Connecting Linear and Rotational Motion - AP Physics 1

1. A rigid platform rotates with angular speed $\omega$ that is constant. Point $U$ is at radius $r$ and point $V$ is at radius $4r$. Both points move in circles about the same axis. Which statement about their centripetal accelerations is correct?
   Answer: $a_{c,V}=4a_{c,U}$ because $a_c=\omega^2 r$ This problem tests connecting linear and rotational motion through centripetal acceleration at different radii. Centripetal acceleration for circular motion is $a_c$ = ω²r, where ω is the angular speed and r is the radius. Points U and V on the same rigid platform share angular speed ω. Point V at radius 4r has centripetal acceleration $a_{c,V}$ = ω²(4r) = 4ω²r, while point U at radius r has $a_{c,U}$ = ω²r. Thus $a_{c,V}$ = $4a_{c,U}$. Choice C incorrectly suggests a quadratic relationship (ωr²), which is not the correct formula for centripetal acceleration. The strategy is to apply $a_c$ = ω²r: when radius increases by a factor of 4, centripetal acceleration also increases by a factor of 4.
2. A wheel rotates at constant angular speed $\omega$. A bug crawls to a point farther from the axle, moving from radius $r_1$ to radius $r_2$ where $r_2>r_1$, while remaining stuck to the wheel. At the moment it reaches $r_2$, what can be inferred about its linear speed compared to at $r_1$?
   Answer: It is larger because $v=\omega r$ and $r$ increased This problem tests connecting linear and rotational motion when a bug changes its radial position. The linear speed of any point on a rotating object is v = ωr, where ω is the wheel's angular speed and r is the distance from the axis. As the bug moves from radius $r_1$ to the larger radius $r_2$ while the wheel maintains constant angular speed ω, its linear speed increases from $v_1$ = ω$r_1$ to $v_2$ = ω$r_2$. Choice A incorrectly suggests the speed decreases due to "sharing" angular speed over more circumference, misunderstanding that each point has its own linear speed. The strategy is to apply v = ωr directly: when ω is constant and r increases, v must increase proportionally.
3. 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?
   Answer: $a_{t,N}=4a_{t,M}$ because $a_t=\alpha r$. 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.
4. A rigid disk spins with constant $\omega$. Point P is at distance $R$ from the center and point Q is at $3R$. How do their centripetal accelerations compare?
   Answer: $a_{c,Q}=3a_{c,P}$ because $a_c=\omega^2 r$ 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.
5. A wheel rotates at angular speed $\omega$. Point W is at radius $r$ and point X at radius $2r$. Which statement about centripetal acceleration is correct?
   Answer: $a_{c,X}=2a_{c,W}$ because $a_c=\omega^2 r$ 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.
6. A rigid platform rotates about a vertical axis with angular speed $\omega$. Point $E$ is at radius $2r$ and point $F$ is at radius $5r$. Which statement about their linear speeds is correct?
   Answer: $v_F=\tfrac{5}{2}v_E$ because $v=\omega r$. This question assesses the skill of connecting linear and rotational motion by evaluating linear speeds on a rotating platform. Linear speed v is calculated as v = ωr, directly proportional to the radius from the axis of rotation. Consequently, points at larger radii experience higher linear speeds while maintaining the same angular displacement over time. For point E at 2r, $v_E$ = ω(2r), and for F at 5r, $v_F$ = ω(5r) = (5/2)ω(2r) = (5/2)$v_E$. Choice A distracts by equating $v_E$ and $v_F$ due to same rotation time, but this overlooks the differing path lengths. A transferable approach is to consider the ratio of radii, as speed ratios match radius ratios for constant ω.
7. A turntable speeds up with constant angular acceleration $\alpha$. Point A is at radius $r$ and point B at radius $3r$. How do their tangential accelerations compare?
   Answer: $a_{t,B}=3a_{t,A}$ because $a_t=\alpha r$ This problem tests connecting linear and rotational motion, specifically the relationship between tangential acceleration and angular acceleration. For any point on a rotating object, the tangential acceleration is given by $a_t$ = αr, where α is the angular acceleration and r is the radius. Since the turntable has constant angular acceleration α for all points, and point B is at radius 3r while point A is at radius r, we find that $a_{t,B}$ = α(3r) = 3αr = $3a_{t,A}$. Choice C incorrectly suggests outer points rotate more slowly, which confuses angular speed (same for all points) with linear speed. The strategy to remember is that tangential acceleration, like tangential speed, scales linearly with radius when angular acceleration is constant.
8. A fan blade rotates with constant angular acceleration $\alpha$ from rest. Point $S$ is at radius $r$ and point $T$ is at radius $3r$. At the same time, how do their tangential accelerations compare?
   Answer: $a_{t,T}=3a_{t,S}$ because $a_t=\alpha r$. This problem tests understanding of connecting linear and rotational motion for tangential acceleration. When a rigid body has angular acceleration α, any point at distance r from the axis experiences tangential acceleration at = αr. Point S at radius r has tangential acceleration at,S = αr, while point T at radius 3r has at,T = α(3r) = 3αr = 3at,S. Choice C incorrectly suggests tangential acceleration scales as r², which would be true for centripetal acceleration at constant ω, not tangential acceleration with constant α. When solving problems with angular acceleration, remember that tangential acceleration represents the rate of change of linear speed and scales linearly with radius.
9. A rigid wheel rotates with angular speed $\omega$. A mark at radius $r$ and a mark at radius $2r$ are observed simultaneously. Which statement about their angular speeds is correct?
   Answer: $\omega_{2r}=\omega_r$ because all points on a rigid body share the same $\omega$. This question tests connecting linear and rotational motion, specifically understanding angular speed for a rigid body. For any rigid body rotating about a fixed axis, all points share the same angular speed ω, regardless of their distance from the axis. This is a fundamental property of rigid body rotation - while linear speeds vary with radius (v = ωr), angular speed is uniform throughout the body. Both marks complete the same angle in the same time, giving them identical angular speeds. Choice A incorrectly tries to relate different linear speeds to different angular speeds, missing this key principle. When analyzing rigid body rotation, remember that angular quantities (ω, α) are shared by all points, while linear quantities (v, at) vary with radius.
10. A rigid rod rotates about one end with angular speed $\omega$. Point E is $L$ from the axis and point F is $\tfrac{L}{2}$. Which is true about their linear speeds?
    Answer: $v_E=2v_F$ because $v=\omega r$ This problem tests connecting linear and rotational motion for a rigid rod rotating about one end. The tangential speed of any point on a rotating rigid body is v = ωr, where ω is the angular speed and r is the distance from the rotation axis. Point E is at distance L from the axis while point F is at distance L/2, and both share the same angular speed ω since they're on the same rigid rod. Therefore, $v_E$ = ωL and $v_F$ = ω(L/2) = ωL/2, which means $v_E$ = 2v_F. Choice B incorrectly suggests smaller radius gives larger speed, reversing the actual relationship. Remember that for rigid body rotation, points farther from the axis always move faster linearly, even though all points complete each revolution in the same time.
11. A carousel rotates with constant angular acceleration $\alpha$ from rest. Rider $1$ stands at radius $r$ and rider $2$ stands at radius $2r$. At a given instant, which statement about their tangential accelerations is correct?
    Answer: $a_{t,2}=2a_{t,1}$ because $a_t=\alpha r$. This question assesses the skill of connecting linear and rotational motion by comparing tangential accelerations on an accelerating carousel. Linear speed connects to rotational variables via v = ωr, illustrating how radius amplifies linear effects from angular motion. Similarly, tangential acceleration $a_t$ = αr scales linearly with radius for constant angular acceleration α. For rider 1 at r, $a_{t,1}$ = αr, and for rider 2 at 2r, $a_{t,2}$ = α(2r) = 2αr, so $a_{t,2}$ = $2a_{t,1}$. Distractor D claims both are zero from rest, but acceleration occurs as speed increases, regardless of starting point. For transferable insight, always relate linear and angular accelerations with $a_t$ = αr and confirm by dimensional analysis.
12. 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?
    Answer: $a_{c,N}>a_{c,M}$ because $a_c=\omega^2 r$ increases with radius. 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.
13. A bicycle wheel rotates at constant $\omega$. A dot on the rim is at radius $R$ and a dot halfway out is at $\tfrac{R}{2}$. Which compares their tangential speeds?
    Answer: $v_{\text{rim}}=2v_{\text{half}}$ because $v=\omega r$ 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. For a rigid wheel rotating at constant ω, all points share the same angular speed, but linear speed increases with distance from the center. Therefore, the dot at R has twice the linear speed of the one at R/2, as v_rim = ω R = 2 (ω (R/2)) = 2 v_half. A common distractor is choice A, which wrongly equates linear speeds because the revolution time is the same, overlooking that path length differs with radius. To approach similar problems, always recall that for rigid bodies, angular quantities are uniform, but linear quantities scale with radius.
14. 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?
    Answer: $a_{t,B}=4a_{t,A}$ because $a_t=\alpha r$. 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.
15. A wheel spins with constant angular speed $\omega$. Two sensors detect points $C$ at radius $r$ and $D$ at radius $3r$. Which statement about their centripetal accelerations is correct?
    Answer: $a_{c,D}=3a_{c,C}$ because $a_c=\omega^2 r$. 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.
16. A rigid disk rotates with angular speed $\omega$ that is increasing in time. Two points $X$ and $Y$ lie on the disk at radii $r_X$ and $r_Y$, with $r_Y>r_X$. At some instant, the disk's angular acceleration is $\alpha$. Which statement about the points' tangential accelerations is correct?
    Answer: $a_{t,Y}>a_{t,X}$ because $a_t=\alpha r$ and $r_Y>r_X$ This problem tests connecting linear and rotational motion for tangential acceleration on an accelerating disk. Tangential acceleration $a_t$ measures how quickly the linear speed changes and equals $a_t$ = αr, where α is angular acceleration and r is radius. Since X and Y are on the same rigid disk, they share the same angular acceleration α at any instant. Point Y, being farther from the axis ($r_Y$ > $r_X$), has tangential acceleration $a_{t,Y}$ = α$r_Y$ > α$r_X$ = $a_{t,X}$. Choice C incorrectly claims points closer to the axis have larger tangential acceleration, reversing the actual relationship. The key principle is that tangential acceleration increases linearly with distance from the rotation axis.
17. 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?
    Answer: $a_{t,B}=3a_{t,A}$ because $a_t=\alpha r$. 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.
18. A rigid disk rotates steadily about a fixed axle through its center. Two small stickers, $P$ and $Q$, are placed on the disk at distances $r_P$ and $r_Q$ from the axis, with $r_Q>r_P$. The disk's angular speed $ omega$ is the same for all points on the disk. Ignore any slipping or deformation. Which statement about the stickers' linear speeds is correct?
    Answer: $v_Q>v_P$ because $v=\omega r$ and $r_Q>r_P$ This problem tests connecting linear and rotational motion for points on a rotating rigid body. For any point on a rigid rotating object, the linear speed v is related to angular speed ω and radius r by v = ωr. Since both stickers P and Q are on the same rigid disk, they share the same angular speed ω. Because Q is farther from the axis ($r_Q$ > $r_P$), and v = ωr, we have $v_Q$ = ω$r_Q$ > ω$r_P$ = $v_P$. Choice A incorrectly assumes equal speeds just because they complete revolutions in the same time, ignoring that Q travels a longer circular path. The key strategy is to recognize that on a rigid body, all points share the same ω, but v increases linearly with distance from the axis.
19. A wheel spins at constant $\omega$. Point M is at radius $3r$ and point N at radius $r$. How do their centripetal accelerations compare?
    Answer: $a_{c,M}=3a_{c,N}$ because $a_c=\omega^2 r$ This problem tests connecting linear and rotational motion, specifically centripetal acceleration in uniform circular motion. The centripetal acceleration for any point in circular motion is $a_c$ = ω²r, where ω is the angular speed and r is the radius. Since the wheel spins at constant ω for all points, and point M is at radius 3r while point N is at radius r, we find $a_{c,M}$ = ω²(3r) = 3ω²r = $3a_{c,N}$. Choice B incorrectly claims smaller radius gives larger centripetal acceleration, which is the opposite of the actual relationship. Remember that centripetal acceleration, like tangential speed, increases with radius when angular speed is constant—points farther out need more acceleration to maintain circular motion.
20. 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?
    Answer: $v_Q=2v_P$ because $v=\omega r$ and $Q$ is twice as far from the axis. 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.

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