5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
Which statement correctly describes what the rotational inertia of a rigid system about a given axis measures?
Answer and reasoning
AHow fast the system is rotating about the axis at a given instant A student who thinks rotational inertia measures how fast a system spins picks this. How fast it rotates is the angular velocity; rotational inertia measures how hard it is to change that angular velocity and does not depend on it.
BThe torque needed to keep the system rotating at a steady rate A student who thinks rotation needs a continual torque to keep going picks this. Rotational inertia describes resistance to a change in rotation and is measured in kg·m², not N·m; a system has the same rotational inertia whether or not any torque is exerted on it.
CHow strongly the system resists a change in its rate of rotationCorrect Rotational inertia is the rotational counterpart of mass as a measure of inertia: the greater it is, the harder it is to start, stop, speed up or slow down the system's rotation about that axis. It depends on the system's mass and on how far that mass is from the axis.
DThe total mass of the system, wherever that mass is located A student who thinks rotational inertia depends only on mass picks this. The distribution of the mass also matters: the same mass placed farther from the axis gives a greater rotational inertia.
A small object of mass m is fastened to a light rod a distance r from the rod's axis of rotation. It is replaced by a small object of mass m/3 fastened a distance 3r from the same axis. The new rotational inertia of the system about the axis is how many times the original rotational inertia?
Answer and reasoning
A×1 A student who takes I to be proportional to r, not r², gets ⅓ × 3 = 1. The distance is squared: tripling it multiplies I by 9, so the overall factor is ⅓ × 9 = 3.
B×3Correct I = mr². Dividing the mass by 3 multiplies I by ⅓, and tripling the distance multiplies it by 3² = 9, so the new rotational inertia is ⅓ × 9 = 3 times the original.
C×⅓ A student who thinks rotational inertia depends only on mass gets a factor of ⅓ from the smaller mass. Moving the object three times as far out multiplies I by 9, which outweighs the smaller mass.
D×9 A student who attends only to the distance gets a factor of 3² = 9. The mass has also been divided by 3, so the factor is 9 × ⅓ = 3.
Working Original: I = mr². New: I′ = (m/3)(3r)² = (m/3)(9r²) = 3mr². I′/I = 3.
The diagram shows three small objects fastened to a light rod, which rotates about an axis through its left end, O, perpendicular to the page. What is the rotational inertia of the system about this axis?
Answer and reasoning
A1.60 kg·m² A student who uses mr without squaring gets 0.20 + 0.80 + 0.60 = 1.60, which is in kg·m, not kg·m². Each distance must be squared: Σ m r² = 0.72 kg·m².
B1.44 kg·m² A student who uses the rod's length for every object puts the total 4.0 kg at 0.60 m: 4.0 × 0.60² = 1.44 kg·m². Each object has its own distance from O; only the right-hand object is 0.60 m away.
C0.64 kg·m² A student who places the total mass at the center of mass finds xcm = 0.40 m and gets 4.0 × 0.40² = 0.64 kg·m². Because each distance is squared, Σ m r² = 0.72 kg·m² is not the same as M rcm².
D0.72 kg·m²Correct Add mr² for each object about O: 1.0 × 0.20² + 2.0 × 0.40² + 1.0 × 0.60² = 0.040 + 0.32 + 0.36 = 0.72 kg·m². The light rod adds nothing.
Working Itot = Σ mi ri² = (1.0 kg)(0.20 m)² + (2.0 kg)(0.40 m)² + (1.0 kg)(0.60 m)² = 0.040 + 0.32 + 0.36 = 0.72 kg·m². The light rod contributes nothing.
A light rod of length L has a small object of mass 3m fastened at its left end and a small object of mass m fastened at its right end. The system can be made to rotate about any one of the axes A, B or C shown in the diagram, each perpendicular to the page. About which axis is the rotational inertia of the system least, and why?
Answer and reasoning
AAxis C, because it passes through the middle of the rod A student who thinks the least rotational inertia is always about the middle picks C. That is true only when the middle is the center of mass. Here more mass is at the left end, so the center of mass is at L/4, and IC = 3m(L/2)² + m(L/2)² = mL², more than IB = 0.75mL².
BAxis B, because it passes through the system's center of massCorrect The center of mass is (3m × 0 + m × L)/(4m) = L/4 from the left end, at axis B. For rotation in a plane, rotational inertia is least about the axis through the center of mass: IB = 3m(L/4)² + m(3L/4)² = 0.75mL², compared with mL² about both A and C.
CAxis A, because the heavier object then adds nothing to the total A student who expects the least value when the heavier object is on the axis picks A. The 3m object then contributes nothing, but the m object is a full L away: IA = mL², more than the 0.75mL² about B.
DIt is the same about all three, because the masses are unchanged A student who thinks rotational inertia is fixed by the masses alone picks this. Rotational inertia depends on the axis: moving the axis changes every object's distance from it, giving mL² about A and about C but only 0.75mL² about B.
Working xcm from the left end = (3m × 0 + m × L)/(4m) = L/4, the position of axis B. IA = 3m(0)² + m(L)² = mL². IB = 3m(L/4)² + m(3L/4)² = (3/16 + 9/16)mL² = 0.75mL². IC = 3m(L/2)² + m(L/2)² = mL². The least is about B.
A uniform rod of mass 2.0 kg and length 1.2 m has a rotational inertia of 0.24 kg·m² about an axis through its center, perpendicular to the rod. What is its rotational inertia about a parallel axis through one end of the rod?
Answer and reasoning
A0.72 kg·m² A student who treats the rod as a single object at its center of mass uses only Md² = 2.0 × 0.60² = 0.72 kg·m². The rod's own rotational inertia about its center, 0.24 kg·m², must be added.
B1.44 kg·m² A student who forgets to square d gets 0.24 + 2.0 × 0.60 = 1.44, which mixes kg·m² with kg·m. The theorem uses d²: 0.24 + 2.0 × 0.36 = 0.96 kg·m².
C3.12 kg·m² A student who takes d to be the rod's full length gets 0.24 + 2.0 × 1.2² = 3.12 kg·m². d is the distance between the two axes, from the center of the rod to its end: 0.60 m.
D0.96 kg·m²Correct The axis through the end is parallel to the axis through the center, a distance d = 1.2/2 = 0.60 m away. I′ = Icm + Md² = 0.24 + 2.0 × 0.60² = 0.24 + 0.72 = 0.96 kg·m².
Working The axis through the end is parallel to the axis through the center of mass, a distance d = 1.2 m/2 = 0.60 m away. I′ = Icm + Md² = 0.24 kg·m² + (2.0 kg)(0.60 m)² = 0.24 + 0.72 = 0.96 kg·m².
In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
5.4.A.1 Rotational inertia (I) Fix
Rotational inertia (I)
A measure of a rigid system's resistance to changes in its rotation about a given axis. It depends on the system's mass and on how that mass is distributed relative to the axis. SI unit: kg·m².
Mass distribution relative to the axis
How far the parts of a system's mass are from the axis of rotation. Of two systems with equal mass, the one with more of its mass farther from the axis has the greater rotational inertia; this is why a hoop has more rotational inertia than a solid disk of the same mass and radius about their central axes.
Students often think Rotational inertia depends only on the mass of a system, so systems of equal mass have equal rotational inertia. In fact No. Rotational inertia depends on how the mass is distributed relative to the axis as well as on how much mass there is. Mass farther from the axis contributes more, so a hoop has more rotational inertia than a solid disk of the same mass and radius.
Students often think Rotational inertia measures how fast a system is rotating, so it increases when the system spins faster. In fact No. How fast a system rotates is its angular velocity. Rotational inertia measures how strongly the system resists a change in its angular velocity, and for a rigid system about a fixed axis it does not change when the system speeds up or slows down.
5.4.A.2 Rotational inertia of a small object Fix
Rotational inertia of a small object
For an object small enough to be treated as a point, a perpendicular distance r from the axis: I = mr². Because r is squared, doubling the distance multiplies the rotational inertia by four.
Perpendicular distance from the axis (r)
The shortest distance from the axis of rotation to an object, measured perpendicular to the axis (SI unit: m). It is measured separately for each object in a system.
Students often think The rotational inertia of an object is proportional to its distance from the axis, I = mr, so doubling the distance doubles I. In fact No. I = mr², so doubling r multiplies I by four.
Students often think Rotational inertia depends only on how far the mass is from the axis, not on how much mass there is. In fact No. I = mr² is proportional to m as well as to r², so two objects at the same distance have rotational inertias in the ratio of their masses.
5.4.A.3 Total rotational inertia of a collection of objects Fix
Total rotational inertia of a collection of objects
The sum of the rotational inertias of the objects about the same axis: Itot = Σ Ii = Σ mi ri². Every term is positive or zero, whichever side of the axis an object is on; an object on the axis contributes zero.
Light rod
A connecting rod whose mass is negligible, so that it adds nothing to the rotational inertia of the system it is part of.
Students often think The rotational inertia of a system equals that of a single object with the system's total mass located at its center of mass, I = M rcm². In fact No. Each object contributes mi ri² about the axis, and because the distances are squared, the sum is not M rcm². Two equal objects on opposite sides of an axis have their center of mass on the axis, yet their rotational inertia about it is not zero.
Students often think Rotational inertia depends on how far apart the objects in a system are from each other, not on their distances from the axis. In fact No. Each object's contribution depends on its own distance from the axis. Objects that are far apart can still have a small rotational inertia about an axis between them, and objects close together have a large one if both are far from the axis.
5.4.B.1 Axis through the center of mass Fix
Axis through the center of mass
For rotation in a given plane, a rigid system's rotational inertia is least about the axis through its center of mass; about any parallel axis it is greater.
Students often think Rotational inertia is a fixed property of an object, like its mass, and is the same about every axis. In fact No. Rotational inertia is always about a particular axis. For rotation in a given plane it is least about the axis through the center of mass and greater about any parallel axis.
Students often think The rotational inertia of a system is least about an axis through its geometric middle, whatever the distribution of its mass. In fact Only if the middle is also the center of mass. For a system with more mass toward one end, the least rotational inertia is about the axis through the center of mass, which is nearer the heavier end.
5.4.B.2 Parallel axis theorem Fix
Parallel axis theorem
I′ = Icm + Md², where Icm is the rotational inertia about an axis through the center of mass, M is the system's total mass and d is the perpendicular distance between that axis and the parallel axis of interest. The Md² term is never negative, so I′ is never less than Icm.
Rotational inertia of an extended rigid object
For an extended object, such as a uniform rod, disk or hoop, the rotational inertia depends on its shape and is given where needed (for example, ½MR² for a uniform disk about its central axis); AP Physics 1 students are not expected to know these values, only to use them.
Students often think In I′ = Icm + Md², d is a size of the object, such as its length, radius or diameter, rather than the distance between the two axes. In fact No. d is the perpendicular distance between the axis through the center of mass and the new, parallel axis. For an axis through the end of a uniform rod, d is half the rod's length; for an axis halfway between a disk's center and its rim, d is half the radius.
5 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 5
Each of the three systems shown in the diagram is a small object fastened to one end of a light rod. Each rod rotates about an axis through its other end, perpendicular to the page. The masses of the objects and their distances from the axes are shown. Which ranking of the systems' rotational inertias, I₁, I₂ and I₃, is correct?
Answer and reasoning
AI₃ > I₁ = I₂Correct Each rotational inertia is I = mr². System 1: 4m × r² = 4mr². System 2: m × (2r)² = 4mr². System 3: 2m × (2r)² = 8mr². Doubling the distance multiplies I by four, which exactly makes up for system 2's mass being a quarter of system 1's.
BI₁ = I₃ > I₂ A student who uses I = mr, without squaring the distance, gets 4mr, 2mr and 4mr. Because the distance is squared, system 2's rotational inertia is 4mr², equal to system 1's, and system 3's is 8mr², twice system 1's.
CI₁ > I₃ > I₂ A student who ranks by mass alone gets 4m > 2m > m. Distance from the axis matters too, and it is squared: system 3's object is twice as far out as system 1's, which more than makes up for its smaller mass.
DI₂ = I₃ > I₁ A student who ranks by distance alone puts systems 2 and 3, whose objects are both 2r from the axis, above system 1. The masses matter too: system 3's object has twice the mass of system 2's, so I₃ = 8mr², while I₂ = 4mr², the same as I₁.
Working The rods are light, so each system's rotational inertia is that of its small object, I = mr². I₁ = (4m)r² = 4mr². I₂ = m(2r)² = 4mr². I₃ = (2m)(2r)² = 8mr². So I₃ > I₁ = I₂.
Two identical small objects, each of mass m, are fastened to a light rod that can rotate about an axis through point O, perpendicular to the rod. In arrangement 1, the objects are on opposite sides of O, each a distance d from O. In arrangement 2, both objects are on the same side of O, one a distance d from O and the other a distance 2d from O. Which claim about the rotational inertias of the two arrangements about O, with its reasoning, is correct?
Answer and reasoning
AArrangement 1's is greater, as its objects are farther from each other. A student who thinks rotational inertia depends on how far apart the objects are picks this: they are 2d apart in arrangement 1 and d apart in arrangement 2. What matters is each object's distance from the axis at O, and in arrangement 2 one object is 2d from O.
BThey are equal, as the two arrangements have the same total mass. A student who thinks rotational inertia depends only on mass picks this. The masses are the same, but in arrangement 2 one object is farther from O, and its contribution, m(2d)², is four times as large as md².
CArrangement 2's is greater, as one object is at d and the other at 2d from O.Correct Each object contributes mr² about O, and every contribution is positive whichever side of O the object is on. I₁ = md² + md² = 2md²; I₂ = md² + m(2d)² = 5md². Moving one object from d out to 2d quadruples its contribution, so arrangement 2's rotational inertia is greater.
DArrangement 1's is zero, as its center of mass is located at O. A student who treats the system as a single object at its center of mass gets zero for arrangement 1, whose center of mass is at O. Each object contributes md² whichever side of O it is on, so I₁ = 2md², not zero.
Working I₁ = md² + md² = 2md². I₂ = md² + m(2d)² = 5md². So arrangement 2 has the greater rotational inertia.
A uniform disk of mass M and radius R has a rotational inertia of ½MR² about an axis through its center, perpendicular to the disk. Which expression gives its rotational inertia about a parallel axis through a point halfway between the center and the rim?
Answer and reasoning
A0.75MR²Correct The new axis is parallel to the central axis, a distance d = R/2 from it, so I′ = Icm + Md² = 0.50MR² + M(R/2)² = 0.50MR² + 0.25MR² = 0.75MR². It is greater than ½MR², as it must be for an axis that does not pass through the center of mass.
B0.50MR² A student who thinks an object has the same rotational inertia about every axis keeps ½MR². The new axis does not pass through the center of mass, so the rotational inertia is greater, by Md² = M(R/2)².
C0.25MR² A student who treats the disk as a single object at its center, R/2 from the axis, gets M(R/2)² = 0.25MR². That leaves out the disk's own rotational inertia about its center, ½MR², which must be added.
D1.50MR² A student who takes d to be the disk's radius, the size of the object, gets ½MR² + MR² = 1.50MR². d is the distance between the two axes, from the center of the disk to the new axis: R/2.
Working The new axis is parallel to the central axis, a distance d = R/2 away. I′ = Icm + Md² = ½MR² + M(R/2)² = 0.50MR² + 0.25MR² = 0.75MR².
A bicycle wheel can rotate about its axle. Which change would increase the wheel's rotational inertia about the axle?
Answer and reasoning
AMoving some of its mass from the rim in toward the hub A student who thinks mass far from the axis is easier to turn, because of leverage, picks this. Leverage applies to where a force is exerted, not to where mass is; moving mass toward the hub reduces the rotational inertia.
BSpinning the wheel at a faster rate about the same axle A student who thinks rotational inertia measures how fast the wheel spins picks this. Spinning faster increases the wheel's angular velocity; its rotational inertia depends only on its mass and how that mass is placed, so it does not change.
CPushing on the rim, rather than near the hub, to turn it A student who thinks rotational inertia depends on how the wheel is pushed picks this. Where the push is exerted changes the torque, not the rotational inertia, which is set by the wheel's mass and its distribution about the axle.
DMoving some of its mass from the hub out to the rimCorrect Rotational inertia depends on how far the mass is from the axis. Moving mass from the hub to the rim puts it farther from the axle, so each piece moved contributes more (mr² for each piece) and the total increases.
Two small objects, each of mass m, are fixed to the ends of a light rod of length 2L. Which expression gives the rotational inertia of the system about an axis perpendicular to the rod through the point halfway between one object and the rod's center?
Answer and reasoning
A0.5mL² A student who puts the whole mass, 2m, at the center of mass, L/2 from the axis, gets 2m(L/2)² = 0.5mL² and picks this. Each object contributes m r² with its own distance; because the distances are squared, the far object's 2.25mL² dominates.
B2.5mL²Correct The axis is L/2 from the near object and 3L/2 from the far one. I = m(L/2)² + m(3L/2)² = 0.25mL² + 2.25mL² = 2.5mL². The parallel axis theorem gives the same result: 2mL² about the center plus 2m(L/2)².
C2.0mL² A student who thinks the system has the same rotational inertia about every axis uses the value about the rod's center, 2mL², and picks this. The axis here is not through the center of mass, so the rotational inertia is greater than 2mL².
D8.0mL² A student who takes r in I = mr² to be the length of the rod, 2L, for each object gets 2m(2L)² = 8.0mL² and picks this. r is each object's own distance from the axis: L/2 for one and 3L/2 for the other.
Working The axis is L/2 from the near object and 2L − L/2 = 3L/2 from the far object. I = Σ m r² = m(L/2)² + m(3L/2)² = 0.25mL² + 2.25mL² = 2.5mL². (Check with the parallel axis theorem: about the center, Icm = 2mL²; the axis is d = L/2 from the center, so I′ = 2mL² + 2m(L/2)² = 2.5mL².) Errors: total mass at the center of mass, L/2 from the axis → 2m(L/2)² = 0.5mL²; value about the center taken for every axis → 2mL²; each object put a rod's length, 2L, from the axis → 2m(2L)² = 8.0mL².
Compiled from the AP Physics 1 Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
