Why would a particle be able to change its angular speed subject to central potential?

Question

Given a central potential $V(r)$, for any particle subject to this potential, we have its angular momentum not changing with respect to time because $\frac{d\mathbf{L}}{dt}=\frac{dm}{dt}\mathbf{x}\times\mathbf{\dot x}+m\mathbf{\dot x}\times\mathbf{\dot x}+m\mathbf{x}\times\mathbf{\ddot x}$, where because of central potential, $\mathbf{\ddot x}$ is in the same direction as particle's position vector $\mathbf{x}$.

On the other hand, we can write equations of motion of the particle using polar coordinates (as $\mathbf{L}$ is constant, motion reduces to 2D and we use polar coordinates) to find that the magnitude of angular momentum per unit mass is constant.

$l=r^2\dot{\theta}$ is constant

Now we know that a central potential is able to exert a force only in radial direction, let us suppose the radial distance now increases from $r_1$ to $r_2$, the constant $l$ immediately implies that we have a change in angular speed of the particle $\dot{\theta}$.

A change in the angular speed $\dot{\theta}$ means at some time during the process of changing radial distance, the particle must have had an angular acceleration. But we know that the particle will not experience any force other than the radial force. What is going on?

Answer 1

The formula: $\tau = I\alpha$ is valid for a rigid body rotating around one of its axis of inertia. It is the derivative of $L= I\omega$.

But if the radius changes, the system is not a rigid body, so the notion of the dependency of torque for an angular acceleration is no longer valid.

In this case, only the definition of $\mathbf L = \mathbf r \times \mathbf p$ can be used. For a planar movement, $|\mathbf L| = |\mathbf r||\mathbf p|sin(\theta)$. The modulus of the angular moment can stay constant, while the 3 variables at the RHS of the equation change.

Answer 2

I think the main question is "how did that radius change?" Whatever force caused that to happen would be responsible for the alteration in angular speed and overall momentum

Answer 3

Few minutes after I posted this question I reminded myself of one simple fact, that a radial force vector is not a fixed vector in space, it changes with changing position vector of the particle. The unit radial direction itself changes with time, and its change would be in direction perpendicular to its original position, namely it changes in $\mathbf{\hat{\theta}}$ direction.

Answer 4

Even if an object's angular coordinate is accelerating ($\ddot{\theta} \neq 0$), this does not necessarily imply that the acceleration vector has an angular component. This is because the acceleration vector expressed in polar coordinates is $$ \ddot{\vec{r}} = (\ddot{r} - r \dot{\theta}^2) \hat{r} + (r \ddot{\theta} + 2 \dot{r} \dot{\theta} ) \hat{\theta}, $$ and so long as $2 \dot{r} \dot{\theta} + r \ddot{\theta} = 0$, we have $a_\theta = 0$ and $F_\theta = 0$.

In fact, it is worth noting that $$ \frac{d l}{dt} = \frac{d}{dt}(r^2 \dot{\theta}) = r^2 \ddot{\theta} + 2 r \dot{r} \dot{\theta} = r(r \ddot{\theta} + 2 \dot{r} \dot{\theta}). $$ This means that so long as $r \neq 0$, $dl/dt = 0$ if and only if $r \ddot{\theta} + 2 \dot{r} \dot{\theta} = 0$, which by the above argument is equivalent to saying that $F_\theta = 0$.