Velocity of a point of a rotating object placed at the end of a rotating link

Question

The figure shows rotating object placed at the end of a rotating link. To find the velocity at the point A,

$\vec{r_p} = \vec{r_{o'}} + \vec{r_{p/o'}}$

So,

$\vec{v_p} = \vec{v_{o'}} + \vec{\omega_{o'}} \times\vec{r_{p/o'}} + \vec{v_{rel}}$

As we have $ \vec{v_{rel}} =\vec{v_{p/o'}} = 0 \qquad, Pure \quad rotation \quad and $

$ \vec{v_{o'}} = \vec{v_{translation}} + \vec{\omega_{o}} \times \vec{r_{o'}}$

As, $\vec{v_{translation}} = 0, pure \quad rotation$

So,

$\vec{v_p} = \vec{\omega_{o}} \times \vec{r_{o'}} + \vec{\omega_{o'}} \times\vec{r_{p/o'}} $

Here

$\vec{v_A} = \vec{\omega_{1}} \times \vec{r_{oo'}} + \vec{\omega_{2}} \times\vec{r_{Ao'}} $

Why is the solution given to be

$\vec{v_A} = \vec{\omega_{1}} \times \vec{r_{oo'}} + (\vec{\omega_{1}} + \vec{\omega_{2}})\times\vec{r_{Ao'}} $.

Answer 1

Suppose the particle at $A$ is stationary with respect the frame centered in $O'$, so is $\vec{\omega_2}=0$. Clearly, the movement wrt the frame centered in $O$ is

$\vec{v_A} = \vec{\omega_{1}} \times(\vec{r_{oo'}}+\vec{r_{Ao'}})$

Now, consider that the particle is observed moving wrt the frame centered in $O'$: $\vec{v_{Ao'}}\ne0$ . In this case the velocities superpose, so is, simply they have to be added to know the movement wrt the frame centered in $O$,

$\vec{v_A} = \vec{\omega_{1}} \times(\vec{r_{oo'}}+\vec{r_{Ao'}})+\vec {v_{Ao'}}$

It's said that the movement is caused by some rotation around the point $O'$: $\vec{v_{Ao'}}=\vec{\omega_2}\times\vec{r_{Ao'}}$

$\vec{v_A} = \vec{\omega_{1}} \times(\vec{r_{oo'}}+\vec{r_{Ao'}})+\vec{\omega_2}\times\vec{r_{Ao'}}$

$\vec{v_A} = \vec{\omega_{1}} \times \vec{r_{oo'}} + (\vec{\omega_{1}} + \vec{\omega_{2}})\times\vec{r_{Ao'}}$