Coriolis force and conservation of angular momentum

Question

I'm trying to understand the relations between the existance of Coriolis force and the conservation of angular momentum. I found this example on Morin, which confuses me.

A carousel rotates counterclockwise with constant angular speed $ω$. Consider someone walking radially inward on the carousel (imagine a radial line painted on the carousel; the person walks along this line), at speed $v$ with respect to the carousel, at radius $r$. [...]

Take $d/dt $ of $L = mr^2ω$, where $ω$ is the person’s angular speed with respect to the lab frame, which is also the carousel’s angular speed. Using $dr/dt =− v$, we have $$dL/dt =− 2mrωv+mr^2(dω/dt)\tag{1}$$

What if the person doesn’t apply a tangential friction force at his feet?

Then the Coriolis force of $2mωv$ produces a tangential acceleration of $2ωv$ in the rotating frame, and hence also in the lab frame (initially, before the direction of the motion in the rotating frame has a chance to change), because the frames are related by a constant $ω$. This acceleration exists essentially to keep the person’s angular momentum (with respect to the lab frame) constant. [...] To see that this tangential acceleration is consistent with conservation of angular momentum, set $dL/dt = 0$ in Eq. (1) to obtain $2ωv = r(dω/dt)$ (this is the person’s $ω$ here, which is changing). The right-hand side of this is by definition the tangential acceleration. Therefore, saying that $L$ is conserved is the same as saying that $2ωv$ is the tangential acceleration (for this situation where the inward radial speed is $v$).

There is no friction force acting here, nor any other real force therefore the angular momentum of the person does not change.

In my view the motion of the person, seen in lab frame, would be linear motion, because, at the beginning, the person has a tangential velocity $\omega r$ and radial velocity $v$, and he will keep these two forever. But then does it makes sense to talk about conservation of angular momentum? I mean it will surely be conserved in the lab frame but the motion is on a straight line (as far as I can see).

The two highlighted parts in text are the most confusing to me.

How is acceleration produced in the lab frame initially? (Coriolis does not act there)

It seems that Coriolis force is there to keep the angular momentum of the person constant in the lab frame. But this cannot be true since Coriolis force is a fictitious force, existing only in the rotating frame. I don't see clearly the link between Coriolis force and conservation of angular momentum in this example.

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So firstly will the angular momentum of the person (who will move on a straight line in the lab) be conserved in the lab frame?

Secondly can anyone give some further explanations about the links between Coriolis force and conservation of angular momentum in this example?

Answer 1

This is indeed confusing. The confusion comes from this very peculiar hypothesis:

What if the person doesn’t apply a tangential friction force at his feet?

It implies there is a radial contact force at the person's feet (I prefer "contact" to "friction", which refers to movement). And, indeed, for the person to move radially inwards, or even to stay immobile in the carousel, they need to at least counterbalance centrifugal acceleration.

So let's imagine how the person could be "frictionless" tangentially yet "frictionful" radially: suppose there are slippery concentric rails all over the carousel, on which the person can lean to move radially, but which prevent them to control rotational speed.

Suppose the person starts immobile with respect to the rail of radius $r$ on which they stand. When the person steps inwards, they undergo the said tangential Coriolis acceleration, which makes them start to glide counterclockwise along the inner rail of radius $r-δr$ on which they now stand, at $δω$ with respect to the carousel. Their rotational speed with respect to the lab is now $ω+δω$, and $δω$ is such that their angular moment has not changed: $rω=(r-δr)(ω+δω)$.

Answer 2

In my view the motion of the person, seen in lab frame, would be linear motion, because, at the beginning, the person has a tangential velocity $ωr$ and radial velocity $v$, and he will keep these two forever. But then does it makes sense to talk about conservation of angular momentum? I mean it will surely be conserved in the lab frame but the motion is on a straight line (as far as I can see).

It does make sense! Remember, to talk about angular momentum we do not necessarily need to talk about rotational motion. Consider a particle in free fall near the Earth, with ititial position $(0,d,h)$ and zero velocity. By the formula $\vec L=\vec r\times\vec p$ one can see it has vanishing angular momentum with respect to any point on the line $(0,d,z)$, but non vanishing angular momentum with respect to the origin. The latter is $$\vec L=m(d\hat j+z\hat k)\times\left(-\sqrt{2g(h-z)} \hat k\right)=-md\sqrt{2g(h-z)}\hat i.$$ Notice that the angular momentum is not conserved through the fall, even though the motion is on a straight line. There is torque about the origin.

Going back to your example, the angular momentum is conserved because there are no torque on the particle and not because it is on a straight line.

Answer 3

After reading your question again I gave a try proving why and how angular momentum is conserved. In the example given by the question one needs to understand how the angular momentum of an object moving in a rotating, non-inertial frame is conserved. I will present in brief, my effort of proving how and why the angular momentum is an integral of motion. The first paragraph below is given for completeness and one may skip to the second, where the problem consider here arises.

Firstly I tried to prove that the angular momentum will be constant with time when the angular velocity of the rotating system is constant. That is, if r,u are the polar coordinates with z=0, then $u=ωt$. By having no potential at all, we build the Lagrangian function:

$$L=T(kinetic energy) = (½) [ \dot r^2 + (r \dot u )^2 ] .$$ One then finds that the angular momentum is $p_u = { \partial L \over \partial \dot u} =r^2 \dot u $. Construct the Hamiltonian function $H=p_r \dot r +p_u \dot u -L $ to find $H=p_r ^2 + {p_u ^2 \over 2r^2} ={ \dot r^2 \over 2} + {r^2 \dot u^2 \over 2} $. Then, obviously, ${\partial H \over \partial u} =0 = \dot p_u $. So, the angular momentum is a constant of motion.

Secondly, and in relationship with your question, I tried proving the same for a non-constant angular velocity, that is $\ddot u \ne 0 $. What one can see is that the the functions (L and H), will be of the same form but with a difference: $ \dot u= \dot u(t) .$ Anyway, if the functions are of the same form, then:

$$ H=p_r ^2 + {p_u ^2 \over 2r^2} ={ \dot r^2 \over 2} + {r^2 \dot u^2 \over 2}, $$

and again, we see that

$${\partial H \over \partial u} = \dot p_u =0 ,$$ that is, the angular momentum is a constant of motion. Then, by taking the time derivative of $p_u$ we have:

$$(d/dt) p_u= \dot p_u ={d \over dt} (r^2 \dot u) = 2r \dot r \dot u +r^2 \ddot u .$$ This is the same result that your example gives. But now, we have proven that the angular momentum is a constant of motion, so $ \dot p_u =0$ and we arrive to the equation given by your example:

$$2r \dot r \dot u =-r^2 \ddot u .$$

But now, what about the inertial frame of reference? We understand that the motion will be a line, but that's something we didn't use in the previous analysis, so why use it here? What we can imagine, to understand the problem better is the following image:

This shows us (I believe...) that both the distance r and the angle u may change during the motion, something that comes closely to what an other answer here, to your post is arguing: that the angular momentum may not be by tautology zero. So, by the same exactly procedure as before, working in poolar coordinate, we find the Hamiltonian. We obviously conclude that the angular momentum will be zero. If we take that the particle moves in path, so that $ \dot u =0$, that is that the angle of the particle is constant, then we see that the Lagrange function is: $L= (½) \dot r^2 $. There is a zero-dependence from the angle, whatsoever.

Hope all this helps. What we see, is that, indeed, even in the inertial frame we can talk about angular momentum that depend solely on the trajectory (linear) of the particle in respect with the origin. In the case of the rotating system, this fact is being translated as a Coriolis force, a way to see for the observer there the conservation of angular momentum. If the angle is constant, we take what you argue as obvious in you example: conservation is a tautology. On all the other cases, we indeed find that conservation implies an acceleration to the system.