# Moment of Inertia of a non-uniform rod about its geometric center

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.

• 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. Commented Aug 19, 2021 at 16:59

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.