# Gyration Sentence Examples

- In the case of an axial moment, the square root of the resulting mean square is called the radius of
**gyration**of the system about the axis in question. - Which we shall meet with presently as the ellipsoid of
**gyration**at G. - Since I~=Ii., I~=o, we deduce 100=3/4Ma2, ~ =4MaZ; hence the value of the squared radius of
**gyration**isfora diameter 3/4ai, and for the axis of symmetry 3/4af. - From its axis (0), if the radius of
**gyration**about a longitudinal axis through G, aiid 0 the inclin - ation of OG to the vertical, FIG. - The square of the radius of
**gyration**with respect to a diameter is ia2. - The formula (16) expresses that the squared radius of
**gyration**about any axis (Ox) exceeds the squared radius of**gyration**about a parallel axis through G by the square of the distance between the two axes. - The squares of the radii of
**gyration**about the principal axes at P may be denoted by k,i+k32, k,f + ki2, k12 + k,2 hence by (32) and (35), they are rfOi, r2Oi, r20s, respectively. - This is called the ellipsoid of
**gyration**at 0; it was introduced into the theory by J. - It possesses thi property that the radius of
**gyration**about any diameter is half thi distance between the two tangents which are parallel to that diameter, In the case of a uniform triangular plate it may be shown that thi momental ellipse at G is concentric, similar and similarly situatec to the ellipse which touches the sides of the triangle at their middle points. - If k be the radius of
**gyration**about p we find k2 =2Xarea AHEDCBAXONap, where a$ is the line in the force-diagram which represents the sum of the masses, and ON is the distance of the pole 0 from this line. - If M be the total mass, k the radius of
**gyration**(~ ii) about the axis, we have sin 0, (3) - If K be the radius of
**gyration**about a parallel axis through G, we have kf=K2+h2 by If (16), and therefore i=h+K1/h, whence GO.GP=K2. - Where K is the radius of
**gyration**about the axis of symmetry, a is the constant distance of G from the plane, and R, F are the normal and tangential components of the reaction of the plane, as shown in fig. - Let W be the weight of a flywheel, R its radius of
**gyration**, ai its maximum, aj its minimum, and A=~1/8(a1+ai) its mean angular velocity. - R is called the radius of
**gyration**of the body with regard to an axi: - The radius of
**gyration**of the section is 2a 2. - After a certain discount for friction and the recoil of the gun, the net work realized by the powder-gas as the shot advances AM is represented by the area Acpm, and this is equated to the kinetic energy e of the shot, in foot-tons, (I) e d2 I + p, a in which the factor 4(k 2 /d 2)tan 2 S represents the fraction due to the rotation of the shot, of diameter d and axial radius of
**gyration**k, and S represents the angle of the rifling; this factor may be ignored in the subsequent calculations as small, less than I %. - The velocity of a liquid particle is thus (a 2 - b 2)/(a 2 +b 2) of what it would be if the liquid was frozen and rotating bodily with the ellipse; and so the effective angular inertia of the liquid is (a 2 -b 2) 2 /(a 2 +b 2) 2 of the solid; and the effective radius of
**gyration**, solid and liquid, is given by k 2 = 4 (a 2 2), and 4 (a 2 For the liquid in the interspace between a and n, m ch 2(0-a) sin 2E 4) 1 4Rc 2 sh 2n sin 2E (a2_ b2)I(a2+ b2) = I/th 2 (na)th 2n; (8) and the effective k 2 of the liquid is reduced to 4c 2 /th 2 (n-a)sh 2n, (9) which becomes 4c 2 /sh 2n = s (a 2 - b 2)/ab, when a =00, and the liquid surrounds the ellipse n to infinity. - The moment of inertia of the body about the axis, denoted by But if is the moment of inertia of the body about a mean diameter, and w the angular velocity about it generated by an impluse couple M, and M' is the couple required to set the surrounding medium in motion, supposed of effective radius of
**gyration**k', If the shot is spinning about its axis with angular velocity p, and is precessing steadily at a rate about a line parallel to the resultant momentum F at an angle 0, the velocity of the vector of angular momentum, as in the case of a top, is C i pµ sin 0- C2µ 2 sin 0 cos 0; (4) and equating this to the impressed couple (multiplied by g), that is, to gN = (c 1 -c 2)c2u 2 tan 0, (5) and dividing out sin 0, which equated to zero would imply perfect centring, we obtain C21 2 cos 0- (c 2 -c 1)c2u 2 sec 0 =o.