Her body continued its momentum down the hill and she fell, twisting so that she wouldn't fall on the kid.
Hence the angular momentum of the part between A and B remains constant, or as much enters at B as leaves at A.
Consider a submarine boat or airship moving freely with the direction of the resultant momentum horizontal, and the axis at a slight inclination 0.
The pressure on CD is equal to the A C momentum which it receives per second.
Since the conditions in the region PQ remain always the same, the momentum perpendicular to AB entering the region at Q is equal to the momentum perpendicular to AB leaving the region at P. But, since the motion at Q is along AB, there is no momentum there perpendicular to AB.
She lit with one foot underneath her body, and the momentum of her fall threw her forward - over the ledge.
If the velocity of a particle at A relative to the undisturbed parts is u from left to right, the velocity of the matter moving out at A is U - u, and the momentum carried out by the moving matter is p(U - u) 2.
There is also the " external " applied pressure X, and the total momentum flowing out per second is X-I-P4-W-1-p(U - u)2.
Equating this to the momentum entering at B and subtracting P' from each X+W+p(U - u)2 =poU 2.
Now the transfer of momentum across a surface occurs in two ways, firstly by the carriage of moving matter through the surface, and secondly by the force acting between the matter on one side of the surface and the matter on the other side.
Consider, for example, a submarine boat under water; the inertia is different for axial and broadside motion, and may be represented by (1) c 1 =W+W'a, c2=W+W'/3' where a, R are numerical factors depending on the external shape; and if the C.G is moving with velocity V at an angle 4) with the axis, so that the axial and broadside component of velocity is u = V cos 0, v =V sin 4), the total momentum F of the medium, represented by the vector OF at an angle 0 with the axis, will have components, expressed in sec. Ib, F cos 0 =c 1 - = (W +W'a) V cos 43, F sin 0 = c 2.11 = (W +W'/3) V sin 4) .
Sulpicius Galba and others, and along with it the development of prose composition, went on with increased momentum till the age of Cicero.
Therefore the momentum entering through a square centimetre at B per second is equal to the momentum leaving through a square centimetre at A.
At B there is only the latter kind, and since the transfer of matter is powoU, where po is the undisturbed density and wo is the undisturbed cross-section, since its velocity is U the passage of momentum per second is powoUo 2.
In fluid movements, he shook out a noose, swung it a few times to get momentum and then threw it at the limb.
Well as of the body from the vector OF to O'F' requires an impulse couple, tending to increase the angle F00', of magnitude, in sec. foot-pounds F.00'.sin FOO'=FVt sin (0-0), (4) equivalent to an incessant couple N=FV sin (0-0) = (F sin 0 cos 0-F cos 0 sin ¢)V = (c 2 -c i) (V /g) sin 0 cos 4) =W'(13-a)uv/g (5) This N is the couple in foot-pounds changing the momentum of the medium, the momentum of the body alone remaining the same; the medium reacts on the body with the same couple N in the opposite direction, tending when c 2 -c 1 is positive to set the body broadside to the advance.
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.
In later memoirs Reynolds followed up this subject by proceeding to establish definitions of the velocity and the momentum and the energy at an element of volume of the molecular medium, with the precision necessary in order that the dynamical equations of the medium in bulk, based in the usual manner on these quantities alone, without directly considering thermal stresses, shall be strictly valid - a discussion in which the relation of ordinary molar mechanics to the more complete molecular theory is involved.
When A is held still, and B rotated, centrifugal action sets up vortex currents in the water in the pockets; thus a continuous circulation is caused between B and A, and the consequent changes of momentum give rise to oblique reactions.
Since the condition of the medium between A and B remains constant, even though the matter is continually changing, the momentum possessed by the matter between A and B is constant.
But the matter to the right of A is also receiving momentum from the matter to the left of it at the rate indicated by the force across A.
Substituting in the momentum equation, we obtain Pv 1 7V + y 2 I V / +PoU 2 I - v) V) = PoU2, whence U 2 = Po (I }-y21 U J .
Since no angular momentum goes out on the whole Z nwra 4 d0/dx -?- 2 pwra 4 Ud0/dt = o.
Lord Kelvin was thereby induced to identify magnetic force with rotation, involving, therefore, angular momentum in the aether.
Long previously Lord Kelvin himself came nearer this view, in offering the opinion that magnetism consisted, in some way, in the angular momentum of the material molecules, of which the energy of irregular translations constitutes.
The equations of motion can be established in a similar way by considering the rate of increase of momentum in a fixed direction of the fluid inside the surface, and equating it to the momentum generated by the force acting throughout the space 5, and by the pressure acting over the surface S.
Taking the fixed direction parallel to the axis of x, the time-rate of increase of momentum, due to the fluid which crosses the surface, is - f'fpuq cos OdS = - f f (lpu 2 -+mpuv+npuw)dS, (1) which by Green's transformation is (d(uiu 2) dy dz dxdydz.
(2) y The rate of generation of momentum in the interior of S by the component of force, X per unit mass, is fffpXdxdydz, f pXdxdydz, (3) and by the pressure at the surface S is -f.
The time rate of increase of momentum of the fluid inside S is )dxdydz; (5) and (5) is the sum of (I), (2), (3), (4), so that /if (dpu+dpu2+dpuv +dpuw_ +d p j d xdyd z = o, (b)` dt dx dy dz dx / leading to the differential equation of motion dpu dpu 2 dpuv dpuv _ X_ (7) dt + dx + dy + dz with two similar equations.
Denoting the effective inertia of the liquid parallel to Ox by aW' the momentum aW'U = 4)0W' (24) _ U i -AO' 25) in this way the air drag was calculated by Green for an ellipsoida pendulum.
The partial differential coefficient of T with respect to a component of velocity, linear or angular, will be the component of momentum, linear or angular, which corresponds.