Angular Momentum and Torque
~5 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable with vector cross products, Newton's second law in vector form, and the concept of a fixed reference frame. Familiarity with the previous lecture on variable-mass systems is helpful but not strictly required.
Why this matters
Angular momentum provides the rotational analogue to linear momentum, unlocking the solution to orbital mechanics and rigid body dynamics.
While linear momentum tells us how an object moves in a straight line, angular momentum describes its tendency to rotate about a point. This quantity is fundamental because it is conserved in the absence of external torque, a principle that explains why planets maintain stable orbits, why ice skaters spin faster when pulling in their arms, and why gyroscopes resist changes in orientation. Understanding this conservation law is the bridge between simple particle dynamics and the complex motion of extended bodies.
The Rotational Analogue
Imagine a ball attached to a string, whirled around your hand. If you let go of the string, the ball flies off in a straight line tangent to its circular path. However, while the string is taut, the ball is constantly changing direction. This change in direction requires a force directed towards the center. In linear mechanics, we track momentum \(\vec{p}\). In rotational mechanics, we track angular momentum \(\vec{L}\). The key insight is that just as force changes linear momentum, torque changes angular momentum. If no external twisting force (torque) acts on a system, its angular momentum remains constant. This is the rotational version of Newton's first law: an object in rotational motion stays in rotational motion with the same angular momentum unless acted upon by an external torque.
Definition and Vector Nature
Angular momentum is defined relative to a specific origin point \(O\). For a particle of mass \(m\) at position \(\vec{r}\) with velocity \(\vec{v}\), the linear momentum is \(\vec{p} = m\vec{v}\). The angular momentum \(\vec{L}\) is the cross product of the position vector and the linear momentum:
The magnitude of this vector is \(L = r p \sin\theta\), where \(\theta\) is the angle between \(\vec{r}\) and \(\vec{p}\). The direction of \(\vec{L}\) is perpendicular to the plane containing \(\vec{r}\) and \(\vec{p}\), determined by the right-hand rule. This vector nature is crucial because it implies that angular momentum is not just a scalar quantity of "amount of rotation" but a directed quantity that specifies the axis of rotation.
Torque and the Rate of Change
To understand how \(\vec{L}\) changes, we differentiate it with respect to time. Using the product rule for cross products:
The first term, \(\frac{d\vec{r}}{dt} \times \vec{p}\), becomes \(\vec{v} \times (m\vec{v})\). Since the cross product of any vector with itself (or a parallel vector) is zero, this term vanishes. The second term remains, and by Newton's second law, \(\frac{d\vec{p}}{dt} = \vec{F}_{\text{net}}\). Thus:
The quantity \(\vec{\tau} = \vec{r} \times \vec{F}\) is defined as the torque. Therefore, the fundamental rotational equation of motion is \(\frac{d\vec{L}}{dt} = \vec{\tau}\). This mirrors the linear form \(\frac{d\vec{p}}{dt} = \vec{F}\), establishing a deep symmetry in Newtonian mechanics.
Conservation in Central Forces
A central force is one that always points directly toward or away from a fixed origin, such as gravity or electrostatic attraction. Mathematically, this means \(\vec{F}\) is parallel or antiparallel to \(\vec{r}\). Consequently, the angle \(\theta\) between \(\vec{r}\) and \(\vec{F}\) is either \(0\) or \(\pi\), making \(\sin\theta = 0\). The torque \(\vec{\tau} = \vec{r} \times \vec{F}\) is therefore zero. Since the torque is zero, the time derivative of angular momentum is zero, implying that \(\vec{L}\) is constant in both magnitude and direction. This conservation law restricts the motion to a fixed plane, a critical step in solving the Kepler problem for planetary orbits.
The derivation
Angular Momentum, Torque, and Central Forces
For a particle of position \(\vec{r}\) and momentum \(\vec{p}=m\vec{v}\), define the angular momentum about a fixed origin \(O\) as \(\vec{L}=\vec{r}\times\vec{p}\). Newton's second law implies that its time rate of change equals the net torque \(\vec{\tau}=\vec{r}\times\vec{F}\) about the same origin, \(\dfrac{d\vec{L}}{dt}=\vec{\tau}\). Consequently, whenever the net force is central — everywhere parallel or antiparallel to \(\vec{r}\), i.e. \(\vec{F}=f(r)\,\hat{r}\) — the torque vanishes and \(\vec{L}\) is conserved.
Angular momentum depends on the chosen origin: move the origin and \(\vec{L}\) changes. The law \(d\vec{L}/dt = \vec{\tau}\) holds when both are taken about a point fixed in an inertial frame, or about the centre of mass of a system. About an accelerating point, extra terms appear (equivalently, the torque of fictitious forces must be included). C
Worked example
Problem. A particle of mass \(2.0\ \mathrm{kg}\) moves with velocity \(\vec{v} = 3.0\ \mathrm{m/s}\ \hat{i} + 4.0\ \mathrm{m/s}\ \hat{j}\) at the instant it is located at position \(\vec{r} = 5.0\ \mathrm{m}\ \hat{i} - 2.0\ \mathrm{m}\ \hat{j}\) relative to the origin. Calculate the angular momentum \(\vec{L}\) about the origin.
- Calculate the linear momentum \(\vec{p} = m\vec{v}\). \(\vec{p} = 2.0\ \mathrm{kg} \times (3.0\ \hat{i} + 4.0\ \hat{j})\ \mathrm{m/s} = 6.0\ \hat{i} + 8.0\ \hat{j}\ \mathrm{kg\cdot m/s}\).
- Compute the cross product \(\vec{L} = \vec{r} \times \vec{p}\). Using the determinant form for the \(\hat{k}\) component: \(L_z = r_x p_y - r_y p_x\).
- Substitute values: \(L_z = (5.0)(8.0) - (-2.0)(6.0) = 40.0 + 12.0 = 52.0\).
- Since \(\vec{r}\) and \(\vec{p}\) are in the \(xy\)-plane, \(\vec{L}\) is along the \(z\)-axis.
Units check. Position is in \(\mathrm{m}\) and momentum in \(\mathrm{kg\cdot m/s}\), so \(\mathrm{m} \cdot (\mathrm{kg \cdot m/s}) = \mathrm{kg \cdot m^2/s}\). This is the standard unit for angular momentum.
Three quick tests
- If a force is applied at the center of mass of a free particle, does it produce torque about that center of mass?
Answer
No, because the position vector \(\vec{r}\) from the center of mass to the point of application is zero, so \(\vec{\tau} = \vec{0} \times \vec{F} = \vec{0}\). - Why is the angular momentum of a planet about the Sun conserved?
Answer
Because the gravitational force is a central force, pointing directly from the planet to the Sun, resulting in zero torque about the Sun. - Can angular momentum be zero if the particle is moving?
Answer
Yes, if the velocity vector is parallel to the position vector (radial motion), the cross product \(\vec{r} \times \vec{p}\) is zero.
Problem set for PU-101 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.