Problem 13 - Entrance Test

A uniform solid disk of mass M and radius R is initially at rest, free to rotate about a fixed axis through its center. A constant tangential force F is applied to the rim of the disk for a time interval Δt. Ignoring friction, what is the final angular speed of the disk?

A. A) ω_f = (F Δt) / (MR)B. B) ω_f = (2F Δt) / (MR)C. C) ω_f = (F Δt) / (2MR)D. D) ω_f = (F R Δt) / (M)E. E) ω_f = (M R F) / (Δt)

Correct: B

This problem can be solved using the angular impulse-momentum theorem, which states that the angular impulse applied to an object is equal to the change in its angular momentum. 1. **Calculate the Torque (τ):** The constant tangential force F is applied at the rim of the disk (distance R from the axis of rotation). τ = F * R 2. **Calculate the Angular Impulse:** Angular Impulse = τ * Δt = (F * R) * Δt 3. **Calculate the Moment of Inertia (I):** For a uniform solid disk rotating about its central axis, I = (1/2)MR². 4. **Calculate the Initial Angular Momentum (L_initial):** The disk is initially at rest, so its initial angular speed ω_initial = 0. L_initial = I * ω_initial = 0. 5. **Calculate the Final Angular Momentum (L_final):** L_final = I * ω_final = (1/2)MR² * ω_final. 6. **Apply the Angular Impulse-Momentum Theorem:** Angular Impulse = ΔL = L_final - L_initial F * R * Δt = (1/2)MR² * ω_final - 0 Solve for ω_final: ω_final = (F * R * Δt) / ((1/2)MR²) ω_final = (2 * F * R * Δt) / (M R²) ω_final = (2F Δt) / (MR) The correct choice is B.