A thin square slab with sides of length 0.550 m and mass 1.80 kg is suspended so it can freely rotate about a horizontal axis of rotation perpendicular to the surface at one corner. The square is held so its top and bottom edges are horizontal, and then is released. What is the angular velocity of the square when the point opposite the pivot is as low as possible?
KE = ½Iω^2
IP = ICM + Mh^2 (parallel axis theorem)
The Attempt at a Solution
I need help setting it up. I feel like I should be determining the angle to find out how far the sheet moved..am I on the right track?