Solution

The phrase "placed on" in this context (and the fact it falls off vertically at high speeds) implies the chain is stretched tightly around the outer cylindrical edge (rim) of the disk.

Consider a small segment of the chain corresponding to a small angle $2\alpha$. Its length is $dl = 2\alpha R$, and its mass is $dm = m \frac{\alpha}{\pi}$.
The forces acting on this segment in the horizontal plane are the tension forces $T$ (directed tangentially) and the normal reaction force $N$ from the edge of the disk (directed radially outward).
The inward radial component of the tension forces is $2T \sin\alpha \approx 2T\alpha$.

According to Newton's second law in the radial direction, the net inward force provides the centripetal acceleration $a_c = \omega^2 R$:
$$2T\alpha - N = dm \cdot a_c = m \frac{\alpha}{\pi} \omega^2 R$$
From this, the normal force exerted by the disk on the chain is:
$$N = 2T\alpha - m \frac{\alpha}{\pi} \omega^2 R$$

The chain starts to fall off (slip downwards) when the maximum static friction force can no longer balance the gravity acting on the segment.
Vertical equilibrium condition:
$$F_{\text{fr}} = dm \cdot g \implies \mu N = m \frac{\alpha}{\pi} g$$

Substitute the expression for $N$:
$$\mu \left( 2T\alpha - m \frac{\alpha}{\pi} \omega^2 R \right) = m \frac{\alpha}{\pi} g$$

Divide the entire equation by $\alpha/\pi$:
$$\mu (2\pi T - m\omega^2 R) = mg$$

From this, we find the coefficient of friction $\mu$:
$$\mu = \frac{mg}{2\pi T - m\omega^2 R}$$