The work done by a torque is given by $$W = \int \vec{\tau} \cdot d\vec{\theta}$$. For a precessing gyroscope, the gravitational torque vector is horizontal. The angular displacement vector for the precession is along the vertical axis of precession. Since the torque vector is always perpendicular to the angular displacement vector, their dot product is zero, and the work done is zero. This torque changes the direction of the angular momentum but not its magnitude.

The work done is the integral of the torque over the angular displacement: $$W = \int_{0}^{\pi} \tau_0 \sin(\theta) d\theta = \tau_0 [-\cos(\theta)]_{0}^{\pi} = -\tau_0 (\cos(\pi) - \cos(0)) = -\tau_0 (-1 - 1) = 2\tau_0$$.

The work done is the area under the torque versus angular position graph. Since the torque decreases linearly, the graph is a triangle. The area of the triangle is $$W = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} (5.0 , \text{rad})(20 , \text{N} \cdot \text{m}) = 50 , \text{J}$$.

The frictional torque opposes the angular velocity, so the angle between the torque vector and the infinitesimal angular displacement vector is $$\pi$$ radians. Thus, the work done by friction is negative. Since friction is the only torque mentioned that causes it to slow down, it is the net torque. Therefore, the net work done is also negative, consistent with the decrease in rotational kinetic energy.

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.

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.

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.

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.

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.

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).