DYNAMICS OF A RIGID BODY. 143 



remarkable form in the present case. The work done 

 in giving the body a small twist is proportional to the 

 vertical distance through which the centre of inertia is 

 elevated. Now, as in the position of equilibrium the 

 centre of inertia is vertically beneath the point of sus- 

 pension, it is obvious from symmetry that the ellipsoid 

 of the potential must be a surface of revolution about a 

 vertical axis. It is further evident that the vertical 

 radius vector of the ellipsoid must be infinite, because 

 no work is done in rotating the body around a vertical 

 axis. 



Let O be the centre of suspension, and / the cen- 

 tre of inertia, and let OP be a radius vector o the 

 ellipsoid of the potential. Let fall IQ perpendicular 

 on OP y and PT perpendicular upon OI. It is extremely 

 easy to show that the vertical height through which / is 

 raised is proportional to /(> x OP* ; whence the area 

 of the triangle OPI is constant, and therefore the locus 

 of P must be a right circular cylinder of which Olis the 

 axis. 



We have now to find the triad of common conjugate 

 diameters of the momental ellipsoid, and the circular 

 cylinder just described. A group of three conjugate dia- 

 meters of the cylinder must consist of the vertical axis, 

 and any two other lines through the origin, which 

 are conjugate diameters of the ellipse in which their 

 plane cuts the cylinder. It follows that the triad required 

 will consist of the vertical axis, and of the pair of 

 common conjugate diameters of the two ellipses in 

 which the plane conjugate to the vertical axis in the 

 momental ellipsoid cuts the momental ellipsoid and 

 the cylinder. These three lines are the three harmonic 

 axes. 



With reference to the vertical axis which appears to 

 be one of the harmonic axes, the time of vibration would 



