Where do you like to go? Up or Down

Write the equations of motion.

Up the incline force of fricion = $\mu_{k} mg \cos \theta$ acting at contact point.

Down the incline force is $mg \sin \theta$

$a = \mu_{k} g \cos \theta - g \sin \theta$

$\Rightarrow v = (\mu_{k} g \cos \theta - g \sin \theta) t$

Angular acceleration $\alpha = \frac{\mu_{k} Mg \cos \theta R}{\frac{MR^2}{2}}$

Hence , $\omega R = \omega_{0} R - \alpha t R = \omega_{0} R - 2 \mu_{k} g \cos \theta t$ ,

Now, velocity will be maximum when rolling without slipping begins, as after that the friction would not be $\mu_{k}M g \cos \theta$ , and will be less than $mg \sin \theta$ , as the angular velocity will decrease, so friction will act such that velocity also decreases and is always $R \omega$ .

At this time, $v = \omega R$ ,

Solving, $t_{*} = \frac{\omega_{0} R}{3 \mu_{k} g\cos \theta - g\sin \theta}$

Hence, $v_{max} = a t_{*} = \boxed{\frac{\mu_{k} - \tan \theta}{3 \mu_{k} - \tan \theta} \omega_{0} R}$

Substitute values to get $v_{max} = 6 cm s^{-1}$