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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?

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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.

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  • $\begingroup$ dont we consider the case where an axis passes thru the point of contact (an instantaneous axis of rotation) for a cylinder performing pure rolling on a surface $\endgroup$ – Schwarz Kugelblitz Jul 31 at 20:11
  • $\begingroup$ @SchwarzKugelblitz He was basically asking about how the center of mass is always the only point in a rotating object that moves with constant velocity. You are right that at a given instant the contact point on the cylinder is stationary and the cylinder rotates about this point, but as you continue watching this point, it traces out a curved path (a cycloid in fact). However, the center of the cylinder moves with constant velocity. $\endgroup$ – Brian Moths Aug 2 at 4:26
  • $\begingroup$ I see. Thanks for the reply! $\endgroup$ – Schwarz Kugelblitz Aug 2 at 12:32
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It happens I think because rotation about axes which do not pass through the center of mass are usually inherently unstable and hence the rotation tends to "decay" to a more stable axis, i.e the one passing through its center of mass. Also, the moment of inertia of a body is usually the lowest through a certain principle axis through its center of mass.(Moment of inertia is analogous to mass in rotation. i.e. more moment of inertia for a certain amount of force force implies lesser angular acceleration) A simple proof for this is:

The moment of inertia $I=mr^2$ for a point particle at a distance r from its axis of rotation. Now consider two particles separated by some distance $l$. Assume an axis of rotation passing through the line joining these both, at a distance $x$ from the first particle, and $l-x$ from the second. then the moment of inertia about this axis is $$I=Mx^2 + M(l-x)^2$$ Differentiating with respect to $x$, we get: $$\frac{dI}{dx}=2Mx-2Ml+2Mx$$ If you want to find the minimum value of x needed for the lowest of moment of inertia, equate the expression above to zero. This gives $I$ to be minimum at $$x=l/2$$ from the first particle which is the position of center of mass of the system. The moment of inertia is always lowest through an axis through the center of mass. (True for all rigid bodies.) and hence the most stable rotation is achieved through an axis through the center of mass.

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    $\begingroup$ The answer I accepted leads me to believe that rotation of an unhinged body about a point other than its centre of mass is impossible(because the velocity of the centre of mass would vary in the absence of an external force). Hence, I would think of rotation around axes other than the COM to be not just unstable but impossible. $\endgroup$ – jerry Dec 9 '14 at 14:57
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    $\begingroup$ Yes, but that is true only for objects which have no constraint on them. for example a door hinged would only rotate about the hinge when a force is applied to it. nonetheless, it is true in an isolated unconstrained object. $\endgroup$ – Hritik Narayan Dec 9 '14 at 15:01
  • $\begingroup$ Exactly. An unhinged body in other words. $\endgroup$ – jerry Dec 9 '14 at 15:02
  • $\begingroup$ Yes. I hope the answer helped though anyway. $\endgroup$ – Hritik Narayan Dec 9 '14 at 15:06
  • $\begingroup$ @HritikNarayan I was also searching for the answer to this question. In fact even I had thought that M.I about C.M should be least and the body should naturally rotate about a point of least M.I , just as in translation the body of greater mass wouldn't be affected as much as a body of less mass by the same force. I was just thinking how to prove it mathematically. And I found it here. Great Answer !! $\endgroup$ – Shashaank Sep 20 '16 at 20:25
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It does happen because, in the free condition only the Centre of mass is the point which can provide the necessary centripetal forces for the rotation of body.

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