3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
Two small spheres of equal mass are fixed to a light rod that rotates about a fixed axis through the rod's center, perpendicular to the rod. The rotational kinetic energy of the rod–spheres system is K0. The spheres are moved to three times their original distances from the axis, and the rod is then set rotating at half its original angular speed. Modeling the spheres as objects, the new rotational kinetic energy equals K0 multiplied by which factor?
Answer and reasoning
A0.25 A student who thinks rotational inertia depends only on mass keeps I the same and multiplies K only by (1/2)² = 1/4. Moving the same masses to three times the distance makes I = Σ m r² nine times as great.
B2.25Correct Each sphere's contribution m r² becomes m(3r)² = 9mr², so I is multiplied by 9. Halving ω multiplies ω² by 1/4. K = (1/2)Iω² is therefore multiplied by 9 × 1/4 = 2.25.
C0.75 A student who takes rotational inertia to be proportional to distance multiplies I by 3 and gets 3 × 1/4 = 0.75. The distance in I = mr² is squared, so tripling it multiplies I by 9.
D4.50 A student who treats kinetic energy as proportional to Iω, like momentum mv, gets 9 × 1/2 = 4.5. Rotational kinetic energy depends on ω², so halving ω divides K by 4, not by 2.
Working I = Σ m r²: r → 3r gives I → 9I. ω → ω/2 gives ω² → ω²/4. K = (1/2)Iω² → (9/4)K0 = 2.25K0. Errors: I unchanged → 1/4 = 0.25; I ∝ r → 3/4 = 0.75; K ∝ Iω → 9/2 = 4.50.
A uniform grinding wheel spins at constant angular velocity on a fixed axle, so its center of mass stays at rest. A student claims: “The velocities of points on opposite sides of the wheel are equal in size and opposite in direction, so they cancel, and the wheel has no kinetic energy.” Which response to the student's claim is correct?
Answer and reasoning
AIt is right: the kinetic energies of points moving in opposite directions cancel. A student who gives kinetic energy the direction of the motion picks this. Velocities cancel because they are vectors; kinetic energies have no direction and are never negative, so they cannot cancel.
BIt is right: a wheel whose center of mass is at rest has no kinetic energy. A student who looks only at the center of mass picks this. The wheel's translational kinetic energy is zero, but every point off the axle is moving and has kinetic energy, so the wheel has rotational kinetic energy.
CIt is wrong: each moving point has positive kinetic energy, so the energies add.Correct The velocity vectors do add to zero, which is why the wheel's linear momentum is zero. Kinetic energy, though, is a scalar: each moving point has (1/2)mv² > 0 whatever its direction, and these add to the wheel's rotational kinetic energy, (1/2)Iω².
DIt is wrong: the wheel's kinetic energy is (1/2)Mv², where v is its rim speed. A student who treats the whole wheel as moving at its rim speed picks this. Points nearer the axle move more slowly; for a uniform wheel K = (1/2)Iω² = (1/2)((1/2)MR²)(v/R)² = (1/4)Mv², only half of (1/2)Mv².
A flywheel with constant rotational inertia turns counterclockwise at 12 rad/s. A motor slows it to a stop and then sets it turning clockwise at 12 rad/s, so that, with counterclockwise taken as positive, its angular velocity changes from +12 rad/s to −12 rad/s. How does the flywheel's final rotational kinetic energy compare with its initial rotational kinetic energy?
Answer and reasoning
AIt is the initial value with a negative sign. A student who gives kinetic energy the sign of the angular velocity picks this. Kinetic energy has no direction and cannot be negative; squaring −12 rad/s gives the same value as squaring +12 rad/s.
BIt equals the initial rotational kinetic energy.Correct K = (1/2)Iω² depends on ω², and (+12 rad/s)² = (−12 rad/s)². Rotational kinetic energy is a scalar with no direction, so reversing the sense of rotation at the same angular speed leaves it unchanged.
CIt is greater, as ω has changed by 24 rad/s in all. A student who finds the change in kinetic energy from the change in angular velocity, (1/2)I(24 rad/s)², picks this. The change in kinetic energy is the difference of the two energies, (1/2)I(12)² − (1/2)I(12)² = 0.
DIt is smaller, as energy was used up in stopping it. A student who thinks energy was used up in stopping the flywheel picks this. Kinetic energy depends only on the angular speed at the instant: it is (1/2)I(12 rad/s)² at the start and at the end, whatever energy was transferred in between.
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 Fix
Rotational kinetic energy
The kinetic energy a rigid system has because it rotates: K = (1/2)Iω², where I is its rotational inertia about the axis of rotation and ω is its angular speed. Unit J. It depends on the square of the angular speed, so doubling ω makes it four times as great.
Rotational inertia (in K = (1/2)Iω²)
A rigid system's resistance to changes in its rotation about an axis, in kg·m². For a collection of objects I = Σ mi ri², with each r measured from the axis, so it depends on how the mass is distributed as well as on the mass; for extended objects such as disks and hoops the value is given.
Angular speed (in K = (1/2)Iω²)
The magnitude of the angular velocity, in rad/s. Every point of a rigid system has the same angular velocity, so one value of ω describes the rotation of the whole system; it must be in radians per second in K = (1/2)Iω².
Rotation about a fixed axis
When an object turns about a fixed axis, (1/2)Iω² is simply another way to calculate the kinetic energy of its moving parts. For a small object of mass m moving in a circle of radius r, (1/2)(mr²)ω² = (1/2)m(rω)² = (1/2)mv²: the rotational and the translational expressions describe the same energy, which is its total kinetic energy, and it is counted once.
Translational kinetic energy of the center of mass
(1/2)Mvcm², where M is the total mass of the rigid system and vcm is the speed of its center of mass. Unit J. It describes only the motion of the system from place to place, not its rotation.
Total kinetic energy of a rigid system
The sum of the translational kinetic energy of its center of mass and its rotational kinetic energy about its center of mass: Ktot = (1/2)Mvcm² + (1/2)Icmω². A disk that slides while it spins has both terms; neither can be left out.
Students often think Rotational inertia depends only on the amount of mass, so moving the same masses to new positions, or comparing objects of equal mass, leaves the rotational inertia unchanged. In fact Yes. I = Σ m r², so the same masses contribute more rotational inertia the farther they are from the axis; moving each mass to three times its distance makes I nine times as great.
Students often think Rotational inertia is proportional to the distance from the axis (I = mr), so tripling the distance triples the rotational inertia. In fact It becomes nine times as great, because the distance is squared. An object three times as far from the axis also moves three times as fast at the same ω, and kinetic energy depends on the square of that speed.
6.1.A.2 Linear speed of a point in a rotating system Fix
Linear speed of a point in a rotating system
A point a distance r from the axis moves with speed v = rω. All points share one ω, but points farther from the axis move faster, and a point on the axis is at rest.
Rotation with the center of mass at rest
A wheel spinning on a fixed axle has its center of mass at rest, so its linear momentum and its translational kinetic energy are zero. Its points off the axis still move, so it has kinetic energy, all of it rotational: K = (1/2)Iω².
Students often think An object whose center of mass is at rest has no kinetic energy, because it is not going anywhere. In fact Yes. Every point off the axle moves with speed v = rω and so has kinetic energy. The wheel's translational kinetic energy is zero, but its rotational kinetic energy, (1/2)Iω², is not.
6.1.A.3 Kinetic energy as a scalar Fix
Kinetic energy as a scalar
Rotational kinetic energy has a size but no direction. It is never negative, it is the same for clockwise and counterclockwise rotation at the same angular speed, and the kinetic energies of the parts of a system add as ordinary numbers.
Students often think Kinetic energy has the sign or direction of the motion: it is negative for clockwise (negative) rotation, and the kinetic energies of parts moving in opposite directions cancel. In fact No. K = (1/2)Iω² contains ω², so it is positive for either sense of rotation, and kinetic energies of parts moving in opposite directions add. Only vector quantities such as velocity, momentum and angular momentum carry a direction or sign that can cancel.
Students often think The kinetic energy at an instant is determined by the slope of the angular velocity–time graph, so where the slope is the same the kinetic energy is the same. In fact No. The slope of an ω–t graph is the angular acceleration. The kinetic energy depends on the VALUE of ω at that instant, read from the height of the graph: K = (1/2)Iω².
6 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 6
The diagram shows, viewed from above, a light rod that rotates in a horizontal plane about a fixed vertical axle at one end. Two small spheres, each of mass 1.0 kg, are fixed to the rod at the distances from the axle shown. The rod rotates with angular speed 4.0 rad/s. Modeling the spheres as objects, what is the rotational kinetic energy of the rod–spheres system?
Answer and reasoning
A3.2 JCorrect I = Σ m r² = (1.0 kg)(0.20 m)² + (1.0 kg)(0.60 m)² = 0.040 + 0.36 = 0.40 kg·m². K = (1/2)Iω² = (1/2)(0.40 kg·m²)(4.0 rad/s)² = 3.2 J.
B6.4 J A student who uses I = Σ m r without squaring the distances gets (1.0)(0.20) + (1.0)(0.60) = 0.80 and (1/2)(0.80)(4.0)² = 6.4 J. Each distance must be squared: I = Σ m r² = 0.40 kg·m².
C2.6 J A student who places the whole 2.0 kg at the center of mass, 0.40 m from the axle, gets I = (2.0)(0.40)² = 0.32 kg·m² and K ≈ 2.6 J. Because distances are squared, the outer sphere contributes more than the average distance allows for; each sphere must be counted at its own distance.
D5.8 J A student who treats all 2.0 kg as moving with the outer sphere's speed, (4.0)(0.60) = 2.4 m/s, gets (1/2)(2.0)(2.4)² ≈ 5.8 J. The inner sphere is only 0.20 m from the axle and moves at 0.80 m/s, so its kinetic energy is much smaller.
Working I = (1.0)(0.20)² + (1.0)(0.60)² = 0.40 kg·m². K = (1/2)(0.40)(4.0)² = 3.2 J. Check: vA = 0.80 m/s, vB = 2.4 m/s; (1/2)(1.0)(0.80)² + (1/2)(1.0)(2.4)² = 0.32 + 2.88 = 3.2 J. Errors: Σmr = 0.80 → 6.4 J; all mass at rcm = 0.40 m → 2.56 ≈ 2.6 J; all mass at 0.60 m → 5.76 ≈ 5.8 J.
A small ball of mass m, attached to a light string, moves in a horizontal circle of radius r about a fixed vertical axis with constant angular speed ω. Modeling the ball as an object, which expression gives the ball's total kinetic energy?
Answer and reasoning
AK = mr²ω² A student who adds (1/2)Iω² = (1/2)mr²ω² to (1/2)mv² = (1/2)m(rω)² picks this. For an object rotating about a fixed axis, both expressions describe the same motion, so adding them counts the kinetic energy twice.
BK = mr²ω A student who treats kinetic energy like momentum, swapping m for I and v for ω in mv, picks Iω = mr²ω. Its unit is kg·m²/s, not J; kinetic energy is (1/2)Iω², with ω squared.
CK = mr²ω²/2Correct About the fixed axis the ball has I = mr², so (1/2)Iω² = (1/2)mr²ω². This is the same energy as (1/2)mv² with v = rω: the rotational expression and the translational expression are two ways of calculating the ball's one kinetic energy, which is its total kinetic energy.
DK = mrω²/2 A student who uses I = mr, without squaring the radius, picks this. For an object a distance r from the axis, I = mr²; the expression chosen, mrω²/2, does not even have the units of energy.
Working I = mr² (object at distance r). K = (1/2)Iω² = (1/2)mr²ω² = (1/2)m(rω)² = (1/2)mv². Errors: (1/2)Iω² + (1/2)mv² = mr²ω²; Iω = mr²ω; I = mr → (1/2)mrω².
A uniform disk of mass 2.0 kg and radius 0.20 m lies flat on horizontal frictionless ice. Its center of mass slides across the ice at 2.0 m/s while the disk spins at 10 rad/s about a vertical axis through its center. The disk's rotational inertia about that axis is 0.040 kg·m². What is the disk's total kinetic energy?
Answer and reasoning
A4.0 J A student who counts only the motion of the center of mass, (1/2)(2.0)(2.0)² = 4.0 J, picks this. The disk also spins about its center of mass, which adds (1/2)Icmω² = 2.0 J of rotational kinetic energy.
B8.0 J A student who treats the whole disk as moving at its rim speed relative to the center, ωR = 2.0 m/s, finds a rotational term of (1/2)(2.0)(2.0)² = 4.0 J and a total of 8.0 J. Most of a uniform disk's mass is closer to the axis than the rim; the given rotational inertia, 0.040 kg·m², gives a rotational term of 2.0 J.
C4.4 J A student who adds Icmω = (0.040)(10) = 0.40 to the translational 4.0 J picks this. Icmω has units kg·m²/s and is not an energy; the rotational kinetic energy is (1/2)Icmω² = 2.0 J.
D6.0 JCorrect Ktot = (1/2)Mvcm² + (1/2)Icmω² = (1/2)(2.0 kg)(2.0 m/s)² + (1/2)(0.040 kg·m²)(10 rad/s)² = 4.0 J + 2.0 J = 6.0 J. The disk moves from place to place and also rotates about its center of mass, so both terms count.
Working Ktrans = (1/2)(2.0)(2.0)² = 4.0 J. Krot = (1/2)(0.040)(10)² = 2.0 J. Ktot = 6.0 J. Errors: translational only → 4.0 J; all mass at rim speed ωR = 2.0 m/s → 4.0 + 4.0 = 8.0 J; 4.0 + Iω = 4.0 + 0.40 = 4.4 J.
A thin hoop and a uniform disk have the same mass and the same radius. Each lies flat on horizontal frictionless ice. They slide with the same center-of-mass speed while spinning with the same angular speed, each about a vertical axis through its own center. How does the hoop's total kinetic energy compare with the disk's?
Answer and reasoning
AThe hoop's is greater, as its rotational inertia is greater.Correct Each total is (1/2)Mvcm² + (1/2)Icmω². The translational terms are equal. All of the hoop's mass is at the rim, so its rotational inertia, and hence its rotational kinetic energy at the same ω, is greater than the disk's.
BThey are equal, as only the motion of the center of mass counts. A student who counts only (1/2)Mvcm² picks this. Both objects also spin about their centers of mass, and that rotational kinetic energy is part of each total.
CThey are equal, as equal masses have equal rotational inertias. A student who thinks rotational inertia depends only on mass picks this. It also depends on how the mass is distributed: the hoop's mass is all at the rim, so its rotational inertia is greater.
DThe disk's is greater, as a solid disk is harder to set spinning. A student who judges rotational inertia by how solid an object looks picks this. For equal mass and radius the hoop, with all its mass far from the axis, is the harder one to set spinning, and it has the greater rotational kinetic energy at the same ω.
The graph shows the angular velocity ω of a flywheel as a function of time t, with counterclockwise rotation taken as positive. The flywheel's rotational inertia does not change. Instants P, Q and R are marked on the graph. Which of the following correctly ranks the flywheel's rotational kinetic energies KP, KQ and KR at these instants?
Answer and reasoning
AKP > KQ > KR A student who gives kinetic energy the sign of ω picks this, placing R, where ω is negative, below Q, where ω = 0. Squaring ω makes K positive for either sense of rotation, and at R the flywheel is turning at 8 rad/s, so it has kinetic energy, while at Q it has none.
BKP = KQ = KR A student who reads the slope of the graph picks this: the slope is the same everywhere on the straight line. The slope is the angular acceleration; the kinetic energy depends on the value of ω at each instant.
CKQ > KR > KP A student who uses the area under the graph up to each instant picks this: the areas are 0 at P, 18 rad at Q and 10 rad at R. That area is the angular displacement since t = 0, not a measure of kinetic energy; K depends on the value of ω at the instant.
DKP > KR > KQCorrect K = (1/2)Iω² depends on the value of ω read from the graph, squared. At P, ω = +12 rad/s; at Q, ω = 0; at R, ω = −8 rad/s. So KP ∝ 144, KR ∝ 64 and KQ = 0. The minus sign at R shows only the sense of rotation; kinetic energy is a scalar and cannot be negative.
Working ωP = +12 rad/s, ωQ = 0, ωR = −8 rad/s. K ∝ ω²: 144 : 0 : 64, so KP > KR > KQ. Areas from t = 0 (for the area distractor): P 0; Q (1/2)(12)(3) = 18 rad; R 18 − (1/2)(8)(2) = 10 rad.
A uniform rod of mass M and length L rotates in a horizontal plane about a fixed vertical axle through one end. The rod's rotational inertia about this axle is (1/3)ML². The free end of the rod moves with speed v. Which expression gives the rod's kinetic energy?
Answer and reasoning
A(1/2)Mv² A student who treats all of the rod's mass as moving with the speed v of its free end picks this. Points nearer the axle move more slowly (speed rω), so the kinetic energy is less than (1/2)Mv².
B(1/8)Mv² A student who places all the mass at the center of mass, L/2 from the axle, uses I = M(L/2)² = (1/4)ML² and gets (1/2)(1/4)ML²(v/L)² = (1/8)Mv². The rotational inertia is not M rcm²; the given value about the axle is (1/3)ML².
C(1/6)Mv²Correct The end is a distance L from the axle, so ω = v/L. K = (1/2)Iω² = (1/2)(1/3)ML²(v/L)² = (1/6)Mv². This is the rod's total kinetic energy: for rotation about a fixed axle, the rotational kinetic energy already counts the motion of every point.
D(2/3)Mv² A student who adds a translational kinetic energy (1/2)Mv², using the one speed given, to the rotational kinetic energy about the axle, (1/6)Mv², picks this. (1/2)Iω² about the fixed axle already counts the motion of every point of the rod, so nothing is added. A translational term belongs only with the rotational inertia about the center of mass: (1/2)M(v/2)² + (1/2)(1/12)ML²(v/L)² = (1/8 + 1/24)Mv², which is the same (1/6)Mv².
Working The free end is a distance L from the axle, so v = Lω and ω = v/L. K = (1/2)Iω² = (1/2)(1/3)ML²(v/L)² = (1/6)Mv². Errors: whole mass moving at the end's speed → (1/2)Mv²; mass taken at the center of mass, I = M(L/2)² = (1/4)ML² → (1/2)(1/4)ML²(v/L)² = (1/8)Mv²; a translational (1/2)Mv² added to the rotational (1/6)Mv² → (2/3)Mv².
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
