Unit 5 · Topic 5.4

5.4 Rotational Inertia

Rotational inertia I is an object's resistance to changes in its spin. It depends on the mass and on how far that mass is from the axis: $I = \sum mr^2$ for particles, $I = \int r^2\,dm$ for solid objects, and $I = I_{\text{cm}} + Md^2$ for an axis that doesn't pass through the center of mass.

Key terms

rotational inertia
moment of inertia
linear mass density
parallel axis theorem
center of mass

What changed in the 2024 update

Before the Fall 2024 course update, this topic was part of one big Rotation unit (old Unit 5). That unit is now split in two: Unit 5 covers spinning motion, torque and rotational inertia, and Unit 6 covers rotational energy, angular momentum and rolling. Older videos and practice labeled "Unit 5: Rotation" still fit, but they may mix in Unit 6 ideas.

Mass and where it sits

In straight-line motion, mass makes an object hard to accelerate. In rotation, the same job is done by rotational inertia (also called moment of inertia). For a small object of mass m at distance r from the axis, $I = mr^2$ . The unit is kg·m².

Because r is squared, mass far from the axis counts much more than mass near it. That's why a hoop has more rotational inertia than a solid disk of the same mass and radius: all the hoop's mass is at the rim.

For a system of several objects, add their rotational inertias about the same axis: $I = \sum m_i r_i^2$ . The same object has a different I about a different axis.

Solid objects: integrate

For a continuous object, split it into tiny pieces of mass dm and add up $r^2\,dm$ : $I = \int r^2\,dm$ where r is each piece's perpendicular distance from the axis.

For a thin rod along the x-axis, use the linear mass density λ (mass per length): $dm = \lambda\,dx$ . If the rod is uniform, $\lambda = \dfrac{M}{L}$ . If it isn't, λ is a function of x, and you find the total mass with $M = \int \lambda\,dx$ .

For a disk or a ring-shaped object around its central axis, use thin rings. A ring of radius r and width dr has area $2\pi r\,dr$ , so $dm = \sigma\,2\pi r\,dr$ , where σ is the mass per area. For a uniform disk, $\int_0^R r^2\,\sigma\,2\pi r\,dr = \dfrac{\pi\sigma R^4}{2} = \dfrac{1}{2}MR^2$ .

Values worth knowing

The calculus derivations you're expected to do are limited to thin rods (uniform or not) about an axis perpendicular to the rod, and to thin hoops, cylindrical shells, disks and annular rings about their central axis. You can use other values, like a solid sphere's, when a problem gives them.

Object and axis	Rotational inertia
point mass at distance r	mr²
thin hoop or cylindrical shell, central axis	MR²
uniform solid disk or cylinder, central axis	½MR²
uniform thin rod, axis through center	(1/12)ML²
uniform thin rod, axis through one end	(1/3)ML²
uniform solid sphere, through center (given when needed)	(2/5)MR²

The parallel axis theorem

Of all the axes pointing in the same direction, the one through the center of mass gives the smallest rotational inertia. For any axis parallel to that one, a distance d away, $I = I_{\text{cm}} + Md^2$ where M is the total mass.

Example: a uniform rod about its center has $I_{\text{cm}} = \tfrac{1}{12}ML^2$ . Move the axis to the end, a distance L/2 away: $\tfrac{1}{12}ML^2 + M\left(\tfrac{L}{2}\right)^2 = \tfrac{1}{3}ML^2$ . This matches the integral.

The two axes must be parallel, and one of them must pass through the center of mass. You can't hop from one off-center axis to another with a single Md² step.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Two masses on a light rod
A light rod has a 2.0 kg ball 0.50 m from its left end and a 3.0 kg ball at its right end, 1.0 m from the left end. Find the rotational inertia about an axis through the left end, perpendicular to the rod.
Show the solutionHide the solution
1. Step 1: Treat each ball as a point mass and add: $I = m_1 r_1^2 + m_2 r_2^2$ .
2. Step 2: $I = (2.0)(0.50)^2 + (3.0)(1.0)^2 = 0.50 + 3.0 = 3.5$ kg·m².
3. Step 3: The ball at the end contributes six times as much, even though it's only 1.5 times as heavy, because r is squared.
Answer: 3.5 kg·m²
Example 2
Deriving I for a uniform rod about its end
Show that a uniform thin rod of mass M and length L has rotational inertia ⅓ML² about an axis through one end, perpendicular to the rod.
Show the solutionHide the solution
1. Step 1: Put the axis at x = 0 and the rod from x = 0 to x = L. The rod is uniform, so $\lambda = \dfrac{M}{L}$ and $dm = \dfrac{M}{L}\,dx$ .
2. Step 2: Each piece is a distance x from the axis: $I = \displaystyle\int_0^L x^2 \dfrac{M}{L}\,dx = \dfrac{M}{L}\cdot\dfrac{L^3}{3} = \dfrac{1}{3}ML^2$ .
3. Step 3: Check with the parallel axis theorem: $\tfrac{1}{12}ML^2 + M\left(\tfrac{L}{2}\right)^2 = \tfrac{1}{3}ML^2$ . It matches.
Answer: I = ⅓ML²
Example 3
Nonuniform rod (exam-level)
A thin rod of length L lies along the x-axis from x = 0 to x = L. Its linear mass density is $\lambda = \beta x$ , where β is a constant. Find its total mass M and its rotational inertia about x = 0, in terms of M and L.
Show the solutionHide the solution
1. Step 1: Total mass: $M = \displaystyle\int_0^L \beta x\,dx = \dfrac{\beta L^2}{2}$ , so $\beta = \dfrac{2M}{L^2}$ .
2. Step 2: Rotational inertia: $I = \displaystyle\int_0^L x^2(\beta x)\,dx = \dfrac{\beta L^4}{4}$ .
3. Step 3: Substitute β: $I = \dfrac{2M}{L^2}\cdot\dfrac{L^4}{4} = \dfrac{1}{2}ML^2$ .
4. Step 4: This is more than ⅓ML² for a uniform rod, because more of the mass sits near the far end. That's a good check on your answer.
Answer: M = βL²/2 and I = ½ML²

Common mistakes

Using the distance to the center of mass instead of each piece's perpendicular distance to the axis.
Applying the parallel axis theorem between two axes when neither passes through the center of mass.
Writing dm = (M/L) dx for a nonuniform rod. Use the given λ(x), and find M by integrating it.
Thinking only the total mass matters. Two objects with the same mass can have very different I.

On the exam

Expect a derivation of I for a rod with a given λ(x). Show dm = λ dx, the integral with limits, and the result in terms of M and L.
Qualitative questions ask what happens to I when mass moves or the axis changes. Mass moving outward raises I; an axis through the center of mass gives the smallest I.

Connected topics

Videos

Topic 5.4 - Rotational Inertia
Lessons With LondotWatch on YouTube (opens in a new tab)
Calculating Rotational Inertia with Integrals | AP Physics C - Unit 5 - Lesson 6C
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Using Integrals to Derive Rotational Inertia of a Long, Thin Rod with Demonstration
Flipping PhysicsWatch on YouTube (opens in a new tab)
AP Physics C - Moment of Inertia
Dan Fullerton (APlusPhysics)Watch on YouTube (opens in a new tab)
29.4 Parallel Axis Theorem
MIT OpenCourseWareWatch on YouTube (opens in a new tab)
Uniform Solid Cylinder Moment of Inertia Derivation
Flipping PhysicsWatch on YouTube (opens in a new tab)

Check yourself

5 questions on 5.4 Rotational Inertia. Pick an answer to see if you got it, and why.

Question 1 of 5Calculator allowed

Three small 0.50 kg balls are fixed to a light rod at distances 0.20 m, 0.40 m and 0.60 m from an axis at one end of the rod. What is the rotational inertia of the system about that axis?

Question 2 of 5Calculator allowed

A uniform rod has mass 3.0 kg and length 2.0 m. About an axis through its center, perpendicular to the rod, its rotational inertia is $\frac{1}{12}ML^2 = 1.0$ kg·m². What is its rotational inertia about a parallel axis 0.50 m from the center?

A thin rod of length L lies along the x-axis from x = 0 to x = L. Its linear mass density is $\lambda(x) = \beta x$ , where β is a positive constant, so the rod gets heavier toward x = L. Its total mass is M.

Described situation

Question 3 of 5Calculator allowed

Which expression gives β in terms of M and L?

Question 4 of 5Calculator allowed

What is the rod's rotational inertia about an axis through x = 0, perpendicular to the rod?

Question 5 of 5Calculator allowed

How does the rod's rotational inertia about a perpendicular axis through x = L compare with its value about x = 0?

0 of 5 answered