Question 11.5: A cube of edge length s and mass M hangs vertically down fro...
A cube of edge length s and mass M hangs vertically down from one of its edges. Find the period for small vibrations about the equilibrium position. How long is the equivalent thread pendulum?
The moment of inertia of the cube about AB is (see Exercise 11.4)
\Theta_{AB} = \frac{2}{3} Ms^{2}.
The center of gravity is in the center of the cube, i.e., for the distance a of the center of gravity S from the axis AB we have
a = \frac{1}{2} s \sqrt{2}.
The equation of motion of the physical pendulum for small angle amplitudes was
\ddot{\varphi}+ \frac{Mga}{\Theta_{AB}} \varphi = 0with the angular frequency
ω =\sqrt{\frac{Mga}{\Theta_{AB}}}and the period
T = \frac{2π}{ω}= 2π \sqrt{\frac{\Theta_{AB}}{Mga}}= 2π \sqrt{\frac{2Ms^{2} · 2}{3Mgs \sqrt{2}}}= 2π \sqrt[4]{2} \sqrt{\frac{2s}{3g}}.The length of the equivalent thread pendulum is calculated as
T = T^{′} = 2π \sqrt{\frac{l}{g}},which just defines the equivalence of the pendulums. By insertion of T one obtains
2π\sqrt[4]{2} \sqrt{\frac{2}{3} \frac{s}{g}}= 2π \sqrt{\frac{l}{g}},or resolved,
l = \frac{2}{3}\sqrt{2}s.This equivalent length of the thread pendulum is also called the reduced pendulum length.
