Rolling Down Rolling Down Rolling Down

First thing to observe is that the maximum friction force is same as the force of gravity along the incline. Hence, as soon as the cylinder is dropped, it will remain stationary and slip until it stops spinning. Then, it will pure roll down to the bottom of the incline.

First, we find time taken in stopping rotation $t_1$ . Friction ( $F_r$ ) is producing the torque. Hence:

$F_rR=I\alpha$ (alpha is angular deceleration, $I$ is moment of inertia)

$\Rightarrow5mR=mR^2\alpha/2$

$\alpha=10/R=2 \Rightarrow t_0=w_0/\alpha=1$

Now we find time taken to pure roll to the bottom $t_1$ . Here, friction acting will not be maximum, and will be some value which allows pure rolling. It's easy make equations:

$mgsin(\theta)-F_r=ma$

$F_rR=I\alpha$

$R\alpha=a$

Solving the above, we get $a=10/3$

Since the cylinder is dropped at height $3R$ , it has to travel a distance $3R/sin(\theta)=6R$ along the incline to reach the bottom. We use equation of motion to get time.

$6R=a(t_1)^2/2$

$\Rightarrow t_1=18^{0.5}$

The final answer is: $t_0+t_1=1+18^{0.5}=\boxed{5.242}$