User:CosineKitty/MathPage

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Point mass ideal pendulum, small angle case:

$t_{c}=2\pi {\sqrt {\frac {L}{g_{c}}}}$

Distributed mass pendulum, with moment of inertia, small angle case:

$t_{c}=2\pi {\sqrt {\frac {I}{mg_{i}L}}}$

Equating the right hand sides of both of these equations, canceling like terms, and squaring both sides:

${\frac {L}{g_{c}}}={\frac {I}{mg_{i}L}}$

The ratio of moment-corrected gravity over experimentally determined gravity is:

${\frac {g_{i}}{g_{c}}}={\frac {I}{mL^{2}}}$

For a cylinder rotating about an axis passing through its diameter at its center of gravity:

$I_{x}={\frac {m(3r^{2}+h^{2})}{12}}$

Using the parallel axis theorem, I move the rotation axis from center of gravity to the pivot point a distance L above the center of gravity:

$I=I_{x}+mL^{2}=m\left({\frac {r^{2}}{4}}+{\frac {h^{2}}{12}}+L^{2}\right)$

Substituting the rightmost expression for I into the g_i/g_c equation, we get:

${\frac {g_{i}}{g_{c}}}={\frac {{\frac {r^{2}}{4}}+{\frac {h^{2}}{12}}+L^{2}}{L^{2}}}$

Buoyancy of air requires a correction of:

${\frac {g_{a}}{g_{i}}}={\frac {m}{m-\rho V}}$

General formula for zero-angle limit of a conical pendulum consisting of a thin rod with an arbitrary linear density:

$g=\left({\frac {2\pi }{t_{c}}}\right)^{2}{\frac {\int _{0}^{L}s^{2}\rho (s)\,ds}{\int _{0}^{L}s\rho (s)\,ds}}$