The horizontally located disk rotates around the vertical axis, making 25 rpm.

The horizontally located disk rotates around the vertical axis, making 25 rpm. At what distance from the axis of rotation of the disc can be kept the body on it if the friction coefficient is 0.2?

The friction force informs the centripetal acceleration of the body. According to Newton’s second law:
F = M * A,
Here
F = μ * m * g – friction force,
μ = 0.2 – friction coefficient,
g = 10m / s ^ 2 – acceleration of free fall,
m – body weight,
a = ω ^ 2 * R – the centripetal acceleration of the body,
ω = 2 * π * F,
F = 25AB / min = (5/12) R / C – rotational speed,
R – the desired distance.
μ * M * G = M * A;
a = μ * g;
ω ^ 2 * r = μ * g;
(2 * π * f) ^ 2 * r = μ * g;
R = (μ * G) / (2 * π * f) ^ 2 = 0.292m = 29.2cm.