Solution

Newton's second law for a rotational movement

$$M = I\varepsilon$$

$$m_1gR - m_2 g R = I\varepsilon\tag{1}$$

Since there is no slippage:

$$\varepsilon = \frac{a}{R}$$

Substituting into $(1)$

$$\boxed{(m_1 - m_2) g R^2 = Ia}\tag{2}$$

Conservation of angular momentum

$$I\omega = m_1vR + m_2vR\tag{3}$$

Let's express velocity through angular velocity

$$v=\omega R\tag{4}$$

After substituting $(4)$ into $(3)$

$$\boxed{I = (m_1+m_2)R^2} \tag{5}$$

Let's equate the expressions $(2)$ and $(5)$

$$(m_1 - m_2) g R^2 = (m_1+m_2)aR^2$$

After mathematical transformations

$$a = g \frac{m_1 - m_2}{m_1 + m_2}\tag{6}$$

Describe the forces acting on the vertical axis

$$N = (m_1+m_2)g - a (m_1-m_2)\tag{7}$$

Substituting the acceleration $(6)$ into expression $(7)$

$$N = g\cdot\left((m_1+m_2)- \frac{(m_1-m_2)^2}{m_1+m_2}\right)$$

From here we find the support reaction force:

$$\boxed{N = \frac{4m_1m_2g}{m_1+m_2}}$$