Rotational inertia

Spin is harder to start or stop for some bodies than for others, even at the same mass. Rotational inertia $I$ (the moment of inertia) measures that resistance to a change in spin about a chosen axis.

For point masses, $I = \sum m_i r_i^2$: each mass times the square of its distance from the axis. For a solid body the sum becomes an integral, $I = \int r^2\,dm$, adding every bit of mass weighted by $r^2$. That $r^2$ weighting is the key idea: mass far from the axis counts far more than mass near it.

Two tools do most of the work. The integral $\int r^2\,dm$ builds $I$ about whatever axis you measure $r$ from; choosing a center-of-mass axis first lets the parallel-axis theorem $I = I_{cm} + Md^2$ shift it to any parallel axis. The calculus is new for Physics C; the payoff: you can find $I$ for any shape.

Rotational inertia about an axis is $I = \int r^2\,dm$. Two pieces have to be right. The weighting is $r^2$, not $r$: squaring the distance is what makes far-out mass count the most. And $dm$ is a slice of mass, written with a density, not a bare $dx$ or $dr$.

For a uniform rod the linear density is $\lambda = M/L$, so $dm = \lambda\,dx$ and $I = \int x^2\,\lambda\,dx$. For a flat plate the area density is $\sigma = M/A$, so $dm = \sigma\,dA$. For a disk built from thin rings of radius $r$, a ring has area $2\pi r\,dr$, so $dm = \sigma\,2\pi r\,dr$ and $I = \int_0^R r^2\,\sigma\,2\pi r\,dr = \tfrac12 MR^2$.

The common slip is to pull $r^2$ outside the integral, writing $I = r^2\int dm = MR^2$. But $r$ is the distance of each element and runs across the whole body, so it cannot leave the integral. Only when every element really is at the same radius, as in a hoop, does $I = MR^2$.

$I = \int r^2\,dm$. Keep the square, and turn mass into $dm$ with a density: $\lambda\,dx$ for a rod, $\sigma\,dA$ for a plate.

The integral gives $I$ about the axis you measured $r$ from; the standard move is to evaluate it about a center-of-mass axis first. To then find $I$ about a different, parallel axis a distance $d$ away, use the parallel-axis theorem: $I = I_{cm} + Md^2$.

Two conditions matter. The base $I_{cm}$ must be the value about the center-of-mass axis, not some other axis. And $d$ is the distance between the two parallel axes, from the center of mass to the new axis. Because $Md^2$ is added, leaving the center-of-mass axis can only raise the inertia, never lower it.

So a disk about its rim is $\tfrac12 MR^2 + MR^2 = \tfrac32 MR^2$, with $d = R$. A rod about its end is $\tfrac1{12}ML^2 + M(L/2)^2 = \tfrac13 ML^2$, with $d = L/2$. The slips are subtracting instead of adding, starting from a non-center value, or using the wrong distance for $d$.

$I = I_{cm} + Md^2$. Start from the center-of-mass value, take $d$ between the parallel axes, and always add.

Rotational inertia is not a single number stamped on a body. It depends on the axis and on how the mass is spread around that axis, weighted by $r^2$. The same rod has $\tfrac1{12}ML^2$ about its center but $\tfrac13 ML^2$ about its end, four times larger, with no change in mass.

The spread of the mass matters just as much. A solid disk and a thin hoop of equal mass and radius are not equal: the hoop is $MR^2$ and the disk only $\tfrac12 MR^2$, because the hoop's mass all sits at $r = R$ while the disk's is spread inward. Moving mass outward raises $I$; a mass element right on the axis ($r = 0$) adds nothing.

The slip is to treat $I$ as fixed by mass alone, as if equal mass meant equal inertia, or to imagine all the mass acting at the center. Ask two questions first: which axis, and where is the mass.

$I$ depends on the axis and the mass distribution, not on total mass alone. Far mass counts more, by $r^2$; mass on the axis counts not at all.

Take a uniform disk with $M = 2.0$ kg and $R = 0.30$ m. About its central axis, $I = \tfrac12 MR^2 = \tfrac12(2.0)(0.30)^2 = 0.090$ kg$\cdot$m$^2$. Set up as an integral, $\int_0^R r^2\,\sigma\,2\pi r\,dr$ gives the same $\tfrac12 MR^2$; the $\tfrac12$ comes from the mass being spread inward, not all at the rim.

Now shift the axis to the rim, a distance $d = R$ away. The parallel-axis theorem gives $I = \tfrac12 MR^2 + MR^2 = \tfrac32 MR^2 = 0.27$ kg$\cdot$m$^2$. The added $Md^2 = (2.0)(0.30)^2 = 0.18$ kg$\cdot$m$^2$ is the cost of leaving the center.

Two traps to avoid: pulling $r^2$ out gives $MR^2 = 0.18$ kg$\cdot$m$^2$ for the central axis, double the right value; and subtracting the shift would give $0.090 - 0.18 < 0$, an impossible negative inertia. Keep the integral honest, and always add the shift.

Build $I$ with $\int r^2\,dm = \tfrac12 MR^2$, then shift with $I_{cm} + Md^2$. Pulling $r$ out or subtracting the shift both break it.

Computing rotational inertia goes sideways in specific, repeatable ways, and every pitfall here is one the diagnostic is built to surface.