A uniform rod of length \(L\) is free to pivot around an axis through its upper end. If it is released from rest when horizontal, at what speed is the lower end moving at its lowest point? [Hint: The gravitational potential energy change is determined by the change in height of the center of gravity.]

For a uniform rod, the center of gravity is located at the midpoint of the rod. When the rod moves from a horizontal position to a vertical position, this midpoint changes in height by a distance equal to half of the rod's length, L/2. Let the mass of the uniform rod be m. The change in gravitational potential energy is given by: $$\Delta U_g = mgh$$ where \(g\) is the acceleration due to gravity, and \(h\) is the change in height. In this case, \(h = \frac{L}{2}\). So, the change in potential energy is: $$\Delta U_g = mg\frac{L}{2}$$

Conservation of energy states that: $$ \Delta U_g = \Delta K $$ where \(\Delta K\) is the change in kinetic energy of the rod. Since the rod is released from rest, its initial kinetic energy is zero. Therefore, the final kinetic energy of the rod is equal to the change in gravitational potential energy: $$\Delta K = mg\frac{L}{2}$$ Now, let's focus on the lower end of the rod when it is in the vertical position. We can calculate the velocity using the relation between kinetic energy and the linear velocity of the lower end: $$\Delta K = \frac{1}{2}m_{lower}v_{lower}^2$$ But, since the whole rod has the mass \(m\), at the lower end we have the same mass: $$\frac{1}{2}mv^2 = mg\frac{L}{2}$$ Solving for the velocity \(v_{lower}\): $$v_{lower}^2 = 2g\frac{L}{2}$$ $$v_{lower} = \sqrt{gL}$$ Thus, the velocity of the lower end of the rod at its lowest point is given by: $$v_{lower} = \sqrt{gL}$$