Hello Bobbin

From pure rolling condition, we have the velocity of the center of the bobbins as $\dfrac {vR}{r+R} .$

Now let $s$ denote the displacement of the center of the bobbin with respect to point $B$ along the horizontal. Then

$\begin{aligned} S &= & R \cot x \\ \frac{ds}{dx} &=& - R \csc^2 x \\ \frac{ds}{dt} \; \frac{dt}{dx} &= & -R \csc^2 x \end{aligned}$

But as $s$ increases, $x$ decreases, i.e.: $dx$ is negative for positive $ds$ . Therefore, the minus sign will be neglected. Also $\dfrac{ds}{dt}$ is the velocity of the center of the bobbin. Hence

$\dfrac{vR}{r+R} \; \dfrac{dt}{dx} = R \csc^2 x \Rightarrow \dfrac{dx}{dt} = \sin^2 x \; \dfrac{v}{r+R} .$

Also 2 $\dfrac {dx}{dt}$ is the required $\omega$ :

$\omega = 2\sin^2 x \; \dfrac v{r+R} = \dfrac{2v \sin^2 x}{r+R}.$

Hence, (k = 2 ).

Rather than going by Differentiation approach .

I Used the fact that as bobbin and board are in contact with each other there relative velocity along the common normal = 0

Call velocity of centre of mass of bobbin be vcm then angular speed of bobbin =vcm/(R+r)

Velocity of point of contact on bobbin along common normal = vcm*sin(2x)

Velocity of point of contact on board along common normal = w*Rcotx

where w is angular velocity of board . equate them and get desired result