Masses m₁ = 3.00 kg and m₂ = 2.00 kg are joined by a light non-slip cord over a uniform-disk pulley of mass 1.00 kg and radius 0.200 m. The axle is frictionless and the system is released from rest. Determine the acceleration, both cord tensions and pulley angular acceleration. After the heavier mass falls 0.800 m, find its speed using energy and show consistency with constant-acceleration kinematics.

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a = 1.78 m s⁻²; T₁ = 24.1 N, T₂ = 23.2 N; α = 8.92 rad s⁻²; speed after 0.800 m = 1.69 m s⁻¹. For the descending mass, m₁g−T₁ = m₁a. For the rising mass, T₂−m₂g = m₂a. The unequal tensions provide the pulley's angular acceleration. Pulley torque is (T₁−T₂)R = Iα with I = ½MR² and α = a/R, hence T₁−T₂ = ½Ma. Combining equations gives (m₁−m₂)g = (m₁+m₂+M/2)a. Thus a = 9.81/5.50 = 1.78 m s⁻². T₁ = m₁(g−a) = 24.1 N and T₂ = m₂(g+a) = 23.2 N. Their difference supplies the required torque; α = a/R = 8.92 rad s⁻². The net lost gravitational energy (m₁−m₂)gh becomes ½(m₁+m₂)v² + ½I(v/R)² = ½(m₁+m₂+M/2)v². This gives v = 1.69 m s⁻¹, equal to √(2ah).

An Atwood machine with pulley inertia

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