Rolling - AP Physics 1

A cylinder rolls without slipping at speed $v$ on a horizontal surface. What is the instantaneous speed of the point in contact with the ground?
Answer: $0$ 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.
A wheel rolls without slipping. At an instant its center moves right at speed $v$ . What is the direction of the wheel's angular velocity?
Answer: Into the page 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.
A wheel rolls without slipping. If $v_{\text{cm}}=3\ \text{m/s}$ , what is the speed of the top point relative to the ground?
Answer: $6\ \text{m/s}$ This question assesses velocity addition in rolling without slipping in AP Physics 1 rotational motion. Rolling without slipping links v_cm = ωR, affecting point velocities. The top point combines v_cm forward with tangential speed ωR forward, totaling 2v_cm. For v_cm = 3 m/s, the top speed is 6 m/s relative to the ground. Choice B, 3 m/s, could be chosen if one neglects the rotational addition. A transferable strategy is to double the center speed for the top point in pure rolling scenarios.
A hoop rolls without slipping. If the hoop's center-of-mass speed doubles, what happens to its angular speed $\omega$ ?
Answer: $\omega$ doubles 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.
A disk rolls without slipping. If its center-of-mass speed doubles, what happens to its angular speed $\omega$ ?
Answer: It doubles 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.
A cylinder of radius $R$ rolls without slipping. If its center moves a distance $x$ along the floor, what is its angular displacement?
Answer: $\theta = x/R$ 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.
A cylinder rolls without slipping. If its center-of-mass speed doubles, how does its angular speed change?
Answer: It doubles This question assesses understanding of rolling without slipping in rotational dynamics. In rolling without slipping, angular speed ω is directly proportional to linear speed v via ω = v/R, with R constant. If v doubles, ω must also double to maintain the no-slip condition. This relationship holds for any rolling object regardless of shape. A common distractor is choice C (increases by 4), which might confuse speed with kinetic energy scaling. A transferable strategy is to use proportional relationships like v = ωR to predict changes in linked variables.
A wheel of radius $R$ rolls without slipping. If the center moves a distance $2\pi R$ , how many radians has the wheel rotated?
Answer: $2\pi$ This question connects linear displacement to angular displacement in rolling motion. When a wheel rolls without slipping, the center's displacement equals the arc length traced by any rim point: s = Rθ. If the center moves distance 2πR, then 2πR = Rθ, giving θ = 2π radians (one complete rotation). This makes physical sense: moving a distance equal to the circumference corresponds to exactly one full turn. Choice C would represent two full rotations, while choice D incorrectly treats the angle as having units of inverse length. To solve rolling displacement problems, use s = Rθ where s is linear displacement and θ is angular displacement in radians.
A wheel of radius $R$ rolls without slipping at speed $v$ . What is the magnitude of the tangential speed of a rim point relative to the center?
Answer: $v$ This problem asks about the tangential speed of a rim point relative to the wheel's center. In the reference frame of the center, rim points move in a circle with tangential speed ωR. Since the wheel rolls without slipping with center speed v, we know ω = v/R. Therefore, the tangential speed relative to the center is ωR = (v/R)R = v. This makes physical sense: rim points rotate around the center at the same speed the center translates. Choice D might come from confusing this with the ground-relative speed of the top point. For rotational motion problems, carefully identify which reference frame is being used.
A wheel of radius $R$ rolls without slipping with angular speed $\omega$ . What is the speed of its center of mass?
Answer: $v=\omega R$ 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.
A hoop rolls without slipping down a ramp and speeds up. Which statement about static friction on the hoop is correct?
Answer: Static friction does zero work on the hoop This question assesses understanding of rolling without slipping in rotational dynamics. Rolling without slipping down a ramp involves static friction providing torque for angular acceleration without slipping at the contact point. The linear acceleration relates to angular acceleration via a = αR, with friction enabling both. Since the contact point is at rest relative to the surface, static friction does no work as there's no displacement at the point of application. A common distractor is choice B, which confuses static friction with kinetic friction that would act if slipping occurred. A transferable strategy is to evaluate work by considering the velocity at the point where the force is applied in rolling scenarios.
A ball rolls without slipping with angular speed $\omega$ . What is the center-of-mass speed in terms of $\omega$ and radius $R$ ?
Answer: $v=\omega R$ 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.
A cylinder rolls without slipping at speed $v$ . What is the instantaneous speed of the point in contact with the ground?
Answer: $0$ 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.
A sphere rolls without slipping at center-of-mass speed $v$ . What is the speed of the bottom point relative to the center of the sphere?
Answer: $v$ 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.
A disk rolls without slipping to the right. Which point on the rim has the greatest speed relative to the ground?
Answer: The point at the top of the disk This question assesses understanding of rolling without slipping in rotational dynamics. Rolling without slipping means v = ωR, with velocities adding vectorially for rim points. The top point has translational v right plus rotational v right, totaling 2v. Other points, like the front, have components yielding less than 2v, such as √2v. A common distractor is choice D (all same), ignoring varying rotational contributions. A transferable strategy is to calculate point speeds using vector addition and compare magnitudes for extrema.
A sphere rolls without slipping down a ramp. Which statement about kinetic energy at any instant is necessarily true?
Answer: Both translational and rotational kinetic energies are nonzero 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.
A disk rolls without slipping. If its radius increases while $\omega$ stays the same, what happens to the center-of-mass speed?
Answer: It increases This question explores how changing radius affects the rolling condition. For rolling without slipping, the relationship v = ωR must be maintained, where v is center-of-mass speed, ω is angular speed, and R is radius. If the radius increases while angular speed ω remains constant, then the center-of-mass speed must increase proportionally: v = ωR increases because R increases. Choice B incorrectly inverts the relationship, choice C fails to recognize the dependence on radius, and choice D is physically impossible for non-zero ω. The strategy is to recognize that for fixed angular speed, linear speed is directly proportional to radius in rolling motion.
A hoop rolls without slipping on a horizontal surface with center-of-mass speed $v$ . What is the speed of the top point relative to the ground?
Answer: $2v$ This question requires applying the rolling without slipping condition to find the velocity of a specific point on a rolling object. For any rolling object, the velocity of a point is the vector sum of the translational velocity (v) and the rotational velocity relative to the center. At the top of a rolling hoop, both velocities point in the same direction: the center moves forward at v, and the top point rotates forward at ωR = v relative to the center. Therefore, the top point's speed relative to the ground is v + v = 2v. Choice A incorrectly assumes the top point is stationary, ignoring that it participates in both translation and rotation. When analyzing point velocities on rolling objects, always add the center's velocity to the point's rotational velocity.
A hoop rolls without slipping along a level surface with center-of-mass speed $v$ . What is the speed of the point at the top of the hoop?
Answer: $2v$ This problem requires analyzing the motion of different points on a rolling object. For an object rolling without slipping, every point participates in both the translational motion of the center (speed v) and the rotational motion about the center. The top of the hoop moves forward with the center's speed v, plus an additional forward speed v from rotation, giving a total speed of 2v relative to the ground. The bottom of the hoop would have speed 0 (v forward from translation minus v backward from rotation). Choice B incorrectly considers only the translational motion. When finding speeds of points on rolling objects, add the translational and rotational velocity components vectorially.
A hoop rolls without slipping along level ground with center-of-mass speed $v$ . What is the speed of the point at the top?
Answer: $2v$ This question requires understanding how different points on a rolling object move. For an object rolling without slipping, the center of mass moves at speed v, and the object rotates with angular speed ω = v/R. Any point on the rim has two velocity components: the translational velocity v of the center of mass and the rotational velocity due to spinning. At the top of the hoop, these velocities add because both point in the same direction (forward), giving a total speed of v + ωR = v + v = 2v. Choice B incorrectly considers only the translational motion, while choice D incorrectly halves the speed. The strategy is to add the center's velocity to the point's velocity relative to the center.

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