AP Physics C Mechanics Quiz: Rolling
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RollingQuestion 1 of 20

A uniform solid sphere of mass MM and radius RR rolls without slipping down an incline that makes an angle θ\theta with the horizontal. What is the magnitude of the linear acceleration of the sphere's center of mass? The rotational inertia of a solid sphere is I=25MR2I = \frac{2}{5}MR^2.

gsin⁡θg \sin\theta
23gsin⁡θ\frac{2}{3} g \sin\theta
57gsin⁡θ\frac{5}{7} g \sin\theta
12gsin⁡θ\frac{1}{2} g \sin\theta
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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Rolling

Practice Rolling in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Rolling, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

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Question 1

A uniform solid sphere of mass MM and radius RR rolls without slipping down an incline that makes an angle θ\theta with the horizontal. What is the magnitude of the linear acceleration of the sphere's center of mass? The rotational inertia of a solid sphere is I=25MR2I = \frac{2}{5}MR^2.

  1. gsin⁡θg \sin\theta
  2. 23gsin⁡θ\frac{2}{3} g \sin\theta
  3. 57gsin⁡θ\frac{5}{7} g \sin\theta (correct answer)
  4. 12gsin⁡θ\frac{1}{2} g \sin\theta
Explanation: The net force down the incline is Mgsin⁡θ−f=MaMg\sin\theta - f = Ma. The torque causing rotation is τ=fR=Iα\tau = fR = I\alpha. For rolling without slipping, a=Rαa = R\alpha. Substituting for ff and α\alpha: f=Iα/R=(25MR2)(a/R)/R=25Maf = I\alpha/R = (\frac{2}{5}MR^2)(a/R)/R = \frac{2}{5}Ma. Now substitute this into the force equation: Mgsin⁡θ−25Ma=MaMg\sin\theta - \frac{2}{5}Ma = Ma. This simplifies to Mgsin⁡θ=75MaMg\sin\theta = \frac{7}{5}Ma, so a=57gsin⁡θa = \frac{5}{7}g\sin\theta.

Question 2

A small solid sphere of radius rr rolls without slipping inside a large fixed hemispherical bowl of radius RR, starting from rest at the same height as the center of the bowl. What is the speed of the sphere's center of mass at the bottom of the bowl?

  1. 2gR\sqrt{2gR}
  2. gR\sqrt{gR}
  3. 107g(R−r)\sqrt{\frac{10}{7}g(R-r)} (correct answer)
  4. 57gR\sqrt{\frac{5}{7}gR}
Explanation: The center of mass of the small sphere falls a vertical distance of h=R−rh = R-r. The initial potential energy is Mg(R−r)Mg(R-r). This is converted into translational and rotational kinetic energy. For a solid sphere rolling, Ktotal=710Mv2K_{total} = \frac{7}{10}Mv^2. By conservation of energy, Mg(R−r)=710Mv2Mg(R-r) = \frac{7}{10}Mv^2. Solving for vv gives v=107g(R−r)v = \sqrt{\frac{10}{7}g(R-r)}.

Question 3

A solid cylinder is placed on a rough incline. The coefficient of static friction is μs\mu_s. What is the maximum angle of inclination θmax\theta_{max} for which the cylinder will roll without slipping? The rotational inertia of a solid cylinder is I=12MR2I=\frac{1}{2}MR^2.

  1. arctan⁡(μs)\arctan(\mu_s)
  2. arctan⁡(2μs)\arctan(2\mu_s)
  3. arctan⁡(3μs)\arctan(3\mu_s) (correct answer)
  4. arctan⁡(23μs)\arctan(\frac{2}{3}\mu_s)
Explanation: For rolling without slipping, the required static friction force is f=13Mgsin⁡θf = \frac{1}{3}Mg\sin\theta. The maximum available static friction force is fmax=μsN=μsMgcos⁡θf_{max} = \mu_s N = \mu_s Mg\cos\theta. To prevent slipping, we need f≤fmaxf \le f_{max}. Therefore, 13Mgsin⁡θ≤μsMgcos⁡θ\frac{1}{3}Mg\sin\theta \le \mu_s Mg\cos\theta. This simplifies to tan⁡θ≤3μs\tan\theta \le 3\mu_s. The maximum angle is thus θmax=arctan⁡(3μs)\theta_{max} = \arctan(3\mu_s).

Question 4

A uniform solid disk of mass 4.0 kg and radius 0.2 m rolls without slipping along a horizontal surface. The speed of its center of mass is 2.0 m/s. What is the total kinetic energy of the disk? The rotational inertia of a solid disk is I=12MR2I = \frac{1}{2}MR^2.

  1. 4.0 J
  2. 8.0 J
  3. 12.0 J (correct answer)
  4. 16.0 J
Explanation: The total kinetic energy is the sum of translational and rotational kinetic energies. Ktotal=Ktrans+Krot=12Mv2+12Iω2K_{total} = K_{trans} + K_{rot} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2. Given M=4.0M=4.0 kg, R=0.2R=0.2 m, v=2.0v=2.0 m/s. I=12(4.0)(0.2)2=0.08I = \frac{1}{2}(4.0)(0.2)^2 = 0.08 kg m2^2. Since it rolls without slipping, ω=v/R=2.0/0.2=10\omega = v/R = 2.0/0.2 = 10 rad/s. Ktotal=12(4.0)(2.0)2+12(0.08)(10)2=8.0+4.0=12.0K_{total} = \frac{1}{2}(4.0)(2.0)^2 + \frac{1}{2}(0.08)(10)^2 = 8.0 + 4.0 = 12.0 J.

Question 5

A uniform disk of mass MM and radius RR rolls without slipping with a center-of-mass speed vv. Which of the following expressions represents the total kinetic energy of the disk? The rotational inertia of a uniform disk is I=12MR2I = \frac{1}{2}MR^2.

  1. 12Mv2\frac{1}{2}Mv^2
  2. 34Mv2\frac{3}{4}Mv^2 (correct answer)
  3. Mv2Mv^2
  4. 14Mv2\frac{1}{4}Mv^2
Explanation: The total kinetic energy is the sum of the translational and rotational kinetic energies: Ktotal=Ktrans+Krot=12Mv2+12Iω2K_{total} = K_{trans} + K_{rot} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2. For a disk, I=12MR2I = \frac{1}{2}MR^2, and for rolling without slipping, ω=v/R\omega = v/R. Substituting these gives Ktotal=12Mv2+12(12MR2)(v/R)2=12Mv2+14Mv2=34Mv2K_{total} = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{1}{2}MR^2)(v/R)^2 = \frac{1}{2}Mv^2 + \frac{1}{4}Mv^2 = \frac{3}{4}Mv^2.

Question 6

A thin hoop of mass MM and radius RR is released from rest at the top of an incline of vertical height hh. It rolls without slipping to the bottom. What is the speed of its center of mass at the bottom of the incline? The rotational inertia of a hoop is I=MR2I = MR^2.

  1. 2gh\sqrt{2gh}
  2. gh\sqrt{gh} (correct answer)
  3. 43gh\sqrt{\frac{4}{3}gh}
  4. 12gh\sqrt{\frac{1}{2}gh}
Explanation: By conservation of energy, the initial potential energy MghMgh is converted into translational and rotational kinetic energy. Mgh=12Mv2+12Iω2Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2. For a hoop, I=MR2I=MR^2, and for rolling without slipping, ω=v/R\omega = v/R. So, Mgh=12Mv2+12(MR2)(v/R)2=12Mv2+12Mv2=Mv2Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}(MR^2)(v/R)^2 = \frac{1}{2}Mv^2 + \frac{1}{2}Mv^2 = Mv^2. Solving for vv gives v=ghv = \sqrt{gh}.

Question 7

A ball is sliding on a rough horizontal surface such that its center of mass speed vcmv_{cm} is greater than RωR\omega, where RR is the ball's radius and ω\omega is its angular speed. Which statement describes the effect of the kinetic friction force on the ball?

  1. The friction force is in the forward direction, increasing vcmv_{cm} and decreasing ω\omega.
  2. The friction force is in the backward direction, decreasing vcmv_{cm} and increasing ω\omega. (correct answer)
  3. The friction force is in the backward direction, decreasing both vcmv_{cm} and ω\omega.
  4. The friction force is in the forward direction, increasing both vcmv_{cm} and ω\omega.
Explanation: Since the ball is sliding, the point of contact is moving forward relative to the surface. The kinetic friction force opposes this relative motion, so it acts in the backward direction. This backward force creates a net force that decreases the translational speed vcmv_{cm}. This force also creates a torque about the center of mass that increases the angular speed ω\omega. This continues until vcm=Rωv_{cm} = R\omega and the ball rolls without slipping.

Question 8

A bowling ball is initially projected with a center of mass speed v0v_0 but zero angular velocity along a rough horizontal lane. The ball eventually begins to roll without slipping. During the initial slipping phase:

  1. both the translational kinetic energy and the angular velocity remain constant.
  2. the translational speed increases, and the angular velocity increases.
  3. the translational speed decreases, and the angular velocity increases. (correct answer)
  4. the translational speed decreases, and the angular velocity decreases.
Explanation: Initially, the ball is sliding, so a kinetic friction force acts opposite to the direction of motion. This force causes a linear deceleration, so the translational speed decreases. The friction force also exerts a torque about the center of mass, causing an angular acceleration, so the angular velocity increases from zero. This process continues until the condition for rolling without slipping, vcm=Rωv_{cm} = R\omega, is met.

Question 9

A uniform solid sphere and a thin hoop, both with the same mass MM and radius RR, are released from rest at the top of the same incline. Both roll without slipping. Which of the following statements is true about their motion?

  1. The hoop reaches the bottom first because it has a larger rotational inertia.
  2. The sphere reaches the bottom first because it has a smaller rotational inertia. (correct answer)
  3. They reach the bottom at the same time because their mass and radius are the same.
  4. They reach the bottom at the same time because the same net force acts on both.
Explanation: For an object rolling down an incline, a smaller rotational inertia means a larger fraction of the initial potential energy is converted into translational kinetic energy, resulting in a larger final linear speed and smaller travel time. A solid sphere (I=25MR2I = \frac{2}{5}MR^2) has a smaller rotational inertia than a hoop (I=MR2I=MR^2). Therefore, the sphere will have a greater acceleration and reach the bottom first.

Question 10

A wheel is projected along a rough horizontal surface with an initial angular speed but zero initial translational speed. It slips on the surface until it begins to roll without slipping. During the slipping phase, the work done by the kinetic friction force on the wheel:

  1. is negative, because it opposes the initial rotation of the wheel.
  2. is zero, because the net displacement of the center of mass is horizontal.
  3. is positive, because it causes the center of mass to accelerate from rest. (correct answer)
  4. is negative with respect to rotation and positive with respect to translation, with a net negative value.
Explanation: The bottom of the wheel is moving backward relative to the surface due to the initial spin. The kinetic friction force opposes this motion, so it acts in the forward direction. Since the friction force is in the same direction as the displacement of the center of mass, the work done by friction is positive. This positive work increases the translational kinetic energy of the wheel. The torque from friction does negative work on the rotation, decreasing rotational kinetic energy. The total mechanical energy decreases, but the work done by the friction force on the center of mass is positive.

Question 11

A solid ball with mass MM and radius RR rolls without slipping along a horizontal surface with a center of mass speed vv. It then encounters a ramp and rolls up to a maximum vertical height hh. What is the value of hh? The rotational inertia of a solid ball is I=25MR2I = \frac{2}{5}MR^2.

  1. v22g\frac{v^2}{2g}
  2. 3v24g\frac{3v^2}{4g}
  3. 5v24g\frac{5v^2}{4g}
  4. 7v210g\frac{7v^2}{10g} (correct answer)
Explanation: The initial total kinetic energy is converted to potential energy. Ktotal=Ktrans+Krot=MghK_{total} = K_{trans} + K_{rot} = Mgh. Ktotal=12Mv2+12Iω2=12Mv2+12(25MR2)(v/R)2=12Mv2+15Mv2=710Mv2K_{total} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{2}{5}MR^2)(v/R)^2 = \frac{1}{2}Mv^2 + \frac{1}{5}Mv^2 = \frac{7}{10}Mv^2. Setting this equal to MghMgh and solving for hh gives h=7v210gh = \frac{7v^2}{10g}.

Question 12

When various objects of the same mass and radius roll without slipping down an incline, they reach the bottom with different speeds. The primary physical principle that explains this difference is that the objects have different:

  1. gravitational potential energies at the top of the incline.
  2. coefficients of static friction with the surface of the incline.
  3. rotational inertias, which affects the energy distribution. (correct answer)
  4. center of mass locations, which alters the effective gravitational force.
Explanation: All objects start with the same potential energy MghMgh. This energy is converted into both translational and rotational kinetic energy. The way this energy is partitioned depends on the rotational inertia, II. An object with a larger II will have more of its energy as rotational kinetic energy, leaving less for translational kinetic energy. This results in a lower final center-of-mass speed. Thus, the differing rotational inertias cause the difference in final speeds.

Question 13

A solid cylinder of mass MM and radius RR rolls without slipping down a ramp inclined at an angle θ\theta to the horizontal. What is the magnitude of the static friction force acting on the cylinder? The rotational inertia of a solid cylinder is I=12MR2I = \frac{1}{2}MR^2.

  1. 13Mgsin⁡θ\frac{1}{3} Mg \sin\theta (correct answer)
  2. 12Mgsin⁡θ\frac{1}{2} Mg \sin\theta
  3. 23Mgsin⁡θ\frac{2}{3} Mg \sin\theta
  4. Mgsin⁡θMg \sin\theta
Explanation: The linear acceleration of a rolling cylinder down an incline is a=gsin⁡θ1+I/MR2=gsin⁡θ1+1/2=23gsin⁡θa = \frac{g\sin\theta}{1+I/MR^2} = \frac{g\sin\theta}{1+1/2} = \frac{2}{3}g\sin\theta. The friction force ff provides the torque fR=IαfR = I\alpha. Since a=Rαa = R\alpha, we have f=Ia/R2f = I a / R^2. Substituting for II and aa: f=(12MR2)(23gsin⁡θ)/R2=13Mgsin⁡θf = (\frac{1}{2}MR^2)(\frac{2}{3}g\sin\theta) / R^2 = \frac{1}{3}Mg\sin\theta.

Question 14

A solid sphere is placed on a rough horizontal surface and is given an initial angular velocity ω0\omega_0 but no initial linear velocity. Which of the following correctly describes the subsequent motion of the sphere?

  1. A kinetic friction force acts backward, causing the sphere to slow its rotation and remain in place.
  2. A kinetic friction force acts forward, causing linear acceleration and rotational deceleration. (correct answer)
  3. No friction force acts, so the sphere continues to spin in place with constant angular velocity.
  4. A kinetic friction force acts forward, causing linear acceleration while the angular velocity remains constant.
Explanation: Due to the initial spin, the point of contact on the sphere moves backward relative to the surface. The kinetic friction force opposes this relative motion, so it acts in the forward direction. This forward force causes a linear acceleration of the center of mass. The friction force also creates a torque that opposes the initial rotation, causing a rotational deceleration. This continues until the sphere rolls without slipping.

Question 15

A small solid sphere is to roll without slipping from rest at a height HH to complete a circular loop-the-loop of radius RR. To just maintain contact with the track at the top of the loop, what is the minimum required height HH? Assume the sphere's radius is negligible compared to RR.

  1. 2.0R2.0 R
  2. 2.5R2.5 R
  3. 2.7R2.7 R (correct answer)
  4. 3.0R3.0 R
Explanation: At the top of the loop, the centripetal force is provided by gravity: Mg=Mvtop2/RMg = Mv_{top}^2/R, so vtop2=gRv_{top}^2 = gR. By conservation of energy from the start to the top of the loop: MgH=Mg(2R)+Ktotal,topMgH = Mg(2R) + K_{total,top}. The total kinetic energy for a rolling sphere is Ktotal=710Mv2K_{total} = \frac{7}{10}Mv^2. So, MgH=2MgR+710Mvtop2=2MgR+710M(gR)=(2+0.7)MgR=2.7MgRMgH = 2MgR + \frac{7}{10}Mv_{top}^2 = 2MgR + \frac{7}{10}M(gR) = (2 + 0.7)MgR = 2.7MgR. Therefore, H=2.7RH = 2.7R.

Question 16

A solid block slides down a frictionless incline of height hh. In a second experiment, a solid sphere of the same mass rolls without slipping down a separate, but identical, incline. Let vslidev_{slide} be the final speed of the block and vrollv_{roll} be the final speed of the sphere. Which of the following correctly compares the two speeds?

  1. vslide>vrollv_{slide} > v_{roll} (correct answer)
  2. vslide<vrollv_{slide} < v_{roll}
  3. vslide=vrollv_{slide} = v_{roll}
  4. The comparison depends on the mass of the objects.
Explanation: For the sliding block, all initial potential energy MghMgh becomes translational kinetic energy: Mgh=12Mvslide2Mgh = \frac{1}{2}Mv_{slide}^2, so vslide=2ghv_{slide} = \sqrt{2gh}. For the rolling sphere, the potential energy becomes both translational and rotational kinetic energy: Mgh=710Mvroll2Mgh = \frac{7}{10}Mv_{roll}^2, so vroll=107ghv_{roll} = \sqrt{\frac{10}{7}gh}. Since 2>10/72 > 10/7, it is clear that vslide>vrollv_{slide} > v_{roll}. Less energy is available for translation in the rolling case.

Question 17

A uniform solid cylinder of mass mm and radius RR is initially at rest on a rough horizontal surface with coefficient of static friction μs\mu_s and kinetic friction μk\mu_k. A horizontal force FF is applied to the cylinder's center of mass. For what range of applied force will the cylinder roll without slipping?

  1. 0≤F≤3μsmg20 \leq F \leq \frac{3\mu_s mg}{2}
  2. 0≤F≤2μsmg0 \leq F \leq 2\mu_s mg
  3. 0≤F≤3μsmg0 \leq F \leq 3\mu_s mg (correct answer)
  4. 0≤F≤μsmg0 \leq F \leq \mu_s mg
Explanation: For rolling without slipping, apply Newton's laws with constraint a=Rαa = R\alpha. Translation: F−f=maF - f = ma. Rotation about center: fR=Iα=12mR2α=12mR2⋅aR=12mRafR = I\alpha = \frac{1}{2}mR^2 \alpha = \frac{1}{2}mR^2 \cdot \frac{a}{R} = \frac{1}{2}mRa, so f=12maf = \frac{1}{2}ma. Substituting into translation equation: F−12ma=maF - \frac{1}{2}ma = ma, giving F=32maF = \frac{3}{2}ma and f=12ma=F3f = \frac{1}{2}ma = \frac{F}{3}. For no slipping, friction force cannot exceed maximum static friction: ∣f∣≤μsmg|f| \leq \mu_s mg. So F3≤μsmg\frac{F}{3} \leq \mu_s mg, giving F≤3μsmgF \leq 3\mu_s mg. Combined with F≥0F \geq 0, the range is 0≤F≤3μsmg0 \leq F \leq 3\mu_s mg. Choice A uses wrong coefficient relationship. Choice B uses wrong moment calculation. Choice D neglects rotational effects.

Question 18

A hollow cylinder and a solid cylinder, both with the same mass MM and radius RR, are released simultaneously from rest at the top of an inclined plane. Both roll without slipping. When the solid cylinder has traveled a distance dd down the incline, what distance has the hollow cylinder traveled?

  1. 2d3\frac{2d}{3}
  2. 3d4\frac{3d}{4} (correct answer)
  3. 4d5\frac{4d}{5}
  4. dd
Explanation: For rolling without slipping, a=gsin⁡θ1+IMR2a = \frac{g\sin\theta}{1 + \frac{I}{MR^2}}. For solid cylinder: I=12MR2I = \frac{1}{2}MR^2, so as=2gsin⁡θ3a_s = \frac{2g\sin\theta}{3}. For hollow cylinder: I=MR2I = MR^2, so ah=gsin⁡θ2a_h = \frac{g\sin\theta}{2}. Using s=12at2s = \frac{1}{2}at^2, when solid travels distance dd: t=3dgsin⁡θt = \sqrt{\frac{3d}{g\sin\theta}}. In this time, hollow travels: sh=12⋅gsin⁡θ2⋅3dgsin⁡θ=3d4s_h = \frac{1}{2} \cdot \frac{g\sin\theta}{2} \cdot \frac{3d}{g\sin\theta} = \frac{3d}{4}. Choice A uses wrong moment ratios. Choice C confuses with sphere values. Choice D assumes equal accelerations.

Question 19

A yo-yo consists of a disk of mass MM and radius RR with a string wound around its edge. The yo-yo is released from rest and falls under gravity while the string unwinds. If the string does not slip on the disk's edge, what is the acceleration of the yo-yo's center of mass?

  1. g3\frac{g}{3}
  2. 2g3\frac{2g}{3} (correct answer)
  3. g2\frac{g}{2}
  4. gg
Explanation: Apply Newton's second law and rotational dynamics. Forces on yo-yo: weight MgMg downward, tension TT upward. For translation: Mg−T=MaMg - T = Ma. For rotation about center: TR=Iα=12MR2αTR = I\alpha = \frac{1}{2}MR^2 \alpha. No-slip condition: a=Rαa = R\alpha, so α=aR\alpha = \frac{a}{R}. Substituting: TR=12MR2⋅aR=12MRaTR = \frac{1}{2}MR^2 \cdot \frac{a}{R} = \frac{1}{2}MRa, giving T=12MaT = \frac{1}{2}Ma. From translation equation: Mg−12Ma=MaMg - \frac{1}{2}Ma = Ma, so Mg=32MaMg = \frac{3}{2}Ma, yielding a=2g3a = \frac{2g}{3}. Choice A uses wrong moment calculation. Choice C ignores rotational inertia. Choice D assumes free fall.

Question 20

A uniform solid sphere rolls down a curved track and then moves along a horizontal loop-the-loop of radius RR. If the sphere is released from rest at height hh above the bottom of the track, what is the minimum value of hh required for the sphere to maintain contact with the track at the top of the loop?

  1. 3R3R
  2. 5R2\frac{5R}{2}
  3. 7R2\frac{7R}{2}
  4. 27R10\frac{27R}{10} (correct answer)
Explanation: This problem combines rotational motion with circular motion dynamics. When you see a rolling object going through a loop, you need to consider both translational and rotational kinetic energy, plus the condition for maintaining contact. At the top of the loop, the minimum condition for contact occurs when the normal force equals zero, meaning gravity alone provides the centripetal force: mg=mvtop2Rmg = \frac{mv_{top}^2}{R}, so vtop=gRv_{top} = \sqrt{gR}. Using energy conservation from release point to loop top: The initial potential energy mghmgh converts to kinetic energy (both translational and rotational) plus potential energy at height 2R2R: mgh=12mvtop2+12Iω2+mg(2R)mgh = \frac{1}{2}mv_{top}^2 + \frac{1}{2}I\omega^2 + mg(2R) For a solid sphere, I=25mr2I = \frac{2}{5}mr^2 and v=ωrv = \omega r, so 12Iω2=15mv2\frac{1}{2}I\omega^2 = \frac{1}{5}mv^2. Therefore: mgh=12mvtop2+15mvtop2+2mgR=710mvtop2+2mgRmgh = \frac{1}{2}mv_{top}^2 + \frac{1}{5}mv_{top}^2 + 2mgR = \frac{7}{10}mv_{top}^2 + 2mgR Substituting vtop2=gRv_{top}^2 = gR: h=7gR10g+2R=7R10+2R=27R10h = \frac{7gR}{10g} + 2R = \frac{7R}{10} + 2R = \frac{27R}{10} Choice A (3R3R) ignores rotational energy entirely. Choice B (5R2\frac{5R}{2}) incorrectly uses the condition for a point mass. Choice C (7R2\frac{7R}{2}) applies the rolling energy ratio incorrectly. Remember: For rolling objects in loops, always account for rotational kinetic energy using 12Iω2\frac{1}{2}I\omega^2, and know the moment of inertia for common shapes—solid spheres have I=25mr2I = \frac{2}{5}mr^2.