1. Center of Mass (G) The combined Center of Mass (G) lies on the line connecting the center of the pipe \(O\) and the mass \(m\). Its distance from the geometric center \(O\) is: $$ r_{com} = \frac{m R}{M + m} $$

2. Condition for Equilibrium For the system to be stable (and not roll down), the torque due to gravity about the point of contact must be zero. This means the Center of Mass (G) must be vertically aligned with the contact point.

From the geometry of the vertical line and the normal to the incline: $$ \sin \theta = \frac{r_{com}}{R} = \frac{m}{M+m} $$

3. Friction Coefficient The force of friction prevents slipping. Using the trigonometric identity \( \tan \theta = \frac{\sin \theta}{\sqrt{1 – \sin^2 \theta}} \) for the condition of sliding without rolling: $$ \mu \leq \tan \theta = \frac{\frac{m}{M+m}}{\sqrt{1 – \left(\frac{m}{M+m}\right)^2}} $$ Simplifying the denominator: $$ \sqrt{1 – \frac{m^2}{(M+m)^2}} = \frac{\sqrt{(M+m)^2 – m^2}}{M+m} = \frac{\sqrt{M^2 + 2Mm}}{M+m} $$ Therefore: $$ \mu \leq \frac{\frac{m}{M+m}}{\frac{\sqrt{M(M+2m)}}{M+m}} $$