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Suppose we have a rod of length $L$, whose mass varies as $kx$ from one of its endpoints. I'm supposed to find the moment of inertia of this rod, and I'm facing a small conceptual problem.

If I'm asked to find the moment of inertia about the center of the rod, what I've done is, I've calculated the $I$ about the endpoint and then used the parallel axis theorem, to find the moment of inertia about the center of mass, which is at a distance $a$ from the end point.

$I_c + (\int\limits_{0}^{L} kxdx).a^2= \int\limits_{0}^{L} kx.x^2dx$

Then I should use the parallel axis theorem again, to find the moment of Inertia about the geometric center of the rod.

$I_g = I_c + (\int\limits_{0}^{L} kxdx)b^2$, where $b$ is the distance from center of mass to geometric center i.e. $a-\frac{L}{2}$.

I'm hoping there is no mistake in the above procedure. If there is one, can someone verify this or point it out for me, where I went wrong ?

I've another question, which might be really dumb, but please help me out. I know we can't use the formula $I_g = (\int\limits_{-L/2}^{L/2} kx.x^2dx)$ by shifting the origin to $L/2$ as that would give us an answer $0$. Why is that so ? Is it because the density starts to vary from $0$ and not $L/2$. Any intuitive explanation on this, would also be highly appreciated.

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  • $\begingroup$ The $I_{small end}$ = $I_{center of mass}$ + M$(d)^2$ where M is the total mass = (1/2)k$L^2$, and (d) is (x) for the center of mass. $\endgroup$
    – R.W. Bird
    Commented Aug 19, 2021 at 16:59

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I've another question, which might be really dumb, but please help me out. I know we can't use the formula Ig=(∫−L/2L/2kx.x2dx) by shifting the origin to L/2 as that would give us an answer 0. Why is that so ? Is it because the density starts to vary from 0 and not L/2. Any intuitive explanation on this, would also be highly appreciated.

You need a concomitant change in your mass density. You originally gave me $\rho = \hat{\rho}(x) = k x$, where $x$ is the distance from one end of the rod to a point in the rod. Note that $x \in [0, L]$. Now consider a coordinate shift $X = x - \frac{L}{2}$. Note that $X \in [-L/2,L/2]$. We get a new function for the mass density $\rho = \tilde{\rho}(X) = k(X+\frac{L}{2})$. The formula

$I_g = \int_{-L/2}^{L/2} \tilde{\rho}(X) X^2 dX$ should now work.

I looked at the first part of your question and didn't notice anything wrong at first glance.

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  • $\begingroup$ Thank you so much ! $\endgroup$ Commented Aug 19, 2021 at 17:23

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