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?
$a_{c,M}>a_{c,N}$ 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.
Both are zero because angular speed is constant.
$a_{c,N}>a_{c,M}$ because $a_c=\omega^2 r$ increases with radius.
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?
$a_{t,B}=0$ and $a_{t,A}=0$ because $\alpha=0$
$a_{t,B}=3a_{t,A}$ because $a_t=\alpha r$
$a_{t,B}=a_{t,A}$ because both riders share the same $\omega$
$a_{t,B}=9a_{t,A}$ 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.
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 $\omega$.
$a_{t,P}=a_{t,Q}$ because all points share the same $\alpha$.
$a_{t,Q}=3a_{t,P}$ because $a_t=\alpha r$.
$a_{t,P}=3a_{t,Q}$ 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?
$a_{c,Q}=\tfrac{1}{3}a_{c,P}$ because the outer point moves farther each revolution
$a_{c,Q}=a_{c,P}$ because both points share the same $\omega$
$a_{c,Q}=9a_{c,P}$ 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 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?
$v_{\text{scratch}}=2v_{\text{dust}}$ because the inner path is shorter.
$v_{\text{dust}}=2v_{\text{scratch}}$ because $v=\omega r$.
Both linear speeds are zero because neither object slips.
$v_{\text{scratch}}=v_{\text{dust}}$ because both have the same angular speed.
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 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?
$a_{t,B}=2a_{t,A}$ because $a_t=\alpha r$.
$a_{t,B}=a_{t,A}/2$ because the outer point has more distance to cover.
$a_{t,A}=a_{t,B}$ because both points share the same angular acceleration.
$a_{t,A}=2a_{t,B}$ because the inner point “turns faster.”
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.
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?
$a_{t,B}=4a_{t,A}$ because $a_t=\alpha r$.
$a_{t,B}=\tfrac{1}{4}a_{t,A}$ because $B$ is farther from the axis.
$a_{t,B}=16a_{t,A}$ because acceleration scales as $r^2$.
$a_{t,A}=a_{t,B}$ because the wheel is rigid.
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.
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?
$a_{c,X}$ and $a_{c,Y}$ are zero because $\omega$ is constant.
$a_{c,X}=4a_{c,Y}$ because the inner bug turns more sharply.
$a_{c,Y}=4a_{c,X}$ because $a_c=\omega^2 r$.
$a_{c,X}=a_{c,Y}$ because both have the same $\omega$.
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.
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 $\omega$.
$a_{t,G}=4a_{t,H}$ because the inner point responds more quickly.
$a_{t,H}$ is smaller because points farther out have greater inertia.
$a_{t,H}=4a_{t,G}$ because $a_t=\alpha r$.
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.
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?
$a_{c,D}=3a_{c,C}$ because $a_c=\omega^2 r$.
$a_{c,C}=3a_{c,D}$ because larger radius reduces inward acceleration.
$a_{c,C}=a_{c,D}$ because $\omega$ is the same everywhere on the wheel.
$a_{c,D}=9a_{c,C}$ because $a_c=\omega r^2$.
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.