Rotational Kinematics & Inertia

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Visualization of a spinning metallic disc with glowing angular velocity and torque vectors extending from the center axis.

In rotational mechanics, mass is not the only thing that resists acceleration. The distribution of that mass is what truly matters.

Welcome to Unit C5! Everything you learned about straight-line motion is about to be applied to objects spinning in circles. The “Big 3” kinematic equations, Newton’s Second Law, and Kinetic Energy all have exact rotational equivalents. You just need to learn the Greek alphabet of motion.

1. The Translation Dictionary

To succeed in rotational physics, you must be able to translate linear variables into rotational variables seamlessly:

Position ( $x$ ) $\rightarrow$ Angle ( $\theta$ ): Measured in radians.
Velocity ( $v$ ) $\rightarrow$ Angular Velocity ( $\omega$ ): Measured in rad/s. ( $\omega = d\theta/dt$ )
Acceleration ( $a$ ) $\rightarrow$ Angular Acceleration ( $\alpha$ ): Measured in rad/s². ( $\alpha = d\omega/dt$ )
Force ( $F$ ) $\rightarrow$ Torque ( $\tau$ ): The twisting force. ( $\tau = \vec{r} \times \vec{F}$ )
Mass ( $m$ ) $\rightarrow$ Moment of Inertia ( $I$ ): Rotational mass. (kg·m²)

2. Moment of Inertia ( $I$ )

In linear motion, a 10 kg block is equally hard to push no matter how it is shaped. But in rotational motion, a 10 kg rod is much harder to spin if the mass is concentrated at the very ends rather than near the center pivot. Moment of Inertia ( $I$ ) is the measure of how much an object resists rotational acceleration.

For a collection of discrete point masses, $I = \sum m_i r_i^2$ . For a solid, continuous object, you must integrate over the volume of the shape:

$I = \int r^2 dm$

Where $r$ is the perpendicular distance from the axis of rotation to the tiny piece of mass $dm$ .

3. The Parallel Axis Theorem

Most AP Physics C problems will not require you to integrate complex shapes from scratch. You are expected to know that the Moment of Inertia for a uniform rod spinning perfectly around its center of mass is $I_{cm} = \frac{1}{12}ML^2$ .

But what if you grab the rod by the very end and swing it like a baseball bat? You don’t need to do the calculus again. You use the Parallel Axis Theorem.

$I_{new} = I_{cm} + Md^2$

This theorem states that the new Moment of Inertia is equal to the inertia through the Center of Mass ( $I_{cm}$ ) PLUS the total mass of the object ( $M$ ) multiplied by the distance ( $d$ ) the axis was moved squared.

Concept First: The minimum possible Moment of Inertia for any object always occurs when it spins around its Center of Mass ( $d=0$ ). Any shift away from the center makes it harder to spin!

4. Quick AP Practice

📚 Unit C5 Mastery Challenge

1. A solid uniform disk has a mass $M$ , radius $R$ , and a center-of-mass inertia of $I_{cm} = \frac{1}{2}MR^2$ . If a nail is driven through the outer edge of the disk (a distance $R$ from the center) and the disk swings like a pendulum, what is its new Moment of Inertia?

Check Answer

Use the Parallel Axis Theorem: $I = I_{cm} + Md^2$ .
Here, the distance $d$ the axis moved is equal to the radius $R$ .
$I = \left(\frac{1}{2}MR^2\right) + M(R)^2$
$I = \frac{1}{2}MR^2 + \frac{2}{2}MR^2 = \mathbf{\frac{3}{2}MR^2}$ .

2. An engine applies a torque given by $\tau(t) = 4t$ . If the flywheel has a moment of inertia of $2 \text{ kg}\cdot\text{m}^2$ and starts from rest, what is its angular velocity $\omega$ at $t = 3 \text{ s}$ ?