In figure, a rod of mass m = 1kg is held vertically by an unknown mass M hanging over a pulley as shown. Assuming h = 50 cm and H = 100 cm, find the minimum value of M (in kg) that makes the equilibrium stable for small displacement.

consider the net torque on the rod corresponding to a small deviation, as shown in figure below. 

The rod will return to the vertical if the torque due to the tension is greater than the torque due to gravity. In the “critical” case, these torques must be equal:

 $\Rightarrow \left(\mathrm{Mg} \mathrm{sin\beta }\right) (h \cos \alpha )+\left(\mathrm{Mg} \mathrm{cos\beta }\right)\left(h \cos \alpha \right) =\mathrm{mg}\frac{\mathrm{h}}{2}\sin \mathrm{\alpha }$  

since, both $\mathrm{\alpha }$ and $\mathrm{\beta }$ are small,

$\Rightarrow Mgh(\alpha +\beta )=mg\frac{h}{2}\alpha$

$\Rightarrow \frac{M}{m}=\frac{\alpha }{2(\alpha +\beta )}$ ....(1)

 From the law of sines, $\frac{\sin \mathrm{\beta }}{\mathrm{h}}=\frac{\sin [\mathrm{\pi }-\left(\mathrm{\alpha } + \mathrm{\beta }\right)]}{\mathrm{H}}$

$\frac{\sin \mathrm{\beta }}{\mathrm{h}}=\frac{\sin \left(\mathrm{\alpha } + \mathrm{\beta }\right)}{\mathrm{H}}$ 

since, both $\mathrm{\alpha }$ and $\mathrm{\beta }$ are small,

 $\Rightarrow \frac{\mathrm{\beta }}{\mathrm{\alpha }}=\frac{h}{\mathrm{H}-\mathrm{h}}$ 

$\Rightarrow \frac{\alpha }{(\alpha +\beta )}=\frac{H-h}{H}$ .....(2) 

$\Rightarrow \mathrm{M}=\frac{\mathrm{m}\left(\mathrm{H}-\mathrm{h}\right)}{2\mathrm{H}}=\frac{1\left(100-50\right)}{2\times 100}=0.25 kg$