# What is the period of a compound pendulum?

ehild
Homework Helper
The centre of the rod from the pivot is half its length? 0.7m
The centre of the disk from the pivot is just d, the length of the rod: 1.4m.

Correct!

For centre of mass for two point masses: X = m1x1+m2+x2

You have to divide by the whole mass. And you meant m2x2 instead of m2+x2.

BvU
Homework Helper
How am I meant to calculate L for T = 2π√(I/mgL) then? I'm only using L=I/mr because that's the formula you gave me.
Oh oh, where did I make that terrible mistake ?

Looks like a mixing up of notation (again! we have to become more careful with that! e.g. : Make a list of known / unknown variable names, with dimension and their meaning)

Correct!

You have to divide by the whole mass. And you meant m2x2 instead of m2+x2.
Okay, so using that equation I get 1.05m for the centre of mass. How does that work into the problem? Can I use that to find L in T = 2π√(I/mgL)?

BvU
Homework Helper
Yes, that is the idea (but you really don't need anyone's permission to do so )

Any questions about these variable names you need clearing up ?

Yes, that is the idea (but you really don't need anyone's permission to do so )

Any questions about these variable names you need clearing up ?
Hmmmm, okay!
So here are the variables I currently have:
d (length of rod) = 1.4m
r = 0.2m
m = 3.9kg
I = 1/3 m d2+ ( 1/2 mr2 + md2 ) = 10.27kg*m2

T = 2π√(I/MgL)
M (combined mass) = 7.8kg
I = 10.27kg*m2
g = 9.81m/s2
L = Wait how do I get this using the centre mass of 1.05m?

BvU
Homework Helper
L = Wait how do I get this using the centre mass of 1.05m?
by

L = 1.05 m
Again, completely in agreement with what's described in the link