OSCILLATION PERIOD ($T$)
Simple (Point Mass)0.00 s
Physical (Uniform Rod)0.00 s
■ Simple Pendulum:
$T_{\text{simple}} = 2\pi\sqrt{\frac{L}{g}}$
Assumes all mass is concentrated at the very tip (distance $L$).
■ Physical Pendulum (Rod):
$T_{\text{phys}} = 2\pi\sqrt{\frac{I}{mgd_{cm}}}$
For a uniform rod pivoted at one end, $I = \frac{1}{3}mL^2$, and the center of mass is at $d_{cm} = L/2$.
Resulting in: $T_{\text{phys}} = 2\pi\sqrt{\frac{2L}{3g}}$
Verdict: A physical rod will swing faster (shorter period $T$) than a point mass on a string of the exact same length, because its mass is distributed closer to the pivot.
$T_{\text{simple}} = 2\pi\sqrt{\frac{L}{g}}$
Assumes all mass is concentrated at the very tip (distance $L$).
■ Physical Pendulum (Rod):
$T_{\text{phys}} = 2\pi\sqrt{\frac{I}{mgd_{cm}}}$
For a uniform rod pivoted at one end, $I = \frac{1}{3}mL^2$, and the center of mass is at $d_{cm} = L/2$.
Resulting in: $T_{\text{phys}} = 2\pi\sqrt{\frac{2L}{3g}}$
Verdict: A physical rod will swing faster (shorter period $T$) than a point mass on a string of the exact same length, because its mass is distributed closer to the pivot.