Problem 7 - Entrance Test

A uniform solid cylinder of mass M and radius R is initially rotating with an angular speed ω₀ about its central axis. A constant tangential force F is applied to its rim, opposite to the direction of rotation, bringing the cylinder to rest after it completes N revolutions. What is the magnitude of the constant force F?

A. A) F = (M R ω₀²) / (4 π N)B. B) F = (M R ω₀²) / (2 π N)C. C) F = (M R ω₀²) / (8 π N)D. D) F = (2 M R ω₀²) / (π N)E. E) F = (M R ω₀²) / (π N)

Correct: C

This problem can be solved using the rotational work-energy theorem, which states that the net work done on a rotating object equals its change in rotational kinetic energy. 1. **Initial Rotational Kinetic Energy (KE_rot_initial):** For a uniform solid cylinder, the moment of inertia about its central axis is I = (1/2)MR². KE_rot_initial = (1/2)Iω₀² = (1/2) * (1/2)MR² * ω₀² = (1/4)MR²ω₀². 2. **Final Rotational Kinetic Energy (KE_rot_final):** The cylinder comes to rest, so KE_rot_final = 0. 3. **Change in Rotational Kinetic Energy (ΔKE_rot):** ΔKE_rot = KE_rot_final - KE_rot_initial = 0 - (1/4)MR²ω₀² = -(1/4)MR²ω₀². 4. **Work Done by the Force (W_F):** The work done by a constant torque (τ) over an angular displacement (Δθ) is W = τΔθ. The constant tangential force F is applied at the rim (distance R from the axis), so the torque it produces is τ = F * R. Since the force opposes the rotation, the work done by this torque is negative. The angular displacement is N revolutions. We need to convert this to radians: Δθ = N revolutions * (2π radians/revolution) = 2πN radians. W_F = -τ * Δθ = -(F * R) * (2πN). 5. **Applying the Work-Energy Theorem:** W_F = ΔKE_rot -(F * R) * (2πN) = -(1/4)MR²ω₀² Cancel the negative signs and solve for F: F * R * (2πN) = (1/4)MR²ω₀² F = ( (1/4)MR²ω₀² ) / (R * 2πN) F = (MRω₀²) / (4 * 2πN) F = (MRω₀²) / (8πN) The correct choice is C.