Newton's Second Law in Angular Form — Question 1

Question 1

The angular equation follows from a product rule

A particle of constant mass $m$ has position $\vec r$, velocity $\vec v$ and momentum $\vec p=m\vec v$, all measured relative to a fixed inertial origin.

1. Derive $d\vec L/dt=\vec\tau$ from $\vec L=\vec r\times\vec p$.

2. Explain why $\vec v\times\vec p$ disappears and why a moving origin needs extra care.

Question 1 - Solution

The angular equation follows from a product rule

(a) Derivation and calculation

Apply the product rule to $\vec L=\vec r\times\vec p$: $$\frac{d\vec L}{dt}=\frac{d\vec r}{dt}\times\vec p+\vec r\times\frac{d\vec p}{dt}.$$ At a fixed inertial origin, $\dot{\vec r}=\vec v$ and $\vec p=m\vec v$, so $$\vec v\times\vec p=m\vec v\times\vec v=\vec0.$$ Newton’s law therefore gives $$\boxed{\frac{d\vec L}{dt}=\vec r\times\vec F=\vec\tau}.$$

(b) Result and interpretation

For an origin $A$ moving at velocity $\vec V_A$, relative position has derivative $\vec v-\vec V_A$. If angular momentum uses inertial momentum, $$\frac{d\vec L_A}{dt}=(\vec v-\vec V_A)\times\vec p+\vec r_{P/A}\times\vec F =\boxed{\vec\tau_A-\vec V_A\times\vec p}.$$ The extra term cannot generally be dropped; the choice of both position origin and momentum frame matters.

Physical check

The fixed-origin law follows when $\vec V_A=0$; the added term also vanishes when $\vec V_A$ is parallel to momentum.