The angular momentum, L, of a point mass is defined as the cross product of the object's linear momentum, p, and its moment arm with respect to a fixed pivot point, r.

 

L = r x p

 

Remember that since L is a vector cross product, all three of these vectors must be mutually perpendicular to each other. That is, L, r and p must be in three separate planes. L is measured in kg m²/sec.

 

 

Using an analogy to torque we will once again use the right hand rule (RHR): r is your fingers, palm is p = mv, thumb is L.

 

 

For example, suppose a mass is falling while attached to the end of string that is unwinding from a pulley.

 

 

Let's look at an example of how we can use a determinant to calculate the angular momentum of the falling mass in the previous diagram with respect to the pulley's axis of rotation.

 

 

 

Continuing, from Newton's Second Law, we know that impulse is equal to the change in an object's momentum

 

F Δt = Δ(mv)
F = Δ(mv)/Δt

 

Multiplying both sides by r

 

Fr = Δ(mvr)/Δt
τ Δt = Δ(mvr)

 

yields an expression about angular impulse, J = τ t.

 

 

 

 

The angular momentum of a point mass is defined as L = mvr. Since all rigid bodies are made up of a collection of point masses in specific configurations, we can derive a general equation for any rigid body by multiplying by r/r, or 1, which yields

 

L = mvr (r/r)

L = mr² (v/r)
L = Iω

 

This equation also follows the previous pattern that we noticed in the sections on rotational kinematics and rotational dynamics; namely, that a analogous rotational equation can be set up when m is replaced by I and v is replaced by ω.

 

 

 

The Law of Conservation of Momentum states that the angular momentum of a system will remain constant unless acted upon by an unbalanced external torque.