Question

An Atwood machine consists of a mass of 3.5 kg connected by a light string to a mass of 6.0 kg over a frictionless pulley with a moment of inertia of 0.0352 kg ∙ m2 and a radius of 12.5 cm. If the system is released from rest, what is the speed of the masses after they have moved through 1.25 m if the string does not slip on the pulley?

Please note: the professor has told us that the correct answer is 2.3 m/s. how does one arrive at this answer?

Answer 1

Given that,

mass of light string, m₁ = 6 kg

mass of an Atwood machine, m₂ = 3.5 kg

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

radius of pulley, r = 0.125 m

Using conservation of energy, we have

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

where, K.E_initial = kinetic energy at rest = 0

then, we get

(0) + m₁ g h = [(1/2) m₁ v2 + (1/2) m₂ v² + (1/2) I $\omega$²] + m₂ g h

we know that, $\omega$ = angular velocity =v / r

m₁ g h = [(1/2) m₁ v² + (1/2) m₂ v² + (1/2) I (v/r)²] + m₂ g h

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

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

v = _$\sqrt{}$2g h (m₁ - m₂) / [m₁ + m₂ + (I / r²)]

v = _$\sqrt{}$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]

v = _$\sqrt{}$(61.2 kg.m²/s²) / (11.7 kg)

v = _$\sqrt{}$5.23 m²/s²

v = 2.28 m/s

v $\approx$ 2.3 m/s