Angular Momentum and Angular Impulse - AP Physics 1

A spinning platform has $L_i=10\,\text{kg}\cdot\text{m}^2/\text{s}$ . A constant torque of $-1.0\,\text{N}\cdot\text{m}$ acts for $6.0\,\text{s}$ . What is $\Delta L$ ?
Answer: $-6.0\,\text{kg·m}^2/\text{s}$ This problem tests understanding of negative torque effects on angular momentum. The angular impulse J = τΔt = (-1.0 N·m)(6.0 s) = -6.0 kg·m²/s. Since angular impulse equals change in angular momentum, ΔL = -6.0 kg·m²/s. The negative value indicates the angular momentum decreases by this amount. Choice C (-60) incorrectly multiplies by an extra factor of 10, suggesting a decimal place error. When calculating change in angular momentum, multiply torque by time and preserve the sign to indicate direction.
A wheel on a low-friction axle experiences a constant torque of $4.0 \text{N}\cdot\text{m}$ for $0.50 \text{s}$ . What is the angular impulse delivered?
Answer: $2.0 \text{N·m·s}$ This question evaluates understanding of angular impulse delivered by a constant torque over time. Angular impulse is the product of constant torque and the time interval, representing the total 'push' in the rotational sense. It quantifies how much the angular momentum changes, but here the question directly asks for the impulse itself. The low-friction axle implies negligible other torques, so the given torque is net. A distractor like $4.0 \text{N}\cdot\text{m}$ could arise from mistaking torque for impulse, a misconception of overlooking the multiplication by time. Remember as a strategy that impulse always involves integrating force or torque over time, ensuring units include seconds.
A wheel experiences a constant torque of $0.40 \text{N}\cdot\text{m}$ for $0.75 \text{s}$ . What is the magnitude of $\Delta L$ ?
Answer: $0.30 \text{ kg·m}^2/\text{s}$ This problem assesses computing change in angular momentum from torque and time. For constant torque, $\Delta L$ equals $\tau$ times $\Delta t$ , embodying the angular impulse. This ties into how sustained torque accumulates change in rotational momentum over time. The wheel experiences this direct proportionality. Choice C ( $1.15 \text{ kg}\cdot\text{m}^2/\text{s}$ ) could be selected by adding torque and time instead of multiplying, revealing a misconception about the multiplicative nature of impulse. A key strategy is to memorize the rotational equivalents: force $\to$ torque, momentum $\to$ angular momentum, impulse $\to$ angular impulse.
A rotor experiences a constant opposing torque of magnitude $0.80\ \text{N}\cdot\text{m}$ for $3.0\ \text{s}$ . What is the magnitude of the angular impulse?
Answer: $2.4\ \text{kg·m}^2/\text{s}$ This scenario examines angular impulse in an opposing torque context. The magnitude of angular impulse is τ Δt, regardless of direction, as it quantifies the total rotational impetus. Qualitatively, it equals the absolute change in angular momentum, opposing the rotor's motion in this case. The constant torque over time delivers this impulse. Choice C (0.80 kg·m²/s) may be picked by mistaking impulse for torque alone, reflecting a misconception that time is not a factor in impulse calculations. A broad strategy is to draw parallels between linear and angular dynamics, treating impulse as the 'push' over time in both realms.
A spinning platform experiences a net torque of $2.5\ \text{N}\cdot\text{m}$ for $0.20\ \text{s}$ . What is the change in angular momentum magnitude?
Answer: $0.50\ \text{kg·m}^2/\text{s}$ This question explores the magnitude of angular momentum change due to net torque over time. Net torque produces impulse, equaling ΔL in magnitude. Longer time or higher torque amplifies the change qualitatively. Only the product matters for magnitude. The distractor 2.5 kg·m²/s might use torque value alone, a misconception of omitting time in impulse. A key strategy is to identify if the question seeks change (τΔt) versus rate (τ), applying the formula accordingly.
A disk's angular momentum decreases by $3.0\ \text{kg}\cdot\text{m}^2/\text{s}$ due to a constant opposing torque acting for $2.0\ \text{s}$ . What is the torque magnitude?
Answer: $1.5\ \text{N·m}$ This question assesses calculating torque magnitude from change in angular momentum and time. The decrease in angular momentum implies an opposing torque, where magnitude of $ΔL$ = $τ Δt$ . Qualitatively, torque acts as the rate of change, so dividing $ΔL$ by time gives $τ$ . The opposing nature determines the sign, but magnitude is absolute. The distractor $6.0 \text{N}\cdot\text{m}$ could result from multiplying instead of dividing, a misconception of inverting the impulse formula. A transferable approach is to rearrange $ΔL$ = $τ Δt$ for the unknown, checking consistency with units.
A motor applies a constant torque $\tau$ to a fan for $0.20\,\text{s}$ , producing angular impulse $0.80\,\text{kg}\cdot\text{m}^2/\text{s}$ . What is $\tau$ ?
Answer: $4.0\,\text{N·m}$ This question involves finding torque from angular impulse and time duration. The angular impulse J = τΔt, so τ = J/Δt = (0.80 kg·m²/s)/(0.20 s) = 4.0 N·m. The torque must be sufficient to produce the given angular impulse in the specified time. Shorter time requires larger torque for the same impulse. Choice A (0.16) incorrectly multiplies the two given values instead of dividing, revealing confusion about the relationship between impulse, torque, and time. When given angular impulse and time, divide impulse by time to find the constant torque.
A wheel experiences a constant torque of $0.25 \text{N}\cdot\text{m}$ for $8.0 \text{s}$ . What is the angular impulse magnitude?
Answer: $2.0 \text{N·m·s}$ This question evaluates computing angular impulse from constant torque and duration. Angular impulse is $τ Δt$ for constant torque, measuring total rotational effect. It relates qualitatively to how much spin is imparted, with units $\text{N}\cdot\text{m}\cdot\text{s}$ . The magnitude ignores direction here. A distractor like 8.0 $\text{N}\cdot\text{m}$ might use time alone, a misconception of swapping torque and time in the product. Strategically, recall impulse as area under torque-time graph; for constant, it's simple multiplication.
A fan initially has $L=0$ . A constant net torque of $0.90\ \text{N}\cdot\text{m}$ acts for $1.0\ \text{s}$ . What is the fan's angular momentum afterward?
Answer: $0.90\ \text{kg·m}^2/\text{s}$ This question probes how angular momentum evolves from zero under constant torque. Starting from rest (L=0), angular impulse from torque over time builds up the final angular momentum. The relationship is linear: more time or torque means higher final L. Since initial is zero, final L equals the impulse τΔt. A distractor like 0.90 N·m might ignore time, a misconception of confusing torque units with momentum units. As a strategy, verify units: angular momentum is kg·m²/s, requiring torque (N·m) times seconds.
A constant torque of $0.40\,\text{N}\cdot\text{m}$ acts on a wheel, changing its angular momentum by $0.80\,\text{kg}\cdot\text{m}^2/\text{s}$ . How long does it act?
Answer: $2.0\,\text{s}$ This problem involves finding time duration from torque and angular momentum change. From ΔL = τΔt, we solve for time: Δt = ΔL/τ = (0.80 kg·m²/s)/(0.40 N·m) = 2.0 s. The time represents how long the torque must act to produce the given momentum change. Smaller torque requires more time for the same momentum change. Choice B (0.50) incorrectly multiplies the values instead of dividing, showing confusion about rearranging the impulse equation. To find time from momentum change and torque, divide the momentum change by the torque.
A rotating rod has $L_i = 7.0 \, \text{kg}\cdot\text{m}^2/\text{s}$ . A net torque of $+0.50 \, \text{N}\cdot\text{m}$ acts for $2.0 \, \text{s}$ . What is $L_f$ ?
Answer: $8.0 \, \text{kg·m}^2/\text{s}$ This question tests applying positive torque to increase angular momentum. The angular impulse $J = \tau \Delta t = (+0.50 \, \text{N}\cdot\text{m})(2.0 \, \text{s}) = +1.0 \, \text{kg}\cdot\text{m}^2/\text{s}$ . The final angular momentum is $L_f = L_i + \Delta L = 7.0 + 1.0 = 8.0 \, \text{kg}\cdot\text{m}^2/\text{s}$ . The positive torque adds to the existing angular momentum in the same direction. Choice A (6.0) incorrectly subtracts instead of adding, treating the positive torque as negative. When torque and initial angular momentum have the same sign, add the angular impulse to find final angular momentum.
A uniform disk initially at rest experiences a constant torque of $0.60\ \text{N}\cdot\text{m}$ for $2.0\ \text{s}$ . What is the magnitude of the disk's change in angular momentum?
Answer: $1.2\ \text{kg·m}^2/\text{s}$ This question assesses the skill of understanding angular momentum and angular impulse in rotational dynamics. Angular impulse is defined as the product of a constant torque and the time interval over which it acts, directly causing a change in angular momentum. Qualitatively, just as linear impulse changes linear momentum, angular impulse changes angular momentum, with the relationship ΔL = τ Δt for constant torque. Since the disk starts at rest, its initial angular momentum is zero, so the change in angular momentum equals the angular impulse applied. One distractor, such as choice A (0.30 kg·m²/s), might be selected if someone mistakenly divides torque by time instead of multiplying, reflecting a misconception about the impulse-momentum theorem. A transferable strategy is to analogize rotational quantities to their linear counterparts, remembering that torque plays the role of force in changing momentum over time.
A disk's angular momentum increases by $6.0\ \text{kg}\cdot\text{m}^2/\text{s}$ due to a constant torque applied for $2.0\ \text{s}$ . What torque magnitude was applied?
Answer: $3.0\ \text{N·m}$ This question probes finding torque from change in angular momentum and time. The relationship ΔL = τ Δt allows solving for τ as ΔL / Δt when torque is constant. Qualitatively, torque drives the rate of angular momentum change, so averaging over time gives the constant value. The disk's increase in L directly informs the torque magnitude. Distractor A (12 N·m) might occur from multiplying instead of dividing ΔL by time, indicating a reversal misconception in the formula application. To generalize, always isolate the unknown variable algebraically before plugging in numbers to avoid arithmetic errors.
A uniform disk rotates about a fixed axle. A constant torque of $0.60\ \text{N}\cdot\text{m}$ acts for $2.0\ \text{s}$ , then stops. What is the magnitude of the disk's change in angular momentum?
Answer: $1.2\ kg\cdot m^2/s$ This question assesses the relationship between torque, time, and change in angular momentum for rotational systems. Angular impulse, which is torque multiplied by the time it acts, equals the change in angular momentum of the disk. Since the torque is constant, the change in angular momentum is directly proportional to both the torque magnitude and the duration. This relationship holds regardless of the disk's initial angular momentum or moment of inertia, as we're only finding the change. One distractor, such as 0.60 kg·m²/s, might be selected by confusing torque with angular impulse, a misconception of ignoring the time factor in impulse calculations. A transferable strategy is to always calculate angular impulse as τΔt when torque is constant, analogous to linear impulse FΔt.
A flywheel's net torque increases its angular momentum by $12\,\text{kg}\cdot\text{m}^2/\text{s}$ over $3.0\,\text{s}$ . What is the torque magnitude?
Answer: $4.0\,\text{N·m}$ This problem requires finding torque from given change in angular momentum and time. Using the angular impulse-momentum theorem: ΔL = τΔt, we can solve for torque: τ = ΔL/Δt = (12 kg·m²/s)/(3.0 s) = 4.0 N·m. The torque represents the rate of change of angular momentum. A larger torque would produce the same momentum change in less time. Choice A (36) incorrectly multiplies instead of dividing, showing confusion about rearranging the impulse equation. To find torque from momentum change and time, divide the momentum change by the time duration.
A wheel experiences a constant net torque of $4.0\,\text{N}\cdot\text{m}$ for $0.50\,\text{s}$ . What is the angular impulse?
Answer: $2.0\,\text{kg·m}^2/\text{s}$ This question requires calculating angular impulse from constant torque and time duration. Angular impulse $J = \tau \Delta t$ , where $\tau$ is the net torque and $\Delta t$ is the time interval. Substituting the given values: $J = (4.0 \, \text{N}\cdot\text{m})(0.50 \, \text{s}) = 2.0 \, \text{N}\cdot\text{m}\cdot\text{s} = 2.0 \, \text{kg}\cdot\text{m}^2/\text{s}$ . The angular impulse represents the total angular effect of the torque over the time period. Choice A (8.0) incorrectly multiplies 4.0 by 2 instead of 0.50, suggesting a calculation error or misreading of the time value. To find angular impulse, always multiply the constant torque by the time duration.
A rotor experiences a net torque of $1.5\ \text{N}\cdot\text{m}$ for $4.0\ \text{s}$ . What is the magnitude of the change in angular momentum?
Answer: $6.0\ \text{kg·m}^2/\text{s}$ This question examines the direct link between net torque, time, and magnitude of change in angular momentum. Net torque over time delivers angular impulse, which equals the absolute change in angular momentum. Qualitatively, longer torque application or stronger torque leads to greater momentum change. No initial conditions are needed since only the change is requested. The distractor 1.5 kg·m²/s may come from using torque alone, a misconception of equating torque to momentum change without time. Strategically, always multiply constant torque by time to find ΔL, mirroring linear dynamics.
A uniform disk spins about a fixed axle. A constant torque of $0.60\,\text{N}\cdot\text{m}$ acts for $3.0\,\text{s}$ . What is $\Delta L$ ?
Answer: $1.8\,\text{kg·m}^2/\text{s}$ This problem tests understanding of angular impulse and its relationship to change in angular momentum. Angular impulse equals the product of torque and time: J = τΔt = (0.60 N·m)(3.0 s) = 1.8 N·m·s. By the angular impulse-momentum theorem, angular impulse equals the change in angular momentum: ΔL = J = 1.8 kg·m²/s. The units N·m·s and kg·m²/s are equivalent because 1 N = 1 kg·m/s². Choice A (0.20) incorrectly divides torque by time instead of multiplying, revealing confusion about the impulse formula. When given constant torque and time, multiply them to find the change in angular momentum.
A fan blade's axle experiences a constant torque of $1.5 \text{N}\cdot\text{m}$ for $0.20 \text{s}$ . What change in angular momentum results?
Answer: $0.30 \text{kg·m}^2/\text{s}$ This question focuses on calculating the change in angular momentum from a given torque and time. Angular momentum changes when a net torque is applied, with the magnitude of change given by $\Delta L = \tau \Delta t$ for constant torque. This qualitative link mirrors the linear case where impulse equals change in momentum. Here, the fan blade's axle torque causes the specified change. Distractor D ( $7.5 \text{kg}\cdot\text{m}^2/\text{s}$ ) could result from multiplying torque by time incorrectly, like using 1.5 * 5 instead of 0.20 s, showing a misconception in reading the time value accurately. For transferable skills, practice dimensional analysis to ensure calculations yield the correct units for angular momentum.
A rotor's angular momentum changes by $+0.40 \text{kg}\cdot\text{m}^2/\text{s}$ when a constant torque acts for $0.80 \text{s}$ . What torque magnitude acted?
Answer: $0.50 \text{N·m}$ This question tests determining torque from angular momentum change and time. The positive change indicates torque direction aligns with increasing L, where $τ = ΔL / Δt$ . Qualitatively, torque accelerates rotation, and its magnitude is the ratio of change to time. The constant nature simplifies to division. The distractor 0.40 N·m·s may confuse impulse with torque, a misconception of using ΔL directly as τ. Always solve for unknowns using $ΔL = τΔt$ , ensuring proper algebraic isolation.

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