Rotational Inertia and Rotational Newton's Second Law — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Definition of rotational inertia, calculation for discrete and continuous systems, parallel axis theorem, rotational Newton’s second law, and torque-angular acceleration analysis for fixed-axis rotation.

You should already know: Linear Newton's second law and net force analysis. Torque calculation about a fixed pivot axis. Kinematics of rotational motion.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Rotational Inertia and Rotational Newton's Second Law?

This is a core topic in Unit 6: Rotational Motion, which accounts for 12-18% of the total AP Physics 1 exam score per the official CED. It appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often combined with torque, energy, or momentum problems. Rotational inertia (also called moment of inertia, symbol $I$) is the rotational analog of mass in linear motion: it quantifies a rigid body’s resistance to angular acceleration about a given axis, just like inertial mass quantifies resistance to linear acceleration. Unlike mass, rotational inertia is not an intrinsic property of the object—it depends on both the object’s mass distribution and the location of the rotation axis. Rotational Newton’s second law extends Newtonian dynamics to rotating rigid bodies, relating net torque about an axis, rotational inertia, and angular acceleration. This topic is the foundation for all rotational dynamics problems you will encounter on the exam, from massive pulleys to rolling motion.

2. Calculating Rotational Inertia

Rotational inertia is defined from first principles for any system of point masses rotating about a fixed axis as the sum of each mass multiplied by the square of its perpendicular distance from the rotation axis: $$I=\sum m_i{r}_i^2$$ For continuous rigid bodies (like a solid disk or rod), this sum becomes an integral, but AP Physics 1 only requires you to remember common results for uniform symmetric objects about their center of mass axes: thin hoop ($M{R}^2$), solid disk ($\frac{1}{2}M{R}^2$), solid sphere ($\frac{2}{5}M{R}^2$), thin rod about center ($\frac{1}{12}M{L}^2$). A key intuition: rotational inertia scales with the square of distance from the axis, so mass far from the axis contributes far more to resistance to angular acceleration than mass close to the axis. For example, a hoop of mass $M$ and radius $R$ has twice the rotational inertia of a solid disk of the same mass and radius, because all of the hoop’s mass sits at distance $R$ from the center.

Worked Example

Three 0.5 kg point masses are mounted on a massless rigid rod: one at $x=-2$ m, one at $x=0$ m, and one at $x=3$ m. What is the rotational inertia of the system about an axis at $x=0$?

1. Identify each mass and its perpendicular distance from the axis: $m_1=0.5$ kg, $r_1=2$ m; $m_2=0.5$ kg, $r_2=0$ m; $m_3=0.5$ kg, $r_3=3$ m.
2. Substitute into the discrete rotational inertia formula: $I=m_1{r}_1^2+m_2{r}_2^2+m_3{r}_3^2$.
3. Calculate: $I=(0.5)(2^2)+(0.5)(0^2)+(0.5)(3^2)=2+0+4.5=6.5$ kg·m².
4. Confirm units: rotational inertia always has units of mass × length squared, which matches the result.

Exam tip: AP Physics 1 will never ask you to integrate to find rotational inertia for an unfamiliar continuous object. If you need a moment of inertia for a non-standard shape, it will always be given to you in the problem stem.

3. Parallel Axis Theorem

If you know the rotational inertia of an object about an axis through its center of mass, you can easily calculate the rotational inertia about any parallel axis without recalculating from scratch, using the parallel axis theorem: $$I=I_{cm}+M{d}^2$$ where $I_{cm}$ is the rotational inertia about the center of mass axis, $M$ is the total mass of the object, and $d$ is the perpendicular distance between the two parallel axes. The intuition for this formula is that rotation about an offset axis can be decomposed into rotation about the center of mass plus translation of the center of mass around the new axis, so the extra $M{d}^2$ term accounts for the translational contribution of the entire body around the new axis. This theorem is particularly useful for AP problems that ask for rotational inertia about an axis at the end of a rod or the edge of a disk.

Worked Example

A uniform solid disk has mass 4 kg and radius 0.5 m. What is its rotational inertia about an axis tangent to the edge of the disk, parallel to the disk’s central axis?

1. Recall the center of mass rotational inertia for a solid disk: $I_{cm}=\frac{1}{2}M{R}^2$.
2. The perpendicular distance between the central axis and the tangent axis equals the disk radius, so $d=R=0.5$ m.
3. Apply the parallel axis theorem: $I=I_{cm}+M{d}^2=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$.
4. Substitute values: $I=\frac{3}{2}(4)(0.5{)}^2=1.5$ kg·m².

Exam tip: You cannot use the parallel axis theorem to shift between two non-center-of-mass axes. Always shift from the center of mass rotational inertia to your target axis, no exceptions.

4. Rotational Newton's Second Law

Rotational Newton’s second law is the direct analog of linear Newton’s second law ($F_{net}=ma$) for fixed-axis rotation. It states that the net torque about a fixed axis equals the product of the rotational inertia about that axis and the resulting angular acceleration: $${\tau }_{net}=I\alpha$$ This is the central governing equation for all rotational dynamics problems. For connected systems (like a block hanging from a massive pulley), you can apply linear Newton’s second law to translating objects and rotational Newton’s second law to the rotating object, linking linear and angular acceleration via the no-slip relation $a=R\alpha$. For a given net torque, a larger rotational inertia produces a smaller angular acceleration, matching the intuitive behavior that more mass far from the axis makes it harder to speed up or slow down rotation.

Worked Example

A 2 kg block hangs from a massless string wrapped around a 1 kg solid disk pulley of radius 0.1 m. The pulley rotates about a fixed frictionless axis through its center. What is the angular acceleration of the pulley after the system is released from rest?

1. Draw free-body diagrams: for the block, downward weight $mg$ and upward tension $T$; for the pulley, only tension produces non-zero torque (weight and axis force act at the center, so their torque is zero).
2. Apply linear Newton’s second law to the block, taking downward as positive: $mg-T=ma$, where $a$ is the linear acceleration of the block.
3. Apply rotational Newton’s second law to the pulley: ${\tau }_{net}=TR=I\alpha$, where $I=\frac{1}{2}M{R}^2$ for the solid disk.
4. Use the no-slip relation $a=R\alpha$ to substitute $\alpha =a/R$ into the torque equation: $TR=\frac{1}{2}M{R}^2(a/R)=\frac{1}{2}MRa$, which simplifies to $T=\frac{1}{2}Ma$.
5. Substitute $T$ back into the block’s equation: $mg-\frac{1}{2}Ma=ma$, so $a=\frac{mg}{m+M/2}=7.84$ m/s².
6. Solve for angular acceleration: $\alpha =a/R=7.84/0.1=78.4$ rad/s².

Exam tip: Always calculate net torque and rotational inertia about the same axis. Mismatching axes is one of the most common errors on AP rotational dynamics problems.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using $I=M{R}^2$ for a solid disk, or $\frac{1}{2}M{R}^2$ for a hoop. Why: Students mix up the coefficients for common rotational inertia formulas because both share the same $M{R}^2$ form. Correct move: Make flashcards for the 5 common rotational inertia formulas, and quiz yourself daily to remember the correct coefficient for each shape.
• Wrong move: Applying the parallel axis theorem with a known rotational inertia about a non-center-of-mass axis. Why: Students misremember that the theorem requires the starting axis to pass through the center of mass. Correct move: Always start from the center of mass rotational inertia when calculating $I$ for a new parallel axis.
• Wrong move: Assuming tension equals the weight of a hanging block when the pulley has mass. Why: Students are used to massless pulleys and carry over the incorrect assumption that $T=mg$. Correct move: Always write a separate Newton’s second law equation for the hanging block to solve for tension, never assume $T=mg$.
• Wrong move: Writing $I=\sum m_i{r}_i$ instead of $I=\sum m_i{r}_i^2$ for discrete systems. Why: Students confuse the formula for center of mass position with the formula for rotational inertia. Correct move: Always explicitly write the exponent 2 for $r_i$ when setting up your calculation to avoid this error.
• Wrong move: Adding rotational inertia directly to mass when solving for acceleration of a block-pulley system. Why: Students take a shortcut and forget that rotational inertia has different units, so it cannot be added directly to mass. Correct move: Always write separate equations for linear and rotational motion, then eliminate variables step-by-step.

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

Two uniform spheres have the same total mass and same outer radius. Sphere A is solid with uniform mass distribution, while Sphere B has most of its mass concentrated near its outer surface. Which statement correctly compares the rotational inertia of the two spheres about an axis through their center of mass? A) $I_A>I_B$, because mass closer to the axis gives larger rotational inertia B) $I_A=I_B$, because they have the same total mass and same radius C) $I_B>I_A$, because mass farther from the axis gives larger rotational inertia D) $I_B>I_A$, because hollow objects always have smaller rotational inertia than solid objects of the same mass and radius

Worked Solution: Rotational inertia depends on the square of the distance of mass from the axis, so mass at larger $r$ contributes more to the total $I$. Both spheres have the same total mass and radius, but Sphere B has more mass at larger radii than the uniform solid Sphere A. Option A reverses the relationship, option B ignores the effect of mass distribution, and option D reverses the relationship between mass distribution and rotational inertia. The correct answer is C.

Question 2 (Free Response)

A uniform thin rod of mass $M=2$ kg and length $L=1.2$ m is pivoted at one end, held at rest in a horizontal position, and released to rotate freely about the pivot. (a) Calculate the rotational inertia of the rod about the pivot axis. (b) Find the net torque on the rod about the pivot immediately after release. (c) What is the angular acceleration of the rod immediately after release?

Worked Solution: (a) Use the parallel axis theorem starting from the center of mass rotational inertia of a rod: $I_{cm}=\frac{1}{12}M{L}^2$. The distance from the center of mass to the pivot is $d=L/2$. Substitute into the parallel axis theorem: $I=\frac{1}{12}M{L}^2+M(L/2{)}^2=\frac{1}{3}M{L}^2=\frac{1}{3}(2)(1.2{)}^2=0.96$ kg·m². (b) Weight acts at the center of mass, $L/2$ from the pivot. Immediately after release, the rod is horizontal, so the angle between weight (vertical) and the rod's position vector (horizontal) is 90°, so $\sin \theta =1$. Net torque: ${\tau }_{net}=(L/2)Mg=(0.6)(2)(9.8)=11.76$ N·m. (c) Apply rotational Newton’s second law: $\alpha =\frac{{\tau }_{net}}{I}=\frac{11.76}{0.96}=12.25$ rad/s² ≈ 12 rad/s².

Question 3 (Application / Real-World Style)

A playground merry-go-round is a uniform solid disk of mass 150 kg and radius 2 m. A 50 kg child stands at the edge of the merry-go-round. A parent pushes the edge of the merry-go-round with a constant tangential force of 100 N. Assuming friction is negligible, what is the tangential acceleration of the child along the edge of the merry-go-round?

Worked Solution: First, calculate total rotational inertia of the system (merry-go-round + child, treated as a point mass). $I_{disk}=\frac{1}{2}M{R}^2=0.5(150)(2^2)=300$ kg·m². $I_{child}=m{R}^2=50(2^2)=200$ kg·m². Total $I_{total}=300+200=500$ kg·m². Net torque from the parent’s force is ${\tau }_{net}=FR=100(2)=200$ N·m. Angular acceleration: $\alpha ={\tau }_{net}/I_{total}=200/500=0.4$ rad/s². Tangential acceleration: $a_t=R\alpha =2(0.4)=0.8$ m/s². In context: This means the child’s tangential speed increases by 0.8 m/s every second, which is a gentle, comfortable acceleration for a child riding the merry-go-round.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the non-negotiable foundation for all rotational dynamics that follows in Unit 6. Next, you will apply rotational Newton’s second law to rolling motion without slipping, where you combine rotational and linear motion for objects moving across a surface. Without mastering how to calculate rotational inertia and relate net torque to angular acceleration, you will not be able to correctly solve rolling motion problems, which frequently appear on the AP Physics 1 FRQ section. This topic also feeds into rotational kinetic energy and angular momentum, two other core concepts that make up a large portion of Unit 6’s exam weight. The follow-on topics to study next are:

• Rolling Motion
• Rotational Kinetic Energy
• Angular Momentum and Conservation