## College Physics (4th Edition)

As the bob passes the lowest point, the tension in the cord is $2mg$
We can use conservation of energy to find the speed at the lowest point: $KE = U_g$ $\frac{1}{2}mv^2 = mg~(\frac{L}{2})$ $v = \sqrt{gL}$ The net force at the bottom provides the centripetal force to keep the bob moving in a circular path. We can find the tension $F_T$ in the cord: $\sum F = \frac{mv^2}{L}$ $F_T-mg = \frac{mv^2}{L}$ $F_T = \frac{m(\sqrt{gL})^2}{L}+mg$ $F_T = \frac{mgL}{L}+mg$ $F_T = 2mg$ As the bob passes the lowest point, the tension in the cord is $2mg$