1. The problem statement, all variables and given/known data The diagram shows a thin rod of uniform mass distribution pivoted about one end by a pin passing through that point. The mass of the rod is 0.380 kg and its length is 2.50 m. When the rod is released from its horizontal position, it swings down to the vertical position as shown. (a) Determine the speed of its center of gravity at its lowest position. (b) When the rod reaches the vertical position, calculate the tangential speed of the free end of the rod. 2. Relevant equations Moment of Inertia for a solid rod: 1/3MR^2 Tangential Acceleration = r * (dω/dt) 3. The attempt at a solution I really doubt this is right, but this is what I have been trying. For part A, this is what I did. To get the center of mass, I tried 2.5/2 for 1.25. For the speed at that point, I tried using mgh for potential/kinetic energy and got (0.38g)(9.81m/s^2)(1.25). This gave me 4.66. Then I square rooted 4.66 over 0.38 (the mass), and got 4.95. My answer was incorrect For part B, I assumed that it was the answer for part A multiplied by two, since A was asking for the center, rather than the full rod. I'll need to figure out part A before part B.