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While calculating moment of inertia for two point particles, we use

$$ I = m_1r_1^2 + m_2r_2^2$$

While calculating moment of inertia of a plank(with mass $m$) around an axis halfway through its length we can split the plank into two point particles having masses $(m/2)$. we can consider these as point particles at their center of mass and apply the above formula.

But this gives the wrong answer. What is wrong in the above method??

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    $\begingroup$ When you take the centre of mass all the mass is concentrated in the point. It has no dimension. In case of plank, you have dimensions and all the mass is distributed. So, you can't consider it as a point to find moment of inertia. $\endgroup$ – Shubham Oct 29 '15 at 8:05
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Suppose you split the plank into small segments of ${\rm d} m=\rho A {\rm d}x$ mass each. If the total length is $\ell$ then the location of each segment is $x = 0 \ldots \ell$.

Summing the contribution of each segment to the total mass moment of inertia is $${\rm I} = \int x^2 {\rm d}m =\rho A \int \limits_{0}^{\ell} x^2 {\rm d}x =\rho A \frac{\ell^3}{3} $$

Using the total mass of $m = \rho A \int \limits_{0}^{\ell} {\rm d}x = \rho A \ell$ makes the above $$\boxed{ I =m \frac{\ell^2}{3} } $$

This quantity is actually the aggregate of two quantities

$$ I = m \frac{\ell^2}{12} + m \frac{\ell^2}{4} $$ The first is what comes out if you pivot about the center of mass with $x = -\frac{\ell}{2} \ldots \frac{\ell}{2}$ and the second is the so called parallel axis theorem. Essentially you lump the entire mass at the center of mass and consider the distance between that and the pivot $d=\frac{\ell}{2}$. The parallel axis theorem states $$I_{pivot} = I_{cm} + m d_{pivot}^2$$

So the problem with lumping the mass at the ends is that you disregard the mass distribution in between.

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You can use this method for things that vary linearly with distance (such as finding the torque, which is $\mathbf{F} \times \mathbf{d}$). For an object with constant density, the pairwise contribution of both ends is the same as the contribution from the middle.

But the moment of inertia varies with the square of distance. The contribution from the far end is more than double the contribution from the middle. So it ends up being larger than the same mass located at the middle.

This method of simplifying the rotation of a plank about the end would yield $\frac{mL^2}{4}$, while the correct answer is the larger $\frac{mL^2}{3}$.

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If that method would work, it would also be possible to split the plank into two planks of the same size but half the thickness (as a sandwich), and concentrate those to their respective centres of mass. Those centres are very close to the centre of the original plank, so one would get very small values of $r$, and almost no moment of inertia. Clearly the operation of concentrating the mass into the centre of mass throws out a lot of moment of inertia. Indeed it precisely throws out the moment of inertia of the half-plank under consideration relative to its own centre of mass.

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