Year 1 Sem 2/LC Classical Mechanics and Relativity 2/Lecture 06

Lecture 6: Mechanics VI: Vectors, Moving Axes and Gyroscopes
Tue 10 Feb 2026 11:00

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Angular Velocity and Momentum as Vectors

We define the angular velocity vector:

$$\vec{\omega} = \omega_x \hat{\imath} + \omega_y \hat{\jmath} + \omega_z \hat{k}$$

For an object which is rotating in a single plane, the angular velocity vector is perpendicular to the plane, i.e. pointing parallel to the axis of rotation. It’s direction can be given by the Right Hand Rule.

We can likewise define the angular momentum vector:

$$\vec{L} = I \vec{\omega}$$

Where $L_x = I_x \omega_x$ etc. As the moment of inertia is different in each axis (often), $\vec{L}$ and $\vec{\omega}$ may not be parallel.

Torque and Changes in Angular Momentum Direction

Torque and angular momentum are fundamentally connected via:

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

If $\tau$ only has a component which is parallel to $\vec{L}$, only the magnitude of $\vec{L}$ changes with no change in direction.

Likewise if the torque only has a perpendicular component, the direction of $\vec{L}$ changes but not the magnitude.

Flywheels

Consider a heavy disk rotating about the horizontal axis, loosely pivoted at $O$.

If $\omega$ is sufficiently large, the flywheel can stay horizontal instead of drooping downwards. As drawn, the $\vec{\omega}$ vector points directly outwards (leftwards).

The angular momentum vector:

$$\vec{L} = I_\text{wheel} \vec{\omega}$$

Where $I_\text{wheel}$ is about the axis of rotation.

$$L = MR^2/2 \omega$$

Which also points directly outwards. The downward force of the weight tries to drag the flywheel downwards. The torque around the origin is:

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

This creates a torque acting into the page, which causes the whole wheel to rotate horizontally around the origin. This results in the flywheel precessing.

Given a particular angular velocity of the wheel, $\omega$, how quickly does this precession take place? Using $\vec{\tau} = d \vec{L} / dt$, we want to find the angular velocity of the precession, $\Omega$.

From a top-down view, suppose the angular momentum vector at $t$ is $\vec{L(t)}$ and after some small $\Delta t$ it has new angular momentum vector $L(t + \Delta t)$, with the vector $\Delta \vec{L}$ between them. The two angular momentum vectors make an angle $\Delta \phi$.

$$\vec{\tau} = \frac{d \vec{L}}{dt} \approx \frac{\Delta \vec{L}}{\Delta t}$$

$$| \Delta L | = |L| \sin \Delta \phi = |L| \Delta \phi \quad \text{for small angles}$$

Hence, considering magnitudes:

$$\tau = \frac{L \Delta \phi}{\Delta t} \tag{1}$$

By definition of the precession angular velocity $\Omega$:

$$\Omega = \frac{d \phi}{dt}$$

Hence from eqn (1):

$$\Omega \approx \frac{\Delta \phi}{\Delta t} = \frac{\tau}{L}$$

We also know that:

$$L = I_\text{wheel} \omega = \frac{mR^2\omega}{2}$$

And:

$$\tau = rF = lmg$$

Hence:

$$\Omega = \frac{2 lmg}{\omega mR^2} = \frac{2gl}{\omega R^2}$$

Why does it not fall over?

This demonstrates precession, but doesn’t explain why it actually stays up. Are there not still gravitational forces that aren’t countered?

This is not a statics problem. This is a dynamics problem with moving motion that doesn’t require balanced forces, as there’s a net acceleration and it’s not in equilibrium.
If the wheel tries to fall, the angular momentum vector will change in a direction where there isn’t an applied torque. This fails conservation.
Torques actually do cause motion perpendicular to the direction of the applied force.
If the wheel isn’t spinning fast enough, it will fall.

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