AP® Physics 1: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics/units/5/5-4)
Unit 5 · Topic 5.4
5.4 Rotational Inertia
Rotational inertia I measures how hard it is to change an object's rotation. It depends on the object's mass and on how far that mass sits from the axis: I = mr² for a small object, added up for a group. Of all the axes pointing in one direction, the one through the center of mass gives the smallest I, and the parallel axis theorem gives I about any axis parallel to it.
Key terms
- rotational inertia
- point object
- mass distribution
- center of mass
- parallel axis theorem
Mass and where it sits
Mass tells you how hard it is to change straight-line motion. Rotational inertia does the same job for rotation, but location matters as much as amount. A small mass far from the axis can resist spinning more than a big mass close to it.
Try spinning a broom about its long handle, then about an axis across its middle. Same broom, same mass, but the second is much harder, because more of the mass sits far from the axis. Rotational inertia always belongs to an object about a particular axis. Its unit is kg·m².
Point objects and collections
A small object of mass m at a perpendicular distance r from the axis has I = mr². The r is squared, so doubling the distance makes I four times as large.
For a collection of objects, add each one's contribution about the same axis: I = Σmr² = m₁r₁² + m₂r₂² + …. An object sitting on the axis has r = 0 and adds nothing. On the exam you'll only calculate this for five or fewer objects laid out in a flat (two-dimensional) arrangement.
Extended objects
For solid shapes like disks and rods, the exam gives you the formula when you need it. You don't need to memorize these, but you should know why they rank the way they do.
A hoop has all its mass at the rim, so I = MR². A solid disk of the same mass and radius has much of its mass near the center, so its I is only half as big. Spread mass outward and I goes up.
| Shape (mass M) | Axis | I (given on the exam) |
|---|---|---|
| Thin hoop, radius R | through center, perpendicular to hoop | MR² |
| Solid disk or cylinder, radius R | through center, along the cylinder's axis | ½MR² |
| Solid sphere, radius R | through center | (2/5)MR² |
| Thin rod, length L | through center, perpendicular to rod | (1/12)ML² |
| Thin rod, length L | through one end, perpendicular to rod | (1/3)ML² |
Changing the axis: the parallel axis theorem
Compare axes that all point the same way (all parallel). Among them, the one through the center of mass gives the smallest rotational inertia. Slide the axis sideways to any other parallel position and, on balance, the mass ends up farther away, so I grows. (An axis pointing a different way is a separate comparison: the broom has a much smaller I about its handle than across its middle, though both axes pass close to its center of mass.)
The parallel axis theorem tells you by how much: I′ = I_cm + Md². I_cm is the rotational inertia about an axis through the center of mass, M is the total mass and d is the distance between that axis and the new parallel one. The rod in the table shows it: (1/12)ML² + M(L/2)² = (1/3)ML².
Measuring rotational inertia
In a lab you can find I from its effect. Apply several known torques to a wheel, measure each angular acceleration (from video or a rotary sensor), and plot torque against angular acceleration. Since τ = Iα (5.6), the slope of the best-fit line is I. Plotting the data this way uses every trial and averages out random error.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Three masses on a rod, two axes
A light rod 1.0 m long holds a 2.0 kg mass at x = 0, a 1.0 kg mass at x = 0.50 m and a 3.0 kg mass at x = 1.0 m. Find the rotational inertia about an axis perpendicular to the rod (a) through x = 0 and (b) through the center of mass. (c) Check your answers with the parallel axis theorem.
Show the solutionHide the solution
- Step 1: (a) I = Σmr² = (2.0)(0)² + (1.0)(0.50)² + (3.0)(1.0)² = 0 + 0.25 + 3.0 = 3.25 kg·m².
- Step 2: (b) Center of mass: x_cm = (2.0·0 + 1.0·0.50 + 3.0·1.0)/6.0 = 3.5/6.0 ≈ 0.583 m. Distances from it: 0.583 m, 0.083 m and 0.417 m.
- Step 3: I_cm = (2.0)(0.583)² + (1.0)(0.083)² + (3.0)(0.417)² ≈ 0.681 + 0.007 + 0.521 ≈ 1.21 kg·m².
- Step 4: (c) I′ = I_cm + Md² = 1.21 + (6.0)(0.583)² ≈ 1.21 + 2.04 = 3.25 kg·m². It matches (a), and I_cm is smaller, as it must be.
Answer: (a) 3.25 kg·m² (b) about 1.21 kg·m² (c) 1.21 + 2.04 = 3.25 kg·m²
- Example 2Calculator allowed
A rod about its end
A uniform rod has mass 1.2 kg and length 0.90 m. Its rotational inertia about its center is I_cm = (1/12)ML². Find its rotational inertia about an axis through one end, perpendicular to the rod.
Show the solutionHide the solution
- Step 1: I_cm = (1/12)(1.2)(0.90)² = 0.081 kg·m².
- Step 2: The end is d = L/2 = 0.45 m from the center.
- Step 3: I′ = I_cm + Md² = 0.081 + (1.2)(0.45)² = 0.081 + 0.243 = 0.324 kg·m².
- Step 4: Check: (1/3)ML² = (1/3)(1.2)(0.81) = 0.324 kg·m².
Answer: About 0.32 kg·m², four times I_cm
- Example 3Calculator allowed
Four masses, three axes (classic trap)
Four 0.50 kg balls sit at the corners of a square with 0.40 m sides, joined by light rods. Find I about (a) an axis through the center, perpendicular to the square, (b) an axis through one corner, perpendicular to the square, and (c) an axis along one side.
Show the solutionHide the solution
- Step 1: (a) Each ball is half a diagonal from the center: r = 0.40/√2 ≈ 0.283 m, so r² = 0.080 m². I = 4(0.50)(0.080) = 0.16 kg·m².
- Step 2: (b) Distances from a corner: 0 (the ball on the axis), 0.40 m, 0.40 m and the diagonal 0.40√2 ≈ 0.566 m. I = 0.50(0 + 0.16 + 0.16 + 0.32) = 0.32 kg·m². Parallel axis check: 0.16 + (2.0)(0.080) = 0.32 kg·m².
- Step 3: (c) The two balls on the axis add nothing. The other two are 0.40 m away (measured perpendicular to the axis): I = 2(0.50)(0.40)² = 0.16 kg·m².
- Step 4: The trap is using the same r for every ball. Always measure each r perpendicular to the axis you're given.
Answer: (a) 0.16 kg·m² (b) 0.32 kg·m² (c) 0.16 kg·m²
Common mistakes
- Forgetting to square r, or measuring r from the end of the object instead of from the axis.
- Thinking rotational inertia depends only on mass. A hoop and a disk of equal mass and radius have different I.
- Using the parallel axis theorem from an axis that isn't through the center of mass. I_cm must be the center-of-mass value.
- Giving an object one fixed I. Always say which axis you mean.
On the exam
- Ranking questions are common: same mass, different shapes or axes. The object with more mass farther from the axis has the larger I.
- In experimental questions, a graph of torque against angular acceleration with slope I is a standard way to find rotational inertia.
Connected topics
Videos
Check yourself
4 questions on 5.4 Rotational Inertia. Pick an answer to see if you got it, and why.
Three small 0.50 kg beads are fixed to a light rod at distances of 0.20 m, 0.40 m and 0.60 m from an axis at one end of the rod. Treat the beads as point objects. What is the rotational inertia of the system about that axis?
Two small 2.0 kg spheres are attached to the ends of a light rod 0.80 m long. Treat the spheres as point objects and ignore the rod's mass. The system can rotate in a horizontal plane about a vertical axis.
Described scenario
What is the system's rotational inertia about an axis through the center of the rod?
What is the system's rotational inertia about an axis through one of the spheres?
A third 2.0 kg sphere is attached at the exact center of the rod. How does this change the rotational inertia about the axis through the center?
0 of 4 answered