The Center of Gravity — Question 10

Question 10

Suspension above the center gives a restoring gravity moment

A rigid object of mass $M$ pivots freely about a fixed point a distance $d$ above its center of gravity in uniform gravity. Let $\theta$ measure tilt from the hanging-down equilibrium.

1. Find its gravity torque and change in potential energy.

2. Compare with the inverted configuration in which the center is directly above the pivot.

Question 10 - Solution

Suspension above the center gives a restoring gravity moment

(a) Derivation and calculation

For a pivot a distance $d$ above $G$, an angular displacement $\theta$ from the hanging orientation gives $$\boxed{\tau=-Mgd\sin\theta},\qquad \boxed{\Delta U=Mgd(1-\cos\theta)}.$$ For small displacement, $$\tau\simeq-Mgd\theta,\qquad\Delta U\simeq\frac12Mgd\theta^2.$$

(b) Result and interpretation

Thus the hanging state has positive potential curvature and restoring torque: $$U''(0)=Mgd>0.$$ For a small displacement $\phi$ from the inverted orientation, $$U\simeq U_{\max}-\frac12Mgd\phi^2,\qquad\tau\simeq+Mgd\phi.$$ The inverted state is unstable because displacement lowers the energy and the torque increases it.

Physical check

The sign of the second derivative distinguishes two orientations that both have zero torque at exact alignment.