To determine the speed of the spool's center, v_O, we will relate it to the given speed v of the string by analyzing the spool's kinematics.

1. Rolling Without Slipping Condition: The spool rolls without slipping on the horizontal surface. This means the point of contact O' is instantaneously at rest. This condition relates the speed of the center, v_O, to the angular speed \omega of the spool:

   $$v_O = \omega R$$

2. Velocity of the Unwinding Point P: The velocity of point P on the inner cylinder, where the string unwinds, is the vector sum of the translational velocity of the center O (\vec{v}_O) and the rotational velocity of P about O (\vec{v}_{P/O} = \vec{\omega} \times \vec{r}_{OP}). Let's define a coordinate system with \hat{i} pointing horizontally to the right and \hat{j} vertically upwards. The velocity of the center is \vec{v}_O = v_O \hat{i}. For the spool to roll to the right, the angular velocity must be clockwise, so \vec{\omega} = -\omega \hat{k}. The string segment PQ is tangent to the inner cylinder at P. Thus, the radius OP is perpendicular to PQ. As PQ makes an angle \varphi with the horizontal, the position vector \vec{r}_{OP} makes an angle of 90^\circ + \varphi with the \hat{i} axis.

   $$\vec{r}_{OP} = r\cos(90^\circ+\varphi)\hat{i} + r\sin(90^\circ+\varphi)\hat{j} = -r\sin\varphi\,\hat{i} + r\cos\varphi\,\hat{j}$$

   The velocity of P is then:

   $$\vec{v}_P = \vec{v}_O + \vec{\omega} \times \vec{r}_{OP} = v_O \hat{i} + (-\omega \hat{k}) \times (-r\sin\varphi\,\hat{i} + r\cos\varphi\,\hat{j})$$ $$\vec{v}_P = v_O \hat{i} + \omega r\sin\varphi\,(\hat{k}\times\hat{i}) - \omega r\cos\varphi\,(\hat{k}\times\hat{j})$$

   Using \hat{k}\times\hat{i} = \hat{j} and \hat{k}\times\hat{j} = -\hat{i}:

   $$\vec{v}_P = v_O \hat{i} + \omega r\sin\varphi\,\hat{j} + \omega r\cos\varphi\,\hat{i}$$ $$\vec{v}_P = (v_O + \omega r\cos\varphi)\hat{i} + (\omega r\sin\varphi)\hat{j}$$

3. String Constraint: The string is inextensible and is pulled with a speed v. This means the component of \vec{v}_P along the direction of the string must be equal to v. The unit vector along the string PQ is \hat{u}_{PQ} = \cos\varphi\,\hat{i} + \sin\varphi\,\hat{j}.

   $$v = \vec{v}_P \cdot \hat{u}_{PQ}$$ $$v = (v_O + \omega r\cos\varphi)\cos\varphi + (\omega r\sin\varphi)\sin\varphi$$ $$v = v_O\cos\varphi + \omega r(\cos^2\varphi + \sin^2\varphi)$$ $$v = v_O\cos\varphi + \omega r$$

4. Solving for the Speed of the Center v_O: We now have a system of two equations:

   1. $v_O = \omega R$
   2. $v = v_O\cos\varphi + \omega r$

   From equation (1), we express \omega in terms of v_O: \omega = v_O / R. Substituting this into equation (2):

   $$v = v_O\cos\varphi + \left(\frac{v_O}{R}\right)r$$

   Factor out v_O:

   $$v = v_O \left(\cos\varphi + \frac{r}{R}\right)$$

   Finally, solving for v_O:

   $$v_O = \frac{v}{\cos\varphi + \frac{r}{R}} = \frac{vR}{R\cos\varphi + r}$$