If the pulley were weightless, we should consider the tension in the rope on both sides of the pulley to be the same (see problem 42). But since the pulley has a moment of inertia, we must take into account that the tension in the rope on the two sides of the pulley will be different and this difference of tension will create the moment which rotates the pulley. Let the tension in the rope on the left of the pulley be T₁ and the tension in that on the right be T₂. Then, applying Newton's second law to the motion of masses m₁ and m₂, we shall obtain the equations
m₁a = m₁g - T₁, (1)
-m₂a = m₂g - T₂. (2)
Here we have considered as positive the direction of acceleration of load m₁, and so acceleration a in the second equation is given the minus sign.
If the angular velocity of the pulley be denoted by ω, so that ω = a/R, where R is the radius of the pulley, then applying Newton's second law to its rotation we shall have
(T₁ - T₂)R = Iω/R = Ia/R². (3)
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Solving equations (1), (2) and (3) simultaneously, we shall have:
T₁ = m₁g(m₂ + I/R²)/(m₁ + m₂ + I/R²),
T₂ = m₂g(m₁ + I/R²)/(m₁ + m₂ + I/R²)
When I = 0 we obtain from these equations that T₁ = T₂, which coincides with the results of problem 42.