A block of mass m is kept on a horizontal ruler. The friction...

The block moving in a circular path experiences two components of acceleration:

Centripetal acceleration towards the center, $a_c = \omega^2 L$

Tangential acceleration along the circular path, $a_t = \alpha L$

Since these two accelerations are perpendicular to each other, the net acceleration of the block is:

$a = \sqrt{a_c^2 + a_t^2} = \sqrt{(\omega^2 L)^2 + (\alpha L)^2} = L \sqrt{\omega^4 + \alpha^2}$

The net force required to keep the block moving with the ruler is provided by the static friction between the block and the ruler. The maximum static friction available is f_\max = \mu mg.

The block will just begin to slip when the required net force equals the maximum static friction:

ma = f_\max

$mL \sqrt{\omega^4 + \alpha^2} = \mu mg$

$\sqrt{\omega^4 + \alpha^2} = \dfrac{\mu g}{L}$

Squaring both sides:

$\omega^4 + \alpha^2 = \left(\dfrac{\mu g}{L}\right)^2$

$\omega^4 = \left(\dfrac{\mu g}{L}\right)^2 - \alpha^2$

$\omega = \left[\left(\dfrac{\mu g}{L}\right)^2 - \alpha^2\right]^{1/4}$