# Question on Angular momentum conservation for changing moment of inertia

Assume there is a rod of negligible mass which can rotate with one of its ends at centre(connected to rigid point). This rod has some regions where a heavy mass let say an iron piece of 1kg mass can be attached or locked.

At starting we attach this iron piece at 1metre distance from centre. And we give it a torque for some time and it gains an angular velocity and angular momentum.

Now the piece of iron is unlocked from its position and pushed or pulled inside and then locked again as it reaches 0.5 metre distance from center.

If angular momentum is conserved then from $$L = r×mv$$, we can say that since $$r$$ is reducing and $$m$$ is constant, $$v$$ must be increasing.

Why is $$v$$ increasing when there is no force component parallel to it. the only force we can see in system is perpendicular to tangential velocity. And a perpendicular force can't change magnitude of velocity. But to have $$L$$ constant,magnitude of $$v$$ must change.

Now it's your turn. Answer for why magnitude of velocity(tangential) changed. Where is that force that caused it to change?

• Imperatives (“Don’t call…”, “Answer…”) from question askers come across as rude. Commented Oct 21, 2022 at 17:48
• Why do you think that there is no force, parallel to the velocity? There is the reaction between the beam and the iron mass Commented Oct 26, 2022 at 12:20
• Ugh! Yes it is but i am talking about a force parallel to tangential velocity. reaction between beam and iron is radial and therefore perpendicular to tangent. Commented Oct 26, 2022 at 12:37
• I added an answer linking the difference in kinetic energy with the work done while moving the mass in the radial position. When you say "a perpendicular force can't change magnitude of the velocity", you're right, but the force and the velocity are not perpendicular while you're changing the radial position of the mass Commented Oct 27, 2022 at 12:30

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 $$t \propto \frac{1}{L}$$.

• $~r(t)=...~$ is not the solution ? of $~\ddot r=...~$
– Eli
Commented Oct 30, 2022 at 13:43
• @Eli - $r(t)$ is the solution of $\ddot{r}$. You can confirm this as $$\ddot{r} = \frac{m L^2 r_0^3}{(L^2 t^2 + m^2 r_0^4)^{3/2}}$$ Commented Oct 30, 2022 at 13:49

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.

• maybe you didn't get my question. I am not talking about why product of momentum and radius is same. I am asking why speed of mass changed. All forces in system are acting perpendicular to iron piece's direction of motion. And we know that speed should not change by this force. Commented Oct 26, 2022 at 3:18
• Your imagined scenario is unclear, as you have given mutually inconsistent descriptions of what happens. However, maybe this will help: just before the mass is locked down in its second position, does it have a nonzero radial velocity component? If so, a force is going to act radially and do work on the mass as it is locked down… Commented Oct 26, 2022 at 4:07
• well for moving it inward we need a force dirwcted toward center. And i also understand that this will do some work during displacememt. But I don't see where is that force that changes tangetial speed. The conversation seems useless. Maybe u are not getting point. Tell me exactly how tangential speed changes Commented Oct 26, 2022 at 4:49
• Now that you have added a radial force to your description, it is very obvious where the change in velocity comes from. As you pull it in, you are pulling inward as the mass moves inward, thus doing positive work and increasing the speed of the mass. It is no surprise that this speed can change from radial to tangential. Imagine firing a bullet 1 meter to the left of someone. The bullet is not being pushed on, yet its speed starts as radial as measured by the person toward whom it is hurtling, becomes tangential as it passes by, and then becomes radial again once it is receding behind. Commented Oct 26, 2022 at 10:42
• "It is no surprise that this speed can change from radial to tangential." Well it is surprising for me. The way i think there should be a force parallel to tangetial velocity to change it. You can't just say that there radial velocity is transferred to tangential velocity by no means. Also it is not necessary that the work done on iron might give it a radial velocity. This velocity can be infinitely small if u use a centripetal force just a little greater than centrifugal force giving a radial velocity nearly of 0 magnitude. The energy gets stored in form of potential energy. Commented Oct 26, 2022 at 12:47

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 $$\dot\theta_1$$

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

• You are thinking that object gets some radial kinetic energy. But you can do the work without giving any radial velocity(actually by using infinitely small velocity). So in this case we don't have any kinetic energy. The work is done against centrifugal force. And as we go closer to center, we get larger and larger centri. force. And energy can be stored as potential energy in beam just like a string stores energy. Obviously there is no need of a large radial velocity to put object closer or farther from centre. So here we don't have a radial velocity and so there is no extra kinetic energy Commented Oct 28, 2022 at 20:25
• no. sum an infinite number of infinitely small changes, and you get a finite displacement. If you perform infinitely slow displacement, the time you need $t_{fin}- t_{in}$ becomes infinitely large, and thus the interval of integration of the power to get the work. As you can see, the expression of the work doesn't depend on the time required to move the mass, but only on it's initial and final state, through the radial position and the angular velocity Commented Oct 28, 2022 at 20:30
• You have to read carefully the answer, and trust physics, and mathematics, especially when they agree with intuition. You don't own the truth, the physics neither, but for sure it holds better model than your intuition, if it's done the right way Commented Oct 28, 2022 at 20:31
• And i was asking about how the object gains large tangetial speed. To change any particular component of velocity we need a force component along it. Also i don't see any explanation for why radial velocity get converted to tangetial velocity if it is really required and really gets converted too. Commented Oct 28, 2022 at 20:33
• Someone else here already asked you not to be rude. Have a think that people try to give you a help for free here Commented Oct 28, 2022 at 20:33

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.

• massless rod would not provide any tangential force. It can only hold axial forces. You can convince yourself just writing the equations of motion of a massless rod (and thus w/o inertia) Commented Oct 30, 2022 at 9:40
• @basics You are right. The massless rod must have net zero force and net zero torque. That restricts it to axial forces. That indicates a mistake in my explanation. However I would like to think it through before editing or deleting my answer. Thanks! Commented Oct 30, 2022 at 10:25
• The first time I read this question I thought at tangential force as well, since I'm not so used to massless elements. But this is a mistake. I really don't like to tell, but take a look at my answer. As I've already told to the OP, I'm 100% sure about the explanation, including formulas (except for typos) and their interpretation. If you can't find there any flaw, I'll take it as another check Commented Oct 30, 2022 at 10:36

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.