Connecting Linear and Rotational Motion
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AP Physics 1 › Connecting Linear and Rotational Motion
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?
Both are zero because angular speed is constant.
$a_{c,N}>a_{c,M}$ because $a_c=\omega^2 r$ increases with radius.
$a_{c,M}>a_{c,N}$ 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.
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.
A carousel rotates at constant $\omega$. Rider A sits at radius $r$ and rider B at radius $3r$. Which compares the magnitudes of their tangential accelerations?
$a_{t,B}=9a_{t,A}$ because tangential acceleration depends on $r^2$
$a_{t,B}=3a_{t,A}$ because $a_t=\alpha r$
$a_{t,B}=0$ and $a_{t,A}=0$ because $\alpha=0$
$a_{t,B}=a_{t,A}$ because both riders share the same $\omega$
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.
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?
$a_{t,Q}=a_{t,P}$ because tangential acceleration depends only on $\omega$.
$a_{t,Q}=3a_{t,P}$ because $a_t=\alpha r$.
$a_{t,P}=a_{t,Q}$ because all points share the same $\alpha$.
$a_{t,P}=3a_{t,Q}$ because the inner point is “closer to the turning.”
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.
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?
$a_{c,Q}=a_{c,P}$ because both points share the same $\omega$
$a_{c,Q}=\tfrac{1}{3}a_{c,P}$ because the outer point moves farther each revolution
$a_{c,Q}=9a_{c,P}$ because $a_c=(\omega r)^2$
$a_{c,Q}=3a_{c,P}$ because $a_c=\omega^2 r$
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.
A fan blade spins with angular speed $\omega$. Point M is at radius $r$ and point N is at radius $4r$. Which compares their angular speeds?
Angular speed cannot be compared without knowing the fan’s radius
$\omega_N=4\omega_M$ because N travels farther each second
$\omega_N=\omega_M$ because all points on a rigid body share the same $\omega$
$\omega_N=\tfrac{1}{4}\omega_M$ because N has greater tangential speed
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.
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?
$v_Y=2v_X$ because $v=\omega r$
$v_Y=\tfrac{1}{2}v_X$ because the outer coin has a longer path
$v_X=2v_Y$ because the inner coin completes more revolutions per second
$v_X=v_Y$ because both have the same $\omega$
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.
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 $a_c=\omega^2 r$.
$a_{c,B}=a_{c,A}$ because both points have the same $\omega$.
$a_{c,A}=0$ and $a_{c,B}=0$ because the wheel’s angular speed is constant.
$a_{c,B}=4a_{c,A}$ because centripetal acceleration scales with $v^2$ and $v$ is the same.
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.
A circular record rotates at constant angular speed $\omega$. A scratch at radius $r$ and a dust speck at radius $2r$ move with the record. Which statement about their linear speeds is correct?
$v_{\text{dust}}=2v_{\text{scratch}}$ because $v=\omega r$.
$v_{\text{scratch}}=v_{\text{dust}}$ because both have the same angular speed.
$v_{\text{scratch}}=2v_{\text{dust}}$ because the inner path is shorter.
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.
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?
Both bolts have zero tangential speed because they are fixed in place on the platform.
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 $v=\omega r$.
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.
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?
$a_{t,B}=\tfrac{1}{3}a_{t,A}$ because larger radius reduces acceleration.
$a_{t,B}=3a_{t,A}$ because $a_t=\alpha r$.
$a_{t,A}=a_{t,B}$ because both points share the same $\alpha$.
$a_{t,A}=3a_{t,B}$ because the inner point changes direction faster.
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.