In our applied mechanics class, we studied about area moment of inertia. Our teacher only explained the mathematical relation of this term that is product of area and square of the perpendicular distance from the axis. But I cannot understand the physical meaning. Help me out.
[Physics] what is the physical meaning of area moment of inertia
applied-physicsclassical-mechanics
Related Solutions
I hope I am interpreting your question right: One of the defining differences between translation and rotation is that translation is commutative. If I move forward 1 and right 1, it is the same as moving right 1 and then forward 1. The same can not be said of rotation. If I rotate my phone clockwise parallel to my body then clockwise perpendicular to my body I end up with a different end position than if I do the same rotations in a different order.
We don't measure translation or rotation though. We can define a translation by measuring distance, speed, direction, etc. We can define a rotation by measuring angles, angular velocity, direction, etc. These measurements are limited by the precision of our measuring equipment.
Any real motion is a combination of many different motions. Pure translation is really how we simplify real motion by excluding the things we don't care about. It is a simplification, and as such, I'm not sure it needs to be "proven".
Why?
Because a moment is a manifestation of a force at a distance, the same way the a velocity is a manifestation of a rotation at a distance. Given two points A and B you know that $$ \vec{M}_A = \vec{r}_{AB} \times \vec{F}_B \\ \vec{v}_A = \vec{r}_{AB} \times \vec{\omega}_B $$
The force at B causes a torque at A, simarly to how a rotation at B causes velocity at A.
So Why is that?
Both forces/torques and velocities/rotations are 3D screws that contain the following properties. a) A line of direction, b) a magnitude, c) a pitch. Forget about the b) and c) for now and focus on the line.
How do you describe a line in 3D. A line has 4 degrees of freedom, and it is usually represented using 6 components with something called Pluecker coordinates. There involve two vectors, each with 3 components. The first vector, I call $\vec{F}$ gives the direction of line line, but its magnitude is not important. So two degrees of freedom are used from the vector. The second vector, I call $\vec{M}$ gives the moment of the line about the origin and it is used to describe the closest point of the line to the origin. It too uses two degrees of freedom because the location along the line is unimportant. It represents either a) The moment of a force along the line, or b) the speed of a rotating body about the line. The location of the line is given by
$$ \vec{r} = \frac{\vec{F} \times \vec{M}}{\vec{F} \cdot \vec{F}} = - \frac{\vec{M} \times \vec{F}}{\vec{F} \cdot \vec{F}} $$ depending on which you like best.
Similarly for motions
$$ \vec{r} = \frac{\vec{\omega} \times \vec{v}}{\vec{\omega} \cdot \vec{\omega}} = - \frac{\vec{v} \times \vec{\omega}}{\vec{\omega} \cdot \vec{\omega}} $$.
So the moment is a manifestation of a line at a distance.
Best Answer
The notion arises fairly naturally when one is thinking about how a constant cross section beam reacts to forces and bending moments.
You imagine a beam to be made up of elastic fibers. When it is bent in a plane, nearest the center of curvature, the fibers must "squash", so they are in compressive strain. On the side outermost from the curvature center, they are in tensile strain. Somewhere in between there is a neutral axis in the cross section that is the intersection between the cross section and the fibers that are neither in compressive nor tensile strain
Indeed, if you sketch the geometry, the amount of strain in the fibers at a normal distance $x$ from the neutral axis is
$$\epsilon(x) = \frac{x}{R} \tag{1}$$
and so the stress normal to the plane is:
$$\sigma(x) = E\,\frac{x}{R} \tag{2}$$
where $R$ is the bending curvature at that cross section and $E$ the Young's modulus. This is because the length of a fiber a normal distance $x$ subtending an angle $\delta\theta$ at the center of curvature is $(R+x)\,\delta \theta$. The same angle is subtended by a length $R\,\delta \theta$ at the neutral axis. Since all the fibers in the arc are the same length when the beam is relaxed, this means that the proportional change in length is $(R+x)/R$, whence (1) (see my sketch below for intuition; the bending is in the plane of the page):
What this means is that if you imagine the beam to be "cut in two" by any cross sectional plane and draw a free body diagram of the two halves, each side exerts an equal and opposite torque on the other.
So, in the light of (2), the contribution to the torque from a fiber with cross-sectional area $\mathrm{d} A$ at distance $x$ from the neutral axis is:
$$\mathrm{d}\tau = \text{lever arm} \times \text{normal force} = x\times E\,\frac{x}{R}\,\mathrm{d} A = \frac{E}{R}\,x^2\,\mathrm{d}A\tag{3}$$
whence one sums up all the contributions to get the total torque of one beam section on the other:
$$\tau = \frac{E\,I}{R}\tag{4}$$
The Punchline
See Euler-Bernoulli or Timoshenko Beam Theory for more information.