# How is this application of the parallel axis theorem wrong?

Let's say I have a rod of length $L$. The moment of inertia for spinning it around some point at a distance $d$ from one end can be calculated using the parallel axis theorem. Specifically, knowing that the moment of inertia when spun around its center is $\frac{1}{12}mL^2$, so the moment of inertia when spun around $d$ is $\frac{1}{2}mL^2 + m(L-d)^2$.

Now instead, let me calculate the moment of inertia when spun around $d$ by using the moment of inertia when spun around the end, which is $\frac{1}{3}mL^2$. Using parallel axis theorem, the moment of inertia when spun around $d$ is $\frac{1}{2}mL^2 + md^2$.

But woah, those are not equal. So one is wrong. I know its the first one that's right, so I'm wondering what I did wrong with the second scenario.

The parallel axis theorem only works when the initial moment is computed about an axis through the object's center of mass. It does not apply when the initial moment is computed about any arbitrary axis. That is, $I = I_{cm} + mr^2$, where $I_{cm}$ is the moment about an axis through the center of mass and $r$ is the perpendicular distance between that axis and the axis about which you wish to calculate $I$. Your second equation does not compute $I$ relative to an axis through the rod's center of mass, so it is not correct. (The first equation is also wrong, because $\frac{1}{2}$ should be $\frac{1}{12}$ and $(L-d)^2$ should be $(\frac{L}{2}-d)^2$, I think.)

I don't know the parallel axis, but here you have an extremely simple test whether your calculi are good or wrong. It just completes the explanation given to you by pwf.

The inertial momentum of a rod of uniform density \mu is

$I = \mu \int_{-d}^{L-d} l^2 dl = \mu \frac {(L-d)^3}{3} + \mu\frac {d^3}{3}$

$= \mu \frac {L^3 -3L^2d +3Ld^2}{3} = \frac {mL^2}{3} - mLd + md^2$.

You can easily rearrange this expression as

$I = \frac {mL^2}{12} + m(\frac {L}{2} -d)^2$.

We have where "d" is the distance of a point from one end of the rod:

$\frac{1}{12}mL^2 + m(\frac{L}{2}-d)^2$ This is the parallel axis formula for our situation.

If "d" is a positive number, it lies within the rod. If "d" is a negative number, it lies outside the rod. Just an FYI.

The parallel axis theorem only works when one considered axis is through the center of mass. Everyone here agrees on that.

When "spun around the end", "d" must be equal to 0. Therefore:

$\frac{1}{12}mL^2 + m(\frac{L}{2}-0)^2 = \frac{1}{12}mL^2 + \frac{1}{4}mL^2 = \frac{1}{12}mL^2 + \frac{3}{12}mL^2 = \frac{1}{3}mL^2$

The center of mass axis MUST be one of the two that are used.