Question

Derive an expression that relates angular momentum with the angular velocity of a rigid body.

Answer 1

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 m₁, m₂, ..., m_n, situated at perpendicular distances r₁, r₂, ..., 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

v₁= r₁ω The magnitude of the linear momentum of the particle is

p₁ = m₁v₁ = m₁r₁ω

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 r₁ and p₁

In this case, θ = 90° ∴ sinθ = 1

∴ L₁ = r₁ p_{1 =} r₁m₁r₁ω = m₁r₁²ω

Similarly, L₂ = m₂r₂²ω, L₃ = m₃r₃²ω etc.

The angular momentum of the body about the given axis is

L= L₁ + L₂ +L_{3 …….} + L_N

= m₁r₁²ω + m₂r₂²ω + m₃r₃²ω …….. m_Nr_N²ω

= ω (m₁r₁² + m₂r₂² + m₃r₃² …….. m_Nr_N²)

\(=(\sum \limits_{i=1}^{N}m_ir_i^2)ω\) 

= 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