Compound pendulums

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.

Reply 1

TSR Jessica

19

Sorry you've not had any responses about this. :frown:

Are you sure you've posted in the right place? :smile:

Here's a link to our subject forum which should help get you more responses if you post there. :redface:

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

Just quoting in Puddles the Monkey so she can move the thread if needed :h:

Spoiler

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

\omega = \sqrt{\frac{mgh}{I}}

where

h

is the distance from the pivot to the center of mass and

I

is the moment of inertia. For a simple pendulum,

I

becomes

mh^2

when the bob has most of the mass in this setup. So square of the period of a simple pendulum is proportional to

h

.

Reply 3

lerjj

13

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:

I\theta '' = -mgh \sin(\theta)

\theta '' ~ -\frac{mgh}{I}\theta

So the pendulum is simple harmonic with period:

T = 2\pi \sqrt{\dfrac{I}{mgh}}

This might look like T^2 is proportional to 1/h - however, you can always rewrite the moment of inertia as:

I = \gamma mh^2

where

\gamma

is purely based on the geometry (and not the scale) of the pendulum.
So the overall equation is actually:

T = 2\pi \sqrt{\dfrac{\gamma h}{g}}

where for a pendulum with all the mass at the end,

\gamma h = l

Reply 4

lerjj

13

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?

Please post this question seperately, and not in another user's thread.

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.

Reply 5

JimmyMcgill

Original post by lerjj

I'm pretty sure that that is still true. The equation for a compound pendulum is:

I\theta '' = -mgh \sin(\theta)

\theta '' ~ -\frac{mgh}{I}\theta

So the pendulum is simple harmonic with period:

T = 2\pi \sqrt{\dfrac{I}{mgh}}

This might look like T^2 is proportional to 1/h - however, you can always rewrite the moment of inertia as:

I = \gamma mh^2

where

\gamma

is purely based on the geometry (and not the scale) of the pendulum.
So the overall equation is actually:

T = 2\pi \sqrt{\dfrac{\gamma h}{g}}

where for a pendulum with all the mass at the end,

\gamma h = l

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.

Reply 6

lerjj

13

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?

Reply 7

JimmyMcgill

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?