Topic 5.4 Rotational Inertia: define rotational inertia as the mass-weighted sum of $r^2$, compute it by integration for continuous bodies, and apply the parallel-axis theorem.

What this topic is asking

The College Board (Topic 5.4) wants you to define rotational inertia (the moment of inertia) as the mass-weighted sum of $r^2$, to compute it by integration for continuous bodies, and to apply the parallel-axis theorem. Rotational inertia is the rotational analogue of mass: it measures resistance to angular acceleration, and unlike mass it depends on the axis and how the mass is distributed. The calculus derivations are a signature AP Physics C skill.

Defining rotational inertia

Rotational inertia plays the role of mass in rotational dynamics: the larger it is, the harder it is to angularly accelerate the body, just as a larger mass resists linear acceleration. The crucial difference is the $r^2$ weighting: a mass element far from the axis contributes much more than one near it. Consequently $I$ depends not only on the total mass but on how that mass is distributed and on which axis you rotate about. The same object has different rotational inertias about different axes.

Computing $I$ by integration

For a continuous body, the sum becomes an integral over the mass elements:

You express $dm$ through the density and geometry, $dm = \lambda\,dx$ for a rod, $dm = \sigma\,dA$ for a plate, $dm = \rho\,dV$ for a solid, and integrate $r^2$ over the body. For a uniform rod of mass $M$, length $L$, about its center, $I = \int_{-L/2}^{L/2}x^2\dfrac{M}{L}\,dx = \tfrac{1}{12}ML^2$; about one end it is $\tfrac{1}{3}ML^2$. A hoop has all its mass at radius $R$, so $I = MR^2$; a solid disk integrates to $\tfrac{1}{2}MR^2$; a solid sphere to $\tfrac{2}{5}MR^2$. These derivations, and reading the right standard result off the equation sheet, are routinely examined.

The parallel-axis theorem

Once you know the rotational inertia $I_{cm}$ about an axis through the center of mass, the parallel-axis theorem gives it about any parallel axis a distance $d$ away:

This saves you from re-integrating for a shifted axis. For example, a rod about its center has $I_{cm} = \tfrac{1}{12}ML^2$; about an end (a distance $L/2$ away) it is $\tfrac{1}{12}ML^2 + M(L/2)^2 = \tfrac{1}{3}ML^2$, matching the direct integral. The theorem also shows that the rotational inertia is smallest about an axis through the center of mass and grows as you move the axis away.

Try this

Q1. A solid disk of mass $4.0$ kg and radius $0.20$ m rotates about its central axis. Calculate its rotational inertia. [2 points]

• Cue. $I = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(4.0)(0.20)^2 = 0.080$ kg m squared.

Q2. A rod has $I_{cm} = \tfrac{1}{12}ML^2$. State its rotational inertia about an axis through one end. [2 points]

• Cue. Parallel axis with $d = L/2$: $I = \tfrac{1}{12}ML^2 + M(L/2)^2 = \tfrac{1}{3}ML^2$.