Angular Momentum and Angular Impulse
Help Questions
AP Physics 1 › Angular Momentum and Angular Impulse
A wheel experiences a constant torque of $0.40 \text{N·m}$ for $0.75 \text{s}$. What is the magnitude of $\Delta L$?
$0.30 \text{ kg·m}^2/\text{s}$
$0.40 \text{ kg·m}^2/\text{s}$
$1.15 \text{ kg·m}^2/\text{s}$
$0.53 \text{ kg·m}^2/\text{s}$
Explanation
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 ($1.15 \text{ kg·m}^2/\text{s}$) 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 wheel’s angular momentum changes by $1.8\ \text{kg·m}^2/\text{s}$ when a constant torque acts. If the torque is $0.60\ \text{N·m}$, how long did it act?
$3.0\ \text{s}$
$2.4\ \text{s}$
$1.2\ \text{s}$
$0.33\ \text{s}$
Explanation
This scenario tests calculating time from change in angular momentum and torque. Rearranging ΔL = τ Δt gives Δt = ΔL / τ for constant torque. This reflects the duration needed for torque to effect the momentum change. The wheel's constant torque determines this time. Choice D (2.4 s) could result from dividing incorrectly, like 1.8 / 0.75, indicating a numerical misconception in division. A transferable approach is to check reasonability: larger ΔL or smaller τ should yield longer times, aiding error detection.
A wheel’s motor applies a constant torque of $4.0\ \text{N·m}$ for $0.50\ \text{s}$. What angular impulse is delivered to the wheel?
$0.50\ \text{kg·m}^2/\text{s}$
$4.0\ \text{kg·m}^2/\text{s}$
$8.0\ \text{kg·m}^2/\text{s}$
$2.0\ \text{kg·m}^2/\text{s}$
Explanation
This question evaluates knowledge of angular impulse in the context of rotational motion. Angular impulse results from a torque applied over a time interval and is calculated as τ Δt for constant torque. It represents the total change in angular momentum imparted to the system, similar to how force over time changes linear momentum. In this case, the motor delivers this impulse directly to the wheel. A common distractor like choice C (4.0 kg·m²/s) could arise from forgetting to multiply by time and just using the torque value, indicating a misconception of impulse as instantaneous rather than time-integrated. To approach such problems effectively, always verify units: angular impulse has units of kg·m²/s, matching angular momentum.
A pulley experiences a net torque of $2.5,\text{N·m}$ for $1.2,\text{s}$. What is the change in angular momentum?
$2.5,\text{kg·m}^2/\text{s}$
$3.0,\text{kg·m}^2/\text{s}$
$1.2,\text{kg·m}^2/\text{s}$
$2.1,\text{kg·m}^2/\text{s}$
Explanation
This question requires calculating change in angular momentum from torque and time. Using the angular impulse-momentum theorem: $ \Delta L = \tau \Delta t = (2.5 , \text{N·m})(1.2 , \text{s}) = 3.0 , \text{kg·m}^2/\text{s} $. The change represents how much the angular momentum increases due to the applied torque. The calculation is straightforward multiplication of the two given values. Choice A (2.1) might result from calculation error or misreading the values, while C (2.5) incorrectly uses just the torque value. Always multiply torque by time to find the change in angular momentum.
A wheel experiences a constant net torque of $4.0,\text{N·m}$ for $0.50,\text{s}$. What is the angular impulse?
$4.0,\text{kg·m}^2/\text{s}$
$8.0,\text{kg·m}^2/\text{s}$
$2.0,\text{kg·m}^2/\text{s}$
$0.50,\text{kg·m}^2/\text{s}$
Explanation
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: $J = (4.0 , \text{N·m})(0.50 , \text{s}) = 2.0 , \text{N·m·s} = 2.0 , \text{kg·m}^2/\text{s}$. 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 spinning platform has $L_i=10,\text{kg·m}^2/\text{s}$. A constant torque of $-1.0,\text{N·m}$ acts for $6.0,\text{s}$. What is $\Delta L$?
$-60,\text{kg·m}^2/\text{s}$
$6.0,\text{kg·m}^2/\text{s}$
$-1.0,\text{kg·m}^2/\text{s}$
$-6.0,\text{kg·m}^2/\text{s}$
Explanation
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 motor applies a constant torque $\tau$ to a fan for $0.20,\text{s}$, producing angular impulse $0.80,\text{kg·m}^2/\text{s}$. What is $\tau$?
$4.0,\text{N·m}$
$0.16,\text{N·m}$
$0.80,\text{N·m}$
$1.0,\text{N·m}$
Explanation
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 rotating rod has $L_i = 7.0 , \text{kg·m}^2/\text{s}$. A net torque of $+0.50 , \text{N·m}$ acts for $2.0 , \text{s}$. What is $L_f$?
$8.0 , \text{kg·m}^2/\text{s}$
$7.5 , \text{kg·m}^2/\text{s}$
$6.0 , \text{kg·m}^2/\text{s}$
$9.0 , \text{kg·m}^2/\text{s}$
Explanation
This question tests applying positive torque to increase angular momentum. The angular impulse $J = \tau \Delta t = (+0.50 , \text{N·m})(2.0 , \text{s}) = +1.0 , \text{kg·m}^2/\text{s}$. The final angular momentum is $L_f = L_i + \Delta L = 7.0 + 1.0 = 8.0 , \text{kg·m}^2/\text{s}$. 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 constant torque of $0.40,\text{N·m}$ acts on a wheel, changing its angular momentum by $0.80,\text{kg·m}^2/\text{s}$. How long does it act?
$3.2,\text{s}$
$0.50,\text{s}$
$2.0,\text{s}$
$1.2,\text{s}$
Explanation
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 rotor experiences a net torque of $1.5\ \text{N·m}$ for $4.0\ \text{s}$. What is the magnitude of the change in angular momentum?
$0.38\ \text{kg·m}^2/\text{s}$
$6.0\ \text{kg·m}^2/\text{s}$
$1.5\ \text{kg·m}^2/\text{s}$
$4.0\ \text{kg·m}^2/\text{s}$
Explanation
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.