Variable Friction on horizontal Rod !!

The torque as a function of P is

$\tau(P)=\frac{Mg\mu}{L}(\int_0^P xdx+\int_0^{L-P}xdx)=\frac{1}{2}Mg\mu(L-2P+2\frac{P^2}{L})$

The rotational inertia as a function of P is

$I(P)=\frac{M}{L}\int_{-P}^{L-P}x^2dx=\frac{M}{3}(L^2+3P^2-3LP)$

The angular acceleration as a function of P is

$\alpha(P)=\frac{\tau(P)}{I(P)}=\frac{3g\mu(L-2P+2P^2/L)}{2(L^2+3P^2-3LP)}$

Differentiating the acceleration with respect to P and setting it equal to zero (to find a minimum or maximum) yields $P=\frac{L}{2}$ which implies that the other (either min or max) is at an endpoint. Plugging in the values you find:

$\alpha_{max}=1.5$ ..... which means ..... $T_{min}=\frac{\omega}{1.5}$

$\alpha_{min}=0.75$ ..... which means ..... $T_{max}=\frac{\omega}{0.75}$

And thus $\frac{T_{min}}{T_{max}}=0.5$