You are Here: Home >< Physics

# Compound pendulums

Announcements Posted on
Last day to win £100 of Amazon vouchers - don't miss out! Take our quick survey to enter 24-10-2016
1. Can you give me a reason as to why for a compound pendulum the (time period)^2 is not directly proportional to h(the distance the pivot is placed from the centre of mass).
For a simple pendulum (time period)^2 is proportional to h.
2. Sorry you've not had any responses about this. Are you sure you've posted in the right place? Here's a link to our subject forum which should help get you more responses if you post there.

You can also find the Exam Thread list for A-levels here and GCSE here.

Just quoting in Puddles the Monkey so she can move the thread if needed
Spoiler:
Show
(Original post by Puddles the Monkey)
x
3. (Original post by hamzah0608)
Can you give me a reason as to why for a compound pendulum the (time period)^2 is not directly proportional to h(the distance the pivot is placed from the centre of mass).
For a simple pendulum (time period)^2 is proportional to h.
It seems to me that you are asking why the period of "a general pendulum" is not following the period of "a specific pendulum" in a specific way.

To me, a compound pendulum can be considered as a general pendulum as compared to a simple pendulum. The angular frequency of both pendulums can be expressed as

where is the distance from the pivot to the center of mass and is the moment of inertia. For a simple pendulum, becomes when the bob has most of the mass in this setup. So square of the period of a simple pendulum is proportional to .
4. (Original post by hamzah0608)
Can you give me a reason as to why for a compound pendulum the (time period)^2 is not directly proportional to h(the distance the pivot is placed from the centre of mass).
For a simple pendulum (time period)^2 is proportional to h.
I'm pretty sure that that is still true. The equation for a compound pendulum is:

So the pendulum is simple harmonic with period:

This might look like T^2 is proportional to 1/h - however, you can always rewrite the moment of inertia as:
where is purely based on the geometry (and not the scale) of the pendulum.
So the overall equation is actually:

where for a pendulum with all the mass at the end,
5. (Original post by JimmyMcgill)
On the topic of compound pendulums I have had to plot a graph of the time period squared against the distance from the pivot. I've found the minimum time period by drawing a tangent to the curve and then a line down to the x-axis to find the distance from the pivot for the that minimum time period. But it says estimate the uncertaintity in this value using the graph and I have no idea how to approach this or what it's even asking for. Could it just be the half the range of values for which the curve touches the tangent as it's quite difficult to observe?

Also, please attach a sketch of the graph you've drawn, because it's much easier to understand what you're meant to be doing from a picture rather than a long description.
6. (Original post by lerjj)
I'm pretty sure that that is still true. The equation for a compound pendulum is:

So the pendulum is simple harmonic with period:

This might look like T^2 is proportional to 1/h - however, you can always rewrite the moment of inertia as:
where is purely based on the geometry (and not the scale) of the pendulum.
So the overall equation is actually:

where for a pendulum with all the mass at the end,
I've made my own thread. Sorry about that. And I don't think this is correct because the total inertia of a compound pendulum is the inertia due to the radius of gyration and then the inertia from the distance so you get two added quantities and hence no direct proportion.
7. (Original post by JimmyMcgill)
I've made my own thread. Sorry about that. And I don't think this is correct because the total inertia of a compound pendulum is the inertia due to the radius of gyration and then the inertia from the distance so you get two added quantities and hence no direct proportion.
I'm not 100% sure what you mean by radius of gyration, but I think the OP means a single rigid body of complex shape when he specified a "compound pendulum". I.e. it's not a double pendulum or anything like that.

In which case, since the centre of mass will hang vertically below the pivot, I think it's fine to simply use the moment of inertia through the CoM. What would you do instead?
8. (Original post by lerjj)
I'm not 100% sure what you mean by radius of gyration, but I think the OP means a single rigid body of complex shape when he specified a "compound pendulum". I.e. it's not a double pendulum or anything like that.

In which case, since the centre of mass will hang vertically below the pivot, I think it's fine to simply use the moment of inertia through the CoM. What would you do instead?
I = IG + mh2 = mk2 + mh2
9. (Original post by JimmyMcgill)
I = IG + mh2 = mk2 + mh2
Is this not the parallel axis formula? In which case, doesn't our axis of rotation go through the CoM? Also, that still gives it just changes the value of
10. ..the graph should be directly proportional, shouldn't it?

T^2 = 4pi^2 * 2l/3g

There's no intercept and everything else is constant.
11. (Original post by Einstein1997)
..the graph should be directly proportional, shouldn't it?

T^2 = 4pi^2 * 2l/3g

There's no intercept and everything else is constant.
The time period of a compound pendulum is T=2pi*sqrt(k^2 + h^2/gh)

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: May 3, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### How does exam reform affect you?

From GCSE to A level, it's all changing

### Q&A with Paralympian Jack Rutter

Poll
Study resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.