What is the period of a compound pendulum?

  • #26
ehild
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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.
 
  • #27
BvU
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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)
 
  • #28
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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)?
 
  • #29
BvU
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Yes, that is the idea (but you really don't need anyone's permission to do so :smile: )

Any questions about these variable names you need clearing up ?
 
  • #30
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Yes, that is the idea (but you really don't need anyone's permission to do so :smile: )

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?
 
  • #31
BvU
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How about replacing
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
 
  • #32
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How about replacing by

L = 1.05 m
Again, completely in agreement with what's described in the link
Okay! That's what I was thinking, I just needed some confirmation just to be super safe. And finally got the right answer! Thank you so much for your help! You and ehild. :smile:
 

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