Why is angular momentum measured at a non inertial CoM conserved?

[Leo Liu]

Homework Statement

Merry

Relevant Equations

Christmas

Question 7.6

Official solution

It seems that the solution uses the conservation of angular momentum to solve this question (τ=0). But the problem is that the frame is set on the centre of mass of the guy, which is non inertial. I would like to know why it is correct to do it this way. My guess is that the author implicitly uses the Konig theorem and measures L of the system at the fixed centre of the circle, and since the force is central, Ls of both CM and the system remain constant. Therefore, L measured at CM is constant ⟹ τ=0. Is my inference correct?

Happy holiday .

 

[PeroK]

Homework Statement:: Merry
Relevant Equations:: Christmas

Question 7.6
View attachment 275090
Official solution
View attachment 275089

It seems that the solution uses the conservation of angular momentum to solve this question (τ=0). But the problem is that the frame is set on the centre of mass of the guy, which is non inertial. I would like to know why it is correct to do it this way. My guess is that the author implicitly uses the Konig theorem and measured L of the system at the fixed centre of the circle, and since the force is central, Ls of both CM and the system remain constant. Therefore, L measured at CM is constant ⟹ τ=0. Is my inference correct?

Happy holiday .

The CM of a rigid body moves according to the net force on the body; and the body rotates according to the torque about its CM.

 

[Leo Liu]

The CM of a rigid body moves according to the net force on the body; and the body rotates according to the torque about its CM.

I understand this. So you have used the Konig theorem, right?
Thanks.

 

[PeroK]

So you have used the Konig theorem, right?

Not explicitly. But, unless those torques are balanced then the man will start to fall over.

 

[haruspex]

It seems that the solution uses the conservation of angular momentum to solve this question...
...measures L of the system at the fixed centre of the circle

I see no reference to angular momentum. It only analyses torques and forces.
The L in the solution is the height of the CoM, not an angular momentum.

 

[etotheipi]

Essentially, the solution manual specifies that they choose a coordinate system with origin at the centre of mass, but they don't specify whether the basis vectors are time dependent and rotate with the cart (in which case the coordinate system is both translating and rotating, and as viewed by this coordinate system he's completely at rest), or if the basis vectors are fixed w.r.t. the ground frame (in which case the coordinate system is just translating, and as viewed by this coordinate system he's just rotating about his own vertical axis). Either choice is fine to begin analysing, but it's worth making that distinction because it does affect which inertial forces you need to include in your analysis.

Luckily, the distinction isn't actually isn't that important here, because inertial forces act through the centre of mass and by extension have zero torque about the centre of mass, so you can basically ignore their effects here when analysing torques.

To get to the point, whichever of these two non-inertial frames you eventually choose, you'll find the man's angular velocity vector is constant, and thus his angular momentum in this coordinate system is constant. So the torque acting on him as measured by such a coordinate system, in this case with origin at the centre of mass, must be zero.

 

[JD_PM]

This is my understanding of the provided solution to the problem.

Fist off, they chose an inertial frame of an external observer (that is, for instance, the frame of someone at rest looking at the man on the merry-go-round system)

We can indeed apply the first part of König's theorem

\begin{equation*}
\vec L_{\text{sys}} = \vec L_{\text{COM}} + \vec L'
\end{equation*}

Where $\vec L_{\text{COM}}$ is the angular momentum of the COM and $\vec L'$ is the angular momentum with respect to the COM.

Taking the time-derivative both sides we get

\begin{equation*}
\vec \tau_{\text{sys}} = \vec \tau_{\text{COM}} + \vec \tau'
\end{equation*}

Next, as etotheipi noted, they picked the COM as the origin of the system. Thus $\tau_{\text{COM}}= 0$.

Assuming that the man does not fall over means that the torque contributions sketched in the solution manual cancel out (i.e. $\tau_{\text{sys}}= 0$), which leads to the solution's manual equation

\begin{equation*}
\tau' = 0
\end{equation*}

Let me stress that the chosen frame is that of an external observer, and not that of the rotating man. If we had picked out the latter we would have encountered fictitious forces.

 

[kuruman]

If we had picked out the latter we would have encountered fictitious forces.

Indeed. In the non-inertial frame there would be a fictitious centrifugal force $F_{\text{cf}}=\dfrac{Mv^2}{R}$ acting on the CoM. The vertical force balance equation would be
$$N_i+N_o-Mg=0.$$If we take as origin the point on the platform directly below the CoM, the torque equation is $$-N_i \frac{d}{2}+ N_o \frac{d}{2}-\frac{Mv^2}{R}L=0.$$These are identical to the ones provided in the solution. Note that in this frame there is no need to bring in the horizontal friction equation in order to answer the question.