Why is moment of inertia minimum about centre of mass of any rigid body?

  • $\begingroup$ because that is where integral for $I$ is minimal. Proof is based on variation principle, I suppose. Intuition can tell you that if you stay in centre of mass, forces acting from elementary volumes of the body will equilibrate each other, hence net momentum will be lowest. $\endgroup$ – aaaaaa Mar 23 '15 at 7:03
  • $\begingroup$ @aandreev, you just wrote "because that is where integral for I is minimal"... that statement is just the question the OP asked, but restated without the question mark... $\endgroup$ – hft Mar 23 '15 at 7:22
  • 2
    $\begingroup$ It's a consequence of the parallel axis theorem. Read the wikipedia article on the parallel axis theorem. The answer to your question, along with a simple derivation, is right there. Come back if you have additional questions. $\endgroup$ – David Hammen Mar 23 '15 at 8:37
  • $\begingroup$ @David Great reference. But I still don't know what is wrong with "compute the integral" answer. It is like asking why 2+2 is 4 $\endgroup$ – aaaaaa Mar 23 '15 at 13:52
  • $\begingroup$ @aandreev Do you have a reference which has the proof? $\endgroup$ – user74261 May 11 '15 at 3:59

Well, to make things more explicit, let's take a generalised case. Suppose there is a frame of reference $A$ in which the motion of a rigid body(made up of many particles) is pure rotation about a fixed axis. Now if the frame $A$ is non inertial, then we cannot just hope $\Gamma ^ {ext} = I \alpha$ to hold, as you may know that the above is derived using $F=ma$ which hold only for inertial frame. But now since the frame is non-inertial, we have to apply a $pseudo$ force $-m \vec{a}$. This pseudo force produces a pseudo torque about the axis.

As a special case, let us calculate the net torque about the centre of mass of the body.

Take the origin at the C.O.M. The total torque of the pseudo forces is, $$\sum \vec{r}_i \times (-m_i \vec{a}) = -\big(\sum m_i \vec{r_i}\big) \times \vec{a} = -M\bigg(\frac{\sum m_i \vec{r_i}}{M}\bigg) \times \vec{a}$$

where $\vec{r_i}$ is the position vector of the $i$th particle as measured from the COM.

But $\dfrac{\sum m_i \vec{r_i}}{M}$ is the position vector of the COM and that is $zero$ as the COM is at the origin. Hence pseudo torque is zero and the net torque comes out to be the same as in the inertial frame, $\Gamma^{ext}=I\alpha$.

So we can conclude that while calculating the net torque about the centre of mass, the pseudo torque comes out to be $zero$ and net torque remains $I\alpha$ while in all the other cases, net torque equals $I\alpha$+ torque due to pseudo forces, hence for any rigid body net torque is minimum about the centre of mass.

Hope it answers your question.

  • $\begingroup$ "hence for any rigid body net torque is minimum about the centre of mass." but he asked "hence for any rigid body net moment of inertia is minimum about the centre of mass."? $\endgroup$ – RE60K Mar 24 '15 at 14:07
  • $\begingroup$ I presume he can relate the torque and the moment of inertia, if he wants to know the physical or intuitive meaning he can look at the answers below. I derived it mathematically. $\endgroup$ – ritvik1512 Mar 24 '15 at 14:23
  • $\begingroup$ he can relate the torque and the moment of inertia, how ? i don't think so. can you explain it to me? $\endgroup$ – RE60K Mar 24 '15 at 14:24
  • $\begingroup$ Sure, but I don't think this is belongs here in the comments, you can drop a email to me if you want. :) $\endgroup$ – ritvik1512 Mar 24 '15 at 14:26
  • $\begingroup$ no don't bother much just give me a little small idea $\endgroup$ – RE60K Mar 24 '15 at 14:28

The other answers are very good so I will concentrate on the more physical meaning, on intuition rather than mathematics. Imagine you have a large rod of a big mass M.it is difficult to rotate it. Now considered the same mass M compressed to nearly a point.Now we have your question about the middle of the rod but on the extreme of it having a huge mass.Well, because all the mass is located at nearly a point, it will be really easy to rotate it, so you conclyde that its moment of inertia is nearly zero.So if we theoretically talk about a single POINT, then the moment of inertia is indeed zero. Hope I helped!

  • $\begingroup$ this assumes that centre of geometry coincides with centre of mass $\endgroup$ – RE60K Mar 24 '15 at 14:48
  • $\begingroup$ Well,how else can you give an intuitive answer?All the other answers give a mathematical meaning.So,if you have a better idea of building him some intuition,then you should just write it on the comments as a supporting comment,not just say the "problem" with my answer. $\endgroup$ – TheQuantumMan Mar 24 '15 at 15:50
  • $\begingroup$ maybe i can't. don't be offended sorry. but i just made you aware of the flaw in your reasoning. $\endgroup$ – RE60K Mar 24 '15 at 16:33
  • $\begingroup$ I am not offended..whatever we can do to help I guess $\endgroup$ – TheQuantumMan Mar 25 '15 at 1:09

By definition, for a 2D object comprising a number of distinct point masses $m_i$ rotating about a point $r_0$, the moment of inertia is given as the sum

$$I = \sum_i{m_i |\vec r_i - \vec{r_0}|^2}$$

If we write the position of the vector $\vec{r}$ as (x,y) and the point $r_0$ as $(x_0, y_0)$ then we can write this as

$$\begin{align} I &= \sum_i{m_i \left((x_i-x_0)^2 + (y_i-y_0)^2\right)}\\ &=\sum_i{m_i \left(x_i^2-2x_i\cdot x_0 + x_0^2 + y_i^2-2y_i\cdot y_0 + y_0^2\right)}\\ \end{align}$$

If we want to minimize this, then we need the partial derivative with respect to $x_0$ and $y_0$ to be zero. This leads to the following equations (I am just showing this for $x$# but the same is obviously true for $y$)

$$\sum_i{m_i \left(-2x_i + 2 x_0\right)}= 0 \implies\\ \sum_i{m_i \cdot x_i} = \sum_i{m_i \cdot x_0}$$

If we divide by the total mass, the expression on the left is the definition of the center of mass in the $x$ direction - and the equation tells us that putting the center of rotation at the center of mass minimizes the moment of inertia.

This result can be expanded with some effort to the 3D case - but the notation becomes messier and I don't think it aids in the understanding.


Here's a mathematical proof to your problem, showing that the polar moment of inertia about the centre of gravity is indeed the minimum, at least for a laminar (2D) rigid body.

For a rigid body of mass $m$:

enter image description here

The polar moment of inertia taken about a general point P is:

$$I_P = \int\limits_m \left( \vec r_P \cdot \vec r_P \right) dm$$

We need to find the minimum value of $I_P$ for this rigid body.

Sub in the following: $\vec r_P = \vec r_G - \vec r_{PG}$

$$I_P = \int\limits_m \left( \vec r_G - \vec r_{PG} \right) \cdot \left( \vec r_G - \vec r_{PG} \right) dm$$

$$I_P = \int\limits_m \left( \vec r_G \cdot \vec r_G \right) dm - \int\limits_m \left( 2 \vec r_{PG} \cdot \vec r_G \right) dm + \int\limits_m \left( \vec r_{PG} \cdot \vec r_{PG} \right) dm$$

$$I_P = \int\limits_m \left( \vec r_G \cdot \vec r_G \right) dm - 2\vec r_{PG} \cdot \int\limits_m \vec r_G dm + \left| \vec r_{PG}\right|^2\int\limits_m dm$$

Now, $\int\limits_m \left( \vec r_G \cdot \vec r_G \right) dm$ is the polar moment of inertia about the centre of gravity, $I_G$. Also, by the definition of the centre of gravity, $\int\limits_m \vec r_G dm = 0$. $\int\limits_m dm$ is equal to the total mass, $m$.

$$I_P = I_G + \left| \vec r_{PG}\right|^2 m$$

Let's try to minimise this expression. $I_G$ is constant, so we cannot minimise this term any further. The term $\left| \vec r_{PG}\right|^2 m$ must be non-negative (squared number multiplied by a mass). This minimum value of this term is zero, where $\vec r_{PG} = 0$.

Therefore, $I_{P_{min}} = I_G$ In other words, the minimum polar moment of inertia occurs about the centre of gravity.


As we know the moment of inertia, I of an object is sum of Mass times distance of the axis from the centre of mass and the moment of inertia about center: $$I=I_{com}+Md^2\\d^2\ge0\implies I\ge I_{com}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy