§ 27. Atwood's Machine: Dynamics of Connected Masses & Pulley Inertia (Grade 9–10)
Motion of two masses on a thread over a pulley with moment of inertia, thread tension. Two weights connected by an inextensible thread over a pulley move with a single common acceleration determined by the difference in weight and total system inertia.
Theory Summary & Scientific Mechanism
Invented by George Atwood in 1784, Atwood’s machine is a classic apparatus for verifying Newton’s laws of motion under “diluted gravity.” Two masses m₁ and m₂ connected by an inextensible cord over a pulley of radius R and mass Mₚ share a common linear acceleration a = α R.
Writing Newton’s second law for both masses and the rotational torque equation for the solid disk pulley (I = (1)/(2)Mₚ R²) gives m₁ g - T₁ = m₁ a, T₂ - m₂ g = m₂ a, and (T₁ - T₂)R = Iα = (1)/(2)Mₚ R a. Summing the three equations yields the system acceleration a = ((m₁ - m₂)g)/(m₁ + m₂ + (1)/(2)Mₚ).
Notice a crucial physical insight: whenever the pulley has inertia (Mₚ > 0), the cord tensions on the two sides are not equal (T₁ ≠ T₂), because a net torque (T₁ - T₂)R is required to angularly accelerate the pulley! Choosing m₁ ≈ m₂ slows the motion (a ≪ g) for accurate timing.
Everyday Analogy
This is exactly how an elevator with a counterweight works! Because masses m₁ and m₂ hang on opposite ends of the same cable, only their weight difference (m₁ - m₂)g drives the motion, slowing gravity down to a gentle pace.
Interactive Question & Derivation
On an ideal massless pulley (mₚulley = 0), two loads m₁ = 3 kg and m₂ = 1 kg are suspended. What is the acceleration a of the system and how does thread tension T compare to the weights of the loads?
- A: Acceleration a = g, and tension equals the heavier load T = m₁ g
- B: Acceleration a = (m₁ - m₂)/(m₁ + m₂)g = 0.5g, and tension satisfies m₂ g < T < m₁ g (T = (2m₁ m₂)/(m₁ + m₂)g)
- C: Tension is zero because the two loads move in opposite directions
Study Hint: The net driving force is (m₁ - m₂)g, while the total accelerated inertia is m₁ + m₂ + I/R². How does acceleration a behave when m₁ and m₂ are close in value?
Proof & Explanation: Writing Newton's second law for both loads: m₁ g - T₁ = m₁ a and T₂ - m₂ g = m₂ a, plus the pulley torque equation (T₁ - T₂)R = Iε = (1)/(2)mₚulleyR²(a/R), yields a = ((m₁ - m₂)g)/(m₁ + m₂ + (1)/(2)mₚulley). For the descending heavier load m₁, tension is less than its weight (T₁ = m₁(g - a) m₂ g).
Key Equations & 3D Simulation Parameters
a = (m₁ - m₂)/(m₁ + m₂ + (1)/(2)mₚulley) g, m₂ g < T₂ ≤ T₁ < m₁ g
- Left load mass m₁: 1 … 5 kg (default: 3.2 kg)
- Right load mass m₂: 1 … 5 kg (default: 1.8 kg)
- Pulley mass Mₚ: 0 … 4 kg (default: 1 kg)
- Pulley radius R: 0.5 … 1.2 m (default: 0.8 m)