Torque and angular momentum

Torque is a cross product, not a plain product. If a force $\vec F$ acts at a point whose position vector relative to a chosen pivot is $\vec r$, only the component of $\vec F$ perpendicular to $\vec r$ actually twists the object — a push straight along $\vec r$ (toward or away from the pivot) produces no rotation at all. The cross product automatically keeps only that perpendicular part:

where: $r$ is the distance from the pivot to the point of application, $F$ is the magnitude of the force, and $\theta$ is the angle between $\vec r$ and $\vec F$. The SI unit is $\text{N m}$ — numerically identical to the joule, but torque is never quoted in joules because it is a vector describing a turning effect, not a scalar energy.

Angular momentum is the rotational twin of momentum. For a particle of mass $m$ moving with velocity $\vec v$ (so momentum $\vec p = m\vec v$) at position $\vec r$ from a chosen origin, its angular momentum about that origin is

where: $\vec r$ locates the particle, $\vec p$ is its linear momentum, and $\vec L$ points along the axis about which the particle appears to circulate, given by the right-hand rule. Like torque, $\vec L$ depends on where the origin is chosen — the same particle has different angular momentum about different points.

For a rigid body spinning about a fixed axis, every particle's individual $\vec L$ adds up into two clean scalar relations that mirror $F=ma$ and $p=mv$ exactly, with mass $m$ replaced by moment of inertia $I$:

where: $I$ is the moment of inertia about the axis, $\alpha$ is angular acceleration, and $\omega$ is angular velocity. The pattern is worth memorising as a translation table: every time you see $m$ in a linear equation, write $I$ instead, and swap $s\to\theta$, $v\to\omega$, $a\to\alpha$, $F\to\tau$, $p\to L$.

The goal of this derivation is to show, using only the product rule for differentiation, that the net external torque on a particle equals the rate at which its angular momentum changes — a genuine derivation, not a stated analogy, and the single equation from which $\tau=I\alpha$ and every conservation-of-angular-momentum result later in the syllabus ultimately follows.

Step 1. Start from the definition $\vec L=\vec r\times\vec p$ and differentiate both sides with respect to time, treating the cross product like an ordinary product so the product rule applies to each factor in turn:

Step 2. Evaluate the first term. Since $\dfrac{d\vec r}{dt}=\vec v$ and $\vec p=m\vec v$, the first term is $\vec v\times m\vec v = m(\vec v\times\vec v)$. The cross product of any vector with itself is zero, because $\sin\theta=\sin 0^\circ=0$ for parallel vectors, so this whole term vanishes.

Step 3. Evaluate the second term using Newton's second law, $\dfrac{d\vec p}{dt}=\vec F$. The second term becomes $\vec r\times\vec F$, which is exactly the definition of torque about the same origin.

Step 4. Combine the two results. The first term is zero and the second term is $\vec\tau$, so

Net torque about a point equals the rate of change of angular momentum about that same point — the exact rotational mirror of $\vec F=d\vec p/dt$, obtained here by direct differentiation rather than by analogy alone.

A force of $15\ \text{N}$ is applied perpendicular to a wrench of length $0.3\ \text{m}$ to tighten a bolt whose fitting has moment of inertia $0.05\ \text{kg m}^2$ about the bolt's axis and is otherwise free to rotate. Find (a) the torque, (b) the angular acceleration produced, and (c) the angular momentum gained after it has turned for $0.2\ \text{s}$ starting from rest.

Step 1. Torque. The force is perpendicular to the wrench, so $\theta=90°$ and $\sin\theta=1$:

Step 2. Angular acceleration, from $\tau=I\alpha$:

Step 3. Angular velocity after $0.2\ \text{s}$ from rest, using the rotational analogue of $v=u+at$:

Step 4. Angular momentum gained, from $L=I\omega$:

This matches $L=\tau t = 4.5\times0.2=0.9\ \text{kg m}^2/\text{s}$, since $d\vec L/dt=\vec\tau$ means a constant torque produces angular momentum at a steady rate. The torque is $4.5\ \text{N m}$, the angular acceleration is $90\ \text{rad/s}^2$, and after $0.2\ \text{s}$ the fitting has gained $0.9\ \text{kg m}^2/\text{s}$ of angular momentum.

• Writing $\tau=rF$ without the $\sin\theta$ factor when the force is not perpendicular to $\vec r$ — only the component of force perpendicular to the position vector (or equivalently the perpendicular "moment arm" distance) produces torque.
• Forgetting that both torque and angular momentum are defined relative to a chosen origin or pivot — the same force or the same moving particle gives different values of $\vec\tau$ or $\vec L$ about different points, so the origin must always be stated.
• Confusing the torque unit $\text{N m}$ with the energy unit $\text{J}$ just because they are numerically the same — torque is a vector turning effect, never added to or substituted for work or energy.
• Applying $L=I\omega$ to a particle that is not in rigid-body rotation or simple circular motion about the axis — for general motion, angular momentum must be computed from the vector definition $\vec L=\vec r\times\vec p$ directly.

Ask the tutor: want to see how $d\vec L/dt=\vec\tau=0$ leads straight into the conservation of angular momentum you meet in the next topic, or work through the right-hand-rule direction of torque with a few more examples?