First and second moments of mass why is the second moment of mass used as moment of inertia, and why is the first moment of mass about an axis not used as moment of inertia?
i understand that first moment is product of mass and its distance from rotational axis, and that second moment is the product of square of distance and mass from the rotational axis, so giving those mathematical formulas will not help me.
please explain why the second instead of the first moment of mass is considered as "moment of inertia" of that mass about the considered axis.
 A: This is a direct consequence of what comes out when you go from equations of motion for single point particles to collective motion of extensive bodies. You start with the second Newton's law:
$$m_i \vec{a}_i = \vec{F}_{i}$$
Summation over the entire system cancels out pairs of mutually opposite internal forces, leaving us with external force and acceleration:
$$\sum m_i \vec{a}_i = \vec{F}$$
Now, by definition, it's $\vec{a}_i = \frac{d^2}{dt^2}\vec{r}_i$. So this can be written as
$$\frac{d^2}{dt^2} \sum m_i \vec{r}_i = \vec{F}$$
notice the sum is exactly the thing that you replace with $m\vec{r}_0$ where $m=\sum m_i$ is the total mass and $\vec{r}_0$ is the center of mass. Why we use the center of mass? Because it turns out equations for linear motion reduce to FORCE = TOTAL MASS * ACCELERATION OF THE CENTER OF MASS.
Now for the rotational part. Of course, if an equation is true, it remains true if you do the same operation on both sides. In this case, according to our (arbitrarily chosen) coordinate frame, the positions of particles were $\vec{r}_i$. Just cross Newton's law with this:
$$m_i \vec{r}_i\times \vec{a}_i = \vec{r}_i\times \vec{F}_{i}$$
Now perform summation over the entire body ... again. In this case, the right-hand side separates into torque of external forces, plus the summation over all pair interactions:
$$\sum_i m_i \vec{r}_i\times \vec{a}_i = \sum_i \vec{r}_{ext,i}\times \vec{F}_{ext,i}+\sum_{i,j}(\vec{r}_i-\vec{r}_j)\times \vec{F}_{ij}$$
Here I took into account that pairs of internal forces are equal in magnitude, but as the positions are different, they wouldn't necessarily cancel out. But classically, they do: forces of one particle on another point along their connecting line so the cross products of the second term are zero (magnetism requires caution here, but I won't get into it).
What's left is
$$\sum_i m_i \vec{r}_i\times \vec{a}_i = \vec{M}$$
You have $\vec{a}_i=d^2 \vec{r}_i/dt^2$. Also, observe:
$$\frac{d}{dt}(\vec{r}\times\frac{d}{dt}\vec{r})=\frac{d}{dt}\vec{r}\times\frac{d}{dt}\vec{r}+\vec{r}\times\frac{d^2}{dt^2}\vec{r}$$
The first term is zero (cross a vector with itself), so you can write
$$\frac{d}{dt}\sum_i m_i \vec{r}_i\times \vec{v}_i = \vec{M}$$
This is the conservation of angular momentum. However... if you have a rigid motion, then you can write the velocities as a sum of translations and rotations! Let's say that we have some arbitrary reference point $\vec{A}$ and a displacement from it: $\vec{r}_i=\vec{A}+\vec{R}_i$. Now, the time derivative must take into account motion of the reference, but also consider rotation about the reference:
$\vec{v}_i=\vec{v}_A+\omega\times\vec{R}_i$. Here the second term is the tangential velocity of rotation around the axis $\omega$. Now, we put this back into the above equation and express all with the relative displacement:
$$\frac{d}{dt}\sum_i m_i (\vec{A}+\vec{R}_i)\times (\vec{v}_A+\omega\times\vec{R}_i) = \vec{M}$$
$$\frac{d}{dt}\left(m\vec{A}\times \vec{v}_A+\vec{A}\times\omega\times\sum_i m_i \vec{R}_i+\sum_i m_i \vec{R}_i\times \vec{v}_A+\sum_i m_i \vec{R}_i\times \omega\times\vec{R}_i\right) = \vec{M}$$
This looks horrible, and it is, unless you specifically put the reference into the center of mass, $\vec{A}=\vec{r}_0$. In that case, the relative positions are expressed in system where $\sum_i m_i \vec{R}_i=0$. The thing reduces further, and let's also recall the double cross product expansion:
$$\frac{d}{dt}\left(m\vec{r}_0\times \vec{v}_0+\sum_i m_i\left( \omega|\vec{R}_i|^2-\vec{R}_i(\omega\cdot\vec{R}_i)\right)\right) = \vec{M}$$
The first term is the angular momentum of the center of mass motion. The second term is $J\vec{\omega}$, where $J$ is the tensor defined as
$$J_{ab}=\sum_{i} \left(|R_i|^2\delta_{ab}-R_{i,a}R_{i,b}\right)$$
In short... the second Newton's law and subsequently conservation of momentum reduces in the case of rigid body motion to an expression that involves the second moment of mass.
