Question

an atwood machine has a mass of 3.50kg connected by a light string to a mass of 6kg over a pulley with a moment of interia of 0.0352 kg m2 and a radius of 12.5 cm . if the system is released from rest, what is the masses after they have moved through 1.25m?

Answer 1

Given that,

mass of the light string, m₁ = 6 kg

mass of the atwood machine, m₂ = 3.5 kg

moment of inertia of pulley, I = 0.0352 kg.m²

Using conservation of energy, we have

K.E_initial + P.E_initial = K.E_final + P.E_final

where, K.E_initial = initial kinetic energy = 0

K.E_final = K.E_trans + K.E_rot

then, we have

m₁ g h = (1/2) m₁ v² + (1/2) m₂ v² + (1/2) I ² + m₂ g h

m₁ g h = (1/2) (m₁ + m₂) v² + (1/2) I (v² / r²) + m₂ g h

(1/2) (m₁ + m₂) v² + (1/2) I (v² / r²) = (m₁ - m₂) g h

(m₁ + m₂) v² + I (v² / r²) = 2 gh (m₁ - m₂)

v² = 2 gh (m₁ - m₂) / [(m₁ + m₂) + I / r²]

where, h = height = 1.25 m

r = radius = 12.5 cm = 0.125 m

v² = 2 (9.8 m/s²) (1.25 m) [(6 kg) - (3.5 kg)] / [(6 kg) + (3.5 kg) + (0.0352 kg.m²) / (0.125 m)²]

v² = (61.2 kg.m²/s²) / [(9.5 kg) + (2.25 kg)]

v² = (61.2 kg.m²/s²) / (11.7 kg)

v = 5.23 m²/s²

v = 2.28 m/s