The first step in solving this problem is to determine the torque acting on the cylinder due to the applied force. The torque is given by:
τ = r × F × sin(θ)
where r is the distance from the point of rotation to the point where the force is applied (0.17 m in this case), F is the magnitude of the force (4.5 N), θ is the angle between the force vector and the lever arm (30 degrees above the horizontal), and the "×" symbol denotes the cross product.
Substituting the given values, we get:
τ = (0.17 m) × (4.5 N) × sin(30°)
τ = 0.147 N·m
Next, we can use the torque to find the angular acceleration of the cylinder using the following equation:
τ = Iα
where I is the moment of inertia of the cylinder and α is the angular acceleration. The moment of inertia of a solid cylinder rotating about its central axis is:
I = (1/2)MR^2
where M is the mass of the cylinder and R is its radius. Substituting the given values, we get:
I = (1/2)(20 kg)(0.4 m)^2
I = 1.6 kg·m^2
Substituting the moment of inertia and the torque into the equation above, we get:
τ = Iα
0.147 N·m = (1.6 kg·m^2)α
α = 0.092 rad/s^2
Finally, we can use the following kinematic equation to find the angular velocity of the cylinder after 5 seconds:
ω = ω0 + αt
where ω0 is the initial angular velocity (0 because the cylinder is released from rest), α is the angular acceleration we just calculated, and t is the time interval (5 seconds in this case). Substituting the given values, we get:
ω = 0 + (0.092 rad/s^2)(5 s)
ω = 0.46 rad/s
Therefore, the angular velocity of the cylinder after 5 seconds is 0.46 rad/s.