 Solution

In a uniform gravitational field, the centre of gravity is the point $G$ through which the resultant gravitational weight acts. It coincides with the center of mass:Indeed the sum of the elemental gravitational torques about any origin isReplacing the distributed gravitational forces by the resultant at $G$ therefore preserves both total force and torque. A strongly nonuniform field need not permit the same identification.

For equilibrium, all external forces and torques must balance. A freely suspended rigid body supported at a pivot requires $G$ on the vertical through that pivot. With $G$ below the pivot, a small angular displacement raises its gravitational potential energy and gives a restoring torque; with $G$ above it, the equilibrium is unstable. A body supported on a horizontal base can remain at rest under its weight only if the vertical projection of $G$ lies within the region that the support forces can sustain. At an edge the overturning torque becomes decisive.

For motion, the center-of-mass equation isThe centre moves as a particle carrying the total mass under the resultant external force, while rotation about it is governed by the external torque and angular momentum. This separates translation from rotation; it does not determine a rigid body's full orientation from its centre's trajectory alone. In uniform gravity and negligible air resistance, a thrown rotating body has a parabolic centre-of-mass path even while its parts follow more complicated paths.