Question regarding calculation of angular velocity connected to coordinate transformation

Question

So i need to calculate the angular velocity of the dark disk, $w^*_3$. In the master solution they state it is : $w^*_3 = w_3 + \overline{w_3}$ with :

$w_3 = \frac{|\dot{r}|}{4a}$

$|\dot{r}| = v_a + \overline{w_3}\cdot a$

$|\dot{r}| = v_b - \overline{w_3}\cdot 2a$

$ \Rightarrow|\dot{r}| = 2a\cdot (w_S+w_R)$

$\overline{w_3} = \frac{1}{a}(|r|-v_1)=\frac{1}{2a}(v_b-|r|)=\frac{1}{3a}(v_b-v_a)$

$ \Rightarrow \overline{w_3} = 2w_r-w_s$

$ \Rightarrow w_3^* = \frac{5}{2}w_r-\frac{1}{2}w_s$

So, the question I have is that if I understood correctly, $w^*_3$ is the angular velocity of the disk itself (in its coordinate system) if you look at it from the disks center. So shouldn't be $\overline{w_3}$ already be my $w^*_3$? Because it is the angular velocity with which my disk is actually spinning?(in it's coordinate system)

Answer 1

$\omega ^*_3$ is the rotation rate of the disk in the frame pictured (equal to $2 * pi *$ frequency point $P$ will appear in the 9 o'clock position). It is the sum of the orbit rate of $\omega _3$'s disc center about the system center in the frame pictured ($\omega_3$), plus the rotation rate about it's own axis in a reference frame where this axis is fixed ($ \bar\omega_3$). The first term considers the contact points with $A$ and $B$ fixed and the system rotating with disk $\omega_3$, so one turn of the center is one rotation of the disk, and point P makes one revolution. The second term considers $\omega_3$'s axis fixed and the contact points free to run.

If we take $\omega_S$ as given in each case, we get two different (linear) relations for $\omega_R$ and $\omega_3$. So $\omega_R = C_1*\omega_S$ and $\bar\omega_R= C_2*\omega_S $ This lets us set up a general solution $\omega^*_3 = f(\omega_S,\omega_R)$, where $C_1$ and $C_2$ are solved using $\omega_S$ and then they are used to compute $\omega^*_3$