Angular-Momentum Theorem for a Mechanical System

The angular momentum of a mechanical system characterizes the rotational aspect of system motion relative to a selected point or axis.

Angular momentum about a point

About a fixed point $O$:

$$\vec K_O=\sum_{i=1}^{n}\vec r_i\times m_i\vec v_i.$$

Angular momentum about an axis is the projection of this vector onto that axis.

Angular-momentum theorem

For a fixed point $O$ in an inertial frame:

$$\frac{d\vec K_O}{dt}=\sum\vec M_O^{e}.$$

The time derivative of system angular momentum equals the resultant moment of external forces about the same point.

Role of internal forces

For central pairwise internal forces, the moments of each interaction pair cancel. Thus only the total moment of external forces remains in the equation for the complete system.

Theorem about an axis

Projecting onto a fixed $z$ axis gives:

$$\frac{dK_z}{dt}=\sum M_z^{e}.$$

Conservation

If $\sum\vec M_O^{e}=0$, then $\vec K_O=const$. If only the resultant external moment about a particular axis is zero, only the corresponding angular-momentum component is conserved.

Rigid-body rotation

For a rigid body rotating about a fixed principal axis $z$, $K_z=I_z\omega$. With constant $I_z$, the equation becomes $I_z\dot\omega=\sum M_z^{e}$.

Example

If a rigid body has $I_z=2$ kg·m² and rotates at $\omega=5$ rad/s, then $K_z=10$ kg·m²/s. With zero external moment about the axis, this value remains constant.

Common mistakes

• confusing angular momentum with moment of force;
• taking force moments about one point and angular momentum about another;
• using $K_z=I_z\omega$ without checking the motion and axis conditions;
• claiming conservation of the full vector when only one external-moment component is zero.