A disc with a flat small bottom beaker placed on it at a distance R...

As the disc is revolving in circular motion about an axis passing through centre and perpendicular to its plane, thus centrifugal force must act on it. The forces acting on disc is shown below.

From FBD

For circular motion, frictional force must balance the required centripetal force.

So, $f = m{\omega}^{2}R$

$⇒ \muN \ge m{\omega}^{2}R$

$⇒ \mumg \ge m{\omega}^{2}R$

$⇒ R \le \dfrac{\mug}{{\omega}^{2}}$