5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
A uniform solid cylinder rolls without slipping across a level floor with total kinetic energy K. A thin hoop with the same mass and the same radius rolls without slipping across the floor at the same speed. What is the total kinetic energy of the hoop?
Answer and reasoning
A1.33KCorrect Each total is translational plus rotational kinetic energy, with ω = v/R. The cylinder has (1/2)Mv² + (1/4)Mv² = (3/4)Mv² = K; the hoop has (1/2)Mv² + (1/2)Mv² = Mv². So the hoop's total is K/(3/4) = 1.33K.
B1.00K A student who gives every round object the rotational inertia MR² treats the cylinder like the hoop and finds equal totals. A uniform solid cylinder has I = (1/2)MR², so less of its energy is rotational.
C1.20K A student who adds the rotational kinetic energy about the contact point to the translational energy finds (1/2 + 3/4)Mv² for the cylinder and (1/2 + 1)Mv² for the hoop, a ratio of 1.20. Rotation about the contact point already includes the translation; use the rotational inertia about the center.
D2.00K A student who takes total kinetic energy to be proportional to rotational inertia doubles K because the hoop's I is twice the cylinder's. Only the rotational part doubles; the translational part, (1/2)Mv², is the same for both, so the ratio is 4/3.
Working With v = Rω: cylinder K = (1/2)Mv² + (1/2)(1/2)MR²(v/R)² = (3/4)Mv². Hoop: (1/2)Mv² + (1/2)MR²(v/R)² = Mv². Ratio = 1/(3/4) = 4/3, so Khoop = 1.33K.
The diagram shows a wheel rolling without slipping to the right along level ground; the velocity of its center C is v. Points P and Q are on the rim at the positions shown. Which statement about the speeds of P and Q relative to the ground at this instant is correct?
Answer and reasoning
AP and Q both move forward at speed v, as C does A student who treats the wheel as moving along without turning gives every point the center's speed. The rotation adds a velocity of magnitude v forward at the top and backward at the bottom, so P moves at 2v and Q is at rest.
BP moves at speed 2v, and Q is momentarily at restCorrect Each point moves with the center's velocity v plus its velocity from rotation, of magnitude Rω = v. At the top, P, both point forward and add to 2v; at the contact point, Q, the rotational velocity points backward and cancels v, as rolling without slipping requires.
CP moves forward and Q moves backward, each at speed v A student who considers only the wheel's rotation about its center picks this; it describes the motion relative to the center. Adding the center's velocity v gives 2v for P and zero for Q.
DP and Q both move at speed 2v, twice that of C A student who adds the speeds from translation and rotation as numbers gets v + v at both points. They must be added as vectors: they point the same way at P but opposite ways at Q, which is instantaneously at rest.
Working vP = v + Rω = 2v (forward); vQ = v − Rω = 0, since v = Rω.
A uniform solid ball rolls without slipping down a rough incline. Air resistance is negligible, and the ball and the incline are rigid. The coefficient of static friction between them is μs, and the normal force on the ball is FN. Which statement about the friction force exerted on the ball by the incline is correct?
Answer and reasoning
AIt acts on a point that is momentarily at rest, so it dissipates no energyCorrect For rolling without slipping the contact point has no velocity relative to the incline, so the static friction force acting there does no work and dissipates no energy. It is not zero: it exerts the torque about the center that gives the ball its angular acceleration.
BIt is zero, because the ball does not slide along the surface of the incline A student who links friction only with sliding picks this. Static friction acts at the contact point; without it there would be no torque about the center, and the ball would slide down without starting to turn.
CIt equals μsFN, because static friction is what keeps the ball from slipping A student who takes static friction to be μsFN picks this. μsFN is the largest value static friction can have; the actual value is whatever rolling without slipping needs, (2/7)Mg sin θ for a uniform ball.
DIt converts some of the ball's mechanical energy into thermal energy as it rolls A student who thinks every friction force dissipates energy picks this. The contact point does not slide, so friction does no work, and the ball's mechanical energy is constant.
A ball of radius 0.10 m is launched along a level floor so that it slips at first. The graph shows, in two panels, the speed v of the ball's center and the ball's angular speed ω as functions of time; the ball moves to the right and spins in the sense that would carry it to the right. At t = 0.50 s, what is the speed, relative to the floor, of the point of the ball in contact with the floor?
Answer and reasoning
A3.5 m/sCorrect At t = 0.50 s the graphs give v = 6.0 m/s and ω = 25 rad/s. The rotation gives the bottom point a backward velocity Rω = (0.10)(25) = 2.5 m/s relative to the center, so relative to the floor it moves forward at 6.0 − 2.5 = 3.5 m/s: the ball is still slipping.
B0.0 m/s A student who applies the rolling-without-slipping condition while the ball slips takes the contact point to be at rest. Here v ≠ Rω (6.0 m/s against 2.5 m/s), so the contact point slides forward at 3.5 m/s.
C6.0 m/s A student who gives every point of the ball the speed of its center ignores the rotation. The spin gives the bottom point a backward velocity of Rω = 2.5 m/s relative to the center.
D2.5 m/s A student who considers only the rotation about the center finds Rω = 2.5 m/s. The center itself moves at 6.0 m/s, which must be added, giving 3.5 m/s forward.
Working At t = 0.50 s: v = 6.0 m/s, ω = 25 rad/s, Rω = (0.10)(25) = 2.5 m/s. The contact point's velocity relative to the floor = v − Rω (rotation carries the bottom point backward) = 6.0 − 2.5 = 3.5 m/s forward.
A 5.00 kg bowling ball, a uniform solid sphere with rotational inertia (2/5)MR² about its center, is launched along a level lane at 7.00 m/s with no rotation. The coefficient of kinetic friction between ball and lane is 0.200. The ball slips until it begins to roll without slipping. How much mechanical energy is dissipated while the ball slips? Use g = 10 m/s².
Answer and reasoning
A60.0 J A student who multiplies the friction force, μkMg = 10.0 N, by the 6.00 m the center travels while slipping gets 60.0 J. That is the loss of translational kinetic energy, but 25.0 J of it becomes rotational kinetic energy; only the sliding at the contact, 3.50 m, dissipates energy.
B0.00 J A student who thinks friction on a rolling ball never dissipates energy picks this. While the ball slips, its surface slides over the lane, so kinetic friction dissipates energy until rolling without slipping begins.
C35.0 JCorrect Slipping ends when v = rω, at v = (5/7)(7.00) = 5.00 m/s. The kinetic energy falls from (1/2)(5.00)(7.00)² = 122.5 J to (1/2)(5.00)(5.00)²(7/5) = 87.5 J, so 35.0 J is dissipated by kinetic friction at the sliding contact.
D79.6 J A student who uses acm = rα while the ball slips finds that slipping stops at v₀/2 = 3.50 m/s, so Kf = (1/2)(5.00)(3.50)²(7/5) = 42.9 J and 79.6 J dissipated. While slipping, α must come from the friction torque, which gives v = (5/7)v₀.
Working Final speed when rolling begins: v = (5/7)v₀ = 5.00 m/s (from v = v₀ − μkgt and Rω = (5/2)μkgt, or from angular momentum about the contact point). Ki = (1/2)(5.00)(7.00)² = 122.5 J. Kf = (1/2)(5.00)(5.00)²(1 + 2/5) = 87.5 J. Dissipated: 122.5 − 87.5 = 35.0 J. (Check: slipping lasts t = 2v₀/(7μkg) = 1.00 s; the surfaces slide v₀t − (1/2)(7/2)μkgt² = 3.50 m; (10.0 N)(3.50 m) = 35.0 J.)
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 of a rolling object Fix
Total kinetic energy of a rolling object
The sum of the translational kinetic energy of the center of mass and the rotational kinetic energy about the center of mass: Ktot = Ktrans + Krot = (1/2)Mvcm² + (1/2)Icmω². Unit: J.
Translational kinetic energy
The kinetic energy the system would have if all its mass moved with the velocity of its center of mass: Ktrans = (1/2)Mvcm². Unit: J.
Rotational kinetic energy about the center of mass
The kinetic energy of the system's rotation about an axis through its center of mass: Krot = (1/2)Icmω². For a given vcm, an object whose mass lies farther from its axis has a larger share of its kinetic energy in rotation. Unit: J.
Students often think Any round object of mass M and radius R has rotational inertia MR², as if all its mass were at its rim. In fact No. MR² holds only for a thin hoop, whose mass is all at radius R. A uniform solid disk or cylinder has (1/2)MR², because much of its mass is near the axis; a uniform solid sphere has (2/5)MR².
Students often think The total kinetic energy of a rolling object is its translational kinetic energy plus the rotational kinetic energy about the contact point. In fact No. A rolling object's kinetic energy can be found either as (1/2)Mvcm² + (1/2)Icmω², or as (1/2)Icontactω² for pure rotation about the contact point, but not as a mixture: (1/2)Icontactω² already includes the translational part, since Icontact = Icm + MR².
6.5.B.1 Rolling without slipping Fix
Rolling without slipping
Motion in which the point of the object touching the surface has no velocity relative to the surface, so the center's motion and the rotation are linked: Δxcm = rΔθ, vcm = rω and acm = rα.
Velocities of points on a rolling object
For rolling without slipping, each point's velocity is v⃗cm plus its velocity due to rotation about the center. The point in contact with the surface is instantaneously at rest, and the top point moves at 2vcm.
Students often think The acceleration (and so the angular acceleration) can be read from the value of a v–t graph instead of from its slope. In fact No. The acceleration is the slope of the v–t graph; the value read off the graph is the speed at that time.
Students often think For a rolling object, the angular acceleration equals the linear acceleration of its center, whatever its radius. In fact No. They are related by acm = rα, so α = acm/r. They have different units (rad/s² and m/s²) and are numerically equal only for a radius of 1 m.
6.5.B.2 Static friction in ideal rolling Fix
Static friction in ideal rolling
The friction force on an object rolling without slipping is static friction, of whatever magnitude (up to μsFN) is needed to keep the contact point from sliding. It acts at a point that is instantaneously at rest, so in the ideal case it does no work and dissipates no energy.
Students often think No friction acts on an object that rolls without slipping, because nothing slides. In fact No. Static friction acts at the contact point; it provides the torque about the center that gives the object its angular acceleration. It does no work in the ideal case, but it is not zero.
Students often think Every friction force dissipates energy as thermal energy, including the static friction on an object rolling without slipping. In fact Not in the ideal case. The friction force acts at the contact point, which is instantaneously at rest, so it does no work and the mechanical energy of the rolling system is constant.
6.5.C.1 Slipping Fix
Slipping
Motion in which the contact point slides relative to the surface, so vcm ≠ rω and acm ≠ rα; the translation and rotation must be found separately, each from its own equation (Newton's second law and its rotational form).
Students often think The linear momentum of a ball sliding on a level lane is conserved, so its center keeps its speed while it starts to spin. In fact No. Kinetic friction from the lane is an external force on the ball, so it exerts an impulse that decreases the speed of the ball's center while its torque increases the ball's angular speed.
Students often think The relations for rolling without slipping hold for any rolling object, so the contact point is always at rest and acm = rα even while the object slips. In fact No. While slipping, the contact point slides over the surface, so vcm ≠ rω and acm ≠ rα. The translation and rotation must be found separately: the speed of the contact point relative to the surface is vcm − rω.
6.5.C.2 Energy dissipated by kinetic friction while slipping Fix
Energy dissipated by kinetic friction while slipping
While an object slips, kinetic friction of magnitude μkFN acts at a point that moves relative to the surface, so it dissipates mechanical energy. The energy dissipated equals the friction force times the distance the contact surfaces slide over each other; for a ball launched without spin, this is less than the distance its center moves. Unit: J.
Students often think Friction on a rolling object never dissipates energy, so a ball's kinetic energy is conserved while it slips as well as when it rolls without slipping. In fact No. While it slips, the kinetic friction force acts at a point that moves relative to the lane, so it dissipates energy. Only after slipping stops does friction cease to dissipate energy.
Students often think The energy dissipated by kinetic friction on a slipping ball equals the friction force multiplied by the distance its center travels. In fact No. For a ball launched without spin, the friction force times the distance the center moves equals the loss of translational kinetic energy only. Part of that energy becomes rotational kinetic energy; the energy dissipated is the friction force times the distance the ball's surface slides over the lane, which is smaller.
7 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 7
The diagram shows three objects, each released from rest at the top of its own incline; the three inclines are identical. Each object rolls without slipping. A uniform solid sphere has rotational inertia (2/5)MR² about its center. The masses and radii are labeled; the objects are not drawn to scale. Which correctly ranks the speeds v₁, v₂ and v₃ of objects 1, 2 and 3 at the bottom of the inclines?
Answer and reasoning
Av₁ > v₃ > v₂ A student who thinks heavier objects roll faster ranks them by mass: 3.0 kg, 2.0 kg, 1.0 kg. Mass cancels from Mgh = (1/2)Mv²(1 + I/(MR²)); only the distribution of mass matters.
Bv₃ > v₂ > v₁Correct With v = Rω, energy conservation gives v = √(2gh/(1 + I/(MR²))): mass and radius cancel, and only the shape matters. I/(MR²) is 2/5 for the sphere, 1/2 for the disk and 1 for the hoop, so the sphere puts the smallest share of its energy into rotation and is fastest, and the hoop is slowest.
Cv₃ > v₁ > v₂ A student who thinks larger objects roll faster ranks them by radius: 0.30 m, 0.20 m, 0.10 m. With v = Rω the radius cancels too; only the ratio I/(MR²) affects the speed.
Dv₁ = v₂ = v₃ A student who ignores rotational kinetic energy uses Mgh = (1/2)Mv² for each and finds the same speed √(2gh). Each object also has rotational kinetic energy, and the share depends on its shape.
Working Energy: Mgh = (1/2)Mv² + (1/2)βMR²(v/R)² ⇒ v = √(2gh/(1 + β)), independent of M and R. Sphere β = 2/5: v₃ = √(10gh/7); disk β = 1/2: v₂ = √(4gh/3); hoop β = 1: v₁ = √(gh). So v₃ > v₂ > v₁.
A uniform solid cylinder is released from rest and rolls without slipping down a ramp, descending a vertical height h. In terms of h and the acceleration due to gravity g, what is the speed of the cylinder's center at the bottom?
Answer and reasoning
A1.41√(gh) A student who ignores the rotational kinetic energy writes Mgh = (1/2)Mv² and gets √(2gh) = 1.41√(gh). The cylinder also turns, so a quarter of Mv² goes into rotation and it moves more slowly.
B1.00√(gh) A student who uses I = MR² for the cylinder writes Mgh = (1/2)Mv² + (1/2)Mv² and gets √(gh). That is the result for a thin hoop; a uniform solid cylinder has I = (1/2)MR².
C0.89√(gh) A student who adds the rotational kinetic energy about the contact point, (1/2)(3/2)MR²ω² = (3/4)Mv², to (1/2)Mv² writes Mgh = (5/4)Mv² and gets √(4gh/5) = 0.89√(gh). Rotation about the contact point already includes the translation.
D1.15√(gh)Correct Mechanical energy is conserved (static friction does no work), and the cylinder's kinetic energy is translational plus rotational: Mgh = (1/2)Mv² + (1/2)(1/2)MR²(v/R)² = (3/4)Mv². So v = √(4gh/3) = 1.15√(gh).
Working Mgh = (1/2)Mv² + (1/2)(1/2)MR²(v/R)² = (3/4)Mv² ⇒ v = √(4gh/3) = 1.15√(gh). (Static friction does no work in rolling without slipping.)
A ball of radius 0.10 m rolls without slipping along a straight, level track while being pushed. The graph shows the speed of the ball's center as a function of time. What is the magnitude of the ball's angular acceleration?
Answer and reasoning
A20 rad/s² A student who reads the graph's final value instead of its slope divides 2.0 m/s by 0.10 m and gets 20, which is the final angular speed in rad/s. The angular acceleration comes from the slope, 0.50 m/s², divided by the radius.
B0.50 rad/s² A student who takes the angular acceleration to equal the linear acceleration of the center picks this. They are related by acm = rα, so α = 0.50/0.10 = 5.0 rad/s².
C5.0 rad/s²Correct The slope of the v–t graph is acm = 2.0/4.0 = 0.50 m/s². For rolling without slipping acm = rα, so α = 0.50/0.10 = 5.0 rad/s².
D0.050 rad/s² A student who forms the angular quantity by multiplying by the radius, α = r·acm, gets (0.10)(0.50) = 0.050. From acm = rα, the radius divides: α = acm/r.
Working acm = slope = (2.0 m/s)/(4.0 s) = 0.50 m/s². Rolling without slipping: α = acm/r = 0.50/0.10 = 5.0 rad/s².
A uniform solid cylinder of mass M rolls without slipping down a rough incline that makes angle θ with the horizontal. The coefficient of static friction between the cylinder and the incline is μs, and g is the acceleration due to gravity. What is the magnitude of the friction force exerted on the cylinder by the incline?
Answer and reasoning
A0.50Mg sin θ A student who uses I = MR² for the cylinder gets f = Ma and a = (1/2)g sin θ, so f = (1/2)Mg sin θ. That is the result for a thin hoop; a uniform solid cylinder has I = (1/2)MR².
B1.00Mg sin θ A student who thinks friction balances the component of gravity along the incline, as for a block at rest, picks this. Then the net force would be zero and the cylinder could not speed up; friction is only (1/3)Mg sin θ.
Cμs Mg cos θ A student who takes static friction to be μsFN picks this. μsFN is only the maximum static friction; the actual value is whatever is needed for rolling without slipping, found from Newton's second law and its rotational form.
D0.33Mg sin θCorrect Newton's second law along the incline gives Mg sin θ − f = Ma, and the rotational form about the center gives fR = (1/2)MR²(a/R), so f = (1/2)Ma. Together, a = (2/3)g sin θ and f = (1/3)Mg sin θ.
Working Along the incline: Mg sin θ − f = Ma. About the center: fR = (1/2)MR²α, with α = a/R ⇒ f = (1/2)Ma. Then Mg sin θ = (3/2)Ma ⇒ a = (2/3)g sin θ and f = (1/3)Mg sin θ = 0.33Mg sin θ.
Two identical uniform solid balls are released from rest at the same height on two ramps that have the same height and slope. Ball 1 rolls without slipping down a rough ramp; ball 2 slides without rotating down a frictionless ramp. Ball 2 reaches the bottom with the greater speed. Which reasoning correctly explains this?
Answer and reasoning
AFriction turns part of ball 1's mechanical energy into thermal energy A student who thinks all friction dissipates energy picks this. Ball 1 rolls without slipping, so the contact point is momentarily at rest and friction does no work; its mechanical energy is conserved.
BGravity exerts a torque on ball 1 about its center that slows its descent A student who thinks gravity exerts a torque about the ball's center picks this. Gravity acts at the center, so its torque about the center is zero; the torque that spins the ball up comes from friction at the contact point.
CPart of ball 1's kinetic energy is rotational, so less of it is translationalCorrect Both balls convert the same gravitational potential energy, Mgh, into kinetic energy, and neither loses any (static friction on ball 1 does no work). Ball 1's kinetic energy is split between translation and rotation, so its translational kinetic energy, and its speed, are smaller.
DThe normal force does negative work on ball 1 as it rolls down the ramp A student who thinks every force on a moving object does work picks this. The normal force is perpendicular to the ball's motion and acts at a point at rest, so it does no work on either ball.
Working Both lose the same Mgh. Ball 2: Mgh = (1/2)Mv₂². Ball 1: Mgh = (1/2)Mv₁² + (1/2)(2/5)MR²(v₁/R)² = (7/10)Mv₁² (friction does no work), so v₁ = √(10gh/7) < v₂ = √(2gh).
A bowling ball, a uniform solid sphere of mass M and radius R with rotational inertia (2/5)MR² about its center, is launched along a level lane with speed v₀ and no rotation. The coefficient of kinetic friction between ball and lane is μk. The ball slips at first and later rolls without slipping. What is the speed of the ball's center when it begins to roll without slipping?
Answer and reasoning
A0.85v₀ A student who keeps the kinetic energy constant writes (1/2)Mv₀² = (7/10)Mv² and gets v₀√(5/7) = 0.85v₀. While the ball slips, kinetic friction acts at a sliding contact and dissipates energy.
B1.00v₀ A student who treats the ball's linear momentum as conserved keeps its center at speed v₀ while it starts to spin. Kinetic friction from the lane is an external force that slows the center as it spins the ball up.
C0.50v₀ A student who uses acm = rα while the ball is slipping takes α = μkg/R and finds v = Rω at t = v₀/(2μkg), when v = 0.50v₀. While slipping, the rotation must be found from the torque: α = 5μkg/(2R).
D0.71v₀Correct While the ball slips, kinetic friction μkMg slows the center, v = v₀ − μkgt, and its torque spins the ball up, Rω = (5/2)μkgt. Slipping stops when v = Rω, at t = 2v₀/(7μkg), when v = (5/7)v₀ = 0.71v₀.
Working While slipping: friction μkMg backward ⇒ v = v₀ − μkgt. Torque μkMgR = (2/5)MR²α ⇒ α = 5μkg/(2R) ⇒ Rω = (5/2)μkgt. Slipping stops when v = Rω: v₀ − μkgt = (5/2)μkgt ⇒ t = 2v₀/(7μkg), v = v₀ − (2/7)v₀ = (5/7)v₀ = 0.71v₀. (Check: angular momentum about the contact point is conserved, Mv₀R = MvR + (2/5)MR²(v/R) ⇒ v = 5v₀/7.)
A uniform solid cylinder rolls without slipping down an incline. It is replaced by a uniform solid cylinder with the same mass but twice the radius, which also rolls without slipping down the same incline. How does the angular acceleration of the new cylinder compare with that of the original cylinder?
Answer and reasoning
AIt is equal to that of the original A student who takes the angular acceleration to equal the center's acceleration, which is the same for both cylinders, picks this. α = acm/R, so the larger cylinder's angular acceleration is smaller.
BIt is half as great as the original'sCorrect For any uniform solid cylinder rolling without slipping, acm = (2/3)g sin θ, independent of mass and radius. With acm = Rα, doubling R halves α. (Equivalently: the friction force is unchanged, its lever arm doubles and I quadruples.)
CIt is twice as great as the original's A student who multiplies by the radius, α = R·acm, picks this. From acm = Rα the radius divides, so doubling R halves α.
DIt is a quarter as great as the original's A student who keeps the friction torque the same while the rotational inertia (1/2)MR² becomes four times as great gets one-fourth. The friction force is unchanged, but it now acts at twice the lever arm, so the torque doubles and α halves.
Working acm = (2/3)g sin θ for any uniform solid cylinder (independent of M and R). α = acm/R, so doubling R halves α. (Friction f = (1/3)Mg sin θ is unchanged, its torque fR doubles and I = (1/2)MR² quadruples: α = fR/I also halves.)
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
