Editing Theory (section)

== The Theory ==
[[File:KaterPendulum.png|thumb|300px]]
When displaced, the restoring force acting on the pendulum is:

: <math>\tau = M_{T} g h_1 \sin \theta</math> 

With <math>\theta</math> being the angle of displacement and the pendulum orientated small mass down (SMD), <math>M_{T}</math> being the total mass of the system, <math>g</math> being local gravity, and <math>h_1</math> being the distance from the pivot to the center of mass of the system. Applying Newton's Second Law rotational analog results in:

: <math> M_{T} g h_1 \sin \theta = - \ddot \theta \mathbb{I}_{smd}</math>

Where <math>\mathbb{I}_{smd}</math> is the moment of inertia. Of course, when <math>\theta</math> is small, the resulting equation can be simplified to:

: <math>\ddot \theta + \frac{M_{T} g h_1 }{\mathbb{I}_{smd}}\theta = 0</math>

Solving the differential equation results in:

: <math> \theta = A \cos{\left(\omega_0 t\right)}+ B \cos{\left(\omega_0 t\right)}</math>

Where:

: <math>\omega_0 = \sqrt{\frac{M_{T} g h_1 }{\mathbb{I}_{smd}}}</math>

resulting in:

: <math>T_{smd} = 2\pi\sqrt{\frac{\mathbb{I}_{smd}}{M_{T} g h_1 }}</math>

Where, <math>T_{smd}</math> is the period in the SMD orientation. Similarly, we can solve for the pendulum in the small mass up (SMU) orientation which results in:

: <math>T_{smu} = 2\pi\sqrt{\frac{\mathbb{I}_{smu}}{M_{T} g h_2 }}</math>

An application of the parallel axis theorm results on the equations:

: <math>\mathbb{I}_{smd} = \mathbb{I}_{cm} + M_{T} h_1^2</math>
and
: <math>\mathbb{I}_{smu} = \mathbb{I}_{cm} + M_{T} h_2^2</math>

Where <math>\mathbb{I}_{cm}</math> is the moment of inertia about the center of mass. By substituting these equations into our previous equations for <math>T</math> we get:

: <math>T_{smd} = 2\pi\sqrt{\frac{\mathbb{I}_{cm} + M_{T} h_1^2}{M_{T} g h_1 }}</math>
and
: <math>T_{smu} = 2\pi\sqrt{\frac{\mathbb{I}_{cm} + M_{T} h_2^2}{M_{T} g h_2 }}</math>

Now, if we can find a small mass position such that our period is the same in both the SMU and SMD orientation, we can set these equations equal to one another and solving for <math>\mathbb{I}_{cm}</math> we find:

: <math>\mathbb{I}_{cm} = M_{T}h_1h_2</math>

Substituting this value into either equation for period results in:

: <math>T = 2\pi\sqrt{\frac{h_1 + h_2}{g}}</math>

And, since the distance, <math>l</math>, between the pivots is the sum of <math>h_1</math> and <math>h_2</math> and we know this distance very well, we can write:

: <math>T = 2\pi\sqrt{\frac{l}{g}}</math>
or
: <math>g = \frac{4\pi^2l}{T^2}</math>

With this relationship, all we need to determine <math>g</math> is, the period of Kater's pendulum when the small mass is adjusted such that the period when swung in the SMU orientation is identical to the period in the SMD orientation.

Some points to consider are if the weights are of different volume -which in our current pendulum, they are- their cross-sectional area will affect results due to air resistance. Also, if the weights are positioned very asymmetrically, the effect of air resistance will affect the results. The small angle approximation has been used to determine the values and the data should be corrected for this.