Answer
The minimum coefficient of static friction to prevent slippage is $~~0.11$
Work Step by Step
In part (a), we found that the centripetal acceleration is $~~a_c = 0.73~m/s^2$
We can express the angular speed in units of $rad/s$:
$\omega = (33\frac{1}{3}~rev/min)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s}) = 3.49~rad/s$
We can find the angular acceleration:
$\alpha = \frac{\omega_f-\omega_0}{t}$
$\alpha = \frac{3.49~rad/s-0}{0.25~s}$
$\alpha = 13.96~rad/s^2$
We can find the tangential acceleration:
$a_t = \alpha~r$
$a_t = (13.96~rad/s^2)(0.060~m)$
$a_t = 0.8376~m/s^2$
We can find the net acceleration:
$a = \sqrt{a_t^2+a_c^2}$
$a = \sqrt{(0.8376~m/s^2)^2+(0.73~m/s^2)^2}$
$a = 1.11~m/s^2$
We can find the minimum coefficient of static friction to prevent slippage:
$F_f = m~a$
$mg~\mu_s = m~a$
$\mu_s = \frac{a}{g}$
$\mu_s = \frac{1.11~m/s^2}{9.8~m/s^2}$
$\mu_s = 0.11$
The minimum coefficient of static friction to prevent slippage is $~~0.11$
