5.4 Rotational Inertia

What is rotational inertia in AP Physics 1?

Rotational inertia tells you how hard it is to change an object's spinning motion. It depends on both mass and how far that mass sits from the axis of rotation, with the formula $I = mr^2$ for a point mass. The farther the mass is from the axis, the bigger the rotational inertia, which is why a hoop is harder to spin up than a solid disk of the same mass.

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Why This Matters for the AP Physics 1 Exam

Rotational inertia is the rotational version of mass, so it shows up anytime you analyze how a system speeds up or slows down its spin. You will use it in later topics when you connect net torque to angular acceleration and when you calculate rotational kinetic energy. On the exam, you often have to predict how a quantity changes when something else changes, like what happens to angular acceleration if you move the masses farther out. That kind of functional-relationship reasoning appears in both multiple-choice and free-response, including the question that asks you to translate between qualitative descriptions and quantitative work.

A few things to know about what you are expected to do:

• You calculate rotational inertia by hand only for systems of five or fewer point objects arranged in two dimensions.
• For extended shapes like rods, disks, and hoops, the rotational inertia values are given to you on the exam.
• You should be able to explain qualitatively how mass distribution changes rotational inertia, even when you are not given a formula.

Key Takeaways

• Rotational inertia measures resistance to changes in rotation and depends on mass and how that mass is spread out from the axis.
• For a single point mass, $I = mr^2$, where $r$ is the perpendicular distance to the axis.
• For several discrete masses, add up the pieces: $I_{tot} = \sum m_i r_i^2$.
• Rotational inertia is smallest when the axis passes through the center of mass.
• The parallel axis theorem, $I' = I_{cm} + Md^2$, gives rotational inertia about any axis parallel to a center-of-mass axis.
• Moving mass farther from the axis raises rotational inertia, which is why a hoop beats a solid disk of equal mass and radius.

Rotational Inertia of Rigid Systems

Rotational inertia measures how much a rigid system resists changes to its rotational motion. It is the rotational analog of mass: just as more mass makes it harder to change linear motion, more rotational inertia makes it harder to change spin.

What sets rotational inertia apart is that distribution matters, not just total mass. Two objects with the same mass can have very different rotational inertias depending on where that mass sits relative to the axis.

• For the same net torque, a system with more mass far from the axis has greater rotational inertia and gets a smaller angular acceleration.
• A hollow cylinder and a solid cylinder of equal mass spin up differently because their mass is arranged differently.
• A figure skater pulling in her arms lowers her rotational inertia because more of her mass moves closer to the axis. (This is an everyday application of the idea, not a required calculation.)

Point Mass and Discrete Systems

For a point mass, or an object treated as concentrated at one point at a perpendicular distance $r$ from the axis:

$I = mr^2$

where:

• $I$ is rotational inertia (kg⋅m²)
• $m$ is mass (kg)
• $r$ is the perpendicular distance to the axis (m)

For a system of several discrete objects, add each object's contribution:

$I_{tot} = \sum I_i = \sum m_i r_i^2$

Notice the $r^2$ term. Distance matters a lot because it gets squared, so doubling how far a mass sits from the axis multiplies its contribution by four.

Rotational Inertia About an Off-Center Axis

Minimum at the Center of Mass

A rigid system has its smallest rotational inertia in a given plane when the axis passes through the center of mass. Any parallel axis that is farther out gives a larger rotational inertia.

• The center-of-mass axis is the "easiest" axis to spin about.
• Shift to a parallel axis and rotational inertia goes up.

Parallel Axis Theorem

The parallel axis theorem connects rotational inertia about a center-of-mass axis to rotational inertia about any parallel axis:

$I' = I_{cm} + Md^2$

where:

• $I'$ is rotational inertia about the parallel axis (kg⋅m²)
• $I_{cm}$ is rotational inertia about the center-of-mass axis (kg⋅m²)
• $M$ is the total mass of the system (kg)
• $d$ is the perpendicular distance between the two parallel axes (m)

If you know the rotational inertia about the center-of-mass axis, this lets you find it about any axis parallel to it.

Example: A 4 kg rod has a rotational inertia of 0.8 kg⋅m² about an axis through its center. About an axis 0.2 m away and parallel to that center axis:

$I' = 0.8\text{ kg}\cdot\text{m}^2 + (4\text{ kg})(0.2\text{ m})^2 = 0.96\text{ kg}\cdot\text{m}^2$

How to Use This on the AP Physics 1 Exam

Problem Solving

When you compute rotational inertia for a group of point masses:

1. Identify the axis and find the perpendicular distance $r$ for each mass.
2. Apply $I = mr^2$ to each mass.
3. Add the contributions: $I_{tot} = \sum m_i r_i^2$.
4. If the axis sits on a mass, that mass has $r = 0$ and contributes nothing.

For the parallel axis theorem, confirm the new axis is parallel to a center-of-mass axis, then add $Md^2$ to $I_{cm}$.

Free Response

When a question asks how rotational inertia or angular acceleration changes, reason through the functional relationships instead of just plugging numbers:

• Because $I = mr^2$, moving a mass twice as far from the axis quadruples its contribution.
• Because angular acceleration is inversely related to rotational inertia, increasing $I$ for the same net torque means a smaller angular acceleration.
• Justify claims with the equation and explain in words, not just with a final number.

Common Trap

The $r$ in $I = mr^2$ is the perpendicular distance to the axis, not just any distance to the mass. Always measure straight to the axis of rotation.

Practice Problem 1: Rotational Inertia Calculation

Use the basic rotational inertia formula and track the distances carefully:

1. For the initial setup, use $I_{tot} = \sum mr^2$:

• $I_{tot} = (3 \text{ kg})(0.4 \text{ m})^2 + (2 \text{ kg})(0.7 \text{ m})^2$
• $I_{tot} = 3 \times 0.16 + 2 \times 0.49 = 0.48 + 0.98 = 1.46 \text{ kg}\cdot\text{m}^2$

1. For the axis through the 3 kg mass:

• The 3 kg mass now has $r = 0$, so it contributes nothing.
• The 2 kg mass is now 1.1 m from the axis (0.7 + 0.4).
• $I_{tot} = 0 + (2 \text{ kg})(1.1 \text{ m})^2 = 2 \times 1.21 = 2.42 \text{ kg}\cdot\text{m}^2$

Practice Problem 2: Parallel Axis Theorem Application

Use the parallel axis theorem:

1. Set up $I' = I_{cm} + Md^2$:

• $I_{cm} = 0.225 \text{ kg}\cdot\text{m}^2$
• $M = 5 \text{ kg}$
• $d = 0.3 \text{ m}$ (center to edge)

1. Calculate:

• $I' = 0.225 + 5 \times (0.3)^2$
• $I' = 0.225 + 5 \times 0.09$
• $I' = 0.225 + 0.45 = 0.675 \text{ kg}\cdot\text{m}^2$

Common Misconceptions

• Rotational inertia is not just total mass. Two objects with the same mass can have very different rotational inertias depending on how the mass is spread out from the axis.
• The $r$ in $I = mr^2$ is the perpendicular distance to the axis, not the size of the object or the distance to its center.
• Rotational inertia is not a fixed property of an object alone. It depends on which axis you choose, so the same object has different values about different axes.
• The parallel axis theorem only works between an axis and a parallel axis through the center of mass. You cannot use it between two random parallel axes directly.
• A heavier object is not automatically harder to spin. A light hoop can have more rotational inertia than a heavier compact object if its mass sits farther from the axis.

Related AP Physics 1 Guides

• 5.1 Rotational Kinematics
• 5.2 Connecting Linear and Rotational Motion
• 5.3 Torque
• 5.5 Rotational Equilibrium and Newton's First Law in Rotational Form
• 5.6 Newton's Second Law in Rotational Form