Two particles of masses m₁ and m₂ are connected by a light rigid rod of length r . The system rotates at an angular speed about an axis passing through the centre of mass of the system and perpendicular to the rod. Determine the angular momentum of the system, where is the reduced mass defined as = m₁ m₂ m₁ + m₂ .

Step-by-step solution

Let the distances of the masses m₁ and m₂ from the centre of mass be r₁ and r₂ respectively. r₁ = m₂ r m₁ + m₂ and r₂ = m₁ r m₁ + m₂ The moment of inertia of the system about the axis passing through the centre of mass and perpendicular to the rod is I = m₁ r₁^2 + m₂ r₂^2 Substituting the values of r₁ and r₂ : I = m₁ ( m₂ r m₁ + m₂ )^2 + m₂ ( m₁ r m₁ + m₂ )^2 I = m₁ m₂^2 r^2 (m₁ + m₂)^2 + m₂ m₁^2 r^2 (m₁ + m₂)^2 I = m₁ m₂ (m₁ + m₂) r^2 (m₁ + m₂)^2 I = m₁ m₂ m₁ + m₂ r^2 = r^2 The angular momentum of the system is gi