A yo-yo is pulled by its string on the floor without slipping. Depending on the angle of the string with respect to the horizontal, the yo-yo will travel towards the puller or away from the puller. Find the angle from the horizontal the string has to be pulled which demarcates the forward movement and backward movement.
Solve for the acceleration, either angular or translational.
Solution
Consider rotation to occur about a pivot at the contact point with the horizontal. Let $F$ be the pulling force on the string. The torque about the that point is given by $\vec r \times \vec F$.
Using some simple trig we can express the vectors for $r$ and $F$ in terms of rectangular coordinates.
\begin{align}
\vec r =& r_1 \sin \theta \hat i + (r_2-r_1\cos \theta) \hat j \\
\vec F =& F \cos \theta \hat i + F \sin \theta \hat j
\end{align}
Now just do the cross product to get the torque and simplify. ( would be out of the paper.)
\begin{align}
\vec \tau =& \vec r \times \vec F = (x F_y - y F_x) \hat k \\
=& [r_1 \sin \theta F \sin \theta - (r_2 -r_1 \cos \theta)(F \cos \theta)] \hat k \\
=& (r_1 F \sin^2 \theta - r_2 F \cos \theta + r_1 F \cos^2 \theta) \hat k\\
=& (r_1 F - r_2 F \cos \theta) \hat k
\end{align}
Let the moment of inertia of the yo-yo be expressed in terms of the outer radius:

Where $c$ is just a constant less than one depending on the mass distribution of the yoyo. Using the parallel axis theorem to get the I about the pivot point at the contact between the surface and the rim:

Thus the angular acceleration is

$$\alpha = \frac{ F(r_1 - r_2 \cos \theta) }{(c+1)mr_2^2}$$
Here positive α would correspond to rolling to the left, away from the puller. This occurs when $r_1/r_2$ is greater than $\cos \theta$.
Another way to solve it
Consider the rotation about the centre of mass and let the static friction force be represented by $f$.
The net force on the object on the c.m. is
$$F_{net} = F \cos \theta - f = ma$$ 
The net torque about c.m. is (magnitude only)

$$\tau = fr_2 - Fr_1 = I_{\rm cm} \alpha$$
Substituting for $f$ and $I$ we get

$$\tau = Fr_1 -(F\cos \theta -ma )r_2 = cmr_2^2(- a/r_2)
\tau = (F\cos \theta -ma )r_2 - Fr_1 = cmr_2^2 (-a/r_2)$$
Notice that  in order for the signs to be consistent.
Solving for $a$

$$a = -\frac{F(r_1 - r_2 \cos \theta)}{(c+1)mr_2}$$
This is consistent with solution 1. When $\cos \theta$ is greater than $r_1/r_2$ then $a$ is positive.
N. Alberding, 2013