1. The problem statement, all variables and given/known data Consider a light rod of negligible mass and length "L" pivoted on a frictionless horizontal bearing at a point "O." Attached to the end of the rod is a mass "m." Also, a second mass "M" of equal size (i.e., m=M) is attached to the rod (0.2L from the lower end). What is the period of this pendulum in the small angle approximation? 2. Relevant equations T=2pi(I/Hgm)^0.5 where H=the length from the center of mass to the point of rotation rcm=(r1m+r2M)/(m+M) Icm=m(L1)^2+M(L2)^2 I=Icm+m(d)^2 3. The attempt at a solution rcm=(r1m+r2M)/(m+M) rcm=[(L+0.8L)m)]/2m rcm=0.9L (from the point of rotation) Icm=mL^2+mL^2 Icm=m[(0.1L)^2+(-0.1L)^2] Icm=0.02mL^2 I=Icm+md^2 I=0.02mL^2+m(0.9L)^2 I=0.83mL^2 T=2pi(I/Hgm)^0.5 T=2pi(0.83mL2/0.9Lgm)^0.5 T=2pi(83L/90g)^0.5 I'm really unsure as to what I'm doing wrong so it'd be great if someone could point out what I'm doing wrong--perhaps I've mistaken what one of the variable is supposed to represent? Thanks! P.S. if it helps, the answer is 2pi(41L/45g)^0.5, I'd just love to know how to get to that. OK so another update. I just realized I'm one digit off--because the answer is 82/90, not 83/90. I think the error must be with finding the Icm, because everything would simplify properly if Icm=0.01 instead of 0.02. I'm just confused as to why Icm would only be with regards to one mass instead of both... FIGURED IT OUT: No parallel axis theorem and now the period equation makes sense because we can put in 2m instead of m--> Solution: rcm=(r1m+r2M)/(m+M) rcm=[(L+0.8L)m)]/2m rcm=0.9L (from the point of rotation) Icm=mr^2+mr^2 Icm=m(1L)^2+m(0.8L)^2 Icm=1.64L^2 T=2pi(I/Hgm)^0.5 T=2pi(1.64mL^2/0.9Lg2m)^0.5 T=2pi(82L/90g)^0.5 T=2pi(41L/45g)^0.5