To find the minimum coefficient of static friction $\mu$ to prevent the object from sliding down, we analyze the forces acting on the object in a stationary reference frame. The object is in uniform circular motion, so the net force must provide the required centripetal acceleration, $a_c = \omega^2 r$, directed horizontally towards the axis of rotation.

The forces acting on the object are:

1. Gravitational force, $F_g = mg$, acting vertically downward.
2. Normal force, $N$, perpendicular to the conical surface.
3. Static friction force, $f_s$, parallel to the surface. To prevent the object from sliding down, this force must be directed up the incline. For the minimum coefficient, the friction is at its maximum value, $f_s = \mu N$.

We apply Newton's second law by resolving forces into horizontal and vertical components.

Vertical forces (no acceleration): The sum of vertical forces is zero. The upward components of the normal force and friction balance the downward gravitational force.

$$ \sum F_y = N \cos\theta + f_s \sin\theta - mg = 0 $$

Substituting $f_s = \mu N$:

$$ N(\cos\theta + \mu \sin\theta) = mg \quad (1) $$

Horizontal forces (centripetal acceleration): The net horizontal force provides the centripetal acceleration, $a_c = \omega^2 r$. The horizontal component of the normal force points towards the center, while the horizontal component of friction points away.

$$ \sum F_x = N \sin\theta - f_s \cos\theta = m a_c = m \omega^2 r $$

Substituting $f_s = \mu N$:

$$ N(\sin\theta - \mu \cos\theta) = m \omega^2 r \quad (2) $$

Now, we solve for $\mu$ by dividing equation (2) by equation (1) to eliminate $N$:

$$ \frac{N(\sin\theta - \mu \cos\theta)}{N(\cos\theta + \mu \sin\theta)} = \frac{m \omega^2 r}{mg} $$ $$ \frac{\sin\theta - \mu \cos\theta}{\cos\theta + \mu \sin\theta} = \frac{\omega^2 r}{g} $$

To isolate $\mu$, we cross-multiply:

$$ g(\sin\theta - \mu \cos\theta) = \omega^2 r (\cos\theta + \mu \sin\theta) $$ $$ g\sin\theta - g\mu\cos\theta = \omega^2 r \cos\theta + \omega^2 r \mu\sin\theta $$

Group the terms containing $\mu$:

$$ g\sin\theta - \omega^2 r \cos\theta = \mu (g\cos\theta + \omega^2 r \sin\theta) $$

Solving for $\mu$:

$$ \mu = \frac{g\sin\theta - \omega^2 r \cos\theta}{g\cos\theta + \omega^2 r \sin\theta} $$

This expression is valid provided that the numerator is non-negative, i.e., $g\sin\theta \ge \omega^2 r \cos\theta$. If the numerator is negative, the object has a tendency to slide up, and no friction is needed to prevent it from sliding down ($\mu=0$).