ok sorry for not being specific.
this question is the second part of the firsst question which is this one.
A solid cylinder is pivoted about a frictionless axle as shown. A rope wrapped around the outer radius of 2 m exerts a downward force of 3N. A rope wrapped around the inner radius of 0.7 m exerts a force of 8 N to the right. The moment of inertia of the cylinder is 8 kg m^2. Find the angular acceleration.
I found the solution to this one by t=+-rFsin(theta) here is the specifics
https://www.physicsforums.com/showthread.php?t=583881
However, my solution to the first part of the problem was the sum of all torque forces Sumt=I(alpha)
-Rfsin(theta)+rFsin(theta)=I(alpha) I had my Moment of inertia giving my the problem so all i had to do was to solve for (alpha) which gave me 0.05 rads/s.
The second part of the problem says,
If the cylinder in problem 10 is initially at rest, how long will it take for the cylinder to turn one revolution??