What is the period of a compound pendulum?

  • #1
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Homework Statement


A compound pendulum consists of a thin rod of length 1.4 m and a disc of radius 0.2 m. The centre of the disc is attached to the end of the rod and the pendulum pivots about the opposite end of the rod. Both the mass of the rod and the mass of the disc are the same, each being 3.9 kg. What is the period of the pendulum?

[g=9.81 ms/2, MI rod about its end 1/3mL[SUP]2[/SUP], MI disc through centre, perpendicular to the plane 1/2mr2]

Homework Equations


Compound pendulum: T = 2π√(I/mgL)
1/3mL2
1/2mr2


The Attempt at a Solution


I tried adding the two inertias together for I = 1/3mL2 + 1/2mr2] but that didn't really work out. Since r<<L, I considered this to be a simple pendulum and used 1/2mr2] to find I in the compound pendulum equation, but this method was also incorrect. Am I supposed to use the parallel axis theorem? Should I have used the simple pendulum equation instead? Please help.
 

Answers and Replies

  • #2
ehild
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Use parallel axis theorem to get the moment of inertia of the disk with respect to the pivot.
 
  • #3
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Use parallel axis theorem to get the moment of inertia of the disk with respect to the pivot.
Would the moment of inertia be: 1/3ml2 + 1/4mr2 then? I'm not 100% sure how to apply the parallel axis theorem correctly.
 
Last edited:
  • #4
ehild
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Would the moment of inertia be: 1/12mr2 + 1/2mr2 then? I'm not 100% sure how to apply the parallel axis theorem correctly.
The total moment of inertia is the sum of that of the rod and that of the disk with respect to the pivot. Apply parallel axis theorem for the disk. http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html
 
  • #6
ehild
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Okay, so I'll say that d2=(r/2)2 and apply it to the disk.

????
The moment of inertia of the disk is its moment of inertia with respect to its centre + md2, with d the distance between the centre of the disk from the pivot.
 
  • #7
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????
The moment of inertia of the disk is its moment of inertia with respect to its centre + md2, with d the distance between the centre of the disk from the pivot.
Hmmmm okay. So the moment inertia of the disk with respect to the pivot is: 1/2+mr2+m(r/2)2? Adding this to the inertia of the rod for the total inertia.
 
  • #8
ehild
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Hmmmm okay. So the moment inertia of the disk with respect to the pivot is: 1/2+mr2+m(r/2)2? Adding this to the inertia of the rod for the total inertia.

NO.Look at parallel axis theorem in your notes or book.What does it say?

compoundpendulum.JPG
 
  • #9
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NO.Look at parallel axis theorem in your notes or book.What does it say?

View attachment 80416
There's nothing in my notes about parallel axis theorem, that's why I'm so confused! I don't know, is it: 1/3mL2+1/2mr2+m(1.4)2???
 
  • #10
BvU
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No grounds for confusion. You did know about this parallel thingy from post #1 and can always google to find out what and why. It's not rocket science (which itself is also overrated :smile:). Simply the moment of inertia of a point mass at distance d - and then if it isn't a point mass you have to add the moment of inertia around its center of mass.

What you have looks good, but your notation is asking for trouble. (L in post #9 is not the L in post #1 equation 1).
 
  • #11
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No grounds for confusion. You did know about this parallel thingy from post #1 and can always google to find out what and why. It's not rocket science (which itself is also overrated :smile:). Simply the moment of inertia of a point mass at distance d - and then if it isn't a point mass you have to add the moment of inertia around its center of mass.

What you have looks good, but your notation is asking for trouble. (L in post #9 is not the L in post #1 equation 1).
Woooo, finally on the right track! Hmmmm I'm looking at L in post #9 and #1 and I see no difference, how are they different?
 
  • #12
BvU
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One is the length of the rod
The other is the distance from the pivot point to the center of mass of the pendulum
 
  • #13
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One is the length of the rod
The other is the distance from the pivot point to the center of mass of the pendulum
Are they not the same value of 1.4?
 
  • #14
BvU
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  • #15
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So I should use L = I/mr to find the L in 1/3mL2+1/2mr2+m(1.4)2?
Which length would I use in T = 2π√(I/mgL)? The rod length of 1.4?
 
  • #16
BvU
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Oh boy, no! Did you check the link I gave ?

Ehild made a clear drawing, let's use that notation: 1.4 m is d, the length of the rod.

If you remember, for the moment of inertia of the whole thing we now have I = 1/3 m d2+ ( 1/2 mr2 + md2 )

That leaves the symbol L for the distance from the pivot point to the center of mass of the pendulum, to be calculated.
By you :smile: .
 
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  • #17
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Oh boy, no! Did you check the link I gave ?

Ehild made a clear drawing, let's use that notation: 1.4 m is d, the length of the rod.

If you remember, for the moment of inertia of the whole thing we now have I = 1/3 m d2+ ( 1/2 mr2 + md2 )

That leaves the symbol L for the distance from the pivot point to the center of mass of the pendulum, to be calculated.
By you :smile: .
Awesome, so the L that I calculate on my own will be used in T = 2π√(I/mgL). Right?
 
  • #18
BvU
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Good plan !
 
  • #19
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Good plan !
Putting everything in:
m=3.9kg
I=10.27
L=13.1666
g=9.81
I get 1.11 which is apparently the incorrect answer. :cry:
 
  • #20
BvU
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Are you saying the center of mass is 13.17 m below the pivot point ?

And how come I has no units ? (the value is OK)

General rule for calculations:
  1. Strike all cancelling quantities
  2. Check dimensions
  3. Estimate order of magnitude -- or approx value if possible
  4. Employ calculator
Caveat:
m is the mass of the rod and the mass of the disc. In your T formula it's also the mass of the compound pendulum. Recipe for disaster.​
 
  • #21
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Are you saying the center of mass is 13.17 m below the pivot point ?

And how come I has no units ? (the value is OK)

General rule for calculations:
  1. Strike all cancelling quantities
  2. Check dimensions
  3. Estimate order of magnitude -- or approx value if possible
  4. Employ calculator
Caveat:
m is the mass of the rod and the mass of the disc. In your T formula it's also the mass of the compound pendulum. Recipe for disaster.​
Ooooooooooh, so m is really 7.8kg since mass of the rod and disc are the same? 13.17 is what I calculated for L strictly using the formula L=I/mr
 
  • #22
BvU
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L=I/mr makes no sense whatsoever to me. To you ?
Look at Ehilds drawing again and locate the center of mass.
 
  • #23
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L=I/mr makes no sense whatsoever to me. To you ?
Look at Ehilds drawing again and locate the center of mass.
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.
 
  • #24
ehild
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You can consider the rod as if its whole mass was concentrated at its centre, and the disk also, as if all its mass was in the centre.

How far is the centre of the rod from the pivot?
How far is the centre of the disk from the pivot?
How do you determine the centre of mass of two point masses?
So how far is the CM of the whole system from the pivot?
 
  • #25
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You can consider the rod as if its whole mass was concentrated at its centre, and the disk also, as if all its mass was in the centre.

How far is the centre of the rod from the pivot?
How far is the centre of the disk from the pivot?
How do you determine the centre of mass of two point masses?
So how far is the CM of the whole system from the pivot?
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.
For centre of mass for two point masses: X = m1x1+m2+x2
 

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