What is the period of a compound pendulum?

In summary: 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:Strike all cancelling quantitiesCheck dimensionsEstimate order of magnitude -- or approx value...In summary, the conversation discussed the calculation of the period of a compound pendulum consisting of a thin rod and a disc attached to one end. The participants considered the moment of inertia of the entire system and used the parallel axis theorem to calculate it. They also discussed the distance from the pivot point to the center of mass of the pendulum, which is necessary for the calculation of the period. After some confusion with notation and units, the final calculation resulted in an answer
  • #1
Dennydont
45
0

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.
 
Physics news on Phys.org
  • #2
Use parallel axis theorem to get the moment of inertia of the disk with respect to the pivot.
 
  • #3
ehild said:
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
Dennydont said:
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
 
  • #5
ehild said:
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
Okay, so I'll say that d2=(r/2)2 and apply it to the disk. I should get: 1/4mr2 (disk) + 1/3ml2. Would you say that this is correct?
 
  • #6
Dennydont said:
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
ehild said:
?
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
Dennydont said:
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
ehild said:
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
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
BvU said:
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
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
BvU said:
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
  • #15
BvU said:
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
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: .
 
Last edited:
  • #17
BvU said:
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
Good plan !
 
  • #19
BvU said:
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
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
BvU said:
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
L=I/mr makes no sense whatsoever to me. To you ?
Look at Ehilds drawing again and locate the center of mass.
 
  • #23
BvU said:
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
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
ehild said:
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
 
  • #26
Dennydont said:
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!

Dennydont said:
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
Dennydont said:
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
ehild said:
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
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
BvU said:
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
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
BvU said:
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:
 

1. What is a compound pendulum?

A compound pendulum is a type of pendulum that consists of a rigid body suspended from a fixed point, allowing it to swing back and forth. It is different from a simple pendulum, which consists of a single mass attached to a string or rod.

2. What is the period of a compound pendulum?

The period of a compound pendulum is the time it takes for one complete swing or oscillation. It is affected by the length of the pendulum, the mass of the pendulum, and the acceleration due to gravity.

3. How is the period of a compound pendulum calculated?

The period of a compound pendulum can be calculated using the equation T = 2π√(l/g), where T is the period, l is the length of the pendulum, and g is the acceleration due to gravity.

4. What factors affect the period of a compound pendulum?

The period of a compound pendulum is affected by the length of the pendulum, the mass of the pendulum, and the acceleration due to gravity. It is also affected by the angle of release, air resistance, and the material of the pendulum.

5. How does the period of a compound pendulum change with different lengths?

The period of a compound pendulum is directly proportional to the square root of its length. This means that as the length of the pendulum increases, the period also increases. Therefore, a longer pendulum will have a longer period than a shorter pendulum.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
632
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
27
Views
648
  • Introductory Physics Homework Help
Replies
9
Views
636
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
741
  • Mechanical Engineering
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Back
Top