Page:EB1911 - Volume 12.djvu/802

Rh (A*)

+ z sin (x −  sin ) − rz cos  = 0, (B*)

sin cos  + 2xz sin2 − rx (x sin  + z cos )−g (x cos −z sin ) = 0. (C*)

The steady motion and nutation superposed may be expressed by

= + L, sin  = sin  + L cos, cos  = cos  − L sin , = + N, r = R + Q,  (1)

where L, N, Q are small terms, involving a factor e nti, to express the periodic nature of the nutation; and then if a, c denote the mean value of x, z, at the point of contact

x = a + L cos, z = c − L sin , (2)

x sin + z cos  = a sin  + c cos  + L (a cos  − c sin ), (3)

x cos − z sin  = a cos  − c sin  − L (a sin  + c cos  − ). (4)

Substituting these values in (C*) with dq/dt = −d2/dt2 = n2L, and ignoring products of the small terms, such as L2, LN, ...

(C**)

which is equivalent to

+ 2 ac sin2 − Ra (a sin  + c cos ) − g (a cos  − c sin ) = 0, (5)

the condition of steady motion; and

DL + EQ + FN = 0, (6)

where

(7)

(8)

+ 2ac sin2 − Ra (a sin  + c cos ). (9)

With the same approximation (A*) and (B*) are equivalent to

(A**)

+ c sin (a −  sin ) − Rc cos  = 0. (B**)

The elimination of L, Q, N will lead to an equation for the determination of n2, and n2 must be positive for the motion to be stable.

If b is the radius of the horizontal circle described by G in steady motion round the centre B,

b = v/ = (cP − aR) / = c sin  − aR / , (10)

and drawing GL vertically upward of length = g/2, the height of the equivalent conical pendulum, the steady motion condition may be written

(CR + K) sin  − 2 sin  cos  = −gM (a cos  − c sin )

+ M (2c sin − Ra) (a sin  + c cos ) (11)

LG produced cuts the plane in T.

Interpreted dynamically, the left-hand side of this equation represents the velocity of the vector of angular momentum about G, so that the right-hand side represents the moment of the applied force about G, in this case the reaction of the plane, which is parallel to GA, and equal to gM·GA/GL; and so the angle AGL must be less than the angle of friction, or slipping will take place.

Spinning upright, with = 0, a = 0, we find F = 0, Q = 0, and

(12)

(13)

(14)

Thus for a top spinning upright on a rounded point, with K = 0, the stability requires that

R > 2k′&radic; {g (c − )} / (k2 + c), (15)

where k, k′ are the radii of gyration about the axis Gz, and a perpendicular axis at a distance c from G; this reduces to the preceding case of § 3 (7) when = 0.

Generally, with = 0, but a ± 0, the condition (A) and (B) becomes

(16)

so that, eliminating Q/L,

(17)

the condition when a coin or platter is rolling nearly flat on the table.

Rolling along in a straight path, with =, c = 0,  = 0, E = 0; and

N/L = (CR + K)/A, (18)

(19)

,  L F

 (20)

(21)

Thus with K = 0, and rolling with velocity V = Ra, stability requires

(22)

or the body must have acquired velocity greater than attained by rolling down a plane through a vertical height (a − ) A/C.

On a sharp edge, with = 0, a thin uniform disk or a thin ring requires

V2/2g > a/6 or a/8. (23)

The gyrostat can hold itself upright on the plane without advance when R = 0, provided

K2/AM − g (a − ) is positive. (24)

For the stability of the monorail carriage of § 5 (6), ignoring the rotary inertia of the wheels by putting C = 0, and replacing K by G′ the theory above would require

For further theory and experiments consult Routh, Advanced Rigid Dynamics, chap. v., and Thomson and Tait, Natural Philosophy, § 345; also Bourlet, Traité des bicycles (analysed in Appell, Mécanique rationnelle, ii. 297, and Carvallo, Journal de l’école polytechnique, 1900); Whipple, Quarterly Journal of Mathematics, vol. xxx., for mathematical theories of the bicycle, and other bodies.

14. Lord Kelvin has studied theoretically and experimentally the vibration of a chain of stretched gyrostats (Proc. London Math. Soc., 1875; J. Perry, Spinning Tops,

for a diagram). Suppose each gyrostat to be equivalent dynamically to a fly-wheel of axial length 2a, and that each connecting link is a light cord or steel wire of length 2l, stretched to a tension T.

Denote by x, y the components of the slight displacement from the central straight line of the centre of a fly-wheel; and let p, q, 1 denote the direction cosines of the axis of a fly-wheel, and r, s, 1 the direction cosines of a link, distinguishing the different bodies by a suffix.

Then with the previous notation and to the order of approximation required,

1 = −dq/dt, 2 = dp/dt, (1)

h1 = A1, h2 = A2, h3 = K, (2)

to be employed in the dynamical equations

(3)

in which 3h1 and 3h2 can be omitted.

For the kth fly-wheel

−Aq .. k + Kp. &#8202;k = Ta (qk − sk) + Ta (qk − s k+1 ), (4)

Ap .. k + Kq. &#8202;k = −Ta (pk − rk) − Ta (pk − r k+1 ); (5)

and for the motion of translation

Mx .. k = T (rk+1 − rk), My .. k = T (sk+1 − sk); (6)

while the geometrical relations are

xk+1 − xk = a (pk+1 + pk) + 2lrk+1, (7)

yk+1 − yk = a (qk+1 + qk) + 2lsk+1. (8)

Putting

x + yi = w, p + qi =, r + si = , (9)