Answer to Question #86987 in Mechanics | Relativity for Maria

Question

An Atwood machine consists of two masses, mA = 63 kg and mB = 72 kg , connected by a massless inelastic cord that passes over a pulley free to rotate (Figure 1). The pulley is a solid cylinder of radius R = 0.41 m and mass 5.0 kg . [Hint: The tensions FTA and FTB are not equal.]
Part A
Determine the acceleration of each mass.
Express your answer to two significant figures and include the appropriate units.
Part B
What % error would be made if the moment of inertia of the pulley is ignored?
Express your answer using two significant figures.

Answer 1

Each mass acceleration a = g(m_B – m_A)/(m_B + m_A + 0.5m) = 9.8*9/(72+63+2.5) = 0.64 (m/s²), cylinder acceleration ε = a/R = 1.56 (s^-2).

In case the moment of inertia of the pulley is ignored mass acceleration a₁ = 9.8*9/(72+63) = 0.65 (m/s²), % error = 100(a₁-a)/a = 1.9%.