I keep seeing that when taking the time derivative of the angular momentum in a rotating reference frame, we get: $$\frac{d\vec{L}}{dt} = \vec{\tau} + \vec{\omega} \times \vec{L}$$ meaning the torque as the rotating frame sees it, plus another term.
Is it derived somewhere in the internet? I couldn't find it. I like to understand what I am doing, but in this instance I have no idea why the derivative gives this extra term.
Best Answer
Consider total derivative of $\vec{A}$ in an inertial frame, $$\frac{d}{dt}\vec{A}=\hat{i}\frac{d}{dt}A_x+\hat{j}\frac{d}{dt}A_y+\hat{k}\frac{d}{dt}A_z+A_x\frac{d\hat{i}}{dt}+A_y\frac{d\hat{j}}{dt}+A_z\frac{d\hat{k}}{dt}.$$ In an inertial system $\frac{d\hat{i}}{dt}=\frac{d\hat{j}}{dt}=\frac{d\hat{k}}{dt}=0$. Thus we have, $$\frac{d\vec{A}}{dt}_{inertial}=\hat{i}\frac{d}{dt}A_x+\hat{j}\frac{d}{dt}A_y+\hat{k}\frac{d}{dt}A_z.$$ Now consider in a rotating frame one has $\vec{A}=A_x^\prime\hat{i}^\prime+A_y^\prime\hat{j}^\prime+A_z^\prime\hat{k}^\prime$. Taking the total derivative, $$\frac{d}{dt}\vec{A}=\hat{i}^\prime\frac{d}{dt}A_x^\prime+\hat{j}^\prime\frac{d}{dt}A_y^\prime+\hat{k}^\prime\frac{d}{dt}A_z^\prime+A_x^\prime\frac{d\hat{i^\prime}}{dt}+A_y^\prime\frac{d\hat{j^\prime}}{dt}+A_z^\prime\frac{d\hat{k}^\prime}{dt}.$$ Note that in a rotating frame the quantities like $\frac{d\hat{i}^\prime}{dt}$ do not vanish. Therefore the total time derivative in a rotating frame is,
$$\frac{d\vec{A}}{dt}=\frac{d\vec{A}}{dt}_{rotating}+A_x^\prime\frac{d\hat{i^\prime}}{dt}+A_y^\prime\frac{d\hat{j^\prime}}{dt}+A_z^\prime\frac{d\hat{k}^\prime}{dt}$$ $\frac{d\vec{A}}{dt}_{rotating}=\hat{i}^\prime\frac{d}{dt}A_x^\prime+\hat{j}^\prime\frac{d}{dt}A_y^\prime+\hat{k}^\prime\frac{d}{dt}A_z^\prime$ is the apparent time derivative of the vector in a rotating frame and the combination $A_x^\prime\frac{d\hat{i^\prime}}{dt}+A_y^\prime\frac{d\hat{j^\prime}}{dt}+A_z^\prime\frac{d\hat{k}^\prime}{dt}$ captures the effects of rotation. Now, we know the relation between the linear velocity and angular velocity, $\vec{v}=\frac{d\vec{r}}{dt}=\omega\times\vec{r}$. Choosing $\vec{r}$ to be equal to $\hat{i}^\prime$, $\hat{j}^\prime$, and $\hat{k}^\prime$ respectively we have, $$\begin{align}\frac{d\hat{i}^\prime}{dt}&=\omega\times\hat{i}^\prime\\\frac{d\hat{j}^\prime}{dt}&=\omega\times\hat{j}^\prime\\\frac{d\hat{k}^\prime}{dt}&=\omega\times\hat{k}^\prime\end{align}$$ Using these we get in a rotating frame, $$\frac{d\vec{A}}{dt}=\frac{d\vec{A}}{dt}_{rotating}+\omega\times\vec{A}$$ Choosing $\vec{A}=\vec{L}$ we get, $$\frac{d\vec{L}}{dt}=\frac{d\vec{L}}{dt}_{rotating}+\omega\times\vec{L}=\vec{\tau}+\omega\times\vec{L}.$$