# Changing angular momentum with no forces to do torque

Suppose a particle of mass $m$ moves in a straight line at velocity $v$ some distance away from an axis, as shown here at times t1 and t2 (where the axis of rotation is located at the origin of the axes shown, oriented out of the page:

Is angular momentum (or $L = r \times p$) conserved here? I assume it should be, because I see no forces acting, so no net torque, but how would that be proven -- that is, that $T = \frac{dL}{dt} = 0$? Specifically, I'm asking how to show from the definition of L as $L = r \times p$ that angular momentum is conserved in this scenario -- looking at it intuitively, it seems to me that $r$ will change so $L$ will change, but that doesn't make sense as there's no torque.

• Angular momentum is conserved here because there's no rotation and $L=0$.
– Gert
Dec 17 '15 at 3:01
• But angular momentum is defined as $r \times p = rmv sin(\theta)$, no? How, mathematically, would you show that that's equal to zero at all times here? Dec 17 '15 at 3:51
• Use your diagram to find $r\sin\theta$ for several ($r,\theta$) position points along a string line. You will see that they are all the same. Dec 21 '15 at 22:46
• @Gert, $L$ about the origin is not zero, but it is constant. If you aren't heading directly toward a point, you are revolving around it. The torque is zero. Dec 21 '15 at 22:48

Let $\mathbf{r}=\mathbf{r_0}+\alpha \frac{\mathbf{p}}{|p|}$ where $\mathbf{r_0}$ is vector from the origin to a fixed point along the straight track. $\mathbf{r}$ a vector from the origin to any point on the track and may be specified by choosing $\alpha$. Then: $$\mathbf{r} \times \mathbf{p}=\mathbf{r_0} \times \mathbf{p} +\alpha \frac{\mathbf{p}}{|p|} \times \mathbf{p}$$ $$\mathbf{r} \times \mathbf{p}=\mathbf{r_0} \times \mathbf{p}$$ So the angular momentum, $L=\mathbf{r} \times \mathbf{p}$ is the same value for all points $\alpha$ along the straight track.