Torque as Rate of Change of Angular Momentum

The relation

$\vec{\tau} = \frac{d\vec{L}}{dt}$

This is the most general form of Newton's Second Law for rotation — more fundamental than $\tau = I\alpha$.

Derivation for a particle

Angular momentum of a particle: $\vec{L} = \vec{r} \times \vec{p}$

Differentiate with respect to time:

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

First term: $\frac{d\vec{r}}{dt} \times \vec{p} = \vec{v} \times m\vec{v} = 0$ (cross product of a vector with itself is zero)

Second term: $\vec{r} \times \frac{d\vec{p}}{dt} = \vec{r} \times \vec{F} = \vec{\tau}$

Therefore:

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

Why this is more general than $\tau = I\alpha$

$\tau = I\alpha$ assumes:

• Rigid body (fixed shape)
• Fixed axis of rotation
• $I$ is constant

$\vec{\tau} = \frac{d\vec{L}}{dt}$ applies even when:

• The body deforms (changes $I$)
• The axis changes direction
• Mass is being added or removed

Example: A spinning top precesses — the axis of rotation itself rotates. $\tau = I\alpha$ doesn't capture this; $\vec{\tau} = \frac{d\vec{L}}{dt}$ does.

Connecting to $\tau = I\alpha$

For a rigid body rotating about a fixed axis:

$L = I\omega \implies \frac{dL}{dt} = I\frac{d\omega}{dt} = I\alpha = \tau$

Consistent. The simpler form $\tau = I\alpha$ is the fixed-axis, rigid-body special case.

Impulse-momentum theorem for rotation

Integrating $\vec{\tau} = \frac{d\vec{L}}{dt}$ over time:

$\int_{t_1}^{t_2} \vec{\tau} \, dt = \Delta\vec{L} = \vec{L}_f - \vec{L}_i$

Angular impulse = change in angular momentum.

For constant torque: $\tau \cdot \Delta t = I\omega_f - I\omega_i$