Recent content by BMZ

Well, if Potential Energy fits the problem by using the 1m change in height in mgh, then what would come to mind is Potential Energy = Linear KE + Angular KE which would be mgh = ½mv2 + ½Iω2 Then I = 0.5mr2 since it's a solid cylinder. equation would then be mgh = ½mv2 + ½(½mr2)ω2 You...

Well, I think the KE would come from the rotation of the cylinder as it unwinds from the string. But at the beginning it doesn't have any if it's not moving right? Then maybe the left side would be zero since the rotational kinetic energy is zero at the beginning, then that would be a solvable...

Homework Statement The problem states the following: One end of a string is attached to the ceiling while the other end is wrapped around a solid cylinder of mass 0.20 kg and radius 0.030m. The cylinder is released from rest and the string unwinds as the cylinder rotates and accelerates...