# Compound Pendulum problem

1. Aug 18, 2008

### EmilyM

I'm having some trouble prooving a basic compound pendulum theory - my company makes a piece of kit designed to do just that.

In a simple experiment involving a long thin bar which is set on a knife edge and allowed to swing freely I have been changing the length of the bar and measuring the time period.

The equation is tau = 2 pi * sqrt((k^2 + h^2) / g * h) designed to enable us to find a local estimate for g (gravity) and k - the radius of gyration of the rod. Plotting a graph and rearranging the eqn into y = mx + c gives an estimation of 9.84 for gravity (very good) and 0.268 for k.

The theory for k is simple, Rouths Rule states k^2 = (L^2) / 3. This gives k = 0.528.

Help. Have redone the expt over and over with increasing accuracy to no avail. Am confident that Rouths Rule holds as is 12.7mm diameter st/st rod with L = 915mm.

2. Aug 18, 2008

### nvn

EmilyM: If you want to use tau = 2*pi*[(k^2 + h^2)/(g*h)]^0.5, then you must use k^2 = (L^2)/12, not k^2 = (L^2)/3. But if you want to use k^2 = (L^2)/3, then you must use tau = 2*pi*[(0.25*k^2 + h^2)/(g*h)]^0.5. See if this resolves your problem.

3. Aug 19, 2008

### EmilyM

Oh dear! Well that's what it is then, works perfectly now - thank you so much!

Ps, tau is the letter representing time period in the text books/refs I've been using...

4. Aug 20, 2008

### nvn

I should also mention, the best form for your tau formula is the one you listed in your first post (which is the same as the first tau formula I listed in my post), because it is general, and is therefore applicable to any object shape. Then, for your current, particular bar shape (a uniform bar), use k^2 = (L^2)/12.

5. Aug 20, 2008

### EmilyM

I agree, and I also now understand that my experiment was finding k at the centre of mass but the original formula I was using to find theoretical k (k^2 = (L^2) / 3) is to find k at the end of the bar.

k^2 = (L^2)/12 finds k at the centre of mass and this is what i wanted. The two are ofcourse related by the parallel axis theorem so that to 'move' from k at the centre of mass to k at the end of the bar you must add on a factor which is the distance squared, in this case (L/2)^2 or (L^2)/4. Then we have k^2 = (L^2)/12 + (L^2)/4 = (L^2)/12 + 3(L^2)/12 = 4(L^2)/12 = (L^2)/3 (most people could probably have skipped a few steps there but i'm happier with the maths if i write it long hand)

6. Aug 20, 2008

### nvn

Very well said, EmilyM.