Motion of a rigid body pivoted on one of its principal axes mounted on a gimballed bowl. Rigid body rotating freely; referred to the 4, 6, 0 co-ordinates (§ 101). Gyroscopes and gyrostats. Gyroscopic pendulum. The kinetic energy of the motion of M relatively to O, its centre of inertia, is (§ 281) } (Aw2 + Bp3 + Co3) ; and (§ 280) its whole kinetic energy is obtained by adding the kinetic energy of a material point equal to its mass moving with the velocity of its centre of inertia. This latter part of the kinetic energy of M is most simply taken into account by supposing n to include a material point equal to M placed at 0; and using the previous notation k, e, ƒ for radii of gyration of n on the understanding that n now includes this addition. Hence for the present example, with the preceding notation G, D, we have T= {(G+D cos 0) i2 + nk2 2} + A (Ö sin p − i sin 0 cos ¢)* + B (Ỏ eos ¢ + sin 0 sin ¢) From this the three equations of motion are easily written down. By putting G = 0, D=0, and k = 0, we have the case of the motion of a free rigid body relatively to its centre of inertia. By putting B = A we fall on a case which includes gyroscopes and gyrostats of every variety; and have the following much simplified formula: T=} {[G+A+(D- 4) cos*0] i + (nh + 4) ở + C (j cos 0 +b), or T = { {(E + F cos2 ◊) †2 + (nk2 + A) Òa + C′ († cos 0 + $)3}, is Example (D).-Gyroscopic pendulum.-A rigid body, P, attached to one axis of a universal flexure joint (§ 109), of which the other is held fixed, and a second body, Q, is supported on P by a fixed axis, in line with, or parallel to, the first-mentioned arm of the joint. For simplicity, we shall suppose to be kinetically symmetrical about its bearing axis, and OB to be a principal axis of an ideal rigid body, PQ, composed of P and a mass so distributed along the bearing axis of the actual body Q as to have the same centre of inertia and the same moments of inertia round axes perpendicular to it. Let 40 be the fixed arm, O the joint, OB the movable arm bearing the body P, and coinciding with, or parallel to, the axis of Q. Let BOA' = 0; let & be the pendulum. angle which the plane AOB makes with a fixed plane of reference, Gyroscopic through OA, chosen so as to contain a second principal axis of the imagined rigid body, PQ, when OB is placed in line with A0; and let be the angle between a plane of reference in Q through its axis of symmetry and the plane of the two principal axes of PQ already mentioned. These three co-ordinates (0, 4, 4) clearly specify the configuration of the system at any time, t. Let the moments of inertia of the imagined rigid body PQ, round its principal axis OB, the other principal axis referred to above, and the remaining one, be denoted by A, B, C respectively; and let A' be the moment of inertia of Q round its bearing axis. B We have seen (§ 109) that, with the kind of joint we have supposed at O, every possible motion of a body rigidly connected with OB, is resolvable into a rotation round OI, the line bisecting the angle AOB, and a rotation round the line through O perpendicular to the plane AOB. The angular velocity of the latter is é, according to our present notation. The former would give to any point in OB the same absolute velocity by rotation round OI, that it has by rotation with angular velocity & round AA'; and is therefore equal to This may be resolved into 24 sin2 10 = $ (1 − cos 0) round OB, and 24 sin 10 cos 10 = 4 sin 0 round the perpendicular to OB, in plane AOB. Again, in virtue of the symmetrical character of the joint with reference to the line OI, the angle 4, as defined above, will be equal to the angle between the plane of the two first-mentioned principal axes of body P, and the plane AOB. Hence the axis of the angular velocity & sin 0, is inclined to the principal axis of moment B at an angle equal to 4. Resolving therefore this angular velocity, and 6, into components round the axes of B and C, we find, for the whole component angular velocities of the imagined rigid body PQ, round these axes, sin cos + sin 4, and 4 sin 0 sin + cos 4, respectively. The whole kinetic energy, T, is composed of that of the imagined rigid body PQ, and that of Q about axes through its centre of inertia we therefore have 2T = A ( 1 − cos 0)2 ¿2 + B (¿ sin @ cos &+ėsinė)2+C (¿sin@sin-cos)2 = dy % (1 – cos 0)° p + 13 (p sin 0 cos ¢ + Ô sin Q) sin 0 cos +C(¿sin@sin-cos p) sin @sin 13 (ở sin 0 cos d + Ô sin Q) ( +C (¿sin 0 sin -'{†—¿(1 − cos 0)}(1 − cos 0), sin 0 sin ( – 0 cos ) - - 0 cos $) ($ sin @ cos &+ėsinė), B(sin cos + sin 4) sin 4-C (¿ sin 0 sin - cos 4) cos & 4 (1 – cos 0) sin 0ộ* + 23 cos 0 cos de ( sin 0 cos p + Ủ sin ) +C cos 0 sin 44 (¿ sin ✪ sin 4–0 cos 4) – A′sin0¿{¿− (1 − cos 0) ¿}. Now let a couple, G, act on the body Q, in a plane perpendicular to its axis, and let L, M, N act on P, in the plane perpendicular to OB, in the plane A'OB, and in the plane through OB perpendicular to the diagram. If is kept constant, and varied, the couple G will do or resist work in simple addition with L. Hence, resolving L + G and N into components round OI, and perpendicular to it, rejecting the latter, and remembering that 2 sin 10 is the angular velocity round OI, we have p=2sin10{−(L+G) sin 10+Ncos {0} = {−(L+G)(1 − cos 0)+Nsin@}. Also, obviously V=G, O=M. Using these several expressions in Lagrange's general equations (24), we have the equations of motion of the system. They will be of great use to us later, when we shall consider several particular cases of remarkable interest and of very great importance. Example (E).-Motion of a free particle referred to rotating axes. Let x, y, z be the co-ordinates of a moving particle referred to axes rotating with a constant or varying angular velocity round the axis OZ. Let x, y, z, be its co-ordinates referred to the same axis, OZ, and two axes OX1, OY,, fixed in the plane per 1 = cos ay sina - (x sin a + y cos a) ȧ, ý, = etc. where a, the angle XX, must be considered as a given function of t. Hence constraint (rotating axes). m (2 – 2 ja – xả - ya) = X, m (j + 2ửa – gả + xa) = Y, m = 2, X, Y, Z denoting simply the components of the force on the particle, parallel to the moving axes at any instant. In this example t enters into the relation between fixed rectangular axes and the co-ordinate system to which the motion is referred; but there is no constraint. The next is given as an example of varying, or kinetic, constraint. varying due to constraint. Example (F).-A particle, influenced by any forces, and at- Example of tached to one end of a string of which the other is moved with any relation constant or varying velocity in a straight line. Let be the kinetic inclination of the string at time t, to the given straight line, and the angle between two planes through this line, one containing the string at any instant, and the other fixed. These two coordinates (0, 4) specify the position, P, of the particle at any instant, the length of the string being a given constant, a, and the distance OE, of its other end E, from a fixed point, Q, of the line in which it is moved, being a given function of t, which we shall denote by u. Let x, y, z be the co-ordinates of the particle referred to three fixed rectangular axes. Choosing OX as the given straight line, and YOX the fixed plane from which & is measured, we have x=u+a cos 0, y = a sin 0 cos p, za sin @ sin o, ¿=¿ – a sin 00; Example of varying relation due to kinetic constraint. and for y, we have the same expressions as in Example (A). Hence where T=T+&m (u - 20a sin 0) denotes the same as the T of Example (A), with → = 0, and r = a. Hence, denoting as there, by G and II the two components of the force on the particle, perpendicular to EP, respectively in the plane of and perpendicular to it, we find, for the two required equations of motion, m {a (Ö – sin 0 cos 0¿a) – sin 0 ü} = G, and ma d (sin❜ 04) = H. These show that the motion is the same as if E were fixed, and a force equal to - mü were applied to the particle in a direction parallel to EX; a result that might have been arrived at at once by superimposing on the whole system an acceleration equal and opposite to that of E, to effect which on P the force - mü is required. Example (F). Any case of varying relations such that in 318 (27) the coefficients (4, 4), (4, 4)... are independent of t. Let denote the quadratic part, L the linear part, and K [as in § 318 (27)] the constant part of T in respect to the velocity components, so that T = } { (4, 4) †2 + 2 (4, 6) 4$ + (4, 6) $2 + L = (4) † + (6) $ + ... K = (4, 4, 0, ...) .......(a), where (4, 4), (4, 4), (4, 4) ... denote functions of the co-ordinates without t, and (4), (4), (y, p, 0, ...) functions of the co-ordinates and, may be also, of t; and Hence the contribution from K to the first member of the |