I have seen a lot of questions here which ask why a free rigid body always rotates about it's centre of mass. The answer in most cases is like a "thought experiment". First, we prove that when a force is applied to a rigid body, it behaves like a point object where the entire mass of the object is concentrated to one point called the "centre of mass". Then we transfer out attention to a co-ordinate system at the centre of mass (so that the centre of mass is at rest, relatively). Then we say that the definition of a rigid body is that the distance between the particles of the rigid body always remain constant. This means that the distance between the centre of mass, and any point in the rigid body also remains constant. So, the only possible motion of any point will be a circular path around the centre of mass : hence, the only possible motion of a rigid body about the centre of mass is a rotation. Also, since the distance between any points in the rigid body must be constant, particles inside the rigid body cannot rotate in opposite directions or different axes, as this would change the distances.

Now, I have also been taught this way. In school and university, even in our Dynamics text book (Meriam & Kraige), the concept of "rotation" and "moment" was just introduced..like its common sense. There was no "mathematical proof" that rotation is the motion around the centre of mass (CM). Rotation and translation are always treated differently, even though its taught that the net motion will be a sum of the two.

I have been wondering whether you can prove that the motion of a particle in a rigid body with respect to the centre of mass is a rotation. I have come up with a kind of half baked derivation below :

First, as always we consider a rigid body as a system of particles connected with massless rigid rods. For simplicity I have considered only the 2D case. In the figure below, I have considered a 3 particle system, with all relevant variables marked.

The red point is the centre of mass (CM) of the system. Here a force $\vec f$ is applied to mass $m_1$ which does not pass through the CM. So, this system would rotate.

To apply the principles of dynamics, we first isolate all masses and draw the free body diagram

Here $\vec f_{12}$ and $\vec f_{13}$ are the reaction forces on $m_1$ from $m_2$ and $m_3$. Applying newtons second law to $m_1$ we have $$\vec f + \vec f_{12} + \vec f_{13} = m_1\ddot{\vec r_1}$$

For mass $m_3$

we have $$\vec f_{31} + \vec f_{32} = m_3\ddot{\vec r_3}$$

and for mass $m_2$

we have $$\vec f_{21} + \vec f_{23} = m_2\ddot{\vec r_2}$$

Now adding up all the above equations and noting that $\vec f_{12}=-\vec f_{21}$ and $\vec f_{13}=-\vec f_{31}$ and $\vec f_{32}=-\vec f_{23}$, we have $$\vec f=m_1\ddot{\vec r_1}+m_2\ddot{\vec r_2}+m_3\ddot{\vec r_3}$$ Introducing the position of centre of mass as $$\vec r_{cm}=\frac{m_1\vec r_1+m_2\vec r_2+m_3\vec r_3}{m_1+m_2+m_3}$$ and differentiating $$\ddot{\vec r_{cm}}=\frac{m_1\ddot{\vec r_1}+m_2\ddot{\vec r_2}+m_3\ddot{\vec r_3}}{m_1+m_2+m_3}$$ Now we can substitute for $m_1\ddot{\vec r_1}+m_2\ddot{\vec r_2}+m_3\ddot{\vec r_3}$ in the dynamic equation to get $$\vec f=(m_1+m_2+m_3)\ddot{\vec r_{cm}}$$ This is nothing but the equation of motion of a point particle whose mass is $m_1+m_2+m_3$ situated at position $\vec r_{cm}$. Thus, the rigid body behaves like the entire mass is concentrated at the centre of mass. Now we turn our attention to the centre of mass co-ordinate system $x_{cm} - y_{cm}$. To do this we note that $\vec r_1=\vec r_{cm}+\vec r_{1c}$ and $\vec r_2=\vec r_{cm}+\vec r_{2c}$ and $\vec r_3=\vec r_{cm}+\vec r_{3c}$ Substituting for $\vec r_1$, $\vec r_2$ and $\vec r_3$ in the dynamic equation for each mass, we have $$\vec f + \vec f_{12} + \vec f_{13} - m_1\ddot{\vec r_{cm}}=m_1\ddot{\vec r_{1c}}\\\vec f_{31} + \vec f_{32} - m_3\ddot{\vec r_{cm}}= m_3\ddot{\vec r_{3c}}\\\vec f_{21} + \vec f_{23} - m_2\ddot{\vec r_{cm}}= m_2\ddot{\vec r_{2c}}$$ Again adding all of the above up, we have $$\vec f-(m_1+m_2+m_3)\ddot{\vec r_{cm}}=m_1\ddot{\vec r_{1c}}+m_2\ddot{\vec r_{2c}}+m_3\ddot{\vec r_{3c}}$$ Now we invoke the definition of rigid body. This means that the distance between any 2 masses is constant. This may be written for our case as $$\frac {d}{dt}\left(\vec r_{12}\cdot\vec r_{12}\right)=0$$ since the magnitude of the vector between any 2 masses is constant. However $\vec r_{12}=\vec r_{2c}-\vec r_{1c}$. So we have $$\frac {d}{dt}\left[\left(\vec r_{2c}-\vec r_{1c}\right)\cdot\left(\vec r_{2c}-\vec r_{1c}\right)\right]=0\\\frac {d}{dt}\left[{\vert\vec r_{2c}\vert}^2+{\vert\vec r_{1c}\vert}^2-2\vec r_{2c}\cdot\vec r_{1c}\right]=0$$ This essentially means that $$\frac {d}{dt}\left[\vec r_{2c}\cdot\vec r_{1c}\right]=0$$ Applying product rule $$\vec r_{2c}\cdot\dot{\vec r_{1c}}+\vec r_{1c}\cdot\dot{\vec r_{2c}}=0$$ Differentiating once again, $$\vec r_{2c}\cdot\ddot{\vec r_{1c}}+\vec r_{1c}\cdot\ddot{\vec r_{2c}}+2\dot{\vec r_{1c}}\cdot\dot{\vec r_{2c}}=0$$ Since the last term is a product of derivatives, we say that it is infinitesimally small, and ignore it. This gives $$\vec r_{2c}\cdot\ddot{\vec r_{1c}}+\vec r_{1c}\cdot\ddot{\vec r_{2c}}=0$$ Applying same treatment for $\vec r_{13}$, we have $$\vec r_{1c}\cdot\ddot{\vec r_{3c}}+\vec r_{3c}\cdot\ddot{\vec r_{1c}}=0$$ From the above 2 equations, we can write $$\ddot{\vec r_{2c}}=\frac{-\vec r_{1c}\cdot\vec r_{2c}\cdot\ddot{\vec r_{1c}}}{{\vert\vec r_{1c}\vert}^2}\\\ddot{\vec r_{3c}}=\frac{-\vec r_{1c}\cdot\vec r_{3c}\cdot\ddot{\vec r_{1c}}}{{\vert\vec r_{1c}\vert}^2}$$ Substituting for $\ddot{\vec r_{2c}}$ and $\ddot{\vec r_{3c}}$ in the summed up dynamic equation, we get $$\vec f-(m_1+m_2+m_3)\ddot{\vec r_{cm}}=m_1\ddot{\vec r_{1c}}+m_2\frac{-\vec r_{1c}\cdot\vec r_{2c}\cdot\ddot{\vec r_{1c}}}{{\vert\vec r_{1c}\vert}^2}+m_3\frac{-\vec r_{1c}\cdot\vec r_{3c}\cdot\ddot{\vec r_{1c}}}{{\vert\vec r_{1c}\vert}^2}$$ Now we focus on the term $(m_1+m_2+m_3)\ddot{\vec r_{cm}}$. From the definition of the centre of mass, we have $$(m_1+m_2+m_3)\ddot{\vec r_{cm}}=m_1\ddot{\vec r_1}+m_2\ddot{\vec r_2}+m_3\ddot{\vec r_3}$$ We will now proceed to invoke the rigid body condition the same way we did above, by noting that $\vec r_{12}=\vec r_2-\vec r_1$ and that $\vec r_{13}=\vec r_1-\vec r_3$. After applying the same treatment as above, we get $$\ddot{\vec r_2}=\frac{-\vec r_1\cdot\vec r_2\cdot\ddot{\vec r_1}}{{\vert\vec r_1\vert}^2}\\\ddot{\vec r_3}=\frac{-\vec r_1\cdot\vec r_3\cdot\ddot{\vec r_1}}{{\vert\vec r_1\vert}^2}$$ Substituting these into the centre of mass definition above, we have $$(m_1+m_2+m_3)\ddot{\vec r_{cm}}=m_1\ddot{\vec r_1}+m_2\frac{-\vec r_1\cdot\vec r_2\cdot\ddot{\vec r_1}}{{\vert\vec r_1\vert}^2}+m_3\frac{-\vec r_1\cdot\vec r_3\cdot\ddot{\vec r_1}}{{\vert\vec r_1\vert}^2}$$. Now, if we take the common term $\ddot{\vec r_1}$ apart, all the other terms on RHS are scalar products. So we may write $$(m_1+m_2+m_3)\ddot{\vec r_{cm}}=K_1\ddot{\vec r_1}$$ where $$K_1=\frac{m_1{\vert\vec r_1\vert}^2-m_2\vec r_1\cdot\vec r_2-m_3\vec r_1\cdot\vec r_3}{{\vert\vec r_1\vert}^2}$$ Now we make the observation $$\vec r_{2c}-\vec r_{1c}=\vec r_2-\vec r_1=\vec r_{12}$$Differentiating twice, we have $$\ddot{\vec r_{2c}}-\ddot{\vec r_{1c}}=\ddot{\vec r_2}-\ddot{\vec r_1}$$ Substituting for $\ddot{\vec r_{2c}}$ in terms of $\ddot{\vec r_{1c}}$ and $\ddot{\vec r_2}$ in terms of $\ddot{\vec r_1}$ as derived above, we have $$\frac{-\vec r_{1c}\cdot\vec r_{2c}\cdot\ddot{\vec r_{1c}}}{{\vert\vec r_{1c}\vert}^2}-\ddot{\vec r_{1c}}=\frac{-\vec r_1\cdot\vec r_2\cdot\ddot{\vec r_1}}{{\vert\vec r_1\vert}^2}-\ddot{\vec r_1}$$ Again, we can notice that after taking term $\ddot{\vec r_{1c}}$ on LHS as common and taking term $\ddot{\vec r_1}$ on RHS as common, what left inside the brackets will be a scalar term. So we write $$\ddot{\vec r_1}=K_2\ddot{\vec r_{1c}}$$ So finally we may write $$(m_1+m_2+m_3)\ddot{\vec r_{cm}}=K_3\ddot{\vec r_{1c}}$$ where $$K_3=K_1K_2$$ Now we can substitute for the term $(m_1+m_2+m_3)\ddot{\vec r_{cm}}$ in the summed dynamic equation which becomes $$\vec f-K_3\ddot{\vec r_{1c}}=m_1\ddot{\vec r_{1c}}+m_2\frac{-\vec r_{1c}\cdot\vec r_{2c}\cdot\ddot{\vec r_{1c}}}{{\vert\vec r_{1c}\vert}^2}+m_3\frac{-\vec r_{1c}\cdot\vec r_{3c}\cdot\ddot{\vec r_{1c}}}{{\vert\vec r_{1c}\vert}^2}$$ Now I am going to do what is called a "pro gamer move". Since scalar product is commutative, I will group the terms in RHS as $$\vec f-K_3\ddot{\vec r_{1c}}=m_1\ddot{\vec r_{1c}}+m_2\frac{-\vec r_{1c}\cdot\left(\vec r_{2c}\cdot\ddot{\vec r_{1c}}\right)}{{\vert\vec r_{1c}\vert}^2}+m_3\frac{-\vec r_{1c}\cdot\left(\vec r_{3c}\cdot\ddot{\vec r_{1c}}\right)}{{\vert\vec r_{1c}\vert}^2}$$ Now, the terms in the brackets are scalar products; which means that the second and third terms in the RHS are vectors in the direction of $\vec r_{1c}$ Now to remove those additional terms, I take a cross product with $\vec r_{1c}$ on both LHS and RHS. $$\vec r_{1c}\times\vec f-K_3\vec r_{1c}\times\ddot{\vec r_{1c}}=m_1\vec r_{1c}\times\ddot{\vec r_{1c}}+m_2\frac{-\vec r_{1c}\times\vec r_{1c}\cdot\left(\vec r_{2c}\cdot\ddot{\vec r_{1c}}\right)}{{\vert\vec r_{1c}\vert}^2}+m_3\frac{-\vec r_{1c}\times\vec r_{1c}\cdot\left(\vec r_{3c}\cdot\ddot{\vec r_{1c}}\right)}{{\vert\vec r_{1c}\vert}^2}$$ In this case, since the second and third terms in RHS before cross product where vectors in the direction of $\vec r_{1c}$, taking cross product means that these terms will be $0$. Thus finally we have $$\vec r_{1c}\times\vec f=\left(m_1+K_3\right)\left(\vec r_{1c}\times\ddot{\vec r_{1c}}\right)$$ Which is nothing but $$\tau=I\alpha$$ where I call the bracketed term as $I$ (moment of inertia). So I have obtained the moment equation in the centre of mass co-ordinate system. I have the following questions :

1. Even though I set out to prove that the motion of $m_1$ will be circular, I didn't quite reach there. Does the moment equation prove that $m_1$ will have circular motion ?
2. Is what I have done correct ?