Torque and Work
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AP Physics C: Mechanics › Torque and Work
A pulley of radius $r=0.10,\text{m}$ is rotated by a constant tangential pull on a rope. The applied force has magnitude $F=60,\text{N}$ and stays perpendicular to the radius as the pulley turns. The axle is frictionless, and the pulley rotates in its plane. The force is applied while the pulley rotates through $\Delta\theta=4.0,\text{rad}$. Assume no slipping and no dissipative losses. The torque is constant with magnitude $\tau=rF$. The work done by the torque is $W=\tau\Delta\theta$. Using the given information, calculate the total work done by the applied force.
$W=24,\text{N,m}$
$W=240,\text{J}$
$W=6.0,\text{J}$
$W=24,\text{J}$
Explanation
This question tests AP Physics C: Mechanics, specifically the concepts of torque and work in rotating systems. Torque is the rotational equivalent of force, calculated as τ = rFsinθ, where r is the distance from the pivot, F is the force, and θ is the angle between force and lever arm. Work in rotation is given by W = τθ, where θ is the angular displacement. In this scenario, a tangential force of 60 N at radius 0.10 m rotates a pulley through 4.0 rad. Choice A is correct because the torque is τ = rF = (0.10 m)(60 N) = 6.0 N·m, and the work is W = τΔθ = (6.0 N·m)(4.0 rad) = 24 J. Choice B incorrectly multiplies by an extra factor of 10, possibly from a decimal error. To help students, emphasize careful unit tracking and decimal placement. Practice with small radii to ensure students don't automatically assume meters when given centimeters.
A pulley of radius $r=0.20,\text{m}$ is turned by pulling on a light rope wrapped around its rim. A constant tangential force $F=50,\text{N}$ is applied to the rope so the force stays perpendicular to the radius. The axle is frictionless, and the pulley rotates in its plane. The force is applied while the pulley turns through an angle $\Delta\theta=2.5,\text{rad}$. Assume the rope does not slip on the rim. The torque from the pull is constant, so rotational work is $W=\tau\Delta\theta$. Using the given information, calculate the total work done by the applied force on the pulley during this rotation.
$W=2.0,\text{N,m}$
$W=125,\text{J}$
$W=25,\text{J}$
$W=5.0,\text{J}$
Explanation
This question tests AP Physics C: Mechanics, specifically the concepts of torque and work in rotating systems. Torque is the rotational equivalent of force, calculated as τ = rFsinθ, where r is the distance from the pivot, F is the force, and θ is the angle between force and lever arm. Work in rotation is given by W = τθ, where θ is the angular displacement. In this scenario, a tangential force of 50 N is applied to a pulley of radius 0.20 m through an angle of 2.5 rad. Choice A is correct because the torque is τ = rF = (0.20 m)(50 N) = 10 N·m (tangential force means sin90° = 1), and the work is W = τΔθ = (10 N·m)(2.5 rad) = 25 J. Choice D incorrectly gives units of N·m instead of J, showing confusion between torque and work units. To help students, stress that work always has units of energy (joules), while torque has units of N·m. Practice dimensional analysis to verify that τΔθ gives units of energy.
A mechanical arm rotates about a pivot, and a force $F=25,\text{N}$ is applied at a point $r=0.40,\text{m}$ from the pivot. The force is applied at an angle $\phi=45^\circ$ relative to the arm (the radius vector), in the plane of rotation. Assume the pivot is frictionless and the force magnitude remains constant at that instant. Only the torque from this applied force is considered. The torque magnitude is $\tau=rF\sin\phi$. Using the given information, determine the torque produced about the pivot.
$\tau=7.1,\text{N,m}$
$\tau=10,\text{N,m}$
$\tau=14,\text{N,m}$
$\tau=0.28,\text{N,m}$
Explanation
This question tests AP Physics C: Mechanics, specifically the concepts of torque and work in rotating systems. Torque is the rotational equivalent of force, calculated as τ = rFsinθ, where r is the distance from the pivot, F is the force, and θ is the angle between force and lever arm. Work in rotation is given by W = τθ, where θ is the angular displacement. In this scenario, a 25 N force is applied at 0.40 m from the pivot at a 45° angle. Choice A is correct because τ = rFsinφ = (0.40 m)(25 N)sin(45°) = (0.40)(25)(0.707) = 7.1 N·m. Choice B incorrectly calculates τ = rF without the angle factor, yielding 10 N·m. To help students, use the mnemonic that torque depends on the 'perpendicular' component of force. Practice with 45° angles specifically, as sin(45°) = cos(45°) = 0.707 is a common value students should memorize.
A balancing beam of negligible mass is pivoted at one end and initially held horizontal. A single weight of magnitude $W=50,\text{N}$ hangs from the beam at a distance $r=0.60,\text{m}$ from the pivot. The weight pulls vertically downward, and the beam rotates in a vertical plane. Consider the instant when the beam is horizontal so the angle between $\vec r$ and the weight is $90^\circ$. Ignore friction at the pivot and assume the weight remains attached at the same point. The torque magnitude from the weight is $\tau=rW\sin 90^\circ$. Using the given information, what is the torque about the pivot due to the hanging weight at this instant?
$\tau=300,\text{N,m}$
$\tau=15,\text{N,m}$
$\tau=30,\text{N,m}$
$\tau=30,\text{J}$
Explanation
This question tests AP Physics C: Mechanics, specifically the concepts of torque and work in rotating systems. Torque is the rotational equivalent of force, calculated as τ = rFsinθ, where r is the distance from the pivot, F is the force, and θ is the angle between force and lever arm. Work in rotation is given by W = τθ, where θ is the angular displacement. In this scenario, a 50 N weight hangs 0.60 m from a pivot on a horizontal beam, creating maximum torque. Choice A is correct because τ = rWsin(90°) = (0.60 m)(50 N)(1) = 30 N·m, since the weight acts vertically and the beam is horizontal. Choice D incorrectly labels the torque with units of joules instead of N·m, confusing torque with work. To help students, emphasize that hanging weights on horizontal beams always produce maximum torque (sin90° = 1). Practice identifying when forces are perpendicular to position vectors for simplified calculations.
A uniform disc of radius $r=0.40,\text{m}$ is mounted on a low-friction axle. A force of magnitude $F=30,\text{N}$ is applied at the rim but not tangentially; instead, it makes an angle of $30^\circ$ with the radius. The force lies in the plane of the disc and stays at this angle with respect to the radius at the point of application. Air resistance is negligible, and the disc remains rigid. The torque magnitude about the center is given by $\tau=rF\sin\phi$, where $\phi$ is the angle between $\vec r$ and $\vec F$. Using the given information, what is the torque exerted by the force about the disc’s center?
$\tau=10.4,\text{N,m}$
$\tau=12,\text{N,m}$
$\tau=24,\text{N,m}$
$\tau=6.0,\text{N,m}$
Explanation
This question tests AP Physics C: Mechanics, specifically the concepts of torque and work in rotating systems. Torque is the rotational equivalent of force, calculated as τ = rFsinθ, where r is the distance from the pivot, F is the force, and θ is the angle between force and lever arm. Work in rotation is given by W = τθ, where θ is the angular displacement. In this scenario, a 30 N force is applied at the rim (0.40 m) at a 30° angle to the radius. Choice C is correct because τ = rFsinφ = (0.40 m)(30 N)sin(30°) = (0.40)(30)(0.5) = 6.0 N·m. Choice A incorrectly uses sin(90°) instead of sin(30°), yielding 12 N·m, which would be the torque if the force were tangential. To help students, use diagrams showing the force vector and its angle relative to the position vector. Emphasize that the angle in the torque formula is between the force and position vectors, not the force and some reference direction.
A disc of radius $r=0.25,\text{m}$ rotates about its center on a frictionless axle. A constant tangential force $F=16,\text{N}$ is applied at the rim so that the force stays perpendicular to the radius. The disc rotates through an angular displacement of $\Delta\theta=\pi,\text{rad}$. Ignore any losses so all work is due to the applied torque. The torque magnitude is constant and equals $\tau=rF$. The rotational work done is $W=\tau\Delta\theta$. Using the given information, how much work is done by the force during this rotation?
$W=4\pi,\text{J}$
$W=4\pi,\text{N,m}$
$W=8\pi,\text{J}$
$W=2\pi,\text{J}$
Explanation
This question tests AP Physics C: Mechanics, specifically the concepts of torque and work in rotating systems. Torque is the rotational equivalent of force, calculated as τ = rFsinθ, where r is the distance from the pivot, F is the force, and θ is the angle between force and lever arm. Work in rotation is given by W = τθ, where θ is the angular displacement. In this scenario, a tangential force of 16 N at radius 0.25 m rotates a disc through π rad. Choice A is correct because the torque is τ = rF = (0.25 m)(16 N) = 4.0 N·m, and the work is W = τΔθ = (4.0 N·m)(π rad) = 4π J. Choice D incorrectly gives units of N·m instead of J, confusing torque with work. To help students, practice problems using π in angular measurements, ensuring comfort with symbolic answers. Emphasize dimensional analysis: torque (N·m) × angle (rad) = work (J).
A robotic arm rotates about a fixed pivot in a horizontal plane. A force of magnitude $F=40,\text{N}$ is applied at a point $r=0.30,\text{m}$ from the pivot. The force makes an angle $\phi=60^\circ$ with the arm (the radius vector from the pivot to the point of application). The applied force remains constant in magnitude and direction relative to the arm at that instant. Frictional effects are negligible, and only the torque from this force is considered. Use the standard torque magnitude relation $\tau=rF\sin\phi$. Using the given information, determine the torque produced about the pivot by the applied force.
$\tau=6.0,\text{J}$
$\tau=24,\text{N,m}$
$\tau=10.4,\text{N,m}$
$\tau=12,\text{N,m}$
Explanation
This question tests AP Physics C: Mechanics, specifically the concepts of torque and work in rotating systems. Torque is the rotational equivalent of force, calculated as τ = rFsinθ, where r is the distance from the pivot, F is the force, and θ is the angle between force and lever arm. Work in rotation is given by W = τθ, where θ is the angular displacement. In this scenario, a 40 N force is applied at 0.30 m from the pivot at a 60° angle to the arm, requiring calculation of torque. Choice B is correct because τ = rFsinφ = (0.30 m)(40 N)sin(60°) = (0.30)(40)(0.866) = 10.4 N·m. Choice A incorrectly calculates τ = rF without considering the angle, yielding 12 N·m. To help students, emphasize the importance of identifying the angle between force and position vectors. Use visual aids showing force decomposition and practice with various angles to build intuition about when torque is maximized (90°) or zero (0° or 180°).
A flywheel experiences a constant torque due to a motor. The motor applies a tangential force $F=80,\text{N}$ at a radius $r=0.15,\text{m}$ from the axis, and the force remains perpendicular to the radius. The flywheel starts from rest and rotates without friction. The motor continues to apply the same torque until the flywheel has rotated through $\Delta\theta=10,\text{rad}$. Treat the torque as constant during this interval. The work done by a constant torque over an angular displacement is $W=\tau\Delta\theta$. Using the given information, determine the work done by the motor during the $10,\text{rad}$ rotation.
$W=800,\text{J}$
$W=12,\text{J}$
$W=60,\text{J}$
$W=120,\text{J}$
Explanation
This question tests AP Physics C: Mechanics, specifically the concepts of torque and work in rotating systems. Torque is the rotational equivalent of force, calculated as τ = rFsinθ, where r is the distance from the pivot, F is the force, and θ is the angle between force and lever arm. Work in rotation is given by W = τθ, where θ is the angular displacement. In this scenario, a tangential force of 80 N at radius 0.15 m rotates a flywheel through 10 rad. Choice A is correct because the torque is τ = rF = (0.15 m)(80 N) = 12 N·m (tangential means sin90° = 1), and the work is W = τΔθ = (12 N·m)(10 rad) = 120 J. Choice C incorrectly calculates only half the work, possibly from an arithmetic error. To help students, practice problems with various angular displacements, emphasizing that work accumulates over the entire rotation. Reinforce unit consistency: rad × N·m = J.
A beam of negligible mass is pivoted at its left end and lies initially horizontal. A weight $W_1=30,\text{N}$ hangs at $r_1=0.40,\text{m}$ from the pivot, and a second weight $W_2=20,\text{N}$ hangs at $r_2=0.70,\text{m}$ from the pivot, both on the same side. The weights act vertically downward, and the beam rotates in a vertical plane. At the instant shown, the beam remains horizontal so each weight produces a moment arm equal to its distance from the pivot. Ignore friction at the pivot and any other forces. The net torque about the pivot is the sum of the individual torques from the two weights. Using the given information, compute the net torque magnitude about the pivot at this instant.
$\tau=52,\text{N,m}$
$\tau=2.6,\text{N,m}$
$\tau=13,\text{N,m}$
$\tau=26,\text{N,m}$
Explanation
This question tests AP Physics C: Mechanics, specifically the concepts of torque and work in rotating systems. Torque is the rotational equivalent of force, calculated as τ = rFsinθ, where r is the distance from the pivot, F is the force, and θ is the angle between force and lever arm. Work in rotation is given by W = τθ, where θ is the angular displacement. In this scenario, two weights hang from a horizontal beam, creating torques that must be summed. Choice A is correct because τ₁ = r₁W₁ = (0.40 m)(30 N) = 12 N·m and τ₂ = r₂W₂ = (0.70 m)(20 N) = 14 N·m, giving a net torque of 12 + 14 = 26 N·m (both rotate the beam in the same direction). Choice C incorrectly calculates only half the total torque. To help students, emphasize that torques add algebraically when they act about the same axis. Practice with multiple forces at different positions, ensuring students understand sign conventions for clockwise vs. counterclockwise torques.
A robotic arm rotates about a fixed pivot in a plane. A force of magnitude $F=30,\text{N}$ is applied at a point $r=0.40,\text{m}$ from the pivot. At the instant considered, the angle between the radius vector and the force is $\phi=60^\circ$. The pivot is frictionless, and only this applied force produces torque about the pivot. Use the standard torque magnitude formula $\tau=rF\sin\phi$. Assume all quantities are constant at this instant and treat the problem as two-dimensional. Ignore any gravitational torque on the arm. Using the given information, determine the torque produced by the mechanical arm.
$\tau=6.0,\text{N!\cdot!m}$
$\tau=10.4,\text{N!\cdot!m}$
$\tau=12,\text{N!\cdot!m}$
$\tau=20.8,\text{N!\cdot!m}$
Explanation
This question tests AP Physics C: Mechanics, specifically the concepts of torque and work in rotating systems. Torque is the rotational equivalent of force, calculated as τ = rFsinφ, where r is the distance from the pivot, F is the force, and φ is the angle between force and lever arm. Work in rotation is given by W = τθ, where θ is the angular displacement. In this scenario, a force of 30 N is applied at 0.40 m from the pivot at an angle of 60° to the radius vector. Choice B is correct because it accurately uses τ = rFsinφ = (0.40)(30)(sin 60°) = (0.40)(30)(0.866) = 10.4 N·m. Choice C is incorrect because it uses sin 30° instead of sin 60°, giving only 6.0 N·m. To help students, emphasize understanding which angle to use in the torque formula - it's the angle between the force vector and the position vector. Practice with various force orientations and ensure students can identify the correct angle for the sine function.