Mistake Master
Mistake Master
Force changes velocity, and the body's mass sets how much velocity-change a given force buys. Torque changes angular velocity, and the body's rotational inertia $I$ sets how much. The catch is that $I$ is not a property of the body alone. Where the mass sits relative to the axis matters at least as much as how much mass there is, and the same body has different $I$ values about different axes.
A door is easy to swing about its hinges and much harder to swing about a hinge moved to the door's center. Same door, same mass, very different effort. The "effort" you feel is the rotational analog of mass, and it depends on two things: how much mass the door has, and where that mass sits relative to the axis you're swinging about.
For a system of point masses, the definition is simple and carries the entire topic:
$$I = \sum_i m_i\, r_i^2,$$where $m_i$ is the mass of particle $i$ and $r_i$ is its perpendicular distance from the rotation axis. Two features of this formula run the rest of the lesson. First, $I$ has both a mass factor and a distance factor: $m$ enters linearly, $r$ enters squared. Far mass counts much more than near mass at fixed total mass. Second, every $r_i$ is measured from the chosen axis. Change the axis and every distance changes, so $I$ changes. The same body has many rotational inertias, one for each axis you might pick.
Topic 5.3 named torque, the rotational input. Topic 5.4 names $I$, the rotational resistance. Topic 5.5 will combine them.
For a system of point masses, $I = \sum_i m_i r_i^2$. The units are kg$\cdot$m$^2$. The quantity is a non-negative scalar: there is no direction attached to $I$.
A massless rod has two beads attached, each of mass $m = 3.0$ kg. The beads sit at the two ends of the rod, separated by $0.80$ m. The rod rotates about an axis through the midpoint, perpendicular to the rod. Find $I$.
Each bead sits at $r = 0.40$ m from the axis. Two terms in the sum: $$I = m_1 r_1^2 + m_2 r_2^2 = (3.0)(0.40)^2 + (3.0)(0.40)^2.$$
$$I = 2 \cdot (3.0)(0.16) = 0.96 \text{ kg} \cdot \text{m}^2.$$
About one kilogram-meter-squared. Note what would happen if you doubled the separation to $1.60$ m: each $r$ becomes $0.80$ m, $r^2$ jumps by a factor of $4$, and $I$ jumps from $0.96$ to $3.84$ kg$\cdot$m$^2$. Same masses, four times the rotational inertia.
For continuous bodies, you don't compute the sum from scratch; the AP equation sheet supplies the answers. Five standard shapes cover almost everything you'll see:
| Shape | Axis | $I$ |
|---|---|---|
| Solid disk or cylinder | through center, perpendicular to face | $\tfrac{1}{2}MR^2$ |
| Thin hoop or ring | through center, perpendicular to plane | $MR^2$ |
| Solid sphere | through any diameter | $\tfrac{2}{5}MR^2$ |
| Thin rod, length $L$ | through center, perpendicular to rod | $\tfrac{1}{12}ML^2$ |
| Thin rod, length $L$ | through one end, perpendicular to rod | $\tfrac{1}{3}ML^2$ |
Each formula carries an implied axis. The hoop's $MR^2$ is for an axis through the center of the hoop, perpendicular to its plane. The rod has two entries in the table because two different rod axes show up so often that both deserve their own line.
One more property of the sum that is easy to overlook: a mass on the axis itself contributes $m \cdot 0^2 = 0$. Sitting on the axis is free. This is why, for a rod rotated about one end, a bead sitting on the rotation axis contributes nothing, while a bead at the other end of the rod, length $L$ away, contributes $mL^2$ all on its own.
Take the same uniform rod, length $L$, mass $M$. Two natural axes, perpendicular to the rod: one through the center, one through one end. From the shape table:
$$I_\text{center} = \tfrac{1}{12}ML^2, \qquad I_\text{end} = \tfrac{1}{3}ML^2.$$Same rod, ratio $4$. Move the axis from the center to the end and the rotational inertia about that axis quadruples. The rod is unchanged. What changed is the axis.
A uniform rod has mass $M = 2.0$ kg and length $L = 1.0$ m. Compute its rotational inertia about an axis through its center, and about an axis through one end, both perpendicular to the rod.
$$I_\text{center} = \tfrac{1}{12}(2.0)(1.0)^2 \approx 0.167 \text{ kg}\cdot\text{m}^2.$$
$$I_\text{end} = \tfrac{1}{3}(2.0)(1.0)^2 \approx 0.667 \text{ kg}\cdot\text{m}^2.$$
Same rod, factor-of-four bigger about the end. The mass farthest from the end-axis sits at the full length $L$ away, contributing the maximum $r^2$ to the sum.
The two axes above are parallel to each other (both perpendicular to the rod), offset by $d = L/2$. For any two parallel axes, one through the center of mass and one shifted by $d$, the two rotational inertias are related by the parallel-axis theorem:
$$I = I_\text{cm} + M d^2.$$$I_\text{cm}$ is the rotational inertia about the parallel axis that goes through the center of mass, $M$ is the total mass, and $d$ is the distance between the two axes. The new $I$ is always at least as large as $I_\text{cm}$, because $Md^2 \geq 0$. Among parallel axes in a given direction, $I$ is smallest about the one through the center of mass.
Check it on the rod. For the center-of-mass axis at the rod's middle, $I_\text{cm} = \tfrac{1}{12}ML^2$. Shift to the end: $d = L/2$, so
$$I_\text{end} = \tfrac{1}{12}ML^2 + M(L/2)^2 = \tfrac{1}{12}ML^2 + \tfrac{1}{4}ML^2 = \tfrac{1}{3}ML^2,$$which matches the shape table. The theorem ties the two entries together.
Three traps. The first two are the same root error coming out of two different mouths, both kinds of single-factor reasoning. The third forgets that $I$ comes paired with an axis.
True for two hoops at the same radius. Not generally true. A $2$ kg solid disk at $R = 0.30$ m has $I = \tfrac{1}{2}(2)(0.30)^2 = 0.09$ kg$\cdot$m$^2$. A $2$ kg thin hoop at the same $R$ has $I = (2)(0.30)^2 = 0.18$ kg$\cdot$m$^2$, twice as much. Same mass, factor-of-two difference. The hoop's mass sits at the maximum distance from the axis; the disk's mass is spread from $r = 0$ outward.
Fix. Both factors in $I = \sum m_i r_i^2$ count. Mass enters linearly, distance enters squared. Asking "is it heavier?" answers half the question. Always pair the mass check with "where does the mass sit relative to the axis?"
The flip side of the same single-factor trap. A $1$ kg thin hoop at $R = 0.50$ m has $I = (1)(0.50)^2 = 0.25$ kg$\cdot$m$^2$. A $10$ kg solid disk at the same $R = 0.50$ m has $I = \tfrac{1}{2}(10)(0.50)^2 = 1.25$ kg$\cdot$m$^2$, five times larger. The disk is more compact in terms of distribution, but the ten-times-more-mass dwarfs the geometric factor.
Fix. Distribution alone does not decide either. Mass and distribution together set $I$. If you find yourself comparing two objects by looking only at how the mass is arranged, stop and check the masses too.
$I$ depends on the axis. Same rod about its center gives $\tfrac{1}{12}ML^2$; about one end gives $\tfrac{1}{3}ML^2$. Factor of four, same rod. The "$I$ of a rod" by itself is meaningless until you specify the axis.
Fix. Treat $I$ the way you learned to treat torque in Topic 5.3: it is a property of a body about an axis, never of the body alone. Every quoted $I$ comes with an implied axis. When the axis shifts between two parts of a problem, the rotational inertia shifts with it; the parallel-axis theorem tells you by how much.