An annular ring with inner and outer radii $R_{1}and{R}_2$ is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring, $\frac{F_1}{F_2}$ is

Let m is the mass of each particle and $\omega$ is the angular speed of the annular ring.

$a_1=\frac{v_1^2}{R_1}=\frac{{\omega }^2{R}_1^2}{R_1}{\omega }^2{R}_1, a_1=\frac{v_2^2}{R_2}={\omega }^2{R}_2$

Taking particle masses equal $\frac{F_1}{F_2}=\frac{m{a}_1}{m{a}_2}=\frac{m{R}_1{\omega }^2}{m{R}_2{\omega }^2}=\frac{R_1}{R_2}$

NOTE:
The force experienced by any particle is only along radial direction.
Force experienced by the particle, $F=m{\omega }^{2}R$

$\therefore \frac{F_1}{F_2}=\frac{R_1}{R_2}$