Rate of change of ##L## in a rotating coordinate system

[Kashmir]

* We've a vector $\mathbf{A}$ lying in space, changing according to some rule.

* We introduce an inertial frame and find $\left(\frac{d}{d t} \mathbf{A} \right)_{i n}$ in it.

* We also introduce a co located frame rotating with $\mathbf{\omega}$. In this rotating frame I find $\left(\frac{d}{d t} \mathbf{A} \right)_{rot}$

* There exists a relationship between the two time rates as
$\left ( \frac{d\mathbf A}{dt} \right )_\text{inertial} = \left ( \frac{d\mathbf A}{dt} \right )_\text{rot} + \boldsymbol \omega \times \mathbf A$

* In all of this derivation it was assumed that the vector $\mathbf{A}$ was independent of the coordinate system. We merely observed the vector in two frames. The vector is independent of the coordinate system.----------------------------------
* Can we use the above equation on angular momentum vector $\mathbf{L}$ i.e $\left ( \frac{d\mathbf L}{dt} \right )_\text{inertial} = \left ( \frac{d\mathbf L}{dt} \right )_\text{rot} + \boldsymbol \omega \times \mathbf L$?

I think no we can't.

In the derivation of the $\left ( \frac{d\mathbf A}{dt} \right )_\text{inertial} = \left ( \frac{d\mathbf A}{dt} \right )_\text{rot} + \boldsymbol \omega \times \mathbf A$ we assumed that the vector $\mathbf{A}$ was independent of the coordinate system, its lengths and direction in space is independent of the coordinate system.

However for $\mathbf{L}$ that isn't the case.
$\mathbf{L}$ has a different length in a stationary frame than in a rotating one. So the derivation doesn't apply.

 

[anuttarasammyak]

$$L=r \times p$$
$$\frac{dL}{dt}=\frac{dr}{dt} \times p + r \ \times \frac{dp}{dt}$$
Why don't you use this relation to check the result ?

 

[vanhees71]

Of course you can use the formula. The formula refers to the time derivative of polar or axial vector components (not vectors!).