Angular momentum L = Iω (fixed axis), L = r × p (general)

1. WHAT is angular momentum?

WHY a cross product? Because only the part of the motion that "goes around" the origin counts. Motion straight toward or away from $O$ produces zero turning — and the cross product automatically kills the parallel component.

$|\vec{L}| = r\,p\,\sin\theta = p \cdot (r\sin\theta) = p\cdot r_\perp$

where $r_\perp = r\sin\theta$ is the perpendicular distance (the "lever arm") from $O$ to the line of motion.

2. HOW $L = I\omega$ comes out of $\vec{L} = \vec{r}\times\vec{p}$ (full derivation)

Setup. Take a rigid body rotating about a fixed axis (say the $z$-axis). Pick particle $i$ at perpendicular distance $r_i$ from the axis. Since it's rigid, every particle has the same angular speed $\omega$.

Step 1 — speed of particle $i$. It travels in a circle of radius $r_i$, so $v_i = \omega\, r_i.$ Why this step? For circular motion, tangential speed $=$ radius $\times$ angular speed.

Step 2 — its momentum is tangential, $p_i = m_i v_i = m_i \omega r_i$. The velocity is perpendicular to the radius vector from the axis, so $\theta = 90^\circ$ and $\sin\theta = 1$.

Step 3 — its angular momentum about the axis. $L_{i,z} = r_i\, p_i \sin 90^\circ = r_i (m_i \omega r_i) = m_i r_i^2\, \omega.$ Why this step? We use the component along the axis, which is the physically conserved/relevant one for fixed-axis spin.

Step 4 — sum over all particles (the axis component adds up): $L_z = \sum_i m_i r_i^2\, \omega = \left(\sum_i m_i r_i^2\right)\omega.$

Step 5 — recognise the bracket as the moment of inertia $I=\sum_i m_i r_i^2$:

3. Connecting to torque (the rotational Newton's law)

Differentiate $\vec L = \vec r\times\vec p$: $\frac{d\vec L}{dt} = \underbrace{\frac{d\vec r}{dt}\times\vec p}_{\vec v\times m\vec v=\,0} + \vec r\times\frac{d\vec p}{dt} = \vec r\times\vec F = \vec\tau.$

4. Worked examples

5. Active recall

Flashcards

Connections

• Moment of Inertia — the $\sum m_i r_i^2$ that turns $\vec r\times\vec p$ into $I\omega$
• Torque — $\vec\tau = \vec r\times\vec F = d\vec L/dt$
• Conservation of Angular Momentum — zero-torque consequence
• Cross Product — why direction & $\sin\theta$ appear
• Linear Momentum — the translational analogue $\vec p=m\vec v$
• Rotational Kinetic Energy — $KE=\tfrac12 I\omega^2 = L^2/2I$
• Kepler's Second Law — equal areas = constant $L$ for orbits

Concept Map

Hinglish (regional understanding)