Slipping a Disk

The key aspect to this problem is realizing that we are dealing with non-uniform circular motion. There are both radial and tangential components to the acceleration. Write an expression for the tangential speed:

$v = r \omega = r \alpha \, t$

Radial force:

$F_r = \frac{m \, v^2}{r} = \frac{m \, ( r \alpha \, t)^2 }{r} = m \, r \alpha^2 \, t^2$

Tangential force (given that within the context of non-uniform circular motion, $\dot{r} = 0$ , radius remains constant)

$F_t = m \frac{dv}{dt} = m \, r \alpha$

Equate the force magnitude to the maximum possible static friction force, $\mu mg$ :

$\begin{aligned} \sqrt{F_r^2 + F_t^2 } &= \mu m g \\ F_r^2 + F_t^2 &= \mu^2 m^2 g^2 \\ m^2 \, r^2 \alpha^4 \, t^4 + m^2 \, r^2 \alpha^2 &= \mu^2 m^2 g^2 \\ r^2 \alpha^4 \, t^4 + r^2 \alpha^2 &= \mu^2 g^2 \\ t^4 &= \frac{\mu^2 g^2 - r^2 \alpha^2}{r^2 \alpha^4}\\ t &= \sqrt[4]{\frac{\mu^2 g^2 - r^2 \alpha^2}{r^2 \alpha^4}} \end{aligned}$

Solving for the critical time value yields $t \approx 0.58175$