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I have the following doubt: suppose the center of mass frame itself of a system of particles is revolving around a given axis in a fixed reference frame with origin $O$ (let the origins of the two system be coincident for simplicity). Let's call this angular velocity $\vec{\Omega}$. Now let's calculate the angular momentum $\vec{L}$ of the system of particle in the fixed frame with respect to the origin $O$. We have (the primed quantities refer to the center of mass frame):

$$\vec{L}=\sum_{i} m_i (\vec{r_{CM}}+\vec{r_i^{\prime}})\times(\vec{\Omega}\times(\vec{r_{CM}}+\vec{r_i^{\prime}})+\vec{v_i^{\prime}})$$ Now expanding the product it looks to me that won't happen that many terms simplify (such as in the case of rectilinear center of mass motion) and thus it appears to me that König theorem for angular momentum (as stated e.g. here https://en.wikipedia.org/wiki/K%C3%B6nig%27s_theorem_(kinetics)) doesn't work in this case. In other words to find $\vec{L}$ we can't just add the two contribution of the center of mass seen as a particle and the angular momentum in the center of mass frame with respect to the center of mass. But if this were the case, then wouldn't it be wrong to find for example the angular momentum of the Earth with respect to the center of the sun (in the sun frame) as the sum of the angular momentum of revolution ($r_{Orbit}M_{Earth}v_{Orbit}$) plus the angular momentum due to rotation ($I_{cm}\omega_{rotation}$)? In this case the center of mass of the Earth is rotating, but if König theorem doesn't apply in this case then this reasoning should be wrong, isn't it?

Please help me clarify these doubts on angular momentum as I find it sometimes very confusing since the majority of textbooks look to me very sloppy on this.

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  • $\begingroup$ The expression for $\vec{L}$ that you have written down seems to be incorrect. Look at the wikipedia page that you have cited. $\endgroup$ – abir Jul 5 at 14:58
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Your CM is revolving around a given axis in a circular motion, but each particle isn't just doing a circular motion, so you can't apply that formula for its velocity. The particles are moving around the CM in some arbitrary fashion. The velocity of each particle, as seen by your observer $O$, will be $\vec{v}_i=\vec{V}^{}_{CM}+\vec{v}_i'$, where $\vec{V}^{}_{CM}=\vec{\Omega}\times \vec{r}^{}_{CM}$, as your CM is in a circular motion around your axis. $\vec{v}_i'$ will be completely unrelated as to what the CM is doing.

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