There's Something Odd About This Yo-Yo

Let $T$ and $f$ represent the Tension in the string (at an angle $\theta$ to the vertical, tilted towards the right) and the frictional force (directed towards left).

Assuming the conditions of pure rolling , $a = R\alpha$

Using Newton's 2nd Law of Motion,

$F\sin\theta-f = Ma$

Writing the Torque equation,

$fR-Fr = I\alpha$

The problem clearly suggests to use $\alpha=0$ and thus $a=0$ .

Simplifying the two equations,

$F\sin\theta=f$
$fR=Fr$

Therefore,

$\sin\theta = \frac{r}{R}$

Substituting the values of $r$ and $R$ ,

$\sin\theta = 0.5$

$\Rightarrow \boxed{\theta = 30^0}$

Firstly , draw the free body diagram of the yoyo. Let us suppose at an angle $\theta$ with respect to vertical The yoyo does not rolls. We are pulling it slowly which indicates that its acceleration is zero.

We have then $T\sin\theta = f$

And net torque is also zero , i.e., $Tr = fR$ Substituting $T$ in first equation . You get $\sin\theta = \frac{r}{R} = \frac{1}{2}$ Hence, $\theta = 30 degrees$

The explanation involves a consideration of the forces and torques on the spool . It is easiest to consider the case where the string is pulled with a force and at an angle such that the spool is just on the verge of slipping without rolling. There are four forces: the weight (mg), the upward normal force of the table (N), the tension in the string (T) and the friction force (µN). If the spool is not yet moving, the net horizontal force is zero, or T cos (theta) = µN. Only two of the forces produce a torque about the center of the spool (T and µN), and these torques must be equal and opposite if the spool is to slip rather than rotate. Equating the torques gives r1T = r2µN. Dividing this equation into the previous one gives cos (theta) = r1/r2. Thus the critical angle that determines which way the spool will rotate depends only of the ratio of the two radii and is independent of the mass of the spool, the tension in the string and the coefficient of friction
so we get angle with horizontal ,cosˉ¹(0.5)=60°
now to get angle with vertical we substract 60 from 90 and we get the answer 30°

0.5sinx = 0.25sin90