Problem 9 - Entrance Test

Let 'a_CM' be the acceleration of the center of mass and 'α' be the angular acceleration of the cylinder. We will use Newton's second law for translation and rotation. Forces acting on the cylinder: 1. Applied force F, acting horizontally to the right (assuming F pulls to the right) at the top of the cylinder. 2. Static friction force f_s, acting horizontally at the point of contact with the ground. 3. Weight Mg, acting downwards at the center of mass. 4. Normal force N, acting upwards at the point of contact with the ground. Since the cylinder rolls without slipping, the relationship between linear and angular acceleration is a_CM = Rα. **1. Translational Equation (ΣFx = Ma_CM):** Let's assume the cylinder accelerates to the right. The applied force F acts to the right. The direction of static friction (f_s) needs to be determined. If F is applied at the top, it causes a clockwise torque. This rotation combined with CM translation means the point of contact would tend to move to the left relative to the ground if friction wasn't present. Therefore, static friction must act to the right to oppose this relative motion and maintain rolling without slipping. So, f_s acts in the same direction as F. F + f_s = M * a_CM (Equation 1) **2. Rotational Equation (Στ_CM = I_CM α):** We take torques about the center of mass (CM). For a uniform solid cylinder, I_CM = (1/2)MR². Torque due to F: τ_F = F * R (clockwise, if F is to the right) Torque due to f_s: τ_fs = f_s * R (counter-clockwise, if f_s is to the right) Net Torque = τ_F - τ_fs (F - f_s)R = I_CM * α (F - f_s)R = (1/2)MR² * α (Equation 2) **3. Substitute a_CM = Rα into Equation 2:** From a_CM = Rα, we have α = a_CM / R. (F - f_s)R = (1/2)MR² * (a_CM / R) (F - f_s)R = (1/2)MR * a_CM Dividing by R (since R ≠ 0): F - f_s = (1/2)M * a_CM (Equation 3) Now we have a system of two equations with two unknowns (a_CM and f_s): (1) F + f_s = M * a_CM (3) F - f_s = (1/2)M * a_CM Add Equation 1 and Equation 3: (F + f_s) + (F - f_s) = (M * a_CM) + ((1/2)M * a_CM) 2F = (3/2)M * a_CM Solve for a_CM: a_CM = 2F / ((3/2)M) a_CM = 4F / (3M) To verify the direction of friction: Substitute a_CM back into Equation 1 or 3. From (1): f_s = M * a_CM - F = M * (4F / (3M)) - F = 4F/3 - F = F/3. Since f_s is positive, our initial assumption that friction acts to the right (in the same direction as F) was correct. The correct choice is B.