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From parallel axis theorem follows that, given a rigid body, the moment of inertia is minimum if calculated with respect to an axis passing through the center of mass. What is the physical meaning of this?

The moment of inertia plays the role of the mass on rotational motion, so it is linked to the reluctance of the body to rotate. Given a torque, if the axis passes through the CM the angular acceleration is maximum, while it decreases increasing the distance from CM. Why does that happen?

I tried to justify it as a consequence of the distribution of mass around the CM, but I do not know if that is right.

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  • $\begingroup$ In some ways this is the definition of CM. The point that minimizes rotational kinetic energy for all rotation axes. $\endgroup$ – ja72 Apr 5 '16 at 20:39
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There are a few ways to justify it.

First, you could look at the motion of the object as it rotates. In 2D, it turns out that all such motion can be decomposed into motion of the CM, and rotation about the CM. Therefore, rotating around the CM itself is the only way to guarantee the CM doesn't move, and hence is the least energetically costly.

Another way is by direct calculus. Suppose some 1D object is made of point masses at positions $x_i$ with masses $m_i$. Then $$I(x) = \sum m_i (x_i-x)^2$$ is the moment of inertia about $x$. The minimum is attained when the derivative is zero, so $$0 = 2 \sum m_i (x_i - x)$$ which implies that $$x = \frac{\sum m_i x_i}{\sum m_i}.$$ This is the definition of the center of mass.

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That the moment of inertia about an axis passing through the CM is minimized, with respect to any other parallel axes, is a consequence of the quadratic (squared) dependence of the moment of inertia on distance. In other words, the ${r^2}$ term in ${I=mr^2}$ makes it so that masses at farther distances are preferentially weighted in their contribution to the overall moment. As you said, for a given object, the moment of inertia will will thus depend on the distribution (distances) of masses about the chosen axis.

For a given torque, one can impart a greater angular acceleration on objects of lesser moment. This is seen in the relation ${\tau=I\alpha}$ (analogous to $F=ma$), where $\alpha$ is angular acceleration. Rearranging, we get $\alpha=\tau/I$, so $\alpha$ is largest when $I$ is smallest.

Intuitively, one can understand how angular acceleration is maximized about the CM by picturing twisting a metal rod. Imagine holding the rod at its end and twisting-- it's difficult. Imagine holding the same rod at its center and twisting-- it's slightly easier.

To describe the above scenario mathematically, we can consider a one dimensional rod of mass $m$ running from ${x=0}$ to $x=l$. The moment of inertia about an axis that runs perpendicular to the rod at $x=0$ (twisting the rod about its end) is given by

$I_{end}=\int_{0}^{l}\rho x^2 dx = \frac{m}{3}l^2$, where $\rho=m/l$ is the mass density of the rod.

The moment about an axis through the middle of the rod, $x=l/2$, is

$I_{mid}=\int_{-l/2}^{l/2}\rho x^2 dx = \frac{m}{12}l^2$.

Note that $I_{mid}<I_{end}$.

Taking a more general approach we could calculate the moment about any perpendicular axis, placed at any value x, as

$I_{x}=\int_{0-x}^{l-x}\rho x^2 dx = \frac{1}{3}\rho(l-x)^3-\frac{1}{3}\rho(0-x)^3$.

This function $I_{x}(x)$ has a minimum at the center of mass when $x=l/2$.

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