Answer:
0.47 m/s²
Explanation:
Assuming the string is inelastic, m₃ will accelerate downward at a rate of -a, and m₄ will accelerate upward at a rate of +a.
Draw a four free body diagrams, one for each hanging mass and one for each wheel.
For m₃, there are two forces: weight force m₃g pulling down, and tension force T₃ pulling up. Sum of forces in the +y direction:
∑F = ma
T₃ − m₃g = m₃(-a)
For m₄, there are two forces: weight force m₄g pulling down, and tension force T₄ pulling up. Sum of forces in the +y direction:
∑F = ma
T₄ − m₄g = m₄a
For m₁, there are two forces: tension force T₃ pulling down, and tension force T pulling right. Sum of the torques in the counterclockwise direction:
∑τ = Iα
T₃r₃ − Tr₃ = (m₁r₃²) (a/r₃)
T₃ − T = m₁a
For m₂, there are two forces: tension force T₄ pulling down, and tension force T pulling left. Sum of the torques in the counterclockwise direction:
∑τ = Iα
Tr₄ − T₄r₄ = (m₂r₄²) (a/r₄)
T − T₄ = m₂a
We now have 4 equations and 4 unknowns. Let's add the third and fourth equations to eliminate T:
(T₃ − T) + (T − T₄) = m₁a + m₂a
T₃ − T₄ = (m₁ + m₂) a
Now let's subtract the second equation from the first:
(T₃ − m₃g) − (T₄ − m₄g) = m₃(-a) − m₄a
T₃ − m₃g − T₄ + m₄g = -(m₃ + m₄) a
T₃ − T₄ = (m₃ − m₄) g − (m₃ + m₄) a
Setting these two expressions equal:
(m₁ + m₂) a = (m₃ − m₄) g − (m₃ + m₄) a
(m₁ + m₂ + m₃ + m₄) a = (m₃ − m₄) g
a = (m₃ − m₄) g / (m₁ + m₂ + m₃ + m₄)
Plugging in values:
a = (1.64 kg − 1.27 kg) (9.8 m/s²) / (2.5 kg + 2.3 kg + 1.64 kg + 1.27 kg)
a = 0.47 m/s²
