Lagrangian in non-inertial frame of reference

I'm in trouble with comprehending derivation of Lagrangian of a particle in non-inertial , translational and rotational frame of reference by Landau's Mechanics.
More precisely I don't understand why the following can happen.

$$L^\prime=\frac{1}{2}m{v^\prime}^2-m\mathbf{W}(t)\cdot\mathbf{r}^\prime-U \tag{39.4}$$

The velocity $$\mathbf{v}'$$ of the particle relative to $$K'$$ is composed of its velocity $$\mathbf{v}$$ relative to $$K$$ and the velocity $$\mathbf{\Omega}\times\mathbf{r}$$ of its rotation with $$K$$: $$\mathbf{v}^\prime=\mathbf{v}+\mathbf{\Omega}\times\mathbf{r}$$ (since the radius vectors $$\mathbf{r}$$ and $$\mathbf{r}'$$ in the frames $$K$$ and $$K'$$ coincide). Substituting this in the Lagrangian (39.4), we obtain

$$L=\frac{1}{2}mv^2+m\mathbf{v}\cdot\mathbf{\Omega}\times \mathbf{r}+\frac{1}{2}m(\mathbf{\Omega}\times \mathbf{r})^2-m\mathbf{W}\cdot \mathbf{r}-U \tag{39.6}$$

(in p.127, L.D. Landau and E.M. Lifshitz Mechanics )

The problem is the rectlinear term. Why the second term in (39.4) $$m\mathbf{W}\cdot\mathbf{r}^\prime$$ simply converts into the third term in (39.6) $$m\mathbf{W}\cdot\mathbf{r}$$. Radius vectors $$\mathbf{r}$$ and $$\mathbf{r}'$$ have rotational relation, hence I guess it can't be just replaced each other.

• It's explained right in the thing you quote. (The part in the parentheses) Jul 6 '19 at 14:13
• @AaronStevens I think the thing they share is only their origin and the watching direction of $K$ and $K'$ are different. How can they be identical? Jul 6 '19 at 14:25
• I don't understand your comment. Jul 6 '19 at 14:27
• @AaronStevens Sorry, English is not my usual language. I meant why does it happen that $r$ and $r^\prime$ coincide in spite that their frame are different. Jul 6 '19 at 14:33
• Because it is the position vector. It points from the origin to the object in question. This doesn't depend on the reference frame. Jul 6 '19 at 14:35

Expanding on AaronStevens's above comment: The reference frames $$K$$ and $$K^{\prime}$$ share the same origin. Therefore position vectors $${\bf r}={\bf r}^{\prime}$$ (measured relative to the origin) are the same.