Timing a pendulum drawn back to the horizontal.

[Grozny]

Does this formula for the period of a pendulum work if the pendulum is drawn all the way back to the horizontal? Or does it only work when the angle off the vertical is small enough that its sine is approximately the angle itself? Note that my pendulum is stiff (a beam, not a weighted wire) so there is no difficulty in holding it out straight.

Period of a pendulum: $$T = 2\pi\sqrt{\frac{I}{mgh}}$$
I: inertia
m: mass
g: accelleration due to gravity, 9.80665
h: distance from the pivot to the center of gravity

I am actually only interested in the fall of the pendulum, from horizontal to vertical. I have figured out I, m, g and h for my device (it uses ball bearings, so friction can probably be ignored) and calculated the time to fall by quartering the period. The answer is close to my observations, but I need it exact, to within +/- about ten microseconds.

Can someone tell me if I am doing this right? If not, then what is the formula for the fall of a pendulum from the horizontal?

 

[rcgldr]

wiki article:

http://en.wikipedia.org/wiki/Pendulum_(mathematics)

 

[Grozny]

wiki article:

http://en.wikipedia.org/wiki/Pendulum_(mathematics)

For a simple pendulum, Wiki considers both the small-angle approximation and the more general case of an arbitrary angle. For the latter, they give this formula for the period:

$$T = \frac{4}{\sqrt{2}}\sqrt{\frac{L}{g}}\int_{0}^{\theta_0}\frac{1}{\sqrt{\cos{\theta} - \cos{\theta_0}}}<br /> d\theta$$

Note: I am using the notation L = length and h = distance from the pivot to the center of gravity, which is different than Wiki.

However, for a compound pendulum, which is what I am concerned with, Wiki only considers the small-angle approximation and gives the same formula that I posted in the OP.

So my question stands.

I need a formula like the one above, but for a compound pendulum.

 

[Grozny]

Simple pendulum, small angle: $$T = 2\pi\sqrt{\frac{L}{g}}$$

Compound pendulum, small angle: $$T = 2\pi\sqrt{\frac{I}{mgh}}$$

Since I have the formula for both simple and compound pendulums given the small-angle approximation, I can set them equal to each other to find the length, L, of a simple pendulum with the same period as the compound pendulum that I am interested in.

If this equivalence held for large angles as well, I would not need a formula for compound pendulums but could just plug L into this formula:

$$T = \frac{4}{\sqrt{2}}\sqrt{\frac{L}{g}}\int_{0}^{\theta_0}\frac{1}{\sqrt{\cos{\theta} - \cos{\theta_0}}}<br /> d\theta$$

So, my question is, if a simple pendulum has the same period as a compound pendulum for small angles, does it also have the same period for large angles?

 

[Grozny]

If a simple pendulum has the same period as a compound pendulum for small angles, does it also have the same period for large angles?

Anybody?

Also, I evaluated this definite integral to (what I think is) five digits:

$$2.6221 = \int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{\cos{\theta}}}<br /> d\theta$$

If someone has access to Mathematica or some similar software, could you do this more accurately for me, please?

 

[Doc Al]

So, my question is, if a simple pendulum has the same period as a compound pendulum for small angles, does it also have the same period for large angles?

Yes. Both a simple pendulum and a physical pendulum have similar equations:

(1) α = - (g/L) sinθ (for a simple pendulum)
(2) α = - (mgh/I) sinθ (for a physical pendulum)

The only difference is the constant in front of the sinθ term. So just replace one constant with the other and you should be OK.

 

[Grozny]

So the period for the compound pendulum is this?

$$T = \frac{4}{\sqrt{2}}\sqrt{\frac{I}{mgh}}\int_{0}^{\theta_0}\frac{1}{\sqrt{\cos{\theta} - \cos{\theta_0}}}<br /> d\theta$$

 

[Doc Al]

So the period for the compound pendulum is this?

$$T = \frac{4}{\sqrt{2}}\sqrt{\frac{I}{mgh}}\int_{0}^{\theta_0}\frac{1}{\sqrt{\cos{\theta} - \cos{\theta_0}}}<br /> d\theta$$

Assuming the wiki equation for the simple pendulum is correct (I didn't check), then yes.