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Problem 10 - Entrance Test

A block of mass m is attached to a light string that is wrapped around a uniform solid cylinder of mass M and radius R. The cylinder pivots without friction about a fixed horizontal axis through its center. The block is released from rest and falls a vertical distance h. Assuming the string does not slip on the cylinder, what is the angular speed of the cylinder just as the block hits the ground?

Correct: B

This problem can be solved using the principle of conservation of mechanical energy, as there are no non-conservative forces doing work (frictionless pivot, light string, string does not slip). **Initial State (block at rest, height h above ground):** Potential Energy (PE_initial) = mgh Kinetic Energy (KE_initial) = 0 (since both block and cylinder are released from rest) Total Initial Energy = mgh **Final State (block at ground, cylinder rotating):** Potential Energy (PE_final) = 0 Kinetic Energy (KE_final) = Translational KE of block + Rotational KE of cylinder KE_final = (1/2)mv² + (1/2)Iω² The cylinder is a uniform solid cylinder, so its moment of inertia I = (1/2)MR². Since the string does not slip, the linear speed of the block (v) is related to the angular speed of the cylinder (ω) by v = Rω. Therefore, v² = R²ω². Substitute I and v into KE_final: KE_final = (1/2)m(R²ω²) + (1/2)(1/2)MR²ω² KE_final = (1/2)mR²ω² + (1/4)MR²ω² KE_final = R²ω² * (m/2 + M/4) KE_final = R²ω² * ( (2m + M) / 4 ) **Apply Conservation of Energy (Total Initial Energy = Total Final Energy):** mgh = R²ω² * ( (2m + M) / 4 ) Solve for ω²: ω² = (mgh * 4) / (R² * (2m + M)) ω² = (4mgh) / (R²(2m + M)) Solve for ω: ω = sqrt [ (4mgh) / (R²(2m + M)) ] ω = (1/R) * sqrt [ (4mgh) / (2m + M) ] The correct choice is B.