Answer:
a = (m₂-m₁) / (m₂ + m₁ + ½ m)
a = (m₂-m₁) / (m₂ + m₁ + 1)
Explanation:
An Atwood machine consists of two masses of different m1 and m2 value that pass through a pulley, in this case with mass. Let's use Newton's second law for this problem.
Assume that m₂> m₁, so the direction of descent of m₂ is positive, this implies that the direction of ascent of m₁ is positive
Equation of the side of m₁
T₁ - W₁ = m₁ a
Equation of the side of m₂
W₂ - T₁ = m₂ a
Mass pulley equation m; by convention the counterclockwise rotation is positive
τ = I α
T₁ R - T₂ R = I (-α)
The moment of inertia of a disk is
I = ½ m R²
Angular and linear acceleration are related
a = α R
α = a / R
The rotation is clockwise, so it is negative
We replace
(T₁ –T₂) R = ½ m R² (-a / R)
T₁ -T₂ = - ½ m a
Let's write our three equations together
T₁ - m₁ g = m₁ a
m₂ g - T₂ = m₂ a
T₁ –T₂ = -½ m a
Let's multiply the last equation by (-1) and add
m₂ g - m₁ g = m₂ a + m₁ a + ½ m a
a = (m₂-m₁) / (m₂ + m₁ + ½ m)
calculate
a = (m₂ - m₁)/ (m₁ +m₂ + 1)