Moment Of Inertia About Centre of Mass Why is moment of inertia minimum about centre of mass of any rigid body?
 A: 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!
A: 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.
A: 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 $m$:

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.
A: 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.
A: 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:
$$I=I_{com}+Md^2\\d^2\ge0\implies I\ge I_{com}$$
