(a) Let us choose the positive direction of the rotation angle $\varphi$, such that $w_{c x}$ and $\beta_{z}$ have identical signs (Fig.). Equation of motion, $F_{x}=m w_{c x}$ and $N_{c z}=I_{c} \beta_{z}$ gives :
$F \cos \alpha-f r=m w_{c x}: f r R-F r=I_{c} \beta_{z}=\gamma m R^{2} \beta_{z}$
In the absence of the slipping of the spool $w_{c x}=\beta_{z} R$ From the three equations $w_{c x}=w_{c}=\frac{F[\cos \alpha-(r / R)]}{m(1+\gamma)}$
(b) As static friction $(f r)$ does not work on the spool, from the equation of the increment of mechanical energy $A_{\text {ex }}=\Delta T$. $A_{e x t}=\frac{1}{2} m v_{c}^{2}+\frac{1}{2} \gamma m R^{2} \frac{v_{c}^{2}}{R^{2}}=\frac{1}{2} m(1+\gamma) v_{c}^{2}$
$=\frac{1}{2} m(1+\gamma) 2 w_{c} x=\frac{1}{2} m(1+\gamma) 2 w_{c}\left(\frac{1}{2} w_{c} t^{2}\right)$
Note |that at $\cos \alpha=r / R$, there is no rolling and for $\cos \alpha<r / R, w_{c x}<0$, i.e. the spool will move towards negative $x$ -axis and rotate in anticlockwise sense.