Consider the system shown in the figure below with m1 230 k

Consider the system shown in the figure below with m_1 = 23.0 kg, m_2 = 12.9 kg, R = 0.290 m, and the mass of the pulley M = 5.00 kg. Object m_2 is resting on the floor, and object mi is 4.70 m above the floor when it is released from rest. The pulley axis is frictionless. The cord is light, does not stretch, and does not slip on the pulley. Calculate the time interval required for m_1 to hit the floor. How would your answer change if the pulley were massless?

Solution

PE₁ = m₁ g h , PE₂ = m₂ g h

KE₁ = 1/2 m₁ * v² , KE₂ = 1/2 m₂ v²

KE_p = 1/2 I w² , I = 1/2 M r² , w = v / r

PE₁ = KE₁ + KE₂ + KE_p + PE₂

m₁ g h = 1/2 m₁ * v²+ 1/2 m₂ * v²+ 1/2 I w²+ m₂ g h

m₁ g h = 1/2 m₁ * v²+ 1/2 m₂ * v²+ 1/2 * 1/2 M r² (v/r)² + m₂ g h

v² = ( m₁ g h - m₂ g h ) / [1/2 m₁ + 1/2 m₂ + 1/4 M]

v = sqrt [g h (m₁ - m₂) / [1/2 m₁ + 1/2 m₂ + 1/4 M] ]

v = sqrt [9.81 * 4.70 * (23.0 - 12.9) / [1/2 * 23 + 1/2 * 12.9 + (1/4) 5.00 ]

v = 4.92 m/s

h = h₀ + (1/2) (v - v₀) * t [ h₀ = 0 m , v₀ = 0 m/s ]

h = v * t / 2

t = 2 * h / v = 2 * 4.70 / 4.92

t = 1.91 s

----------------------------------------------

PE₁ = KE₁+ KE₂ + KE_p + PE₂

v = sqrt [g h (m₁ - m₂) / [ (1/2) m₁ + (1/2) m₂ ] ]

= sqrt [ 9.81 * 4.70 * (23.0 - 12.9) / [1/2 * 23 + 1/2 * 12.9 ] ]

v = 5.09 m/s

t = 2 * h / v  = 2 * 4.70 / 5.09

t = 1.85 s