Chapter 11 - Rolling, Torque, and Angular Momentum - Problems - Page 321: 15c

$\alpha = -47~rad/s^2$

We can write an expression for the rotational inertia of a solid sphere: $I = \frac{2}{5}MR^2$ We can find the angular acceleration: $f_k~R = I~\alpha$ $-Mg~\mu_k~R = (\frac{2}{5}MR^2)~\alpha$ $\alpha = \frac{-5~g~\mu_k}{2~R}$ $\alpha = \frac{-(5)(9.8~m/s^2)(0.21)}{(2)(0.11~m)}$ $\alpha = -47~rad/s^2$