# A Compound Pendulum's Period

1. Jan 12, 2010

### dgl7

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

Last edited: Jan 12, 2010
2. Jan 12, 2010

### ideasrule

Is there a diagram that goes with this? Where's O?

3. Jan 16, 2010

### dgl7

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