3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
A uniform solid disk of mass 4.00 kg and radius 0.500 m rotates at 12.0 rad/s about a fixed axle through its center, perpendicular to its face. What is the rotational kinetic energy of the disk?
Answer and reasoning
A36.0 JCorrect For a uniform solid disk about its central axis, I = (1/2)MR² = (1/2)(4.00 kg)(0.500 m)² = 0.500 kg·m². Then Krot = (1/2)Iω² = (1/2)(0.500 kg·m²)(12.0 rad/s)² = 36.0 J.
B72.0 J A student who treats all of the disk's mass as if it were at the rim picks this: I = MR² = 1.00 kg·m² gives (1/2)(1.00)(12.0)² = 72.0 J. Much of a solid disk's mass is closer to the axle than R, so I = (1/2)MR².
C3.00 J A student who leaves out the square of the angular speed picks this: (1/2)(0.500)(12.0) = 3.00 J. Rotational kinetic energy is proportional to ω², so the angular speed must be squared.
D0.00 J A student who thinks an object whose center of mass is at rest has no kinetic energy picks this. The disk's center does not move, but every other point of the disk moves in a circle, so the disk has kinetic energy (1/2)Iω².
Working I = (1/2)MR² = (1/2)(4.00 kg)(0.500 m)² = 0.500 kg·m². Krot = (1/2)Iω² = (1/2)(0.500 kg·m²)(12.0 rad/s)² = 36.0 J.
The diagram is a top view of a light rod that carries two small beads and rotates about a fixed vertical axle at the angular speed shown. The axle passes through the center of mass of the rod–bead system. What is the kinetic energy of the system?
Answer and reasoning
A0.00 J A student who thinks a system whose center of mass is at rest has no kinetic energy picks this. The center of mass stays on the axle, but both beads move in circles, and each has kinetic energy (1/2)mv².
B0.90 JCorrect Each bead moves in a circle with v = rω: the 0.30 kg bead at (0.40 m)(5.0 rad/s) = 2.0 m/s and the 0.60 kg bead at (0.20 m)(5.0 rad/s) = 1.0 m/s. Their kinetic energies, 0.60 J and 0.30 J, add to 0.90 J, the same as (1/2)Iω² with I = 0.072 kg·m². The center of mass is at rest, but the beads are moving.
C0.30 J A student who thinks the beads' kinetic energies cancel because the beads move in opposite directions picks this: 0.60 J − 0.30 J = 0.30 J. Kinetic energy is a scalar and never negative, so the two values add.
D1.80 J A student who puts the system's whole mass at the far end of the rod picks this: I = (0.90 kg)(0.40 m)² = 0.144 kg·m² gives (1/2)(0.144)(5.0)² = 1.80 J. Each bead contributes mr² at its own distance from the axle.
Working v₁ = (0.40 m)(5.0 rad/s) = 2.0 m/s, K₁ = (1/2)(0.30 kg)(2.0 m/s)² = 0.60 J. v₂ = (0.20 m)(5.0 rad/s) = 1.0 m/s, K₂ = (1/2)(0.60 kg)(1.0 m/s)² = 0.30 J. K = 0.90 J. Check: I = (0.30)(0.40)² + (0.60)(0.20)² = 0.072 kg·m², (1/2)(0.072)(5.0)² = 0.90 J. Center of mass: (0.30)(0.40) = (0.60)(0.20), so it is on the axle.
A motor turns a flywheel of rotational inertia I counterclockwise with angular speed ω. The motor is then reversed, and a short time later the flywheel turns clockwise with the same angular speed ω. Which claim about the flywheel's rotational kinetic energy is correct?
Answer and reasoning
AIt is the same before and after, as rotational kinetic energy has no direction.Correct Krot = (1/2)Iω² depends on the square of the angular speed, which is the same before and after, and kinetic energy is a scalar with no direction. Reversing the sense of rotation does not change it.
BIt changes sign, as a clockwise rotation gives the flywheel negative kinetic energy. A student who gives kinetic energy a sign set by the sense of rotation picks this. Kinetic energy is a scalar and is never negative; (1/2)Iω² is the same for either sense.
CIt increases by 2Iω², as the flywheel's angular velocity has changed by a total of 2ω. A student who finds the change in kinetic energy from the change in angular velocity picks this: (1/2)I(2ω)² = 2Iω². The kinetic energy before and after is (1/2)Iω² each time, so the change is zero.
DIt is greater after, as the motor has done work on the flywheel to reverse it. A student who treats all work as adding energy picks this. While slowing the flywheel the motor does negative work on it, and while speeding it up again it does positive work; with the same final angular speed the total is zero.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
6.1.A.1 Rotational kinetic energy, KrotFix
Rotational kinetic energy, Krot
The kinetic energy a rigid system has because it rotates: Krot = (1/2)Iω², where I is the system's rotational inertia about the axis of rotation and ω is its angular speed about that axis. Unit: joule (J).
Rotational inertia, I
A rigid system's resistance to changes in its rotation about a given axis, set by its mass and by how that mass is distributed relative to the axis: I = Σmi ri² for a collection of objects, or I = ∫r² dm for a continuous solid. Its value depends on the axis chosen. Unit: kg·m².
Angular speed, ω
The magnitude of a rigid system's angular velocity, the rate at which its angular position changes, ω = |dθ/dt|. Every point of a rigid system has the same angular speed. Unit: rad/s.
Kinetic energy of an object rotating about a fixed axis
Each piece of mass mi at distance ri from a fixed axis moves with linear speed vi = riω, so the object's total kinetic energy, the sum of the pieces' translational kinetic energies Σ(1/2)mi vi², equals (1/2)(Σmi ri²)ω² = (1/2)Iω². For rotation about a fixed axis, (1/2)Iω² with I about that axis is the object's total kinetic energy.
Total kinetic energy of a rigid system
The sum of the translational kinetic energy of the motion of its center of mass, (1/2)Mvcm², and its rotational kinetic energy about its center of mass, (1/2)Icm ω²: K = (1/2)Mvcm² + (1/2)Icm ω². Together the two terms are the whole kinetic energy, so no further translational or rotational term is added.
Rotational inertia about a pivot and the split of kinetic energy
For a rigid body rotating about a fixed pivot a distance d from its center of mass, the parallel axis theorem, Ipivot = Icm + Md², together with vcm = ωd, makes (1/2)Ipivot ω² equal to (1/2)Icm ω² + (1/2)Mvcm²: either expression gives the total kinetic energy, but not both added together.
Students often think The rotational inertia of any object is its total mass times the square of its size (its radius or length), as if all the mass were at the rim or far end. In fact No. That is true of a thin hoop about its central axis, whose mass is all at distance R, giving I = MR². In other shapes much of the mass is closer to the axis and contributes less, so a solid disk has I = (1/2)MR² and a uniform rod about one end has I = (1/3)Mℓ².
Students often think Kinetic energy is proportional to speed, so the square in Krot = (1/2)Iω² can be left out and doubling ω doubles the energy. In fact No. Krot = (1/2)Iω² is proportional to the square of the angular speed: doubling ω, with I unchanged, makes Krot four times as great.
6.1.A.2 Rotational kinetic energy with the center of mass at rest Fix
Rotational kinetic energy with the center of mass at rest
A rigid system spinning about an axis through its center of mass has zero center-of-mass velocity but nonzero kinetic energy, (1/2)Icm ω², because its individual points move with linear speeds rω.
Students often think An object whose center of mass is at rest has no kinetic energy, because the object as a whole is not going anywhere. In fact Yes. Every point not on the axis moves with linear speed rω and has kinetic energy. The object's kinetic energy is (1/2)Icm ω², even though its center of mass does not move.
6.1.A.3 Rotational kinetic energy as a scalar Fix
Rotational kinetic energy as a scalar
Rotational kinetic energy has magnitude only and is never negative. The kinetic energies of parts of a system add as plain numbers, whatever their senses of rotation or directions of motion.
Students often think Kinetic energy has a direction or sign set by the direction of motion, so parts moving in opposite directions, or turning in opposite senses, have kinetic energies that cancel. In fact No. Kinetic energy is a scalar and is never negative, so the kinetic energies of all parts add, whatever their directions of motion or senses of rotation.
Students often think The change in kinetic energy equals the kinetic energy calculated from the change in velocity, (1/2)m(Δv)² or (1/2)I(Δω)². In fact No. The change in kinetic energy is K₂ − K₁ = (1/2)Iω₂² − (1/2)Iω₁². When a wheel reverses from ω to the same angular speed in the opposite sense, Δω has magnitude 2ω, but K is the same before and after, so ΔK = 0.
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
A thin uniform rod of mass M and length ℓ, whose rotational inertia about an axis through its center and perpendicular to it is (1/12)Mℓ², rotates about that axis with angular speed ω. The same rod is then made to rotate about a parallel axis through one end, with angular speed ω/2. What is the ratio of the rod's new rotational kinetic energy to its original rotational kinetic energy?
Answer and reasoning
A0.25 A student who uses the same rotational inertia about both axes picks this: only ω² changes, by a factor of 1/4. Rotational inertia depends on the axis; about the end it is (1/3)Mℓ², four times the value about the center.
B2.00 A student who finds the rotational inertia correctly (four times as great) but takes kinetic energy to be proportional to ω picks this: 4 × 1/2 = 2.00. Krot depends on ω², so halving ω divides the energy by 4.
C1.00Correct About one end, the parallel axis theorem with d = ℓ/2 gives I = (1/12)Mℓ² + M(ℓ/2)² = (1/3)Mℓ², four times the value about the center. Halving ω multiplies ω² by 1/4. The ratio is 4 × 1/4 = 1.00.
D3.25 A student who uses the rod's length as d in the parallel axis theorem picks this: I = (1/12)Mℓ² + Mℓ² = (13/12)Mℓ², 13 times the value about the center, and 13 × 1/4 = 3.25. The distance between the axes is ℓ/2, so I = (1/3)Mℓ².
A light rod carries two small beads, of masses m₁ and m₂, on opposite sides of a fixed axle, at distances r₁ and r₂ from it, and rotates about the axle with angular speed ω. Student X finds the system's kinetic energy as ½m₁v₁² + ½m₂v₂², using each bead's speed. Student Y finds it as ½Iω², with I = m₁r₁² + m₂r₂². Which reasoning correctly supports a claim about the two results?
Answer and reasoning
AThey must be added: X's sum is the beads' energy of translation and Y's is their extra energy of rotation. A student who thinks a rotating object has rotational kinetic energy in addition to the translational kinetic energy of its parts picks this. The beads' motion is the rotation: (1/2)Iω² is the same energy as X's sum, and adding the two counts it twice.
BX's result is smaller: the beads move in opposite directions, so their energies partly cancel. A student who gives kinetic energy a direction picks this. Kinetic energy is a scalar and never negative, so the beads' kinetic energies add however they move, and X's sum equals (1/2)Iω².
CY's result is wrong: I is the total mass times the square of the center of mass's distance. A student who replaces the beads by their total mass at the center of mass picks this. Each bead contributes mr² at its own distance from the axle, so Y's I = m₁r₁² + m₂r₂² is correct, and (m₁ + m₂)rcm² is not the rotational inertia.
DThey are equal: each bead's speed is its distance from the axle times ω, so X's sum equals Y's result.Correct Each bead moves in a circle with v = rω, so (1/2)m₁v₁² + (1/2)m₂v₂² = (1/2)m₁r₁²ω² + (1/2)m₂r₂²ω² = (1/2)(m₁r₁² + m₂r₂²)ω² = (1/2)Iω². For rotation about a fixed axis, the sum of the parts' translational kinetic energies and the rotational kinetic energy are the same quantity.
Working v₁ = r₁ω and v₂ = r₂ω, so (1/2)m₁v₁² + (1/2)m₂v₂² = (1/2)(m₁r₁² + m₂r₂²)ω² = (1/2)Iω²: the two methods give the same kinetic energy.
A uniform disk of mass M and radius R swings in its own plane about a fixed horizontal axle through a point on its rim. The disk's rotational inertia about its center is (1/2)MR². At an instant when the disk's angular speed is ω, what is its total kinetic energy?
Answer and reasoning
A1.25MR²ω² A student who adds the translational kinetic energy of the center of mass to the rotational kinetic energy about the axle picks this: (1/2)(3/2)MR²ω² + (1/2)MR²ω² = 1.25MR²ω². With I about the axle, (1/2)Iaxle ω² is already the total; the rotational term that goes with (1/2)Mvcm² uses I about the center.
B0.25MR²ω² A student who uses the rotational inertia about the center for rotation about the axle on the rim picks this: (1/2)(1/2)MR²ω² = 0.25MR²ω². This leaves out the motion of the center of mass, which moves with speed ωR.
C2.25MR²ω² A student who measures d from the axle across the disk to its far edge, the diameter 2R, picks this: I = (1/2)MR² + M(2R)² = 4.5MR², so (1/2)Iω² = 2.25MR²ω². The distance from the center to the axle is R, so Iaxle = (3/2)MR².
D0.75MR²ω²Correct The center of mass is a distance R from the axle, so vcm = ωR and the translational kinetic energy is (1/2)M(ωR)² = 0.50MR²ω². The rotational kinetic energy about the center of mass is (1/2)(1/2)MR²ω² = 0.25MR²ω². The total is 0.75MR²ω², the same as (1/2)Iaxle ω² with Iaxle = (1/2)MR² + MR² = (3/2)MR².
A thin uniform rod of mass M and length ℓ can rotate with negligible friction about a fixed horizontal axle through one end. Its rotational inertia about its center is (1/12)Mℓ². The rod is released from rest in a horizontal position. What is its angular speed as it swings through the vertical position? (g is the magnitude of the acceleration due to gravity.)
Answer and reasoning
A√(g/ℓ) A student who puts all of the rod's mass at its far end picks this: (1/2)Mℓ²ω² = Mgℓ/2 gives ω² = g/ℓ. The mass is spread along the rod, so I about the end is (1/3)Mℓ².
B√(12g/ℓ) A student who uses the rotational inertia about the center for rotation about the end picks this: (1/2)(1/12)Mℓ²ω² = Mgℓ/2 gives ω² = 12g/ℓ. About the end, the parallel axis theorem gives (1/3)Mℓ².
C√(3g/ℓ)Correct The rod's center of mass falls ℓ/2, so the rod–Earth system loses gravitational potential energy Mgℓ/2. About the end, I = (1/12)Mℓ² + M(ℓ/2)² = (1/3)Mℓ², and (1/2)(1/3)Mℓ²ω² = Mgℓ/2 gives ω² = 3g/ℓ.
D√(3g/(2ℓ)) A student who drops the ½ from the rotational kinetic energy picks this: (1/3)Mℓ²ω² = Mgℓ/2 gives ω² = 3g/(2ℓ). The rod's kinetic energy is (1/2)Iω².
Working ΔUg = −Mg(ℓ/2) (center of mass falls ℓ/2). Iend = (1/12)Mℓ² + M(ℓ/2)² = (1/3)Mℓ². Energy: (1/2)(1/3)Mℓ²ω² = Mgℓ/2 → ω² = 3g/ℓ → ω = √(3g/ℓ). Distractors: I = Mℓ² → ω² = g/ℓ; I = (1/12)Mℓ² → ω² = 12g/ℓ; K = Iω² → ω² = 3g/(2ℓ).
The diagram shows three objects, each rotating about a fixed axis through its center, perpendicular to the page. Objects A and B are thin hoops and object C is a uniform solid disk. Which correctly ranks the rotational kinetic energies KA, KB and KC of the three objects?
Answer and reasoning
AKC > KA > KB A student who takes rotational inertia to depend only on mass gives A and C equal rotational inertias and B twice as much; then KA : KB : KC = 2 : 1 : 4.5, and the student picks this. Rotational inertia also depends on the square of the radius and on the shape, so B, with twice the mass at twice the radius, has eight times A's rotational inertia.
BKB > KC > KACorrect With a hoop's I = MR² and a disk's I = (1/2)MR²: KA = (1/2)(MR²)(2ω)² = 2MR²ω²; KB = (1/2)(2M)(2R)²ω² = 4MR²ω²; KC = (1/2)(1/2)MR²(3ω)² = 2.25MR²ω². So KB > KC > KA.
CKC > KB > KA A student who treats the disk as if all its mass were at its rim, I = MR², gets KC = 4.5MR²ω², larger than KB = 4MR²ω², and picks this. A solid disk's rotational inertia is (1/2)MR², so KC = 2.25MR²ω².
DKB > KA > KC A student who uses the correct rotational inertias but takes kinetic energy to be proportional to ω gets KA = MR²ω, KB = 4MR²ω and KC = 0.75MR²ω and picks this. Krot depends on ω², which makes C's three-times angular speed count nine times.
Working KA = (1/2)MR²(2ω)² = 2MR²ω²; KB = (1/2)(2M)(2R)²ω² = 4MR²ω²; KC = (1/2)(1/2)MR²(3ω)² = (9/4)MR²ω². KB > KC > KA.
Compiled from the AP Physics C: Mechanics 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
