Question

Obtain an expression for the angular momentum of a body rotating with uniform angular velocity.

Answer 1

1. Consider a rigid object rotating with a constant angular speed ‘ω’ about an axis perpendicular to the plane of the paper.
   

   A body of N particles

2. Let us consider the object to consist of N number of particles of masses m₁, m₂, …..m_N at respective perpendicular distances r₁, r₂, …..r_N from the axis of rotation.
3. As the object rotates, all these particles perform UCM with the same angular speed ω, but with different linear speeds
   v₁ = r₁ ω, v₂ = r₂ ω, .....v_N = r_N ω.  
   Directions of individual velocities `vecv_1, vecv_2, .......vecv_N,` are along the tangents to the irrespective tracks.
4. Linear momentum of the first particle is of magnitude p₁ = m₁v₁ = m₁r₁ω. Its direction is along that of `vecv_1.`
   Its angular momentum is thus of magnitude L₁ = p₁r₁ = `m_1r_1^2omega`
   Similarly, `L_2 = m_2r_2^2 ω`, `L_3 = m_3r_3^2ω`, …., `L_N = m_Nr_N^2 ω`.
5. For a rigid body with a fixed axis of rotation, all these angular momenta are directed along the axis of rotation, and this direction can be obtained by using the right-hand thumb rule.
   As all of them have the same direction, their magnitudes can be algebraically added.
6. Thus, the magnitude of angular momentum of the body is given by 
   `L = m_1r_1^2omega + m_1r_2^2omega + .... + m_Nr_N^2omega`
   = `(m_1r_1^2 + m_2r_2^2 + ...... + m_Nr_N^2)omega = Iomega`
   where `I = m_1r_1^2 + m_2r_2^2 + .... + m_Nr_N^2` is the moment of inertia of the body about the given axis of rotation.