Answer
The minimum coefficient of static friction to prevent slippage is $~~0.075$
Work Step by Step
In part (a), we found that the centripetal acceleration is $~~a_c = 0.73~m/s^2$
Note that the frictional force provides the centripetal force.
We can find the minimum coefficient of static friction to prevent slippage:
$F_f = m~a_c$
$mg~\mu_s = m~a_c$
$\mu_s = \frac{a_c}{g}$
$\mu_s = \frac{0.73~m/s^2}{9.8~m/s^2}$
$\mu_s = 0.075$
The minimum coefficient of static friction to prevent slippage is $~~0.075$
