AP Physics 1 Quiz: Rolling

What this quiz covers

This quiz focuses on Rolling, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\theta = xR$
2. $\theta = x/R$ (correct answer)
3. $\theta = R/x$
4. $\theta = 2\pi x$

Explanation: This question examines angular displacement in rolling without slipping for AP Physics 1. Rolling without slipping equates linear distance x to arc length θR, yielding θ = x/R. Here, θ is angular displacement and R is radius. This relationship ensures consistent motion without slippage. Choice A, θ = xR, might be selected by swapping the formula incorrectly. A transferable strategy is to relate linear and angular displacements via radius in rolling problems.

Question 2

1. $v$
2. $2v$
3. $0$ (correct answer)
4. $\dfrac{v}{R}$

Explanation: This problem examines the instantaneous velocity of the contact point in rolling without slipping motion. The key insight is that rolling without slipping means the contact point has zero velocity relative to the ground at each instant. This occurs because the point's rotational velocity (ωR = v backward) exactly cancels its translational velocity (v forward). The velocities add vectorially: v forward + v backward = 0. Choice A incorrectly considers only the translational motion, while choice B considers only the rotational motion. To understand rolling motion, remember that the contact point acts as an instantaneous pivot point with zero velocity.

Question 3

1. $0$ (correct answer)
2. $v$
3. $2v$
4. It depends on the coefficient of kinetic friction

Explanation: This question tests the fundamental principle of rolling without slipping. When an object rolls without slipping, the point in contact with the ground must be instantaneously at rest - this is the definition of "no slipping." The contact point has two velocity components that exactly cancel: the forward velocity v from the center's motion and the backward velocity -v from rotation about the center. These add to give zero instantaneous velocity at the contact point. Choice B incorrectly assumes the contact point moves with the center, choice C doubles the speed incorrectly, and choice D confuses static friction (which enables rolling) with kinetic friction. Remember that "rolling without slipping" means the contact point has zero velocity relative to the ground.

Question 4

1. It is unchanged
2. It doubles (correct answer)
3. It halves
4. It quadruples

Explanation: This question examines the relationship between angular and linear speeds in rolling motion. For rolling without slipping, the condition v = ωR must always be satisfied, where v is the center-of-mass speed, ω is angular speed, and R is radius. If the angular speed doubles from ω to 2ω while the radius R remains constant, then the center-of-mass speed must change from v = ωR to v' = (2ω)R = 2(ωR) = 2v. Choice A incorrectly assumes the speeds are independent, choice C reverses the relationship, and choice D incorrectly squares the factor. The key insight is that v and ω are directly proportional for rolling without slipping, so doubling one doubles the other.

Question 5

1. $v$ must increase proportionally with $\omega$ (correct answer)
2. $v$ must decrease as $\omega$ increases
3. $v$ can stay zero while $\omega$ increases
4. $v$ must equal $\dfrac{\omega}{R}$

Explanation: This question tests understanding of the constraint imposed by rolling without slipping. The fundamental relationship v = ωR must be maintained at all times during rolling motion. If ω increases while R remains constant, then v must increase proportionally to maintain this equality. For example, if ω doubles, v must also double. Choice C violates the rolling constraint by suggesting v can remain zero while ω increases - this would cause slipping. The key insight is that rolling without slipping creates a rigid mathematical relationship between translational and rotational motion that cannot be violated.

Question 6

1. It stays the same because rolling implies constant $\omega$
2. It doubles (correct answer)
3. It quadruples
4. It halves

Explanation: This question explores how changes in linear speed affect angular speed in rolling motion. Since the rolling constraint v = ωR must always hold (with constant R), if v doubles, then ω must also double to maintain the relationship. Mathematically, if v₂ = 2v₁ and v = ωR, then ω₂R = 2ω₁R, giving ω₂ = 2ω₁. Choice A incorrectly assumes rolling motion requires constant angular speed, when it actually requires a constant ratio between v and ω. When one quantity in the rolling relationship changes, the other must change proportionally to maintain the constraint.

Question 7

1. $v=\omega R$ (correct answer)
2. $v=\dfrac{\omega}{R}$
3. $v=2\omega R$
4. $v=0$ because it does not slip

Explanation: This problem requires applying the rolling without slipping condition when angular speed is given. For rolling without slipping, the linear speed v of the center of mass equals the product of angular speed ω and radius R: v = ωR. This relationship ensures that the distance traveled by the center equals the arc length swept by any radius during rotation. The contact point travels zero distance relative to the ground, maintaining the no-slip condition. Choice B incorrectly inverts the relationship, while choice C doubles it unnecessarily. To convert between linear and angular quantities in rolling motion, always use v = ωR as your fundamental equation.

Question 8

1. $\omega=\dfrac{v}{2R}$
2. $\omega=\dfrac{v}{R}$ (correct answer)
3. $\omega=\dfrac{2v}{R}$
4. $\omega=0$ because there is no slipping

Explanation: This problem tests understanding of the rolling without slipping condition for rotational motion. When an object rolls without slipping, the linear speed of its center of mass is directly related to its angular speed by the equation v = ωR, where v is the center-of-mass speed, ω is the angular speed, and R is the radius. This relationship exists because the arc length traveled by a point on the rim during one rotation must equal the linear distance traveled by the center. Solving for ω gives ω = v/R. Choice D incorrectly assumes that no slipping means no rotation, when actually it means the rotation and translation are perfectly synchronized. To solve rolling problems, always start with the fundamental constraint v = ωR.

Question 9

1. It triples (correct answer)
2. It is unchanged
3. It becomes one-third as large
4. It doubles because the rim speed is always $2v$

Explanation: This problem examines how radius affects linear speed when angular speed is constant. From v = ωR, we see that linear speed is directly proportional to radius when angular speed is fixed. Tripling the radius triples the linear speed because each point on the larger rim must travel three times as far in each rotation. The wheel covers three times the distance per revolution while maintaining the same rotation rate. Choice B incorrectly assumes v is independent of R, while choice C suggests an inverse relationship. When analyzing rolling motion with changing parameters, identify which variables are held constant and apply v = ωR to find the proportional changes.

Question 10

1. $\alpha=\dfrac{v}{R}$
2. $\alpha=\dfrac{v^2}{R}$
3. $\alpha=0$ (correct answer)
4. $\alpha=\dfrac{g}{R}$

Explanation: This question tests understanding of angular acceleration in rolling motion. Angular acceleration α is the rate of change of angular velocity with respect to time: α = dω/dt. Since the ball rolls with constant center-of-mass speed v, and the rolling condition gives ω = v/R, the angular velocity ω must also be constant (as both v and R are constant). When angular velocity is constant, its time derivative is zero, so α = 0. Choice A incorrectly confuses angular acceleration with angular velocity, choice B incorrectly uses a centripetal acceleration formula, and choice D introduces gravity unnecessarily. Remember that constant linear speed in rolling motion means constant angular speed, which means zero angular acceleration.

Question 11

1. Out of the page
2. Into the page (correct answer)
3. To the right
4. To the left

Explanation: This question requires understanding the vector nature of angular velocity. Angular velocity is a vector quantity whose direction is determined by the right-hand rule: curl your fingers in the direction of rotation, and your thumb points in the direction of the angular velocity vector. For a wheel rolling to the right, the top of the wheel moves forward (right) and the bottom moves backward relative to the center, meaning the wheel rotates clockwise when viewed from the side. Using the right-hand rule for clockwise rotation, the angular velocity vector points into the page. Choice A would correspond to counterclockwise rotation, while choices C and D incorrectly treat angular velocity as pointing in the direction of linear motion. Always use the right-hand rule to determine angular velocity direction.

Question 12

1. $\omega=\dfrac{R}{v}$
2. $\omega=\dfrac{v}{R}$ (correct answer)
3. $\omega=vR$
4. $\omega=0$ because it is not slipping

Explanation: This question tests understanding of the rolling without slipping condition. When an object rolls without slipping, the linear speed of its center equals the tangential speed at its rim, giving us the relationship v = ωR. Solving for angular speed ω, we divide both sides by R to get ω = v/R. This fundamental relationship connects translational and rotational motion for any rolling object. Choice A incorrectly inverts the relationship, while choice D misunderstands that rolling objects do rotate even without slipping. To solve rolling problems, always start with v = ωR and manipulate algebraically to find the unknown quantity.

Question 13

1. $v$
2. $2v$
3. $\dfrac{v}{2}$
4. $0$ (correct answer)

Explanation: This question examines the velocity of the bottom rim point in rolling motion. The bottom point of a rolling wheel moves forward at speed v due to the center's translation, but backward at speed v due to rotation about the center. These two velocity components have equal magnitudes but opposite directions, resulting in zero net velocity relative to the ground. This zero velocity at the contact point is the defining characteristic of rolling without slipping. Choice A incorrectly assumes the bottom point moves with the center. Remember that the instantaneous contact point must have zero velocity to prevent slipping against the surface.

Question 14

1. Zero (correct answer)
2. $v$ forward, equal to the center-of-mass speed
3. $v$ backward
4. $2v$ forward

Explanation: This question examines the instantaneous velocity at the contact point in rolling motion. The rolling without slipping condition requires that the point touching the ground has zero velocity relative to the ground - this prevents sliding. At the contact point, the forward translational velocity v is exactly canceled by the backward rotational velocity v, resulting in zero net velocity. This zero velocity at the contact point is what defines rolling without slipping. Choice B incorrectly ignores the rotational component, while choice C considers only rotation. Remember that rolling without slipping means the contact point is momentarily at rest relative to the ground.

Question 15

1. $0$
2. $v$ (correct answer)
3. $2v$
4. $\tfrac{1}{2}v$

Explanation: This question examines relative velocities in rolling without slipping for AP Physics 1 rotational motion. In rolling without slipping, v = ωR relates center-of-mass speed to angular speed. Relative to the center, rim points have tangential speed ωR, which equals v. For the bottom point, this tangential speed is opposite to the center's motion direction but equals v in magnitude. Choice A, 0, could be confused with the bottom point's speed relative to the ground, not the center. A transferable strategy is to specify the reference frame clearly when calculating relative speeds in rotating systems.

Question 16

1. $0$
2. $v$ (correct answer)
3. $2v$
4. $\dfrac{v}{2}$

Explanation: This question asks about the relative velocity between two points on a rolling sphere. In the reference frame of the sphere's center, all points on the surface move with tangential speed ωR. Since the sphere rolls without slipping at center speed v, we have ω = v/R, making the tangential speed ωR = v. The bottom point moves backward at speed v relative to the center (in the opposite direction of the center's motion). Choice A would only be correct if asking about the bottom point's speed relative to the ground. When finding relative velocities, always clearly identify your reference frame and use the rolling constraint v = ωR.

Question 17

1. $v=\dfrac{\omega}{R}$
2. $v=\omega R$ (correct answer)
3. $v=2\omega R$
4. $v=0$ because rolling without slipping implies no translation

Explanation: This problem tests the fundamental relationship between angular and linear speeds in rolling motion. For rolling without slipping, the center of mass travels a distance equal to the arc length traced by the rim, giving v = ωR. This relationship holds for all shapes rolling without slipping - spheres, cylinders, hoops, or disks. The linear speed equals the angular speed multiplied by the radius. Choice A inverts this relationship, while choice D misunderstands that rolling objects both translate and rotate. To remember this relationship, think of the wheel's circumference 2πR being covered in one full rotation of 2π radians, confirming v = ωR.

Question 18

1. All kinetic energy is translational because the center of mass moves
2. All kinetic energy is rotational because the sphere spins
3. Both translational and rotational kinetic energies are nonzero (correct answer)
4. Kinetic energy is zero at the point of contact with the ramp

Explanation: This question probes kinetic energy distribution in rolling without slipping down an incline in AP Physics 1. Rolling without slipping connects linear velocity v and angular velocity ω via v = ωR, leading to both translational and rotational kinetic energies. Translational KE is (1/2)Mv² from center-of-mass motion, while rotational KE is (1/2)Iω² from spinning. Both components are present and nonzero as the sphere moves and rotates simultaneously. Choice A might be selected if one ignores the rotational contribution entirely. A transferable strategy is to always calculate total kinetic energy as the sum of translational and rotational parts for rolling objects.

Question 19

1. $v=\omega/R$
2. $v=\omega R$ (correct answer)
3. $v=2\omega R$
4. $v=\omega R^2$

Explanation: This question tests the relationship between linear and angular speeds in rolling without slipping for AP Physics 1. Rolling without slipping imposes the condition that the distance traveled linearly equals the arc length rotated, giving v = ωR. Here, v is the center-of-mass speed, ω is the angular speed, and R is the radius. This direct proportionality ensures synchronization between translation and rotation. Choice A, v = ω/R, could be picked if one inverts the formula mistakenly. A transferable strategy is to remember that linear quantities often equal angular quantities multiplied by radius in rolling scenarios.

Question 20

1. $\omega$ is unchanged because rolling is constant
2. $\omega$ doubles (correct answer)
3. $\omega$ halves
4. $\omega$ becomes zero at the contact point

Explanation: This question assesses how changes in linear speed affect angular speed in rolling without slipping in AP Physics 1. Rolling without slipping maintains v = ωR, where v is center-of-mass speed and R is constant. If v doubles, ω must also double to preserve the relationship. This direct proportionality ensures no slipping occurs regardless of speed changes. Choice A might be chosen if one thinks angular speed is independent of linear speed. A transferable strategy is to use the rolling condition v = ωR to predict changes in related variables.