When to use L = Iω and Ecin = ½Iω²

The compact formulas $L = I\omega$ and $E_\text{cin} = \tfrac{1}{2}I\omega^2$ look simple, but hide a trap: which $I$? The answer depends on how the body moves, and there are two distinct situations that must never be confused.

Principle — Two use cases

Case A — body pinned at a fixed point $P$. If the rod is hinged and rotates about the hinge $P$, use: $L = I_P\,\omega \qquad E_\text{cin} = \tfrac{1}{2}I_P\,\omega^2$ where $I_P$ is the moment of inertia about $P$ (not the CM). For a uniform bar about one end, from the Huygens–Steiner theorem: $I_P = \tfrac{1}{12}ML^2 + M(\tfrac{L}{2})^2 = \tfrac{1}{3}ML^2$.

Case B — body that translates and rotates (rolling, collision with a free rod). Use König’s theorem, separating the CM contribution from the internal one: $L_\text{tot} = \underbrace{\vec{r}_\text{CM}\times\vec{p}_\text{tot}}_{L_\text{CM}} + \underbrace{I_\text{CM}\,\omega}_{L_\text{int}} \qquad E_\text{cin,tot} = \underbrace{\tfrac{1}{2}M v_\text{CM}^2}_{E_\text{cin,CM}} + \underbrace{\tfrac{1}{2}I_\text{CM}\omega^2}_{E_\text{cin,int}}$ where $I_\text{CM}$ is the moment of inertia about the CM.

The golden rule: if the body is fixed at a point, use $I_P$; if the body translates and rotates, use $I_\text{CM}$ and separately add the centre-of-mass term.

Don't mix up the two cases

Common mistake: using $I_\text{CM}$ when the body rotates about a hinge, or forgetting the CM term in case B. In the hinged-rod case, if you mistakenly used $I_\text{CM} = \tfrac{1}{12}ML^2$ instead of $\tfrac{1}{3}ML^2$, you would get an $\omega$ four times too large and a kinetic energy sixteen times too large. The lost-energy check (which must be $\geq 0$ in an inelastic collision) would fail, flagging the error.

Collegamenti

Argomenti: Dinamica rotazionale Concetti: Momento angolare · Energia cinetica rotazionale · Momento d’inerzia Metodi: Teorema di König

Esercizi collegati: Problema — Asta incernierata che cade dall’orizzontale · Problema — Momento angolare di rotazione propria della Terra · Problema — Ranking di ω a parità di momento angolare