Answer
a) $\omega=\frac{2\omega_0}{7}$
b) $t=\frac{2R\omega_0}{\mu_k g}$
Work Step by Step
We first find torque:
$\tau = R\mu_k mg$
Thus, we find alpha:
$\alpha=\frac{\tau}{I}=\frac{R\mu_kmg}{\frac{2}{5}mR^2}=\frac{5mu_kg}{2R}$
Thus, because $v=\mu_kgt$, it follows:
$v=\omega_0R+\alpha t$
$\mu_kgt=\omega_0R+\frac{5mu_kg}{2R} t$
$t=\frac{2R\omega_0}{\mu_k g}$
We now find the angular velocity. Using $v=\omega_0R+\alpha t$ and plugging in the values that we have solved for, we obtain:
$\omega=\frac{2\omega_0}{7}$
