Pendulums — Question 3

Question 3

Moving the pivot too close to the center makes a pendulum slow

A rigid body has mass $M$ and radius of gyration $k_g$ about its center of mass, so $I_{\mathrm{CM}}=Mk_g^2$. A horizontal pivot is a distance $d>0$ above the center of mass; parallel-axis inertia is $I_P=M(k_g^2+d^2)$.

1. Find the small-angle period as a function of $d$ and the value of $d$ minimizing it.

2. Find the minimum period and explain the limits $d\to0$ and $d\to\infty$, assuming those pivot locations are mechanically possible.

Question 3 - Solution

Moving the pivot too close to the center makes a pendulum slow

(a) Derivation and calculation

$$I_P=M(k_g^2+d^2),\quad T^2=\frac{4\pi^2I_P}{Mgd}=\frac{4\pi^2}{g}\left(d+\frac{k_g^2}{d}\right).$$$$\frac{dT^2}{dd}=\frac{4\pi^2}{g}\left(1-\frac{k_g^2}{d^2}\right)=0\Rightarrow\boxed{d=k_g}.$$$$\frac{d^2T^2}{dd^2}=\frac{8\pi^2k_g^2}{gd^3}>0.$$

(b) Result and interpretation

$$\boxed{T_{\min}=2\pi\sqrt{2k_g/g}}.$$$$d\to0:\quad T\sim\frac{2\pi k_g}{\sqrt{gd}}\to\infty;\qquad d\to\infty:\quad T\sim2\pi\sqrt{d/g}.$$Near the center, the gravitational restoring moment vanishes; far away, the pivot inertia grows quadratically.

Physical check

The minimum is global for positive $d$, since the period diverges at both ends of the domain.