Torque and Angular Momentum

Pushing on a door near its hinge does little. Push at the far edge, and it swings easily. The same force gives different rotational effect depending on where it acts. The rotational analogue of force is torque, and the analogue of linear momentum is angular momentum.

Concept

Torque (about a chosen origin) due to force $\vec{F}$ applied at position $\vec{r}$:

$\vec{\tau} = \vec{r}\times\vec{F}$

Magnitude: $|\tau| = rF\sin\theta = F\cdot d$, where $d = r\sin\theta$ is the perpendicular distance from the line of force to the origin (the moment arm).

SI unit: N·m.

Angular momentum of a particle (about same origin):

$\vec{L} = \vec{r}\times\vec{p} = m(\vec{r}\times\vec{v})$

SI unit: kg·m²/s (or J·s).

For a rigid body rotating about a fixed axis:

$L = I\omega$

where $I$ is the moment of inertia and $\omega$ angular speed.

Rotational Newton's second law:

$\vec{\tau}_{\text{net}} = \frac{d\vec{L}}{dt}$

(About a fixed point or about the CM.) For a rigid body about a fixed axis, $\tau = I\alpha$, where $\alpha$ is angular acceleration.

Derivation

Start from $\vec{L} = \vec{r}\times\vec{p}$. Differentiating with respect to time:

$\frac{d\vec{L}}{dt} = \frac{d\vec{r}}{dt}\times\vec{p} + \vec{r}\times\frac{d\vec{p}}{dt}$

The first term: $\frac{d\vec{r}}{dt} = \vec{v}$, so $\vec{v}\times\vec{p} = \vec{v}\times m\vec{v} = \vec{0}$.

The second term: $\frac{d\vec{p}}{dt} = \vec{F}_{\text{net}}$, so $\vec{r}\times\vec{F}_{\text{net}} = \vec{\tau}_{\text{net}}$.

Therefore

$\boxed{\vec{\tau}_{\text{net}} = \frac{d\vec{L}}{dt}}$

This is the rotational analogue of $\vec{F} = d\vec{p}/dt$.

Worked Example

A particle of mass 2 kg moves with velocity $\vec{v} = 3\hat{i}$ m/s at position $\vec{r} = 4\hat{j}$ m from the origin. (a) Find its angular momentum. (b) A force $\vec{F} = 5\hat{i}$ N is applied at that point. Find the torque.

Solution:

(a) $\vec{p} = m\vec{v} = (2)(3\hat{i}) = 6\hat{i}$ kg·m/s.

$\vec{L} = \vec{r}\times\vec{p} = (4\hat{j})\times(6\hat{i}) = 24(\hat{j}\times\hat{i}) = -24\hat{k} \, \text{kg·m}^2\text{/s}$

(b) $\vec{\tau} = \vec{r}\times\vec{F} = (4\hat{j})\times(5\hat{i}) = -20\hat{k}$ N·m.

Both are in the $-\hat{k}$ direction (into the page in standard orientation) — consistent with clockwise rotational sense.

Common Confusions

• Torque depends on the chosen origin. Same force, different origin gives different torque (because $\vec{r}$ changes). But the change in $\vec{L}$ about that origin is consistent.
• Force through the origin: zero torque. A force whose line of action passes through the chosen point produces no torque about that point.
• Angular momentum is not necessarily along the axis. For asymmetric bodies, $\vec{L}$ and $\vec{\omega}$ can point in different directions. For symmetric bodies and fixed axes, they align.
• A particle moving in a straight line has nonzero $\vec{L}$ about any point not on the line. The magnitude is $mvd$ where $d$ is the perpendicular distance.

Key Takeaways

• $\vec{\tau} = \vec{r}\times\vec{F}$, $\vec{L} = \vec{r}\times\vec{p}$.
• Magnitude of torque: $\tau = F \cdot d$, where $d$ is the moment arm.
• Rotational second law: $\vec{\tau}_{\text{net}} = d\vec{L}/dt$.
• For rigid body about fixed axis: $L = I\omega$, $\tau = I\alpha$.
• A particle moving in a straight line has nonzero angular momentum about any external point.