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If an object is freely suspended from a pivot, why does the centre of gravity fall directly below that pivot? Would this be the same in non-uniform gravitational fields?

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  • $\begingroup$ We define "below" by the direction of gravitational attraction. The center of gravity by definition is along that direction. $\endgroup$ – Lewis Miller Feb 28 '16 at 22:55
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When an object is freely suspended to a pivot, the centre of gravity of the object does come below the pivot. However, note that here 'below' is defined in terms of the gravitational field itself, as pointed out by Lewis Miller in his comment.

Take for example a long rod suspended from a pivot. The gravitational force on the rod can be considered to act through its centre of gravity. If the centre of gravity is not exactly below the pivot then a torque is going to act due to the gravity, as in figure. figure

F is the force due to gravity. r is the distance between the point of application of the force and the pivot. $\theta$ is the angle between the direction of F and r.

The torque $\tau$ is then $$\tau = \mathbf {Fr}\sin\theta$$ When the centre of gravity is exactly below the pivot then the angle, $\theta=0$ so $\tau=0$. The body stays in equilibrium due to absence of net rotational force and net translational force(the weight of the rod is counteracted by the reaction force of the pivot).

As regards to your second question, this would be the same in non uniform gravitational fields as then the position of centre of gravity would change but the same logic applies.

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Because the pivot cannot exert a torque to counteract gravity. So the only stable configuration is such that the force-line of the weight passes through the pivot having no net torque about the pivot.

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