Compound pendulum In the case of a compound pendulum, we can proceed in a similar manner. O Conservation of angular momentum about o gives lo mgr sin 8 Expressing Io in terms of the radius of gyration, ko we have sin 6=0 (5) We note that this equation is the same as equation 1 for the simple pendulum, if we identify the term g/Lin equation I with the term grG/k2 in equation 5. This leads to the definition of an equivalent length, lequiv k3忌+r Here we have used the fact that lo =IG+mrG, and therefore ko = kG+rG. Thus, we have that the motion of a compound pendulum is identical to that of a simple pendulum of equivalent length Lequiv, given by equation 6. Using the small amplitude approximation, the period of the compound pendulum will be One question we may want to ask is whether, for a given body(kG fixed), we can make the period (or Lequiv) rbitrarily small by choosing rG (or the hinge point O)appropriately. From equation 6, we can write which shows that, when we try to reduce Lequiv/kG by reducing rG, the term rG/kG is reduced, but the term 1/rG/kG)increases. This situation is shown in the graph, which also shows that the minimum value for Lequiv/kG is 2, a value which is attained for rG=kGCompound Pendulum In the case of a compound pendulum, we can proceed in a similar manner. Conservation of angular momentum about O gives, IO ¨θ = −mgrG sin θ . Expressing IO in terms of the radius of gyration, kO, IO = mk2 O we have, ¨θ + grG k 2 O sin θ = 0 . (5) We note that this equation is the same as equation 1 for the simple pendulum, if we identify the term g/L in equation 1 with the term grG/k2 O in equation 5. This leads to the definition of an equivalent length, Lequiv, as, Lequiv = k 2 O rG = k 2 G + r 2 G rG . (6) Here we have used the fact that IO = IG + mr2 G, and therefore k 2 O = k 2 G + r 2 G. Thus, we have that the motion of a compound pendulum is identical to that of a simple pendulum of equivalent length Lequiv, given by equation 6. Using the small amplitude approximation, the period of the compound pendulum will be T = 2π s Lequiv g . One question we may want to ask is whether, for a given body (kG fixed), we can make the period (or Lequiv) arbitrarily small by choosing rG (or the hinge point O) appropriately. From equation 6, we can write Lequiv kG = rG kG + 1 (rG/kG) , which shows that, when we try to reduce Lequiv/kG by reducing rG, the term rG/kG is reduced, but the term 1/(rG/kG) increases. This situation is shown in the graph, which also shows that the minimum value for Lequiv/kG is 2, a value which is attained for rG = kG. 3