5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
A solid sphere of mass M and radius R rolls without slipping along a level floor. Its center of mass moves with speed v, and its rotational inertia about an axis through its center is (2/5)MR². Which expression gives the sphere's total kinetic energy?
Answer and reasoning
A0.50Mv² A student who counts only the motion of the center of mass picks this. That is Ktrans alone; the sphere also rotates about its center, which adds Krot = (1/5)Mv².
B0.90Mv² A student who takes the rotational kinetic energy to be (2/5)Mv², the fraction in the rotational inertia with no factor ½, gets (1/2 + 2/5)Mv². Krot = (1/2)Iω² = (1/2)(2/5)MR²(v/R)² = (1/5)Mv².
C0.20Mv² A student who treats rolling as rotation only gives just Krot = (1/5)Mv². The center of the sphere also moves, so Ktrans = (1/2)Mv² must be added.
D0.70Mv²Correct The sphere's kinetic energy is the sum of its translational and rotational kinetic energies. Ktrans = (1/2)Mv². With ω = v/R, Krot = (1/2)(2/5)MR²(v/R)² = (1/5)Mv². The total is (1/2 + 1/5)Mv² = 0.70Mv².
Working Ktrans = (1/2)Mv². Rolling without slipping: ω = v/R, so Krot = (1/2)Iω² = (1/2)(2/5)MR²(v/R)² = (1/5)Mv². Ktot = (1/2)Mv² + (1/5)Mv² = (7/10)Mv² = 0.70Mv².
Wheel 1, of radius R, rolls without slipping along a level road with center-of-mass speed v. Wheel 2, of radius 2R, rolls without slipping along the same road with center-of-mass speed 2v. What is the ratio ω₂/ω₁ of the wheels' angular speeds?
Answer and reasoning
A2.0 A student who takes the angular speed to be set by the speed alone doubles ω because the speed doubles. ω = vcm/r: wheel 2's speed is doubled, but so is its radius, and each turn carries it twice as far.
B1.0Correct For rolling without slipping vcm = rω, so ω = vcm/r. ω₁ = v/R and ω₂ = 2v/(2R) = v/R: doubling the speed and doubling the radius cancel, and the wheels turn at the same angular speed.
C4.0 A student who thinks a larger wheel must turn faster, as well as a faster one, multiplies the factors, 2 × 2. A larger wheel covers more distance per turn, so at a given speed it turns more slowly: ω = vcm/r.
D0.5 A student who compares only the sizes of the wheels halves ω for the wheel of twice the radius. That would be true at the same speed; wheel 2 also moves twice as fast, which doubles ω again.
Working Rolling without slipping: ω = vcm/r. ω₁ = v/R; ω₂ = (2v)/(2R) = v/R. ω₂/ω₁ = 1.0.
A solid ball is released from rest and rolls without slipping down a ramp fixed to the ground. Air resistance is negligible. Consider the system consisting of the ball and Earth. Which statement about the energy of this system as the ball rolls down is correct?
Answer and reasoning
AUg decreases by more than the kinetic energy increases, since friction dissipates energy. A student who thinks every friction force dissipates energy picks this. The ball does not slip, so the friction force is exerted at a point that is instantaneously at rest and dissipates no energy.
BThe decrease in Ug equals the increase in the ball's total kinetic energy.Correct Without slipping, the contact point is instantaneously at rest, so the static friction force dissipates no energy, and no external work is done on the ball–Earth system. Its mechanical energy is constant: every joule of Ug lost appears as translational plus rotational kinetic energy.
CThe decrease in Ug equals the increase in the ball's translational kinetic energy. A student who counts only the motion of the ball's center picks this. The ball also rotates faster as it rolls down, so part of the decrease in Ug becomes rotational kinetic energy.
DUg decreases by less than the kinetic energy increases, since friction adds energy. A student who thinks friction supplies the energy of rotation picks this. Friction's torque makes the ball turn, but friction is exerted at a point at rest and adds no energy; all the kinetic energy comes from the decrease in Ug.
A wheel spins clockwise about its axle, as seen from the side, while its center is held at rest just above a level floor. The wheel is lowered onto the floor and released, so at first it slips on the floor. Which statement describes the wheel's motion while it slips?
Answer and reasoning
AIts center stays at rest while its spin slows until it stops turning. A student who thinks friction can only slow motion down picks this. The bottom of the spinning wheel slides backward over the floor, so kinetic friction on the wheel points forward and speeds up its center from rest, while its torque slows the spin.
BIts center speeds up from rest while its angular speed decreases, until vcm = rω.Correct The bottom of the clockwise-spinning wheel slides backward over the floor, so kinetic friction on the wheel points forward. That force speeds up the center, and its torque about the center slows the spin. The slipping stops when vcm has risen and ω has fallen to the point where vcm = rω.
CIt moves off at once with vcm = rω₀, where ω₀ is its initial angular speed. A student who thinks vcm = rω holds as soon as a wheel touches the ground picks this. While the wheel slips, vcm and rω are different: vcm starts at zero and grows, and ω starts at ω₀ and falls.
DIts angular speed stays constant while its center speeds up from rest until vcm = rω. A student who thinks friction affects only the motion of the center picks this. Friction is exerted at the rim, so it also exerts a torque about the center that slows the spin; ω decreases while vcm increases.
A bowling ball is launched along a level lane so that at first it slides without rotating. While it slips, its rotation speeds up and its center slows down. Which claim about the ball's total kinetic energy while it slips, with its reasoning, is correct?
Answer and reasoning
AIt decreases, since the ball's surface slides over the lane where kinetic friction acts.Correct The point where kinetic friction is exerted moves relative to the lane, so friction dissipates energy, transforming some of the ball's kinetic energy into thermal energy. The total kinetic energy decreases until the ball rolls without slipping.
BIt stays the same, since the kinetic energy lost by translation all becomes rotational. A student who thinks friction only moves kinetic energy from translation to rotation picks this. The ball's surface slides over the lane, so kinetic friction also transforms some kinetic energy into thermal energy; the rotational kinetic energy gained is less than the translational kinetic energy lost.
CIt increases, since friction does work on the ball to set it spinning. A student who thinks friction supplies the energy of rotation picks this. Friction's torque does speed up the rotation, but friction slows the center more than enough to pay for it, and it dissipates energy as the surfaces slide; the total kinetic energy decreases.
DIt stays the same, since friction on a turning ball does not dissipate energy. A student who applies the rolling rule without its condition picks this. Friction dissipates no energy only in rolling without slipping, when the contact point is at rest; here the ball slips, so its contact point slides over the lane.
In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
6.5.A.1 Total kinetic energy Fix
Total kinetic energy
The sum of a system's translational and rotational kinetic energies, Ktot = Ktrans + Krot. For a rigid object, Ktrans comes from the motion of its center of mass and Krot from its rotation about its center of mass. SI unit: joule (J).
Translational kinetic energy
Ktrans = (1/2)Mvcm², the kinetic energy associated with the motion of a system's center of mass, where M is the total mass and vcm the speed of the center of mass. Unit: J.
Rotational kinetic energy
Krot = (1/2)Iω², the kinetic energy associated with a rigid system's rotation about its center of mass, where I is its rotational inertia about that axis and ω its angular speed. For the same M, R and ω it is larger when more of the mass is far from the axis. Unit: J.
Rotational inertia of a rolling object
A measure of how the object's mass is distributed about the axis through its center of mass, written I = kMR²: k = 1 for a thin hoop, 1/2 for a solid cylinder, 2/5 for a solid sphere. The value for an extended object is given in the problem. Unit: kg·m².
Students often think The kinetic energy of a rolling object is only its translational kinetic energy, (1/2)Mvcm². In fact No. A rolling object also rotates about its center of mass, so its total kinetic energy is (1/2)Mvcm² + (1/2)Iω². Leaving out the rotational part underestimates its kinetic energy and overestimates its speed at the bottom of a ramp.
Students often think For a rotational inertia I = kMR², the rotational kinetic energy of the rolling object is kMv²: the fraction in I is the whole coefficient, and the ½ of (1/2)Iω² is left out. In fact No. Krot = (1/2)Iω² and ω = v/R, so Krot = (1/2)(2/5)MR²(v/R)² = (1/5)Mv². The ½ of the kinetic energy formula multiplies the fraction in the rotational inertia.
6.5.B.1 Rolling without slipping Fix
Rolling without slipping
Rolling in which the point of the object touching the surface does not slide over it, so the motion of the center of mass is tied to the rotation: Δxcm = rΔθ, vcm = rω and acm = rα, with r the radius of the rolling surface.
Contact point
The point of a rolling object that touches the surface. In rolling without slipping it is instantaneously at rest relative to the surface: its velocity relative to the center, rω backward, cancels the center's velocity vcm forward. The top point moves forward at vcm + rω = 2vcm.
Distance per revolution
From Δxcm = rΔθ, one revolution (Δθ = 2π rad) of an object rolling without slipping moves its center forward one circumference, 2πr. A larger wheel therefore turns through a smaller angle, at a smaller angular speed, to cover the same distance at the same speed.
Students often think The angular and linear quantities of a rolling object are interchangeable: its angular displacement in radians is its displacement in meters, and its angular speed is set by its speed alone, whatever its radius. In fact No. They are related through the radius: Δxcm = rΔθ, vcm = rω and acm = rα. An angle in radians is turned into a distance by multiplying by r, and a larger wheel turns through a smaller angle to cover the same distance.
Students often think A larger wheel must rotate faster than a smaller one to move along with it, so a wheel's angular speed increases with both its radius and its speed. In fact No. For rolling without slipping ω = vcm/r. At the same speed, a wheel with twice the radius turns at half the angular speed, because each turn carries it twice as far.
6.5.B.2 Static friction on a rolling object Fix
Static friction on a rolling object
The friction force exerted by the surface at the contact point of an object rolling without slipping. The contact point does not slide, so in ideal cases this force dissipates no energy: the mechanical energy of an object–Earth system rolling down a ramp is constant. Its torque about the center affects how the kinetic energy is shared between translation and rotation.
Students often think Any friction force dissipates mechanical energy, including the static friction exerted on an object that rolls without slipping. In fact No, in the ideal case. The part of the ball touching the ramp is instantaneously at rest, so the static friction force there transforms no mechanical energy into thermal energy; the mechanical energy of the ball–Earth system is constant.
Students often think Friction from the surface adds energy to a rolling object: its rotational kinetic energy is supplied by friction, on top of the energy from the decrease in Ug. In fact No. In rolling without slipping the static friction force is exerted at a point that is instantaneously at rest, so it adds no energy to the object–Earth system. Its torque makes the object turn, so part of the kinetic energy gained from the decrease in Ug is rotational, but the total kinetic energy gained still equals the decrease in Ug.
6.5.C.1 Rolling while slipping Fix
Rolling while slipping
Motion in which the contact point slides over the surface, as when a bowling ball is launched without spin or a spinning wheel is set down on the floor. Then vcm ≠ rω: the translational and rotational motions cannot be directly related, and kinetic friction changes both until, possibly, vcm = rω.
Students often think An object in contact with a surface always has vcm = rω: the rolling relationship holds whenever it touches the ground. In fact No. vcm = rω holds only when the object rolls without slipping. A ball launched sliding, or a spinning wheel set down on the floor, slips at first, and then vcm and rω differ.
Students often think Friction always slows down whatever motion an object has; it cannot start an object moving or make any of its motion speed up. In fact Yes. Friction opposes the sliding of the two surfaces over each other, not the object's motion as a whole. A spinning wheel set down on the floor is pushed forward by kinetic friction, so its center speeds up from rest while the same force slows its spin.
6.5.C.2 Energy dissipated by kinetic friction Fix
Energy dissipated by kinetic friction
When a rotating system slips on a surface, the point of application of the kinetic friction force moves relative to the surface, so kinetic friction transforms some of the system's kinetic energy into thermal energy. The total kinetic energy of a slipping object on a level surface therefore decreases while it slips.
Students often think When friction makes a slipping object start to roll, it only moves kinetic energy from translation into rotation, so the total kinetic energy stays the same. In fact No. While the ball slips, its surface slides over the lane, so kinetic friction dissipates some energy as thermal energy. The total kinetic energy decreases, and the rotational kinetic energy gained is less than the translational kinetic energy lost.
Students often think Friction on a round object that is turning never dissipates energy, whether or not it slips. In fact No. Friction dissipates no energy only when the object rolls without slipping, because the contact point is then at rest. While it slips, the contact point moves relative to the surface, so kinetic friction dissipates energy.
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
A solid cylinder, with rotational inertia (1/2)MR² about its axis, is released from rest on a ramp and rolls down without slipping. Its center of mass descends a vertical height of 1.2 m. Use g = 10 m/s². What is the speed of the cylinder's center of mass at the bottom of the ramp?
Answer and reasoning
A4.9 m/s A student who ignores the rotation sets MgΔh = (1/2)Mv² and gets √(2gh) = 4.9 m/s, the speed of an object sliding without friction. Part of the decrease in Ug becomes rotational kinetic energy, so the center moves more slowly.
B3.5 m/s A student who writes the rotational kinetic energy as (1/2)Mv², using the fraction in I without the ½ of (1/2)Iω², gets MgΔh = Mv² and v = √(gh) = 3.5 m/s. In fact Krot = (1/4)Mv².
C4.0 m/sCorrect The mechanical energy of the cylinder–Earth system is constant, because friction does not dissipate energy when there is no slipping. MgΔh = (1/2)Mv² + (1/2)(1/2)MR²(v/R)² = (3/4)Mv², so v = √(4gh/3) = √(4 × 10 × 1.2/3) m/s = 4.0 m/s.
D6.9 m/s A student who puts all of the decrease in Ug into rotation, MgΔh = (1/4)Mv², gets √(4gh) = 6.9 m/s. The cylinder's center also moves, so the energy must provide (1/2)Mv² of translational kinetic energy as well.
Working System: cylinder and Earth. Rolling without slipping, friction dissipates no energy, so MgΔh = (1/2)Mv² + (1/2)Iω² with ω = v/R: (1/2)Mv² + (1/2)(1/2)MR²(v/R)² = (3/4)Mv². v = √(4gh/3) = √(4 × 10 m/s² × 1.2 m / 3) = √(16 m²/s²) = 4.0 m/s.
A thin hoop (rotational inertia MR²) and a solid cylinder (rotational inertia (1/2)MR²) have the same mass M and the same radius R. They are released from rest side by side at the top of a ramp and roll down without slipping. Which object's center of mass is moving faster at the bottom, and why?
Answer and reasoning
AThe cylinder, since a smaller share of its kinetic energy is rotational.Correct Both lose the same Ug, and without slipping no energy is dissipated, so both gain the same total kinetic energy. For the hoop Krot = Ktrans; for the cylinder Krot is only half of Ktrans. More of the cylinder's kinetic energy is translational, so its center moves faster: v = √(4gh/3) for the cylinder and √(gh) for the hoop.
BNeither, since they have the same mass and each loses the same amount of Ug. A student who thinks rolling speed depends only on mass and height picks this. The decrease in Ug is equal, but the hoop, with all its mass at the rim, puts half its kinetic energy into rotation, leaving less for the motion of its center.
CThe hoop, since more of its kinetic energy is rotation that drives it on. A student who thinks rotation pushes a rolling object along picks this. Rotational kinetic energy does not add to the speed of the center; the more of the fixed total that goes into rotation, the less is left for translation, so the hoop is slower.
DThe cylinder, since all its lost Ug spins it and its I is smaller. A student who treats rolling as rotation only sets MgΔh = (1/2)Iω² and concludes that the cylinder, with the smaller rotational inertia, spins faster. The object is right but the reason is not: part of the decrease in Ug becomes translational kinetic energy, and the cylinder is faster because a smaller share of its kinetic energy is rotational.
A wheel of radius 0.25 m starts from rest and rolls without slipping in a straight line along level ground. The graph shows the wheel's angular velocity ω as a function of time t. What is the displacement of the wheel's center from t = 0 to t = 2.0 s?
Answer and reasoning
A2.0 mCorrect The angular displacement is the area under the ω–t graph: (1/2)(2.0 s)(8.0 rad/s) = 8.0 rad. Because the wheel rolls without slipping, its center moves Δxcm = rΔθ = (0.25 m)(8.0 rad) = 2.0 m.
B4.0 m A student who multiplies the final angular velocity by the time, (8.0 rad/s)(2.0 s) = 16 rad, and then by r gets 4.0 m. The wheel speeds up from rest, so its angular displacement is the triangle's area, 8.0 rad, not 16 rad.
C1.0 m A student who uses the slope of the graph, α = 8.0/2.0 = 4.0 rad/s², and multiplies it by r gets 1.0 m. The slope gives the angular acceleration; the angular displacement is the area under the graph.
D8.0 m A student who reports the angular displacement, 8.0 rad, as the distance moved picks this. An angle is turned into a distance by multiplying by the radius: Δxcm = rΔθ = 0.25 m × 8.0 rad = 2.0 m.
Working Angular displacement = area under the ω–t graph from 0 to 2.0 s: Δθ = (1/2)(2.0 s)(8.0 rad/s) = 8.0 rad. Rolling without slipping: Δxcm = rΔθ = (0.25 m)(8.0 rad) = 2.0 m.
The diagram shows a wheel rolling without slipping along level ground; the arrow shows the velocity v of its center C. Points P and B are on the rim, as shown. Which statement describes the velocities of P and B relative to the ground at the instant shown?
Answer and reasoning
AP and B both move to the right at v. A student who thinks every point of a rolling wheel moves with its center picks this. The rotation about C adds to the center's velocity at the top and cancels it at the bottom, so P moves at 2v and B is at rest.
BP moves right at v, and B moves left at v. A student who uses only the rotation about C picks this. Those are the velocities relative to the center; relative to the ground the center's velocity v must be added, giving 2v for P and 0 for B.
CP and B both move to the right at 2v. A student who adds v and rω as numbers at every point of the rim picks this. At B the rotation's velocity points backward, so it cancels v rather than adding to it; B is at rest.
DP moves right at 2v, and B is at rest.Correct Each point's velocity is the center's velocity plus its velocity from rotation about the center, which has magnitude rω = v. At P the rotation carries the point forward: v + v = 2v to the right. At B it carries the point backward: v − v = 0. That is what rolling without slipping means: the contact point does not slide.
Working Rolling without slipping: v = rω. Take right as positive. Velocity of a rim point relative to the ground = velocity of C + velocity relative to C. For clockwise rotation, P moves right relative to C at rω = v, and B moves left relative to C at rω = v. P: v + v = +2v (right). B: v − v = 0 (at rest).
A bowling ball is launched along a level lane so that at first it slides without rotating. The graph shows the speed vcm of its center of mass and the product rω of its radius and angular speed as functions of time t. Which claim is supported by the graph?
Answer and reasoning
AIt rolls without slipping throughout, as it touches the lane. A student who thinks vcm = rω holds whenever a ball touches a surface picks this. The graph shows vcm and rω differing before t₁, which is exactly what slipping means.
BBefore t₁ friction slows the ball down but does not set it spinning. A student who thinks friction can only slow things down picks this. The graph shows rω increasing before t₁: friction on the slipping ball points backward, slowing the center but exerting a torque that speeds up the rotation.
CBefore t₁ the ball slips on the lane, since vcm and rω differ.Correct Rolling without slipping requires vcm = rω. Before t₁ the two lines differ, so the ball's surface slides over the lane; kinetic friction slows the center and speeds up the rotation until, at t₁, vcm = rω and slipping stops.
DAfter t₁ friction still dissipates energy, as the ball touches the lane. A student who thinks any friction dissipates energy picks this. After t₁, vcm = rω, so the contact point no longer slides; the graph shows vcm and rω constant, with no further loss of kinetic energy.
A solid cylinder of mass M and radius R, with rotational inertia (1/2)MR² about its axis, rolls without slipping down a ramp inclined at angle θ to the horizontal. Which expression gives the magnitude of the acceleration of the cylinder's center of mass?
Answer and reasoning
A(g sinθ)/2 A student who uses energy with the rotational kinetic energy written as (1/2)Mv², taking the ½ in I as the whole coefficient, gets Mgd sinθ = Mv² for a distance d along the ramp. With v² = 2ad this gives a = (1/2)g sinθ. In fact Krot = (1/2)(1/2)MR²(v/R)² = (1/4)Mv².
B(2g sinθ)/3Correct Along the ramp, Mg sinθ − f = Macm. Friction's torque about the center is fR = (1/2)MR²α, and rolling without slipping gives α = acm/R, so f = (1/2)Macm. Then Mg sinθ = (3/2)Macm and acm = (2/3)g sinθ, written (2g sinθ)/3.
Cg sinθ A student who thinks the way an object moves down a ramp does not depend on how its mass is distributed treats the cylinder like a block sliding without friction and picks this. Friction's torque makes the cylinder turn, so part of the energy goes into rotation and the center accelerates less.
D2g sinθ A student who thinks friction adds to the cylinder's motion takes f down the ramp, writes Mg sinθ + (1/2)Ma = Ma and picks this. Friction's torque must turn the cylinder in the sense of its rolling, so friction acts up the ramp and reduces acm.
Working Forces along the ramp: the component Mg sinθ down the ramp and static friction f up the ramp (the normal force has no component along the ramp). Mg sinθ − f = Macm. About the center only friction exerts a torque: fR = Iα = (1/2)MR²α. Rolling without slipping: acm = Rα, so fR = (1/2)MR(acm) and f = (1/2)Macm. Then Mg sinθ = (3/2)Macm, acm = (2/3)g sinθ = (2g sinθ)/3. (Energy check: Mgd sinθ = (3/4)Mv², v² = (4/3)gd sinθ = 2acm d.) Errors: rolling treated like frictionless sliding → g sinθ; energy route with Krot = (1/2)Mv² (the ½ of (1/2)Iω² left out) → Mgd sinθ = Mv², a = (1/2)g sinθ; friction taken down the ramp → Mg sinθ + (1/2)Ma = Ma, a = 2g sinθ.
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
