AP Physics 1 Quiz: Connecting Linear And Rotational Motion

20 questions · exam conditions

0:00

Connecting Linear And Rotational MotionQuestion 1 of 20

A ceiling fan blade rotates with angular speed $\omega$ . Point $M$ is located near the hub and point $N$ is at the tip, farther from the axis. Consider the fan at an instant when it is spinning steadily. Which statement correctly compares the magnitudes of the points' centripetal accelerations?

a_{c,M}>a_{c,N}

because the hub region rotates more times per second.

a_{c,M}=a_{c,N}

because

\omega

is the same everywhere on a rigid body.

a_{c,N}>a_{c,M}

because

a_c=\omega^2 r

increases with radius.

Both are zero because angular speed is constant.

← Back to quizzes

AP Physics 1 Quiz

AP Physics 1 Quiz: Connecting Linear And Rotational Motion

Practice Connecting Linear And Rotational Motion in AP Physics 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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.

All questions

Question 1

$a_{c,M}>a_{c,N}$ because the hub region rotates more times per second.
$a_{c,M}=a_{c,N}$ because $\omega$ is the same everywhere on a rigid body.
$a_{c,N}>a_{c,M}$ because $a_c=\omega^2 r$ increases with radius. (correct answer)
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

A rigid disk rotates with constant angular acceleration $\alpha$ about its center. Two points, $P$ at radius $r$ and $Q$ at radius $3r$ , are marked. At a given instant, which statement correctly compares their tangential accelerations?

$a_{t,P}=a_{t,Q}$ because all points share the same $\alpha$ .
$a_{t,Q}=3a_{t,P}$ because $a_t=\alpha r$ . (correct answer)
$a_{t,P}=3a_{t,Q}$ because the inner point is "closer to the turning."
$a_{t,Q}=a_{t,P}$ 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 3

A rigid disk starts from rest and speeds up with constant angular acceleration $\alpha$ . Two points, $A$ at radius $r$ and $B$ at radius $2r$ , are painted on the disk. At the same instant during the spin-up, which statement correctly compares the magnitudes of their tangential accelerations?

$a_{t,A}=a_{t,B}$ because both points share the same angular acceleration.
$a_{t,B}=2a_{t,A}$ because $a_t=\alpha r$ . (correct answer)
$a_{t,A}=2a_{t,B}$ because the inner point "turns faster."
$a_{t,B}=a_{t,A}/2$ 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 4

A wheel speeds up with constant angular acceleration $\alpha$ about its center. Point $A$ is at radius $r$ and point $B$ is at radius $4r$ . At the same instant, how do their tangential accelerations compare?

$a_{t,A}=a_{t,B}$ because the wheel is rigid.
$a_{t,B}=4a_{t,A}$ because $a_t=\alpha r$ . (correct answer)
$a_{t,B}=\tfrac{1}{4}a_{t,A}$ because $B$ is farther from the axis.
$a_{t,B}=16a_{t,A}$ 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 5

A wheel rotates at constant angular speed $\omega$ . A bug sits at point $X$ a distance $r$ from the center, and another bug sits at point $Y$ a distance $4r$ from the center. Which statement about their centripetal accelerations is correct?

$a_{c,Y}=4a_{c,X}$ because $a_c=\omega^2 r$ . (correct answer)
$a_{c,X}=4a_{c,Y}$ because the inner bug turns more sharply.
$a_{c,X}=a_{c,Y}$ because both have the same $\omega$ .
$a_{c,X}$ 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 6

A fan blade rotates with angular speed $\omega$ that is increasing at a constant rate $\alpha$ . Point $G$ is at radius $r$ and point $H$ is at radius $4r$ . Which statement about their tangential accelerations is correct?

$a_{t,G}=a_{t,H}$ because both points have the same change in $\omega$ .
$a_{t,H}=4a_{t,G}$ because $a_t=\alpha r$ . (correct answer)
$a_{t,G}=4a_{t,H}$ because the inner point responds more quickly.
$a_{t,H}$ 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 7

A bicycle wheel rolls without slipping while the bike moves at constant speed. Consider point $T$ at the top of the rim and point $C$ at the wheel's center. At an instant when the wheel's angular speed is $\omega$ and radius is $R$ , which statement about their speeds relative to the ground is correct?

$v_T=v_C$ because all points on a rigid body have the same speed.
$v_T>v_C$ because the rim's rotation adds to the translational motion at the top. (correct answer)
$v_T<v_C$ because points farther from the axis have smaller linear speed.
$v_T=0$ 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 8

A rigid fan blade rotates with constant angular speed $\omega$ . Point $M$ is at radius $r$ from the center; point $N$ is at radius $2r$ . Both points rotate with the blade. Which relationship between their centripetal accelerations is correct?

$a_{c,N}=a_{c,M}$ because both points share the same $\omega$
$a_{c,N}=2a_{c,M}$ because $a_c=\omega^2 r$ (correct answer)
$a_{c,N}=4a_{c,M}$ because $a_c=v^2/r$ and $v$ doubles
$a_{c,N}=\tfrac{1}{2}a_{c,M}$ 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 9

A rigid disk rotates with constant angular acceleration $\alpha$ about its center. Two embedded LEDs at radii $r$ and $3r$ flash simultaneously at a particular instant. At that instant, which comparison of their tangential accelerations is correct?

They are equal because both LEDs have the same angular acceleration
The LED at $r$ has greater tangential acceleration because it is closer to the axis
The LED at $3r$ has three times the tangential acceleration because $a_t=\alpha r$ (correct answer)
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 10

A horizontal turntable rotates at constant angular speed $\omega$ . Two coins are taped down: coin X at radius $r$ and coin Y at radius $2r$ . Which statement about their linear speeds is correct?

$v_X=v_Y$ because both have the same $\omega$
$v_Y=2v_X$ because $v=\omega r$ (correct answer)
$v_X=2v_Y$ because the inner coin completes more revolutions per second
$v_Y=\tfrac{1}{2}v_X$ 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 11

A rigid platform rotates about a vertical axis with angular speed $\omega$ . Two bolts are fixed to the platform: bolt 1 at radius $r$ and bolt 2 at radius $4r$ . Assume the platform spins without changing $\omega$ . Which statement correctly compares the bolts' tangential (linear) speeds?

Bolt 1 has greater tangential speed because it is closer to the axis.
Both bolts have the same tangential speed because they share the same angular speed.
Bolt 2 has greater tangential speed because $v=\omega r$ . (correct answer)
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 12

A rigid wheel rotates steadily with angular speed $\omega$ . Point $A$ is at radius $r$ and point $B$ is at radius $2r$ . At the same instant, which statement correctly compares the magnitudes of their centripetal accelerations?

$a_{c,B}=2a_{c,A}$ because $a_c=\omega^2 r$ . (correct answer)
$a_{c,B}=a_{c,A}$ because both points have the same $\omega$ .
$a_{c,B}=4a_{c,A}$ because centripetal acceleration scales with $v^2$ and $v$ is the same.
$a_{c,A}=0$ 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 13

A turntable speeds up with constant angular acceleration $\alpha$ . Two dots, $A$ at radius $r$ and $B$ at radius $3r$ , are painted on the turntable. At the same instant, which comparison of their tangential accelerations is correct?

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

Question 14

A rigid turntable spins at constant angular speed $\omega$ . Two small stickers are placed at radii $r_1$ and $r_2$ from the center, with $r_2>r_1$ . The turntable completes each revolution in the same time throughout the motion. Neglect slipping and assume both stickers move in perfect circles about the axis. Which statement correctly compares the stickers' tangential (linear) speeds?

The sticker at $r_2$ has greater tangential speed because $v=\omega r$ . (correct answer)
Both stickers have the same tangential speed because they share the same $\omega$ .
The sticker at $r_1$ has greater tangential speed because it travels a smaller circle each revolution.
Both stickers have zero tangential speed because the motion is rotational, not linear.

Explanation: This question tests the skill of connecting linear and rotational motion, specifically how tangential speed relates to angular speed and radius. For any point on a rotating rigid body, the tangential (linear) speed is given by v = ωr, where ω is the angular speed and r is the distance from the axis of rotation. Since both stickers are on the same turntable rotating at the same angular speed ω, and r₂ > r₁, the sticker at r₂ must have a greater tangential speed than the sticker at r₁. Choice B incorrectly suggests equal tangential speeds by ignoring the radius dependence. To solve problems like this, remember that while all points on a rigid rotating object share the same angular speed, their linear speeds increase proportionally with distance from the axis.

Question 15

A bicycle wheel rotates with constant angular speed $\omega$ . Point $P$ is on the rim and point $Q$ is halfway between the center and rim. Both points move in circles about the axle. Which statement correctly compares the magnitudes of their centripetal accelerations?

$a_{c,P}=a_{c,Q}$ because both points have the same angular speed.
$a_{c,P}<a_{c,Q}$ because the rim point has a larger radius.
$a_{c,P}>a_{c,Q}$ because $a_c=\omega^2 r$ . (correct answer)
$a_{c,P}=0$ because constant $\omega$ implies zero acceleration.

Explanation: This question tests the skill of connecting linear and rotational motion, specifically how centripetal acceleration depends on angular speed and radius. For circular motion, centripetal acceleration is given by ac = ω²r, where ω is the angular speed and r is the radius. Since both points are on the same wheel rotating at the same angular speed ω, and point P is on the rim (larger r) while point Q is halfway to the center (smaller r), point P must have a greater centripetal acceleration than point Q. Choice A incorrectly suggests equal accelerations by ignoring the radius dependence. To solve problems involving centripetal acceleration in rigid body rotation, remember that ac increases linearly with radius when angular speed is constant.

Question 16

A disk rotates at constant $\omega$ . Point $M$ is at radius $r$ and point $N$ is at radius $2r$ . If both points have the same angular speed, what is the ratio of their centripetal accelerations $a_{c,N}/a_{c,M}$ ?

$\tfrac{1}{2}$
$1$
$2$ (correct answer)
$4$

Explanation: This problem tests understanding of connecting linear and rotational motion for centripetal acceleration ratios. Centripetal acceleration for circular motion is ac = ω²r, where ω is angular speed and r is radius. Point M is at radius r with acceleration ac,M = ω²r, while point N is at radius 2r with acceleration ac,N = ω²(2r) = 2ω²r. The ratio is ac,N/ac,M = 2ω²r/ω²r = 2. Choice D might result from incorrectly thinking acceleration scales as r², but that would apply to comparing different angular speeds, not different radii at the same ω. When finding ratios in rotational motion, set up the complete expressions first, then simplify to see how the scaling works.

Question 17

A turntable rotates with constant angular speed $\omega$ . A coin at radius $r$ and a second coin at radius $3r$ do not slip. How do their centripetal accelerations compare?

$a_{c,3r}=3a_{c,r}$ because $a_c=\omega^2 r$ . (correct answer)
$a_{c,3r}=a_{c,r}$ because both share the same $\omega$ .
$a_{c,3r}=\tfrac{1}{3}a_{c,r}$ because the outer coin takes longer to go around.
$a_{c,3r}$ depends on the coins' masses, so it cannot be compared.

Explanation: This question tests connecting linear and rotational motion, specifically for centripetal acceleration. For circular motion, centripetal acceleration is given by ac = ω²r, where ω is angular speed and r is radius. Since both coins are on the same turntable, they share the same angular speed ω. The coin at radius 3r has centripetal acceleration ac,3r = ω²(3r) = 3ω²r, while the coin at radius r has ac,r = ω²r. Therefore, ac,3r = 3ac,r. Choice B incorrectly assumes centripetal acceleration is independent of radius, ignoring the fundamental relationship ac = ω²r. To solve problems involving centripetal acceleration on rotating platforms, remember that ac scales linearly with radius when angular speed is constant.

Question 18

A rigid carousel rotates counterclockwise with constant angular speed $\omega$ . Two riders stand on the platform at distances $r_1$ and $r_2$ from the center, with $r_2>r_1$ . Assume both riders rotate with the platform without slipping. Which statement about their linear speeds is correct?

$v_1>v_2$ because the inner rider has a shorter path and must move faster to keep up
$v_1=v_2$ because both riders have the same angular speed
$v_2>v_1$ because $v=\omega r$ and $r_2>r_1$ (correct answer)
$v_2=v_1$ because linear speed depends only on $\omega$ , not on $r$

Explanation: This problem tests connecting linear and rotational motion for a carousel system. The linear speed v of any point on a rigid rotating body is given by v = ωr, where ω is the angular speed and r is the distance from the rotation axis. Since both riders are on the same rigid carousel rotating at angular speed ω, they share this angular speed. The rider at r_2 (where $r_2$ > $r_1$ ) has linear speed v_2 = ωr_2 > ωr_1 = v_1. Choice B incorrectly claims equal linear speeds, confusing the shared angular speed with linear speed. The strategy is to recognize that farther points on a rotating rigid body move faster linearly, even though all points complete rotations in the same time.

Question 19

A rigid disk rotates counterclockwise about a fixed axle with constant angular speed $\omega$ . Two small dots are painted on the disk: dot $P$ at radius $r$ and dot $Q$ at radius $2r$ . After the disk has been spinning steadily for several seconds, the dots pass a mark on the rim once per revolution. Neglect slipping and wobble. Which statement about the dots' instantaneous linear speeds is correct?

$v_P=v_Q$ because both dots complete each revolution in the same time.
$v_Q=2v_P$ because $v=\omega r$ and $Q$ is twice as far from the axis. (correct answer)
$v_P=2v_Q$ because points closer to the axis move faster.
$v_Q=v_P/2$ because angular speed decreases with radius.

Explanation: This question tests the relationship between linear speed and radius in rotational motion. For a rigid rotating object, all points share the same angular speed ω, meaning they complete each revolution in the same time period. The linear speed v of any point is given by v = ωr, where r is the distance from the axis of rotation. Since dot Q is at radius 2r while dot P is at radius r, and both have the same ω, we get v_Q = ω(2r) = 2ωr = 2v_P. Choice A incorrectly assumes equal linear speeds just because the revolution time is the same, failing to account for the different path lengths. The key strategy is to remember that linear speed increases proportionally with radius when angular speed is constant.

Question 20

A rigid disk rotates about its central axis with angular speed increasing at a constant rate $\alpha$ . Point $M$ is at radius $r$ and point $N$ is at radius $4r$ . At a particular instant, which comparison of their tangential accelerations is correct?

$a_{t,M}=a_{t,N}$ because $\alpha$ is the same everywhere on the disk.
$a_{t,N}=4a_{t,M}$ because $a_t=\alpha r$ . (correct answer)
$a_{t,M}=4a_{t,N}$ because the inner point has greater angular acceleration.
$a_{t,N}=16a_{t,M}$ because tangential acceleration scales as $r^2$ .

Explanation: This question tests understanding of tangential acceleration in rotational motion with angular acceleration. For a rigid disk with angular acceleration α, the tangential acceleration at any point is a_t = αr, where r is the radius. Point M at radius r has tangential acceleration a_{t,M} = αr, while point N at radius 4r has a_{t,N} = α(4r) = 4αr. Therefore, a_{t,N} = 4a_{t,M}. Choice D incorrectly suggests tangential acceleration scales as r², confusing it with centripetal acceleration's dependence on angular speed. The key concept is that tangential acceleration varies linearly with radius when angular acceleration is uniform across the rigid body.