AP Physics 1 Quiz: Connecting Linear And Rotational Motion

What this quiz covers

This quiz focuses on Connecting Linear And Rotational Motion, 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. $a_{c,M}>a_{c,N}$ because the hub region rotates more times per second.
2. $a_{c,M}=a_{c,N}$ because $\omega$ is the same everywhere on a rigid body.
3. $a_{c,N}>a_{c,M}$ because $a_c=\omega^2 r$ increases with radius. (correct answer)
4. Both are zero because angular speed is constant.

Explanation: This question tests the skill of connecting linear and rotational motion, specifically how centripetal acceleration varies with position on a rotating rigid body. The centripetal acceleration for circular motion is ac = ω²r, where ω is the angular speed and r is the distance from the axis. Since the entire fan blade rotates as a rigid body with the same angular speed ω, and point N at the tip is farther from the axis than point M near the hub, point N must have a greater centripetal acceleration. Choice D incorrectly suggests zero acceleration, confusing constant angular speed with the absence of centripetal acceleration. To solve such problems, remember that even at constant angular speed, points in circular motion always experience centripetal acceleration directed toward the center.

Question 2

1. $a_{t,B}=3a_{t,A}$ because $a_t=\alpha r$
2. $a_{t,B}=a_{t,A}$ because both riders share the same $\omega$
3. $a_{t,B}=0$ and $a_{t,A}=0$ because $\alpha=0$ (correct answer)
4. $a_{t,B}=9a_{t,A}$ because tangential acceleration depends on $r^2$

Explanation: This question assesses the skill of connecting linear and rotational motion in AP Physics 1. Linear speed, or tangential speed, for a point on a rotating object is given by v = ω r, where ω is the angular speed and r is the radius from the axis of rotation. Tangential acceleration a_t = α r, where α is angular acceleration, relates linear and rotational motion similarly. For constant ω, α = 0, so tangential acceleration is zero everywhere, regardless of radius. A common distractor is choice A, which applies a_t = α r but forgets that α = 0 for constant speed. To approach similar problems, always recall that for rigid bodies, angular quantities are uniform, but linear quantities scale with radius.

Question 3

1. $a_{t,P}=a_{t,Q}$ because all points share the same $\alpha$.
2. $a_{t,Q}=3a_{t,P}$ because $a_t=\alpha r$. (correct answer)
3. $a_{t,P}=3a_{t,Q}$ because the inner point is “closer to the turning.”
4. $a_{t,Q}=a_{t,P}$ because tangential acceleration depends only on $\omega$.

Explanation: This question tests the skill of connecting linear and rotational motion, specifically the relationship between tangential acceleration and position on a rotating disk. The tangential acceleration for any point on a rotating rigid body is at = αr, where α is the angular acceleration and r is the radius. Since both points are on the same disk with constant angular acceleration α, and point Q is at radius 3r while point P is at radius r, point Q must have three times the tangential acceleration of point P. Choice C incorrectly reverses this relationship with a vague notion about being "closer to the turning." To solve problems involving tangential acceleration, remember that it scales linearly with distance from the rotation axis for a rigid body.

Question 4

1. $a_{c,Q}=3a_{c,P}$ because $a_c=\omega^2 r$ (correct answer)
2. $a_{c,Q}=a_{c,P}$ because both points share the same $\omega$
3. $a_{c,Q}=9a_{c,P}$ because $a_c=(\omega r)^2$
4. $a_{c,Q}=\tfrac{1}{3}a_{c,P}$ because the outer point moves farther each revolution

Explanation: This question assesses the skill of connecting linear and rotational motion in AP Physics 1. Linear speed, or tangential speed, for a point on a rotating object is given by v = ω r, where ω is the angular speed and r is the radius from the axis of rotation. Since points on a rigid disk share the same ω, the point farther out has greater linear speed proportional to r. Centripetal acceleration, however, is a_c = ω² r, so it also increases with radius for constant ω. A common distractor is choice C, which mistakenly uses a_c = (ω r)² instead of a_c = ω² r or v²/r, leading to a quadratic instead of linear dependence on r. To approach similar problems, always recall that for rigid bodies, angular quantities are uniform, but linear quantities scale with radius.

Question 5

1. $v_{\text{scratch}}=v_{\text{dust}}$ because both have the same angular speed.
2. $v_{\text{dust}}=2v_{\text{scratch}}$ because $v=\omega r$. (correct answer)
3. $v_{\text{scratch}}=2v_{\text{dust}}$ because the inner path is shorter.
4. Both linear speeds are zero because neither object slips.

Explanation: This question assesses the skill of connecting linear and rotational motion by analyzing linear speeds on a rotating record. The relationship v = ωr demonstrates that linear speed increases with distance from the center for constant angular speed. This explains why outer points move faster linearly, covering larger circumferences in the same angular time. For the scratch at r, v_scratch = ωr, and dust at 2r, v_dust = ω(2r) = 2ωr, so v_dust = 2v_scratch. Choice A is a distractor, wrongly equating speeds due to same ω, disregarding the radius factor. A key strategy is to visualize the circular paths and compute speeds using v = 2πr / T, where T is the period, applicable to any uniform circular motion.

Question 6

1. $a_{t,A}=a_{t,B}$ because both points share the same angular acceleration.
2. $a_{t,B}=2a_{t,A}$ because $a_t=\alpha r$. (correct answer)
3. $a_{t,A}=2a_{t,B}$ because the inner point “turns faster.”
4. $a_{t,B}=a_{t,A}/2$ because the outer point has more distance to cover.

Explanation: This question tests the skill of connecting linear and rotational motion, specifically how tangential acceleration relates to angular acceleration and radius. For any point on a rotating rigid body, the tangential acceleration is given by at = αr, where α is the angular acceleration and r is the distance from the axis. Since both points are on the same disk with the same angular acceleration α, and point B is at radius 2r while point A is at radius r, point B must have twice the tangential acceleration of point A. Choice C incorrectly reverses the relationship, perhaps confusing the concept with angular quantities. To solve problems involving tangential acceleration, remember that it increases linearly with distance from the rotation axis when angular acceleration is constant.

Question 7

1. $a_{t,A}=a_{t,B}$ because the wheel is rigid.
2. $a_{t,B}=4a_{t,A}$ because $a_t=\alpha r$. (correct answer)
3. $a_{t,B}=\tfrac{1}{4}a_{t,A}$ because $B$ is farther from the axis.
4. $a_{t,B}=16a_{t,A}$ because acceleration scales as $r^2$.

Explanation: This problem tests understanding of connecting linear and rotational motion for tangential acceleration. When a rigid body undergoes angular acceleration α, the tangential acceleration at any point is given by at = αr, where r is the distance from the rotation axis. Since the wheel has constant angular acceleration α, point A at radius r has tangential acceleration at,A = αr, while point B at radius 4r has at,B = α(4r) = 4αr = 4at,A. Choice D incorrectly suggests acceleration scales as r², confusing tangential acceleration with centripetal acceleration relationships. When analyzing rotational motion with angular acceleration, remember that tangential acceleration increases linearly with radius, just like linear speed does with angular speed.

Question 8

1. $a_{c,Y}=4a_{c,X}$ because $a_c=\omega^2 r$. (correct answer)
2. $a_{c,X}=4a_{c,Y}$ because the inner bug turns more sharply.
3. $a_{c,X}=a_{c,Y}$ because both have the same $\omega$.
4. $a_{c,X}$ and $a_{c,Y}$ are zero because $\omega$ is constant.

Explanation: This question assesses the skill of connecting linear and rotational motion by evaluating centripetal accelerations on a rotating wheel. Linear speed v is given by v = ωr, showing dependence on both angular speed ω and radius r for points on rigid bodies. This foundation extends to centripetal acceleration a_c = ω²r, or equivalently v²/r, highlighting greater inward acceleration for larger radii at constant ω. For bug X at r, a_{c,X} = ω²r, and for Y at 4r, a_{c,Y} = ω²(4r) = 4ω²r, so a_{c,Y} = 4a_{c,X}. Distractor C claims a_{c,X} = a_{c,Y} because ω is the same, but this ignores the radius in the formula. Remember to use a_c = v²/r as a strategy, calculating v first if needed, for problems involving circular motion.

Question 9

1. $a_{t,G}=a_{t,H}$ because both points have the same change in $\omega$.
2. $a_{t,H}=4a_{t,G}$ because $a_t=\alpha r$. (correct answer)
3. $a_{t,G}=4a_{t,H}$ because the inner point responds more quickly.
4. $a_{t,H}$ is smaller because points farther out have greater inertia.

Explanation: This question assesses the skill of connecting linear and rotational motion by comparing tangential accelerations on an accelerating fan blade. Although linear speed follows v = ωr, tangential acceleration a_t = αr mirrors this dependence on radius and angular acceleration α. Points farther from the axis thus have greater linear acceleration magnitudes. For point G at r, a_{t,G} = αr, and for H at 4r, a_{t,H} = α(4r) = 4αr, so a_{t,H} = 4a_{t,G}. Distractor A equates them based on same change in ω, but α is the rate of change, and linear effects scale with r. Use the strategy of converting angular to linear via multiplication by r for accelerations in rotational dynamics problems.

Question 10

1. $a_{c,C}=a_{c,D}$ because $\omega$ is the same everywhere on the wheel.
2. $a_{c,D}=3a_{c,C}$ because $a_c=\omega^2 r$. (correct answer)
3. $a_{c,D}=9a_{c,C}$ because $a_c=\omega r^2$.
4. $a_{c,C}=3a_{c,D}$ because larger radius reduces inward acceleration.

Explanation: This question assesses the skill of connecting linear and rotational motion by examining centripetal accelerations on a spinning wheel. Linear speed v depends on radius and angular speed through v = ωr, providing a basis for acceleration relations. Centripetal acceleration a_c = ω²r increases with radius, as outer points require greater inward force to maintain circular paths. For point C at r, a_{c,C} = ω²r, and for D at 3r, a_{c,D} = ω²(3r) = 3ω²r, so a_{c,D} = 3a_{c,C}. Distractor C claims a_{c,D} = 9a_{c,C} using a wrong formula a_c = ω r², confusing it with moment of inertia or other concepts. To tackle similar questions, derive a_c from v²/r after finding v = ωr, ensuring conceptual linkage.

Question 11

1. $v_T=v_C$ because all points on a rigid body have the same speed.
2. $v_T>v_C$ because the rim’s rotation adds to the translational motion at the top. (correct answer)
3. $v_T<v_C$ because points farther from the axis have smaller linear speed.
4. $v_T=0$ because the top point is instantaneously at rest like the contact point.

Explanation: This question tests understanding of rolling motion and the superposition of translational and rotational velocities. For a wheel rolling without slipping, the center C moves at speed v_C = ωR relative to the ground. The top point T has both the translational velocity of the center (v_C) plus the rotational velocity due to spinning (ωR at the rim), giving v_T = v_C + ωR = 2ωR = 2v_C. Since v_C = ωR, we have v_T = 2v_C, making v_T > v_C. Choice D incorrectly assumes the top point is at rest like the contact point, failing to recognize that only the bottom contact point has zero velocity. The strategy is to add the translational and rotational components of velocity, remembering they add at the top and subtract at the bottom.

Question 12

1. $a_{t,R}=a_{t,S}$ because both start from rest
2. $a_{t,S}=2a_{t,R}$ because $a_t=\alpha r$ (correct answer)
3. $a_{t,R}=2a_{t,S}$ because inner points have greater angular acceleration
4. $a_{t,S}$ is greater, but only if the mass at S is larger

Explanation: This problem tests connecting linear and rotational motion during angular acceleration from rest. The tangential acceleration at any point on a rotating platform is a_t = αr, where α is the angular acceleration and r is the radius. Since the platform has constant angular acceleration α for all points, and point S is at radius 2r while point R is at radius r, we calculate a_{t,S} = α(2r) = 2αr = 2a_{t,R}. Choice C incorrectly suggests inner points have greater angular acceleration, when actually all points on a rigid body share the same α. The strategy is to recognize that tangential acceleration depends on both the angular acceleration (same for all points) and the radius (different for each point).

Question 13

1. $a_{c,W}=a_{c,X}$ because both points have the same $\omega$
2. $a_{c,X}=4a_{c,W}$ because $a_c=\omega^2 r^2$
3. $a_{c,X}=2a_{c,W}$ because $a_c=\omega^2 r$ (correct answer)
4. $a_{c,W}=2a_{c,X}$ because the inner point turns more quickly

Explanation: This problem tests connecting linear and rotational motion, specifically centripetal acceleration relationships. For circular motion, centripetal acceleration is a_c = ω²r, where ω is the angular speed and r is the radius. Since both points W and X are on the same wheel with angular speed ω, and X is at radius 2r while W is at radius r, we find a_{c,X} = ω²(2r) = 2ω²r = 2a_{c,W}. Choice B incorrectly includes r² in the formula, suggesting quadratic scaling, while choice D reverses the relationship entirely. The strategy is to remember that centripetal acceleration scales linearly with radius (not quadratically) when angular speed is constant for all points on a rigid body.

Question 14

1. $a_{c,N}=a_{c,M}$ because both points share the same $\omega$
2. $a_{c,N}=2a_{c,M}$ because $a_c=\omega^2 r$ (correct answer)
3. $a_{c,N}=4a_{c,M}$ because $a_c=v^2/r$ and $v$ doubles
4. $a_{c,N}=\tfrac{1}{2}a_{c,M}$ because the larger radius reduces centripetal acceleration

Explanation: This problem tests connecting linear and rotational motion through the relationship between centripetal accelerations at different radii. Centripetal acceleration for circular motion is a_c = ω²r, where ω is angular speed and r is radius. Since points M and N are on the same rigid fan blade, they share angular speed ω. Point N at radius 2r has centripetal acceleration a_{c,N} = ω²(2r) = 2ω²r = 2a_{c,M}, where a_{c,M} = ω²r. Choice C incorrectly uses a_c = v²/r and claims the acceleration quadruples, forgetting that v also depends on r. The key is to use a_c = ω²r directly when ω is shared: doubling the radius doubles the centripetal acceleration.

Question 15

1. They are equal because both LEDs have the same angular acceleration
2. The LED at $r$ has greater tangential acceleration because it is closer to the axis
3. The LED at $3r$ has three times the tangential acceleration because $a_t=\alpha r$ (correct answer)
4. The LED at $3r$ has nine times the tangential acceleration because $a_t=\alpha r^2$

Explanation: This problem tests connecting linear and rotational motion for tangential acceleration at different radii. Tangential acceleration a_t represents the rate of change of linear speed and equals a_t = αr for rotational motion. Since both LEDs are embedded in the same rigid disk, they experience the same angular acceleration α. The LED at radius 3r has tangential acceleration a_t = α(3r) = 3αr, which is three times that of the LED at radius r with a_t = αr. Choice D incorrectly suggests a quadratic relationship (αr²), confusing this with other rotational formulas. The strategy is to remember that tangential acceleration varies linearly with radius when angular acceleration is uniform.

Question 16

1. $\omega_N=4\omega_M$ because N travels farther each second
2. $\omega_N=\omega_M$ because all points on a rigid body share the same $\omega$ (correct answer)
3. $\omega_N=\tfrac{1}{4}\omega_M$ because N has greater tangential speed
4. Angular speed cannot be compared without knowing the fan’s radius

Explanation: This question assesses the skill of connecting linear and rotational motion in AP Physics 1. Linear speed, or tangential speed, for a point on a rotating object is given by v = ω r, where ω is the angular speed and r is the radius from the axis of rotation. Although linear speed increases with radius for constant ω, the angular speed itself is the same for all points on a rigid body, regardless of position. Therefore, points M and N on the fan blade share the same ω, even though their linear speeds differ by a factor of 4. A common distractor is choice A, which incorrectly assumes greater distance means higher angular speed, mixing up linear and angular quantities. To approach similar problems, always recall that for rigid bodies, angular quantities are uniform, but linear quantities scale with radius.

Question 17

1. $v_X=v_Y$ because both have the same $\omega$
2. $v_Y=2v_X$ because $v=\omega r$ (correct answer)
3. $v_X=2v_Y$ because the inner coin completes more revolutions per second
4. $v_Y=\tfrac{1}{2}v_X$ because the outer coin has a longer path

Explanation: This question assesses the skill of connecting linear and rotational motion in AP Physics 1. Linear speed, or tangential speed, for a point on a rotating object is given by v = ω r, where ω is the angular speed and r is the radius from the axis of rotation. Since both coins share the same angular speed ω due to the rigid turntable, the coin at larger radius has greater linear speed proportional to its radius. Thus, for coin Y at 2r, v_Y = ω (2r) = 2 (ω r) = 2 v_X. A common distractor is choice A, which incorrectly assumes linear speeds are equal because angular speeds are the same, ignoring the role of radius. To approach similar problems, always recall that for rigid bodies, angular quantities are uniform, but linear quantities scale with radius.

Question 18

1. Bolt 1 has greater tangential speed because it is closer to the axis.
2. Both bolts have the same tangential speed because they share the same angular speed.
3. Bolt 2 has greater tangential speed because $v=\omega r$. (correct answer)
4. Both bolts have zero tangential speed because they are fixed in place on the platform.

Explanation: This question tests the skill of connecting linear and rotational motion, specifically the relationship between tangential speed and radial position. For any point on a rotating rigid body, the tangential speed is v = ωr, where ω is the angular speed and r is the distance from the axis. Since both bolts are fixed to the same platform rotating at angular speed ω, and bolt 2 is at radius 4r while bolt 1 is at radius r, bolt 2 must have four times the tangential speed of bolt 1. Choice D incorrectly suggests zero speed because the bolts are "fixed in place," misunderstanding that being fixed to a rotating platform means moving in a circle. To solve these problems, remember that "fixed" points on rotating objects still have tangential speeds proportional to their distances from the axis.

Question 19

1. $a_{c,B}=2a_{c,A}$ because $a_c=\omega^2 r$. (correct answer)
2. $a_{c,B}=a_{c,A}$ because both points have the same $\omega$.
3. $a_{c,B}=4a_{c,A}$ because centripetal acceleration scales with $v^2$ and $v$ is the same.
4. $a_{c,A}=0$ and $a_{c,B}=0$ because the wheel’s angular speed is constant.

Explanation: This question tests the skill of connecting linear and rotational motion, specifically how centripetal acceleration scales with radius in rigid body rotation. The centripetal acceleration for circular motion is ac = ω²r, where ω is the angular speed and r is the radius. Since both points are on the same wheel rotating at angular speed ω, and point B is at radius 2r while point A is at radius r, point B must have twice the centripetal acceleration of point A. Choice C incorrectly suggests a factor of 4, perhaps confusing the v² relationship in ac = v²/r with the direct application here. To solve problems involving centripetal acceleration in rigid rotation, use ac = ω²r directly, which shows linear scaling with radius.

Question 20

1. $a_{t,A}=a_{t,B}$ because both points share the same $\alpha$.
2. $a_{t,B}=3a_{t,A}$ because $a_t=\alpha r$. (correct answer)
3. $a_{t,A}=3a_{t,B}$ because the inner point changes direction faster.
4. $a_{t,B}=\tfrac{1}{3}a_{t,A}$ because larger radius reduces acceleration.

Explanation: This question assesses the skill of connecting linear and rotational motion by comparing tangential accelerations on an accelerating turntable. While linear speed v depends on radius r and angular speed ω via v = ωr, tangential acceleration a_t similarly relates as a_t = αr, where α is angular acceleration. For point A at r, a_{t,A} = αr, and for B at 3r, a_{t,B} = α(3r) = 3αr, so a_{t,B} = 3a_{t,A}. This arises because points farther out cover greater linear distances while accelerating angularly at the same rate. Choice A is a distractor that wrongly equates a_t since α is shared, overlooking the radius factor in the linear quantity. A transferable strategy is to derive linear quantities from angular ones using radius and double-check by considering the path circumference.