When will you go away?

Set up a coordinate system with the origin at the meeting point of the wall and the floor. The endpoints $A$ and $B$ of the rod have coordinates $(0,\ell\sin\theta)$ and $(\ell\cos\theta,0)$ respectively, and hence have velocities $\mathbf{v}_A \; = \; \left(\begin{array}{c} 0 \\ \ell\cos\theta \dot\theta \end{array} \right) \qquad \mathbf{v}_B \; = \; \left(\begin{array}{c} -\ell\sin\theta \dot\theta \\ 0 \end{array} \right)$ respectively. Thus the rod has kinetic energy $T \; = \; \tfrac16m\big[|\mathbf{v}_A|^2 + |\mathbf{v}_B|^2 + \mathbf{v}_A \cdot\mathbf{v}_B\big] \; = \; \tfrac16m\ell^2 \dot\theta^2$ On the other hand, the potential energy of the rod is $V \; = \; \tfrac12mg\ell\sin\theta$ Apart from gravity, the only forces acting are the reaction forces $X$ and $Y$ which, since the surfaces are smooth, act horizontally and vertically respectively. Consequently energy is conserved, so we have $T + V \,=\, \tfrac12mg\ell\sin\theta_0$ , where $\theta_0 = 75^\circ$ is the initial angle, and hence $\dot\theta^2 \; = \; \tfrac{3g}{\ell}(\sin\theta_0 - \sin\theta)$ which, on differentiating, gives $\ddot{\theta} \; = \; -\tfrac{3g}{2\ell}\cos\theta$ The centre of mass $G$ of the rod has coordinates $(x,y) = \big(\tfrac12\ell\cos\theta,\tfrac12\ell\sin\theta\big)$ . Resolving forces horizontally, we see that $X \; = \; m\ddot{x} \; = \; m\big[-\tfrac12\ell\sin\theta \ddot\theta - \tfrac12\ell\cos\theta \dot\theta^2\big] \; = \; \tfrac34mg\cos\theta\big[3\sin\theta - 2\sin\theta_0\big]$ Thus the rod loses contact with the wall when $\sin\theta = \tfrac23\sin\theta_0$ , or when $\theta = \boxed{40.08703691^\circ}$ .

Note that similar calculations show that $Y \; = \; \tfrac14mg + \tfrac34mg\sin\theta(3\sin\theta - 2\sin\theta_0)$ so that $Y > 0$ throughout the motion.