newtonian mechanics - Why does an unhinged body rotate about its centre of mass?

If a force is applied to a body which does not act through its centre of mass, it rotates about its centre of mass and not any other point. Why?

Answer

The idea is that if there are no forces on an object, then no matter how it rotates, its center of mass must move at a constant velocity. Then in the frame of the object, the center of mass appears stationary and everything else rotates around it. In general this cannot be said of any other point in the object.

To see that the center of mass moves at a constant velocity, remember that the center of mass $\renewcommand{\r}{\mathbf{r}} \renewcommand{\rcm}{\r_\textrm{cm}} \rcm$ is defined by $\rcm = \frac{1}{M}\int \r \rho d \r$, where $\rho$ is the mass density distribution of the object and $M = \int \rho d \mathbf{r}$ is the total mass of the object. Then the center of mass velocity $\renewcommand{\a}{\mathbf{a}} \renewcommand{\acm}{\a_\textrm{cm}} \acm$ can be found by taking two time derivatives: $\acm = \frac{1}{M}\int \a \rho d \r$.

Now the integrand $\a(\r) \rho(\r)$, by newton's second law must be the total force $\renewcommand{\F}{\mathbf{F}}\F$ at the point $\r$. This force has two contributions: an external force $\renewcommand{\Fext}{\F_\mathrm{ext}}\Fext$ and an internal force $\renewcommand{\Fint}{\F_\mathrm{int}}\Fint$. Thus we have that $\a(\r) \rho(\r) = \F(\r) = \Fint(\r) + \Fext(\r)$. Plugging this back into our expression for $\acm$, we find $\acm = \frac{1}{M} \int \Fint(\r)+\Fext(\r) d \r = \frac{1}{M} \Fext + \frac{1}{M} \int \Fint(\r) d\r$, where $\Fext$ is the total external force.

Now the internal force comes from pairwise interactions with other parts of the object. So if $\Fint(\r, \r')$ is the force of the piece of the object at $\r'$ on the piece of the object at $\r$, then the total internal force at $\r$ is given by $\Fint(\r)=\int \Fint(\r,\r') d \r'$. Then the total contribution of $\Fint$ to $\acm$ can be written $\frac{1}{M} \int \Fint(\r) d\r = \frac{1}{M} \int \Fint(\r,\r') d\r' d\r$. But switching the order of integration, we find $\frac{1}{M} \int \Fint(\r,\r') d\r' d\r = \frac{1}{M} \int \Fint(\r,\r') d\r d\r'$ by Newton's third law we have $\Fint(\r,\r')=-\Fint(\r',\r)$. Combining this with the previous equation we find $\frac{1}{M} \int \Fint(\r,\r') d\r' d\r = -\frac{1}{M} \int \Fint(\r',\r) d\r d\r'$. But then relabelling the dummy variables on the right hand side, we find that $\frac{1}{M} \int \Fint(\r,\r') d\r' d\r = -\frac{1}{M} \int \Fint(\r,\r') d\r' d\r$. which implies that $\frac{1}{M} \int \Fint(\r,\r') d\r' d\r =0$. Therefore $\frac{1}{M} \int \Fint(\r) d\r =0$, and so $\acm = \frac{1}{M} \Fext $. Therefore if the external force is zero, the center of mass moves with constant velocity.