A uniform disc of radius is mounted on a low-friction axle. A force of magnitude is applied at the rim but not tangentially; instead, it makes an angle of 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 , where is the angle between and . Using the given information, what is the torque exerted by the force about the disc’s center?
- (correct answer)
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.