Problem 20 - Entrance Test

To determine the minimum coefficient of static friction, we need to analyze the forces and torques acting on the disk as it rolls without slipping. Forces on the disk: 1. Weight (Mg) acting vertically downwards. 2. Normal force (N) perpendicular to the incline, upwards. 3. Static friction (f_s) parallel to the incline, upwards (opposing the tendency to slide down). Let the x-axis be parallel to the incline (downwards) and the y-axis perpendicular to the incline (outwards). 1. **Translational Motion (along the incline):** The component of gravity down the incline is Mg sinθ. The friction force is f_s (up the incline). ΣFx = Ma_CM Mg sinθ - f_s = Ma_CM (Equation 1) 2. **Translational Motion (perpendicular to the incline):** ΣFy = 0 N - Mg cosθ = 0 N = Mg cosθ (Equation 2) 3. **Rotational Motion (about the center of mass):** The only force causing a torque about the CM is the static friction f_s. Its lever arm is R. τ_CM = I_CM * α For a uniform disk, I_CM = (1/2)MR². f_s * R = (1/2)MR² * α (Equation 3) 4. **Rolling Without Slipping Condition:** a_CM = Rα => α = a_CM / R Substitute α into Equation 3: f_s * R = (1/2)MR² * (a_CM / R) f_s * R = (1/2)MR * a_CM f_s = (1/2)M * a_CM (Equation 4) Now, we have a system of equations for a_CM and f_s. Substitute f_s from Equation 4 into Equation 1: Mg sinθ - (1/2)M * a_CM = Ma_CM Mg sinθ = (3/2)M * a_CM a_CM = (2/3)g sinθ (Equation 5) Now, substitute a_CM back into Equation 4 to find f_s: f_s = (1/2)M * ((2/3)g sinθ) f_s = (1/3)Mg sinθ (Equation 6) For the disk to roll without slipping, the static friction force required (f_s) must be less than or equal to the maximum possible static friction (μs * N). f_s ≤ μs * N Substitute f_s from Equation 6 and N from Equation 2: (1/3)Mg sinθ ≤ μs * Mg cosθ To find the minimum μs (μs_min) needed, we set the inequality to equality: (1/3)Mg sinθ = μs_min * Mg cosθ Solve for μs_min: μs_min = (1/3) (Mg sinθ) / (Mg cosθ) μs_min = (1/3) (sinθ / cosθ) μs_min = (1/3)tanθ The correct choice is A.