Answer:
3.53 rad/s² counterclockwise
Explanation:
I assume the figure is the one attached, and that the question is as follows: "What is the angular acceleration of the cylinders?"
From the figure, the hanging masses will move in opposite directions. Let's assume that the 2.5 kg block will move down and the 4.0 kg block will move up, such that the cylinders turn counterclockwise.
Draw a free body diagram for each block and the cylinder.
For the 2.5 kg block, there are 2 forces:
Weight force m₁g pulling down,
Tension force T₁ pulling up.
For the 4.0 kg block, there are 2 forces:
Weight force m₂g pulling down,
Tension force T₂ pulling up.
For the cylinders, there are 2 forces:
Tension force T₁ pulling counterclockwise at radius r₁ = 10 cm,
Tension force T₂ pulling clockwise at radius r₂ = 5 cm.
Sum of forces on the 2.5 kg block in the down direction:
∑F = ma
m₁g − T₁ = m₁ αr₁
Sum of forces on the 4.0 kg block in the up direction:
∑F = ma
T₂ − m₂g = m₂ αr₂
Sum of torques on the cylinders in the counterclockwise direction:
∑τ = Iα
T₁r₁ − T₂r₂ = Iα
Multiply the first equation by r₁ and the second equation by r₂.
m₁gr₁ − T₁r₁ = m₁ αr₁²
T₂r₂ − m₂gr₂ = m₂ αr₂²
Add both to the third equation:
m₁gr₁ − m₂gr₂ = Iα + m₁ αr₁² + m₂ αr₂²
g (m₁r₁ − m₂r₂) = α (I + m₁r₁² + m₂r₂²)
α = g (m₁r₁ − m₂r₂) / (I + m₁r₁² + m₂r₂²)
For a solid cylinder, I = ½ mr². Therefore, the total moment of inertia of the cylinders is:
I = ½ (3.0 kg) (0.05 m)² + ½ (20 kg) (0.10 m)²
I = 0.104 kg m²
Plugging in values:
α = 9.8 (2.5 × 0.10 − 4.0 × 0.05) / (0.104 + 2.5 × 0.10² + 4.0 × 0.05²)
α = 3.53 rad/s²
