rc/kc) rG/kG In conclusion, we have that for a given solid, the shortest equivalent length, and hence the fastest pendulum occurs when it is suspended from a point which is at a distance rG= kG from the center of mass. In this ase, the minimum equivalent length is 2kG Example Minimum period pendulum Consider a uniform rod of length L. In this case, IG m12/12, and kG=1/v12 a0.291.Therefore, the fastest pendulum is obtained when the bar is suspended from a distance 0. 29l away from the center of mass. We note that if the suspension point is moved slightly, the period of the pendulum will increase. However moving the suspension point will make practically no difference in the frequency, because the tangent at point A(see graph) is horizontal. This fact has been used in the construction of extremely accurate pendulums O k 0.29l0 1 2 3 4 0 1 2 3 4 In conclusion, we have that for a given solid, the shortest equivalent length, and hence the fastest pendulum, occurs when it is suspended from a point which is at a distance rG = kG from the center of mass. In this case, the minimum equivalent length is (Lequiv)min = 2kG . Example Minimum period pendulum Consider a uniform rod of length l. In this case, IG = ml2/12, and kG = l/√ 12 ≈ 0.29l. Therefore, the fastest pendulum is obtained when the bar is suspended from a distance 0.29l away from the center of mass. We note that if the suspension point is moved slightly, the period of the pendulum will increase. However, moving the suspension point will make practically no difference in the frequency, because the tangent at point A (see graph) is horizontal. This fact has been used in the construction of extremely accurate pendulums for clocks. 4