Question on Angular momentum conservation for changing moment of inertia

Take a free sliding bead on a perfectly rigid massless rod that is pivoted on one end. The two degrees of freedom here are $r$ the distance from the pivot to the bead and $\theta$ the angular position of the rod.

A torque is applied on the pivot (or not).

Define the equations of motion along the radial and tangential direction and you will find the following two expressions

$$ \begin{aligned} F_r & = m \left( \ddot{r} - r \dot{\theta}^2 \right) \\ F_n & = m \left( r \ddot{\theta} + 2 \dot{r} \dot{\theta} \right) \end{aligned} $$

Now if the torque on the pivot is known (zero or non-zero) then you can relate it to the tangential force as such $\tau = r\,F_n$.

Together with the equations of motion yields the following accelerations for the two DOF variables

$$ \begin{aligned} \ddot{r} & = \frac{F_r}{m} + r \dot{\theta}^2 \\ \ddot{\theta} & = \frac{\tau}{m r^2} - \frac{ 2\, \dot{r} \dot{\theta}}{r} \end{aligned}$$

Also note the speed at any point being $v = \sqrt{ \dot{r}^2 + r^2 \dot{\theta}^2 }$

Take angular momentum $L = r ( r \dot{\theta})$ and its first derivative $\dot{L} = m r \left( r \ddot{\theta} + 2 \dot{r} \dot{\theta} \right)$ and notice that using $\ddot{\theta}$ from above you have

$$ \dot{L} = \tau $$

or angular momentum is conserved when no torque is applied.

The change in speed is evaluated with the following differentiation $$ \dot{v} = \frac{ \dot{r} \ddot{r} + r^2 \dot{\theta} \ddot{\theta} + r \dot{r} \dot{\theta}^2}{\sqrt{\dot{r}^2 + r^2 \dot{\theta}^2}} = \frac{F_r \dot{r} + \tau \dot{\theta}}{m v} = \frac{\text{power}}{\text{momentum}} $$

Which states that with no power added to the system ($F_r = 0$ and $\tau = 0$) speed remains a constant.

Taking a look at speed again, but use $L = m r^2 \dot{\theta}$ as a constant to get $$v = \sqrt{ \dot{r}^2 + \frac{L}{m^2 r^2} }$$

and knowing that $\ddot{r} = r \dot{\theta}^2 > 0$ at all times, it means that $\dot{r}>0$ at all times. So $r$ increases all the time. Also, because $1/r^2$ decreases all the time, $\dot{r}$ should increase to keep $v$ constant.

Solving $\dot{v} =0$ for $\ddot{r}$ yields the solution

$$ \ddot{r} = \frac{L}{m^2 r^3} $$

The above is sufficient to fully solve the problem as

$$ r(t) = \sqrt{ r_0^2 + \frac{ L^2 t^2}{m^2 r_0^2} } $$

where $r_0$ is the initial radius at $t=0$. See how angular momentum $L$ and $t$ are combined here. The time needed to reach a set radius is $ t \propto \frac{1}{L}$.

There are at least 2 major issues with your logic here:

1. Why do you assume the mass will approach the center and reach a smaller radius? (it won’t, given the initial conditions you stipulate)

2. Remember that momentum and displacement are vectors! So their cross product could remain constant while only one of their magnitudes changes…

There is no paradox to be found here. Carefully consider point number 2, and you will see why. Review the definition of a cross product if necessary.

There are several detailed answers here, which I haven't followed, but I think there is a simpler, qualitative answer to your question. You ask why 𝑣 increases if there is no force component parallel to it but there is such a component. As the mass is being pushed in, the radius of its motion about the pivot point is decreasing - it's not purely radial motion - and the force has a component along the direction of the tangential velocity (dotted line in diagram):

If I got your question right, I can translate it as "why does the kinetic energy of the mass increase, in the conditions described where angular momentum is conserved, since I applied only radial forces to the mass?"

The answer is that, when you move the mass from the initial radial position, $R_0$, to the final radial position $R_1$, you're doing work $W$ on the mass since the velocity of the mass has non-zero radial component, and so the kinetic energy changes,

$\Delta K = K_1 - K_0 = W$.

Details

The dynamical equation of the mass can be projected in radial and azimutal directions

$\hat{r}: m( \ddot r - r \dot\theta^2 ) = F_r$
$\hat{\theta}: m( 2 \dot{r}\dot{\theta} + r \ddot{\theta} ) = F_{\theta}$

being $F_r$, $F_{\theta}$ the components of the reaction with the rod, and whatever keeps the mass fixed or make it move. Before going on, if the rod is massless, $F_{\theta} = 0$ from the dynamical equation of the rod.

The second equation reduces to the conservation of angular momentum

$0 = 2 \dot{r} \dot{\theta} + r \ddot{\theta}$$\qquad \text{for $r \ne 0$} \rightarrow \qquad$$ 0 = r \left( 2 \dot{r} \dot{\theta} + r \ddot{\theta} \right) = \dfrac{d}{dt} \left( r^2 \dot\theta \right)$$\quad \rightarrow \quad$ $\Gamma = m r^2(t) \dot \theta(t) = m r_1^2 \dot\theta_1 = m r_0^2 \dot\theta_0$

so that

$\dot\theta(t) = \dfrac{r_0^2}{r^2(t)} \dot\theta_0$

Knowing the law of motion $r(t)$ describing the change of the radial position of the mass, starting and ending at rest $\dot{r}(t_A) = \dot{r}(t_B) = 0$, we can find:

• the force you need to prescribe that motion

  $\mathbf{F} = m \left[ \ddot r(t) - r(t) \dot\theta(t)^2 \right] \mathbf{\hat{r}} = m \left[ \ddot r(t) - r(t) \left(\dfrac{r_0}{r(t)}\right)^4\dot\theta^2_0 \right] \mathbf{\hat{r}}$

• the power of that force

  $P(t) = \mathbf{F} \cdot \mathbf{v} = m \left[ \ddot r(t) - r(t) \left(\dfrac{r_0}{r(t)}\right)^4\dot\theta^2_0 \right] \mathbf{\hat{r}} \cdot \left( \dot r \mathbf{\hat{r}} + r \dot\theta \mathbf{\hat{\theta}} \right) = $
  $\qquad \qquad \qquad = m \left[ \ddot r(t) - r(t) \left(\dfrac{r_0}{r(t)}\right)^4\dot\theta^2_0 \right] \dot r = $
  $\qquad \qquad \qquad = \dfrac{d}{dt} \left( \dfrac{1}{2} m \dot r^2(t) + \dfrac{1}{2} m \dfrac{r_0^4}{r^2(t)} \dot\theta^2_0 \right)$

• and the work done, from starting conditions with radial position $r_0$ and angular velocity $\dot\theta_0$ to final radial position $r_1$ and angular velocity $\dot\theta_1$

  $W = \displaystyle \int_{t_{in}}^{t_{fin}} P(t) dt = \left[ \dfrac{1}{2} m \dot r^2(t) + \dfrac{1}{2} m \dfrac{r_0^4}{r^2(t)} \dot\theta^2_0 \right]\bigg|_{t_{in}}^{t_{fin}} =$
  $\qquad \qquad \qquad \qquad = \underbrace{\dfrac{1}{2} m \dot r_1^2 - \dfrac{1}{2} m \dot r_0^2}_{\dot{r}_{in, fin} = 0} + \dfrac{1}{2} m \dfrac{r_0^4}{r_1^2} \dot\theta^2_0 - \dfrac{1}{2} m r_0^2 \dot\theta^2_0 = $
  $\qquad \qquad \qquad \qquad = \dfrac{1}{2} m r_0^2 \dot \theta_0 \left( \underbrace{\dfrac{r_0^2}{r_1^2}\dot\theta_0}_{=\dot\theta_1} - \dot\theta \right) = \dfrac{1}{2} \Gamma (\dot\theta_1 - \dot\theta_0) = \dfrac{1}{2} \Gamma \left[\left(\dfrac{r_0}{r_1}\right)^2 - 1\right]\dot\theta_0$
  $\qquad \qquad \qquad \qquad = \dfrac{1}{2} m r_1^2 \dot \theta^2_1 - \dfrac{1}{2} m r_0^2 \dot \theta^2_0 = \dfrac{1}{2} m v_{\theta,1}^2 - \dfrac{1}{2} m v_{\theta,0}^2 = K_1 - K_0$

There are two ways to analyze the system. In both ways, you are not considering all forces acting on your system.

Consider the block alone as the system. Apart from the force pulling it (or pushing it) along radial direction, it is connected to rod. The rod would provide the necessary tangential force to the block.

Consider the block and rod system as your system. This system is connected to the hinge. The hinge would provide the required force to accelerate center of mass of the system.

Lets rotate the rod with constant angular velocity $~\omega~$ instead of applying a torque, which give you the same "situation"

the position vector to the mass is: \begin{align*} &\vec{R}(\tau)= \left[ \begin {array}{ccc} \cos \left( \omega\,\tau \right) &-\sin \left( \omega\,\tau \right) &0\\ \sin \left( \omega \,\tau \right) &\cos \left( \omega\,\tau \right) &0 \\ 0&0&1\end {array} \right]\, r\,\begin{bmatrix} \sin(\varphi) \\ -\cos(\varphi) \\ 0 \\ \end{bmatrix}\quad,\text{the velocity}\\\\ &\vec{v}(\tau)=\left[ \begin {array}{c} {\dot{r}}\,\sin \left( \omega\,\tau+\varphi \right) + \left( r\,\dot{\varphi} +r\omega \right) \cos \left( \omega\,\tau+ \varphi \right) \\ \left( r\,\dot{\varphi} +r\omega \right) \sin \left( \omega\,\tau+\varphi \right) -{\dot{r}}\,\cos \left( \omega\,\tau+\varphi \right) \\ 0 \end {array} \right]\\ \end{align*} with $~T=\frac{m}{2}\,\vec{v}\cdot\vec{v}~$ you obtain the EOM's \begin{align*} &m{\ddot r}-mr{\dot\varphi }^{2}-mr{\omega}^{2}-2\,mr \dot\varphi \,\omega=0\quad\quad (1)\\ &m{r}^{2}\ddot \varphi+2\,{\dot{r}}\,mr \dot\varphi +2\,{\dot{r}}\,mr\omega =0\quad,\text{or}\quad\frac{d}{d\tau}\left[ m\,\underbrace{\,r^2\,(\dot\varphi+\omega)}_{=L}\right]=0 \end{align*} the angular momentum \begin{align*} \vec{L}=\vec{R}\times\,m\vec{v}= m\, L\,\begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix}\quad,\text{is conserved. } \end{align*} the magnitude of the velocity $~\vec v$ \begin{align*} &v^2= \vec{v}\cdot\vec{v}=r^2\,(\omega+\dot{\varphi})^2+\dot{r}^2= \dot{r}^2+\left(\frac{L}{r}\right)^2\\ &v(\tau)=\sqrt{(\dot{r}(\tau))^2+\left(\frac{L}{r(\tau)}\right)^2} \end{align*} thus $~v(\tau)~$ is increasing because the tangential velocity $~\dot{r}(\tau)~$ is increasing and $~\dot{r}(\tau)~$ is increasing because the centrifugal force $~m\,\,\omega^2\,r~$ in equation (1).

The way I understand what you are asking is why the force in the $\theta$ direction $F_{\theta }$, which is zero is not the same as m $\frac{dv_\theta}{dt}$ which is increasing as the orbital radius $r$ decreases.

Let's look at a mass on a frictional surface with a hole in the center. The mass is connected to a string which extends down through the hole and is undergoing circular motion. We pull on the string below the surface to reduce the radius. In this way there is no confusion about a rod producing any torque on the mass, but angular momentum is still conserved and $v_\theta$ still increases.

The total acceleration of the system is still radially inward toward the center, just as is the force so that $F=m\ a$ for the system. For circular motion with constant radius, the speed is constant, but $v$ is accelerating radially inward due to its direction change and the fact that $v$ is a vector. There are physics equations that do not hold for angular coordinate systems. They only hold for Cartesian coordinates. The electric vector potential is one of these as pointed out by J. D. Jackson in his Electrodynamics book. In our case the situation is a little more complicated, since the $x$ and $y$ coordinates for $F$, $v$ and $a$ change with $\theta$ and therefore with time. If you assign the rate at which $r$ decreases with time, or the magnitude of $F$ that decreases $r$, you could theoretically compute, at least numerically, the forces, accelerations, and velocities in the $x$ and $y$ directions. This would show that $F_x=m\ a_x$ and the same for $y$ components.

The forces you are looking for are called the Coriolis and Euler force. Like the centrifugal force, they are described as fictitious forces that can be used to apply Newton's laws in non-inertial frames of reference such as circular motion.

In a linear (inertial) coordinate system, we are used to being able to decompose forces along axes independently. For example, $F=m \cdot a$ might become :

$$ \begin{aligned} F_x & = m \cdot a_x = m \ddot{x} \\ F_y & = m \cdot a_y = m \ddot{y} \end{aligned} $$

In rotating systems, however, things are more complicated.

For the radial part, you probably know that we cannot simply write $F_r = m \ddot{r}$, because even in the absence of any actual radial movement ($\ddot{r}=0$) there is still a force on the system which we call the centrifugal force. Mechanical engineers write this force as $F_{\it\text{Centrifugal}} = m{\omega^2}{r} $, physicists will prefer using $\dot{\theta}$, yielding :

$$ \begin{aligned} F_r & = F_{\ddot{r}}+F_{\it\text{Centrifugal}}\\ & = m \ddot{r} - m{\dot\theta^2}{r} \end{aligned} $$

But what about the angular part? In problems concerning uniform circular motion this doesn't play a role, but in the problem at hand we need to write out the precise terms:

$$ \begin{aligned} F_{\theta} & = F_{\it\text{Euler}}+F_{\it\text{Coriolis}}\\ & = mr \ddot{\theta} + 2 m\dot{r}\dot{\theta} \end{aligned} $$

Here we can separate the tangential force into a term related to the increase in angular speed ($\ddot{\theta}$) called the Euler force and another term called the Coriolis force, which depend on $r$ and $\dot{r}$ respectively. When you change $r$ by pulling your mass inwards, these terms become non-zero even though the external angular force $F_{\theta}$ remains zero. The interplay of these two tangential forces gives you the tangential acceleration.