There are no forces, or torques!

I'm probably missing something big but I cannot see the mistake

• What is special about your point O? Why would you think that anything to do with O should have any impact on some random particle sailing by? Apr 18, 2016 at 18:18

If you draw similar triangles, then you'll find that $r_A/r_B = v_y/v_x$, and so the product $r_A v_x$ is equal to $r_B v_y$. Try drawing a line from the tip of your lower $\vec{v}$ vector to the tip of your lower $v_y$ component to see this.
• @MichaelSeifert Thanks for the reply! If I may ask, so the fact that $L$ is conserved does not imply that $v_{\theta}$ (i.e. the velocity perpendicular to $r$) is constant, right? Apr 18, 2016 at 21:53
• @Sørën: Only if $r$ is constant as well. This is the case, for example, for a planet orbiting the sun in a circle (which has a constant tangential speed at all times). On the other hand, a comet orbiting the sun in an elliptical orbit has a varying $r$ and a varying $v_\theta$, but it has a constant angular momentum $L$. (This is why the comet is moving faster when it's closer to the Sun.) Apr 18, 2016 at 22:00