Answer
$\mu = 0.482$
Work Step by Step
We can find the rate of the angular deceleration.
$\alpha = \frac{0-\omega_0}{t}$
$\alpha = \frac{-(850~rev/min)(1~min/60~s)(2\pi~rad/rev)}{7.50~s}$
$\alpha = -11.87~rad/s^2$
We can use the magnitude of angular deceleration to find the coefficient of friction.
$\tau = I\alpha$
$F_f~R = \frac{1}{2}MR^2\alpha$
$F_N~\mu~R = \frac{1}{2}MR^2\alpha$
$\mu = \frac{MR~\alpha}{2~F_N}$
$\mu = \frac{(50.0~kg)(0.260~m)(11.87~rad/s^2)}{(2)(160~N)}$
$\mu = 0.482$
