The FBD and inertia diagram is
Where \(M=m_{disk}+m_{bar}\) and \(H\) is location of system center of mass. Total mass is \(M=15+10=25\) kg.\begin{align*} H & =\frac{m_{sphere}\left ( L+R\right ) +m_{rod}\left ( \frac{L}{2}\right ) }{m_{sphere}+m_{rod}}\\ & =\frac{15\left ( 0.6+0.1\right ) +10\left ( 0.3\right ) }{15+10}\\ & =0.54\text{ m} \end{align*}
And \begin{align*} I_{o} & =I_{sphere_{0}}+I_{bar_{0}}\\ & =\left ( \frac{2}{5}m_{sphere}R^{2}+m_{sphere}\left ( L+R\right ) ^{2}\right ) +\frac{1}{3}m_{bar}L^{2}\\ & =\left ( \frac{2\left ( 15\right ) \left ( 0.1\right ) ^{2}}{5}+15\left ( 0.6+0.1\right ) ^{2}\right ) +\frac{1}{3}\left ( 10\right ) \left ( 0.6\right ) ^{2}\\ & =8.61\text{ kg-m}^{2} \end{align*}
From FBD we obtain 3 equations.\begin{align*} F_{x} & =Ma_{x}\\ F_{y}-Mg & =Ma_{y}\\ -\tau -\left ( Mg\cos \theta \right ) H & =I_{o}\alpha \end{align*}
Or\begin{align*} F_{x} & =25a_{x}\\ F_{y}-\left ( 25\right ) \left ( 9.81\right ) & =25a_{y}\\ -50-\left ( \left ( 25\right ) \left ( 9.81\right ) \cos \left ( 45\left ( \frac{\pi }{180}\right ) \right ) \right ) \left ( 0.54\right ) & =\left ( 8.61\right ) \alpha \end{align*}
Or\begin{align} F_{x} & =25a_{x}\tag{1}\\ F_{y}-245.25 & =25a_{y}\tag{2}\\ -16.684 & =\alpha \tag{3} \end{align}
\(3\) equations with \(4\) unknowns: \(F_{x},F_{y},a_{x},a_{y}\). But looking at this diagram, which relates \(a_{x},a_{y}\) to \(\alpha \).
We see that\begin{align*} a_{y} & =H\omega ^{2}\sin \theta +H\alpha \cos \theta \\ & =\left ( 0.54\right ) \left ( 3\right ) ^{2}\sin \left ( 45\left ( \frac{\pi }{180}\right ) \right ) +\left ( 0.54\right ) \left ( -16.684\right ) \cos \left ( 45\left ( \frac{\pi }{180}\right ) \right ) \\ & =-2.934\text{ m/s}^{2} \end{align*}
And\begin{align*} a_{x} & =H\alpha \sin \theta -H\omega ^{2}\cos \theta \\ & =\left ( 0.54\right ) \left ( -16.684\right ) \sin \left ( 45\left ( \frac{\pi }{180}\right ) \right ) -\left ( 0.54\right ) \left ( 3^{2}\right ) \cos \left ( 45\left ( \frac{\pi }{180}\right ) \right ) \\ & =-9.807\text{ m/s}^{2} \end{align*}
Using these in (1,2), we find the reaction forces \begin{align*} F_{x} & =25a_{x}\\ & =25\left ( -9.807\right ) \\ & =-245.175\text{ N} \end{align*}
And
\begin{align*} F_{y}-245.25 & =25a_{y}\\ F_{y}-245.25 & =25\left ( -2.934\right ) \\ F_{y} & =171.9\text{ N} \end{align*}
Total reaction force is \(\sqrt{F_{y}^{2}+F_{x}^{2}}=\sqrt{\left ( 171.9\right ) ^{2}+\left ( -245.175\right ) ^{2}}=299.4334\) N