The angular momentum \( \mathbf{L} \) of a particle about a point \( O \) is defined as the cross product of the particle’s position vector \( \mathbf{r} \) (from point \( O \) to the particle) and its linear momentum \( \mathbf{p} = m\mathbf{v} \):
\[ \mathbf{L} = \mathbf{r} \times \mathbf{p} = \mathbf{r} \times (m\mathbf{v}) \]
Angular momentum has SI units of \( \mathrm{kg \cdot m^2/s} \).
The direction of \( \mathbf{L} \) follows the right-hand rule: Curl the fingers of your right hand from \( \mathbf{r} \) to \( \mathbf{v} \); the thumb points in the direction of \( \mathbf{L} \).
The magnitude of angular momentum is given by:
\[ L = |\mathbf{r}| \cdot |\mathbf{p}| \cdot \sin\theta = mvr\sin\theta \]
where \( \theta \) is the angle between \( \mathbf{r} \) and \( \mathbf{v} \). Angular momentum is maximum when \( \theta = 90^\circ \).
For a particle of mass \( m \) moving in a circle of radius \( r \) with angular velocity \( \omega \), the angular momentum is:
\[ L = mvr = mr^2\omega \]
The figure below shows a particle in motion. The position vector \( \mathbf{r} \), velocity vector \( \mathbf{v} \), and angular momentum direction \( \mathbf{L} \) are shown. The local path of the particle is indicated as a curved trajectory.
Figure: Angular momentum \( \mathbf{L} \) is perpendicular to the plane formed by \( \mathbf{r} \) and \( \mathbf{v} \).
The diagram illustrates the angular momentum of a single particle in two-dimensional space. The particle is shown as a large black dot located at position vector \( \vec{r} \) from the origin \( O \).
The particle follows a curved trajectory (dashed arc), and its instantaneous velocity vector \( \vec{v} \) is drawn tangential to this path.
The angular momentum vector \( \vec{L} = \vec{r} \times m\vec{v} \) is perpendicular to both \( \vec{r} \) and \( \vec{v} \), and is depicted as a circled dot \( \odot \), indicating a direction **out of the plane** of motion.
This representation follows the right-hand rule: if the fingers of your right hand curl from \( \vec{r} \) to \( \vec{v} \), your thumb points in the direction of \( \vec{L} \).
The time rate of change of angular momentum gives the torque:
\[ \boldsymbol{\tau} = \frac{d\mathbf{L}}{dt} \]
For a continuous mass distribution, the total angular momentum about the origin is given by:
\[ \mathbf{L} = \iiint_V \mathbf{r} \times \rho(\mathbf{r}) \mathbf{v}(\mathbf{r}) \, dV \]
This form reduces to a summation for discrete particles: \[ \mathbf{L} = \sum_{i} \mathbf{r}_i \times m_i \mathbf{v}_i \] which is useful in analyzing systems of particles.
In a continuous mass distribution, each infinitesimal volume element \( dV \) contains a small amount of mass:
\[ dm = \rho(\mathbf{r}) \, dV \]
Substituting into the volume integral form of angular momentum: \[ \mathbf{L} = \iiint_V \mathbf{r} \times \rho(\mathbf{r}) \mathbf{v}(\mathbf{r}) \, dV \quad \Rightarrow \quad \mathbf{L} = \int \mathbf{r} \times \mathbf{v} \, dm \]
This formulation is useful for integrating over arbitrary mass distributions, especially when the velocity and position vectors of mass elements are known.
Consider a thin ring of mass \( M \) and radius \( R \), rotating in the XY-plane about its center with constant angular velocity \( \omega \).
The position vector of a mass element at angle \( \theta \) is: \[ \mathbf{r} = R \cos\theta\, \hat{\mathbf{i}} + R \sin\theta\, \hat{\mathbf{j}} \] The velocity of the mass element (tangential to the ring) is: \[ \mathbf{v} = -R \omega \sin\theta\, \hat{\mathbf{i}} + R \omega \cos\theta\, \hat{\mathbf{j}} \]
The angular momentum of a small element \( dm \) is: \[ d\mathbf{L} = \mathbf{r} \times \mathbf{v} \, dm = R^2 \omega \, dm \, \hat{\mathbf{k}} \] Since this value is constant for all \( \theta \), the total angular momentum is: \[ \mathbf{L} = \int d\mathbf{L} = R^2 \omega \int dm \, \hat{\mathbf{k}} = M R^2 \omega \, \hat{\mathbf{k}} \]
This result matches \( \mathbf{L} = I \omega \) for a ring, where \( I = M R^2 \) is the moment of inertia.
Suppose an extended rigid body is mounted on a fixed axis and is free to rotate. When a force F acts at a point with position vector r (from the axis), the torque N is defined as:
\( \mathbf{N} = \mathbf{r} \times \mathbf{F} \)
Torque depends on the force, the position of application, and the choice of axis. However, when the net torque is zero, the torque is independent of the choice of origin.
For a body acted on by multiple forces, if the net torque about point \( O \) is zero, then the torque about any other point \( O' \) is also zero, provided the net force is zero. This supports rotational equilibrium.
For a particle with angular momentum \( \mathbf{L} = \mathbf{r} \times \mathbf{p} \), we differentiate with respect to time:
\( \frac{d\mathbf{L}}{dt} = \frac{d}{dt}(\mathbf{r} \times \mathbf{p}) = \mathbf{v} \times \mathbf{p} + \mathbf{r} \times \frac{d\mathbf{p}}{dt} \)
The first term vanishes because \( \mathbf{v} \times \mathbf{p} = m \mathbf{v} \times \mathbf{v} = 0 \). Hence:
\( \frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F} = \mathbf{N} \)
So the rate of change of angular momentum is equal to the torque. If \( \mathbf{N} = 0 \), then \( \frac{d\mathbf{L}}{dt} = 0 \), and angular momentum is conserved.
For a system of \( N \) particles, each with angular momentum \( \mathbf{L}_i = \mathbf{r}_i \times \mathbf{p}_i \), the total angular momentum is:
\( \mathbf{L} = \sum_{i=1}^N \mathbf{r}_i \times \mathbf{p}_i \)
From Newton’s third law, internal forces between particles are equal and opposite and directed along the line connecting the particles. Therefore, internal torques cancel out in pairs:
\( \sum_{i=1}^N \sum_{j \neq i} \mathbf{r}_i \times \mathbf{F}_{ij} = 0 \)
Hence, the rate of change of total angular momentum equals the net external torque:
\( \frac{d\mathbf{L}}{dt} = \sum_{i=1}^N \mathbf{r}_i \times \mathbf{F}_i^{\text{(ext)}} = \mathbf{N}_{\text{ext}} \)
Step 1: Define total angular momentum
For a system of \( N \) particles: \[ \mathbf{L} = \sum_{i=1}^N \mathbf{r}_i \times \mathbf{p}_i = \sum_{i=1}^N \mathbf{r}_i \times m_i \mathbf{v}_i \]
Step 2: Differentiate with respect to time
\[ \frac{d\mathbf{L}}{dt} = \sum_{i=1}^N \left( \frac{d\mathbf{r}_i}{dt} \times m_i \mathbf{v}_i + \mathbf{r}_i \times m_i \frac{d\mathbf{v}_i}{dt} \right) \] Since \( \frac{d\mathbf{r}_i}{dt} = \mathbf{v}_i \), and \( \mathbf{v}_i \times \mathbf{v}_i = 0 \), we get: \[ \frac{d\mathbf{L}}{dt} = \sum_{i=1}^N \mathbf{r}_i \times m_i \mathbf{a}_i \]
Step 3: Apply Newton's Second Law
\[ m_i \mathbf{a}_i = \mathbf{F}_i = \mathbf{F}_i^{\text{(ext)}} + \sum_{j \ne i} \mathbf{F}_{ij} \] Therefore, \[ \frac{d\mathbf{L}}{dt} = \sum_{i=1}^N \mathbf{r}_i \times \left( \mathbf{F}_i^{\text{(ext)}} + \sum_{j \ne i} \mathbf{F}_{ij} \right) \]
Step 4: Split the summation
\[ \frac{d\mathbf{L}}{dt} = \sum_{i=1}^N \mathbf{r}_i \times \mathbf{F}_i^{\text{(ext)}} + \sum_{i=1}^N \sum_{j \ne i} \mathbf{r}_i \times \mathbf{F}_{ij} \]
Step 5: Internal torques cancel (Contact me if you have any doubt)
By Newton’s third law, \( \mathbf{F}_{ij} = -\mathbf{F}_{ji} \), and since the internal forces are collinear (along \( \mathbf{r}_j - \mathbf{r}_i \)), we have: \[ (\mathbf{r}_i - \mathbf{r}_j) \times \mathbf{F}_{ij} = 0 \] So all internal torques cancel pairwise: \[ \sum_{i=1}^N \sum_{j \ne i} \mathbf{r}_i \times \mathbf{F}_{ij} = 0 \]
Final result:
\[ \frac{d\mathbf{L}}{dt} = \sum_{i=1}^N \mathbf{r}_i \times \mathbf{F}_i^{\text{(ext)}} = \mathbf{N}_{\text{ext}} \] This proves that the time derivative of the total angular momentum equals the total external torque acting on the system.
If \( \mathbf{N}_{\text{ext}} = 0 \), then \( \frac{d\mathbf{L}}{dt} = 0 \), implying:
\( \mathbf{L}_{\text{initial}} = \mathbf{L}_{\text{final}} \)
This is the conservation of angular momentum: when no external torque acts, the total angular momentum of a system remains constant.
Total Angular Momentum:
For a system of \( N \) particles, the angular momentum is: \[ \mathbf{L} = \sum_{i=1}^N \mathbf{r}_i \times m_i \mathbf{v}_i \]
Internal Forces and Torques:
Internal forces occur in equal and opposite pairs due to Newton's third law: \[ \mathbf{F}_{ij} = -\mathbf{F}_{ji} \] Hence, \[ \mathbf{r}_i \times \mathbf{F}_{ij} + \mathbf{r}_j \times \mathbf{F}_{ji} = (\mathbf{r}_i - \mathbf{r}_j) \times \mathbf{F}_{ij} \] If \( \mathbf{F}_{ij} \) acts along the line joining \( i \) and \( j \), the torque cancels: \[ (\mathbf{r}_i - \mathbf{r}_j) \times \mathbf{F}_{ij} = 0 \] Therefore, internal torques sum to zero.
Time Derivative of Angular Momentum:
Using Newton's second law: \[ m_i \ddot{\mathbf{r}}_i = \mathbf{F}_i^{\text{ext}} + \sum_{j \ne i} \mathbf{F}_{ij} \] Taking cross product with \( \mathbf{r}_i \), summing over all particles: \[ \frac{d\mathbf{L}}{dt} = \sum_i \mathbf{r}_i \times \mathbf{F}_i^{\text{ext}} + \sum_i \sum_{j \ne i} \mathbf{r}_i \times \mathbf{F}_{ij} \] The second term vanishes, so: \[ \frac{d\mathbf{L}}{dt} = \sum_i \mathbf{r}_i \times \mathbf{F}_i^{\text{ext}} = \boldsymbol{\tau}_{\text{ext}} \]
Conservation of Angular Momentum:
If \( \boldsymbol{\tau}_{\text{ext}} = 0 \), then \[ \frac{d\mathbf{L}}{dt} = 0 \quad \Rightarrow \quad \mathbf{L} = \text{constant} \]
Let \( \mathbf{r}_i = \mathbf{r}_c + \mathbf{r}_i' \), where:
Then total angular momentum becomes: \[ \mathbf{L} = \sum_i m_i (\mathbf{r}_c + \mathbf{r}_i') \times (\mathbf{v}_c + \mathbf{v}_i') \] Expanding and using \( \sum m_i \mathbf{r}_i' = 0 \): \[ \mathbf{L} = M \mathbf{r}_c \times \mathbf{v}_c + \sum_i m_i \mathbf{r}_i' \times \mathbf{v}_i' \] Therefore, \[ \mathbf{L} = \mathbf{L}_{\text{CM}} + \mathbf{L}' \] where: