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??
Best Answer
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.