Consider a rigid body rotating with a constant angular velocity w about an axis through the point O and perpendicular to the plane of the figure. All the particles of the body perform uniform circular motion about the axis of rotation with the same angular velocity ω
Suppose that the body consists of N particles of masses m_{1}, m_{2}, ..., m_{n}, situated at perpendicular distances r_{1}, r_{2}, ..., r_{n}, respectively from the axis of rotation.
The particle of mass m, revolves along a circle of radius r,, with a linear velocity of magnitude
v_{1}= r_{1}ω The magnitude of the linear momentum of the particle is
p_{1} = m_{1}v_{1} = m_{1}r_{1}ω
The angular momentum of the particle about the axis of rotation is by definition,
\(\vec{L_1}=\vec{r_1}\vec{p_1}\)
where θ is the smaller of the two angles between r_{1} and p_{1}
In this case, θ = 90° ∴ sinθ = 1
∴ L_{1} = r_{1} p_{1 = }r_{1}m_{1}r_{1}ω = m_{1}r_{1}^{2}ω
Similarly, L_{2} = m_{2}r_{2}^{2}ω, L_{3} = m_{3}r_{3}^{2}ω etc.
The angular momentum of the body about the given axis is
L= L_{1 }+ L_{2 }+L_{3 ……. }+ L_{N}
= m_{1}r_{1}^{2}ω + m_{2}r_{2}^{2}ω + m_{3}r_{3}^{2}ω …….. m_{N}r_{N}^{2}ω
= ω (m_{1}r_{1}^{2} + m_{2}r_{2}^{2} + m_{3}r_{3}^{2} …….. m_{N}r_{N}^{2})
\(=(\sum \limits_{i=1}^{N}m_ir_i^2)ω\)
= moment of inertia of the body about the given axis.
In vector form, L⃗ =Iω⃗
Thus, angular momentum = moment of inertia x angular velocity.
Note: Angular momentum is a vector quantity. It has the same direction as ω⃗