
#1
Jan1210, 06:46 PM

P: 8

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 wrongperhaps 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 offbecause 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 



#3
Jan1610, 08:21 AM

P: 8

It is a pendulum, thus O is at the opposite end of the rod.



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