Considering conservation of angular momentum of an object, originally in uniform circular motion, moving with constant $v$ towards the center

Say an object originally undergoing uniform circle motion has some constant radial velocity towards the center. Is its angular momentum actively changing because it's net velocity is now not along its original circular path? Or is it not changing because no torque is acting on the object? Or is this where we talk about conservation of angular momentum - where since no external torque is present but we did work to change our radius our energy loss must've been transferred?

I don't see $v_{\theta}$ being able to change by moving towards the center in hindsight, but I have a feeling angular momentum changes by doing this. And if it does, how can it while $v_{\theta}$ remains constant? Do we basically argue $L = mv_{\theta}r$ can still change if with our only variable as $r$, so this is indeed a change in angular momentum but not the obvious external torque kind?

• If the object has a radial velocity towards the center, it is not undergoing uniform circular motion. It is not clear to me what you're asking. – ACuriousMind Nov 14 '17 at 11:50
• I have written my answer based on uniform circular motion, I don't exactly know what you mean btw. – ubuntu_noob Nov 14 '17 at 12:41

If I understand your question correctly, I will try to answer with an example. First, note that the statement about the conservation of angular momentum is

$$\frac{d\vec{L}}{dt}=\vec{\tau}$$

where $\tau=\vec{r}\times\vec{F}$ is the torque applied on the system. Imagine now a ball attached to a rope rotating in a plane, and suddenly you start to pull the rope towards the center. This scenario is depicted in the following image. Observe that there is no external torque in this system, since the force is central. Thus angular momentum is conserved, and if $r=R-v_{r}t$ (someone pulls the rope with constant velocity towards the center) then

$$v_{\theta}=\frac{L}{mr}=\frac{L}{m(R-v_{r}t)}$$

As you can see, the result of the conservation of angular momentum is that the velocity in the angular direction increases as the rope radius decreases. This is perfectly okay! The increases in velocity means that the one who pulls the rope must do work that inject energy into the system.

In summary, angular momentum doesn't have to change in the case you've described - and the result is an increase in the angular velocity with a decrease in the radius for example. You can, however, think about a case where $L$ does change, but then it is clear that there must be some torque acting on the object.

Suppose the particle is connected to a string at the centre of the circle having Tension $\vec{T}=\sum\vec{F}=\frac{mv^2}{r}\hat{r}$ From newton's second law of motion $$\vec{F}=\frac{\mathrm{d}\vec{p}}{\mathrm{d}t}$$ $$\frac{mv^2}{r}\hat{r}=\frac{\mathrm{d}\vec{p}}{\mathrm{d}t}$$ $$\vec{r}\times\frac{mv^2}{r}\hat{r}=\vec{r}\times\frac{\mathrm{d}\vec{p}}{\mathrm{d}t}$$

Now, let angular momentum be defined as $\vec{L}=\vec{r}\times\vec{p}$ $$\implies \frac{\mathrm{d}\vec{L}}{\mathrm{d}t}=\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\times\vec{p}+\vec{r}\times\frac{\mathrm{d}\vec{p}}{\mathrm{d}t}=\vec{v}\times m\vec{v}+\vec{r}\times\frac{\mathrm{d}\vec{p}}{\mathrm{d}t}=\vec{r}\times\frac{\mathrm{d}\vec{p}}{\mathrm{d}t}$$ Hence, from the previous equations $$\vec{r}\times\frac{mv^2}{r}\hat{r}=0=\frac{\mathrm{d}\vec{L}}{\mathrm{d}t}$$ Hence, for this case, $$\vec{L}=const$$