Problem 18 - Entrance Test

This problem involves both conservation of energy and Newton's second law for circular motion. Assume the string is wrapped around a small, massless axle so that the string length does not change significantly during the swing, but the sphere itself might be rotating as it swings. Let's assume the question implies a simple pendulum where the sphere acts as a point mass for the length calculation, and the internal rotation of the sphere is not initiated or relevant (e.g., if the string is just attached to the top of the sphere, not wrapped around it to cause rotation). *Self-correction*: The phrasing "The string is wrapped around a horizontal axle (massless, negligible radius)" usually implies that the string is fixed to the axle and the pendulum oscillates. However, the mention of a 'solid sphere' rather than a 'point mass' might suggest that the sphere itself undergoes rotation, if the string were to unroll from it, or if it's rolling along a surface. Given the wording 'suspended as a pendulum', the typical interpretation is a simple pendulum of length L (which would be the string length from pivot to CM of sphere). If the sphere itself *rotates* as it swings (e.g., due to friction at its top where the string is attached), that would be an advanced scenario not usually assumed for AP Physics 1 without explicit clarification. Let's assume the simpler case: The string is attached to the center of mass of the sphere, and the sphere acts as a point mass in its pendulum motion. The phrase 'string is wrapped around a horizontal axle' just means the pivot point for the pendulum is the axle. The length of the pendulum is L (length of the string). 1. **Conservation of Mechanical Energy:** Let the lowest point be the reference for potential energy (PE=0). Initial height of the sphere's center of mass (CM) = L(1 - cos θ). Initial Energy (at angle θ, released from rest): E_initial = PE_initial = MgL(1 - cos θ). Final Energy (at the lowest point, height 0): E_final = KE_final = (1/2)Mv² (where v is the linear speed at the lowest point). So, MgL(1 - cos θ) = (1/2)Mv² v² = 2gL(1 - cos θ) (Equation 1) 2. **Newton's Second Law at the Lowest Point:** At the lowest point, the sphere is moving in a circular path. The forces acting on it are gravity (Mg downwards) and tension (T upwards). ΣFy = Ma_centripetal = Mv²/L T - Mg = Mv²/L T = Mg + Mv²/L (Equation 2) 3. **Substitute v² from Equation 1 into Equation 2:** T = Mg + M[2gL(1 - cos θ)] / L T = Mg + 2Mg(1 - cos θ) T = Mg + 2Mg - 2Mg cos θ T = 3Mg - 2Mg cos θ T = Mg(3 - 2cos θ) If the problem *intended* for the solid sphere to also have rotational kinetic energy (as if the string unrolls from the surface of the sphere, causing it to spin while swinging), this would be significantly more complex and would require a different setup (e.g., a compound pendulum or rolling pendulum). The phrase 'suspended as a pendulum' usually rules this out for AP Physics 1. If it were a compound pendulum, the moment of inertia would be (1/3)ML² for a rod or (2/5)MR²+ML² for a sphere and the length to CM. Given the choices, the simple pendulum approach (no rotation of the sphere itself, just CM motion) is most likely. The only slight ambiguity is "string is wrapped around a horizontal axle". This merely sets the pivot point. It does not imply string unrolls from the sphere itself. The choices are also typical for a simple pendulum. Therefore, the interpretation of a simple pendulum with the sphere's mass concentrated at its CM, undergoing only translational motion along the arc, is appropriate. The correct choice is B.