A thin rod of length = 2.00m and negligible mass can pivot about one end to rotate in a vertical circle. A ball of mass m = 5.00 kg is attached to the other end. The rod is pulled aside to angle [tex]\theta[/tex]= 30 degrees and released with initial velocity v = 0 m/s. As the ball descends to its lowest point, (a) How much work does the gravitational force do on it? (b) What is the change in the gravitational potential energy of the ball-Earth system? (c) If the gravitational potential energy is taken to be zero at the lowest point, what is its value just as the ball is released? (d) Doe the magnitude of the answers to (a) through (c) increase, decrease, or remain the same if the angle is increased? Relevant Equations: W (gravity) = mgh W (gravity) = -[tex]\Delta[/tex]U So, theoretically, I know what I need to do, but when it comes to plugging in the numbers, I get the incorrect answer. I know that I'm given m, and g is the gravitational constant. What I'm having trouble doing is manipulating the arc of the circle to solve for h, and the h that I keep finding (roughly 0.27 m) gives me the incorrect answer for (a), which should be 20 J. If I'm struggling with part (a), I assume I will have similar issues with the rest of it, so any help would be greatly appreciated.