Rotational Kinematics and Moment of Inertia, Explained

Rotational kinematics reuses the constant-acceleration equations with angle, angular velocity, and angular acceleration in place of x, v, and a. Moment of inertia (sum of mr squared) measures resistance to spin, and angular momentum L = I times omega is conserved when net external torque is zero.

AP Physics: Unit 5 (topics 5.1 Rotational Kinematics, 5.4 Rotational Inertia, 6.3 Angular Momentum and Angular Impulse, 6.4 Conservation of Angular Momentum). Maps to Topics 5.1 and 5.4 in AP Physics 1 Unit 5 (10 to 15% of the multiple-choice section) and Topics 6.3 and 6.4 in Unit 6 (5 to 8%). AP Physics C: Mechanics covers the same rotational topics in its Units 5 and 6 with calculus added.

Angular position, velocity, and acceleration

Rotational kinematics describes spinning motion with three quantities that map one-to-one onto the linear ones you already know: angular position $\theta$ (measured in radians), angular velocity $\omega$ (in $\text{rad/s}$ ), and angular acceleration $\alpha$ (in $\text{rad/s}^2$ ). Every point on a rigid object sweeps the same angle in the same time, so one set of angular values describes the whole object, even though a point near the rim moves faster than a point near the axle.

The radius ties each linear quantity to its angular partner:

Linear	Angular	Connection
Position $x$	Angle $\theta$	$\Delta x = r\Delta\theta$
Velocity $v$	Angular velocity $\omega$	$v = r\omega$
Acceleration $a$	Angular acceleration $\alpha$	$a = r\alpha$ (tangential)

These connections only work in radians. One revolution is $2\pi$ radians, about 6.28 rad, and 1 rad is about 57.3 degrees.

The three rotational kinematics equations

Take the three kinematic equations printed on the AP equation sheet, swap in the angular symbols, and you have the whole rotational set:

\omega = \omega_0 + \alpha t

\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2

\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)

They apply only while angular acceleration is constant, the same restriction the linear set carries. Choose the equation that skips the variable you neither know nor need: no $\theta$ ? Use the first. No $t$ ? Use the third.

A fourth relation, $\Delta\theta = \frac{\omega_0 + \omega}{2}t$ , follows from average angular velocity. Its linear counterpart is not printed on the AP equation sheet, so treat it as a shortcut you derive rather than one you look up.

The solving routine is identical to linear problems: list knowns with signs, name the target, pick the equation, substitute with units. If you can handle a braking car, you can handle a slowing wheel.

Moment of inertia: where the mass sits

Moment of inertia (rotational inertia in the AP course outline) is the rotational counterpart of mass: it measures how strongly an object resists angular acceleration. For point masses,

I = \sum mr^2

where $r$ is each mass's distance from the axis. The $r^2$ is the whole story: a mass twice as far out counts four times as much. That is why a skater's arm position matters and why flywheels put their mass in the rim.

For solid shapes, the sum becomes standard moment of inertia formulas:

Object (mass $M$ )	Axis	Formula
Thin hoop, radius $R$	through center	$I = MR^2$
Solid disk or cylinder, radius $R$	through center	$I = \frac{1}{2}MR^2$
Solid sphere, radius $R$	through center	$I = \frac{2}{5}MR^2$
Thin rod, length $L$	through center	$I = \frac{1}{12}ML^2$
Thin rod, length $L$	through end	$I = \frac{1}{3}ML^2$

The closer the mass sits to the axis, the smaller the coefficient, and moving the axis away from the center of mass always increases $I$ .

From torque to angular acceleration

Rotational inertia earns its keep in the rotational form of Newton's second law:

\alpha = \frac{\tau_{net}}{I}

A net torque produces angular acceleration, and a larger rotational inertia means less angular acceleration from the same torque, just as a larger mass means less linear acceleration from the same force. Torque itself is $\tau = r_{\perp}F = rF\sin\theta$ : force times lever arm. You can check your numbers with the torque calculator.

Two useful limits. If the net torque is zero, $\alpha = 0$ and the object keeps spinning at constant $\omega$ (or stays at rest), which is the rotational version of Newton's first law. If the net torque is constant, $\alpha$ is constant, which is exactly when the kinematics equations above apply.

That is the workflow for a huge class of problems: use torque and $I$ to find $\alpha$ , then use the kinematics equations to find $\omega$ and $\Delta\theta$ at any later time.

Angular momentum and the spinning skater

Angular momentum measures spin content the way linear momentum measures straight-line motion:

L = I\omega

which appears on the AP Physics 1 equation sheet in exactly that form. Units are $\text{kg}\cdot\text{m}^2/\text{s}$ .

Angular momentum conservation parallels conservation of linear momentum: if the net external torque on a system is zero, total angular momentum cannot change. A net torque acting for a time delivers an angular impulse, $\Delta L = \tau_{net}\Delta t$ , so no net torque means no change in $L$ .

The spinning skater is the standard example. On nearly frictionless ice there is almost no external torque about her vertical spin axis, so $L$ is locked in. Pulling her arms toward her body shrinks $I$ (that $r^2$ dependence again), and since $I\omega$ must stay constant, $\omega$ climbs. Extending her arms reverses it. Her muscles are internal to the system, so they can trade $I$ for $\omega$ but never change the product.

Where this sits on the AP exam

Rotational kinematics is Topic 5.1 and rotational inertia is Topic 5.4, both in AP Physics 1 Unit 5, Torque and Rotational Dynamics, which carries 10 to 15% of the multiple-choice section. Angular momentum and angular impulse (Topic 6.3) and conservation of angular momentum (Topic 6.4) sit in Unit 6, Energy and Momentum of Rotating Systems, at 5 to 8%. AP Physics C: Mechanics has units with the same names, each weighted 10 to 15%.

Common question patterns:

Read an $\omega$ vs. $t$ graph: the slope is $\alpha$ and the area under the curve is $\Delta\theta$ .
Rank the rotational inertias of objects with equal mass but different shapes or axis choices.
Conserve angular momentum for a skater, a shrinking spinning object, or a mass dropped onto a rotating disk, then compare kinetic energy before and after.

A calculator (four-function, scientific, or graphing) is allowed on both sections of the exam, so carry exact values through your algebra and round to 3 sig figs at the end.

Ceiling fan spin-up

A ceiling fan starts from rest and reaches an angular velocity of 12.0 rad/s after 4.0 s of constant angular acceleration. Find the angular acceleration, the total angle turned, and the number of revolutions completed.

List knowns: $\omega_0 = 0$ , $\omega = 12.0\ \text{rad/s}$ , $t = 4.0\ \text{s}$ . Targets: $\alpha$ and $\Delta\theta$ .
Angular acceleration from the first equation: $\alpha = \frac{\omega - \omega_0}{t} = \frac{12.0 - 0}{4.0} = 3.0\ \text{rad/s}^2$ .
Angle from the second equation: $\Delta\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 0 + \frac{1}{2}(3.0)(4.0)^2 = \frac{1}{2}(3.0)(16) = 24\ \text{rad}$ .
Convert to revolutions: one revolution is $2\pi = 6.283$ rad, so $24 \div 6.283 = 3.82$ revolutions.
Check with the third equation: $\omega^2 = 0 + 2(3.0)(24) = 144$ , so $\omega = 12.0\ \text{rad/s}$ . Consistent.

$\alpha = 3.0\ \text{rad/s}^2$ ; the fan turns through 24 rad, which is 3.82 revolutions.

Rotational inertia of a dumbbell, two ways

Two 0.50 kg point masses sit at the ends of a light rod 0.80 m long. Find the rotational inertia about (a) an axis through the rod's center and (b) an axis through one end, both perpendicular to the rod. Then find the angular acceleration if a net torque of $0.24\ \text{N}\cdot\text{m}$ acts about the center axis.

(a) Each mass sits $r = 0.40\ \text{m}$ from the center: $I = \sum mr^2 = 2 \times (0.50)(0.40)^2 = 2 \times (0.50)(0.16) = 0.16\ \text{kg}\cdot\text{m}^2$ .
(b) About one end, one mass is at $r = 0$ and contributes nothing; the other is at $r = 0.80\ \text{m}$ : $I = (0.50)(0.80)^2 = (0.50)(0.64) = 0.32\ \text{kg}\cdot\text{m}^2$ . Same object, double the rotational inertia, purely from moving the axis.
Apply the rotational second law about the center axis: $\alpha = \frac{\tau_{net}}{I} = \frac{0.24}{0.16} = 1.5\ \text{rad/s}^2$ .

$I_{center} = 0.16\ \text{kg}\cdot\text{m}^2$ , $I_{end} = 0.32\ \text{kg}\cdot\text{m}^2$ , and the torque produces $\alpha = 1.5\ \text{rad/s}^2$ about the center.

The spinning skater

A skater spins at 2.0 rad/s with arms extended, giving a rotational inertia of $4.8\ \text{kg}\cdot\text{m}^2$ . She pulls her arms in, dropping her rotational inertia to $1.2\ \text{kg}\cdot\text{m}^2$ . Ice friction is negligible. Find her new angular velocity and compare her rotational kinetic energy before and after.

The ice exerts negligible torque about her spin axis, and her muscles are internal forces, so the net external torque is zero and angular momentum is conserved: $I_i\omega_i = I_f\omega_f$ .
Initial angular momentum: $L = I_i\omega_i = (4.8)(2.0) = 9.6\ \text{kg}\cdot\text{m}^2/\text{s}$ .
New angular velocity: $\omega_f = \frac{L}{I_f} = \frac{9.6}{1.2} = 8.0\ \text{rad/s}$ . Cutting $I$ to one quarter multiplies $\omega$ by four.
Kinetic energy before: $K_i = \frac{1}{2}I_i\omega_i^2 = \frac{1}{2}(4.8)(2.0)^2 = \frac{1}{2}(4.8)(4.0) = 9.6\ \text{J}$ .
Kinetic energy after: $K_f = \frac{1}{2}(1.2)(8.0)^2 = \frac{1}{2}(1.2)(64) = 38.4\ \text{J}$ . The extra $38.4 - 9.6 = 28.8\ \text{J}$ is work her muscles do pulling her arms inward.

$\omega_f = 8.0\ \text{rad/s}$ . Angular momentum stays at $9.6\ \text{kg}\cdot\text{m}^2/\text{s}$ , but kinetic energy rises from 9.6 J to 38.4 J.

Frequently asked questions

Do the rotational kinematics equations require constant angular acceleration?

Yes. Like the linear versions, all three assume angular acceleration stays constant over the interval you analyze. If it changes (a motor ramping up unevenly, for example), split the motion into intervals where it is roughly constant, or work from a graph: the slope of an omega versus time graph gives the angular acceleration, and the area under it gives the angle swept.

Why does moment of inertia use r squared instead of r?

A mass farther from the axis must move faster to keep up with the same rotation rate (v equals r times omega), and kinetic energy depends on speed squared. Combining the two gives each mass an effective contribution of m times r squared. Doubling the distance quadruples the contribution, which is why arm position changes a skater's spin so dramatically.

When is angular momentum conserved?

Whenever the net external torque on the system is zero. Internal forces, like a skater's muscles, can rearrange mass and trade rotational inertia for angular velocity, but they cannot change the total angular momentum. On the AP exam, phrases like frictionless axle or the skater pulls in her arms signal a conservation problem.

If a skater spins faster after pulling in her arms, where does the extra kinetic energy come from?

From the skater. Angular momentum is conserved, but rotational kinetic energy is not: pulling mass inward takes muscular work, and that work becomes the added kinetic energy. In the worked example above, kinetic energy rises from 9.6 J to 38.4 J, and the 28.8 J difference is exactly the work done by the skater's muscles.

Do angles have to be in radians?

For anything connecting linear and angular quantities (v equals r omega, arc length, tangential acceleration), yes, radians only. The three rotational kinematics equations work in any single consistent angle unit, but mixing units is the most common way to lose points, so convert everything to radians at the start. One revolution is 2 pi, about 6.28 radians.