z = S mi zi /S mi.B3. Motion on a prescribed trajectory
Material body. System of material points. Conservation of momentum. Centre of mass
We explain the diversity of the mechanical properties of bodies, that is, of their mode of reaction to external effects (forces from outside a system) by the fact that the individual material points of different kinds of bodies have different interactions (inside their systems); rigid ones differ from liquid or gaseous ones, hard ones from soft ones, etc. However, to start with, we will not postulate regarding these inner forces more than that pairs of mass points interact with equally large and mutually opposite forces, that is, if the mass point m acts on the mass point m1 with the force q, then m1 acts on m with the force -q. Then the internal forces do not affect the state of motion of the system of points, just as the internal forces in the systems, illustrated by Fig.38 or Fig. 15. Only external forces, that is those arising from points outside a system, influence it.
If the mass system consists of two mass points m1 and m2 and there act in the direction of the straight line linking them forces, which arise from masses outside the system, then m1 and m2 obtain in the time t the velocities v1 and v2. Then p1t = m1v1 and p2t = m2v2 , whence (p1 + p2)t = m1v1 + m2v2 - This quantity is called the momentum of the system. If, in addition, there act on m1v1 + m2 internal forces, that is, forces which the masses exert on each other, these forces are equal and opposite to one another: q = -q. The sum of the driving forces is then (p1 + p2 + q1 + q2)t = (p1 + p2)t, as before, and the momentum of the system is the same. Hence only external forces determine the momentum of a system, that is, only those arising from masses outside the system.
The same considerations can be made for arbitrarily many free masses m1m2m3 ···, forces p1p2p3 ··· and velocities n1v2v3 ···. We decompose the forces and the velocities in three orthogonal directions x, y, z. If there act between the masses pairwise equal, but oppositely directed forces q and -q, r and -r, s and -s, they yield in each of the three directions pairs of equal, but opposite components and do not affect the sum of the driving forces. Again, the momentum is only determined by the external forces.This if the Law of Conservation of the Momentum of a System.
This law receives a different form, if the centre of mass of the system is introduced, which in a certain sense takes the place of the masses (it is sometimes also called, not quite correctly, the centre of gravity). We define the centre of mass as follows: Given two points in space with the coordinates x1, y1, z1 and x2, y2, z2, then the coordinates of the centre of the line linking them are x0 = (x1+x2)/2, y0 = (y1+y2)/2, z0 = (z1+z2)/2. In general, it turns out that, given n points xi, yi, zi (i = 1,2,3 ··· n), then the centre of the system of points has the coordinates:
x0 = S xi/n, y0 = S yi/n ,z0 = S zi/n.
Beside this geometrical centre, we also have the mass centre. If we imagine masses mi fixed at the points xi, yi, zi, then we can define the coordinates of the point
x =S mixi /S mi,
h = S mi yi /S mi,
z = S mi zi /S mi.
If we now set S mi = m, then
mx = S mi xi, mh =S mi
yi, mz = S mi
zi.
This is the definition of the centre of mass. It seems to be linked to a definite reference system, that is, it seems to change with the system; however, in fact, it does not. (We omit the proof.) The equation mx = S mi xi establishes a relationship of the mass system to the yz-plane and is valid for every system, that is, also for every yz-plane.
If we place the origin of the coordinate system at the centre of mass, then x=0, h= 0, z = 0, and since in the equations for x h z the numeratorS mi does not vanish, one has then S mi xi = S mi yi = S mi zi = 0. Hence the numerical treatment of problems involving the centre of gravity is greatly simplified, if the origin of the coordinate system is placed there .
In the expressions for x, h, z , the coordinates x y z can change uniformly or uniformly accelerated, depending on whether a force acts on the corresponding mass; the centre of mass (x, h, z) moves accordingly with uniform velocity or acceleration or not at all. If now internal forces arise, which act between pairs of masses m1 and m2, then there arise mutually opposite displacements and therefore only additions in the expressions for x h z, which cancel one another. The motion of the centre of mass (centre of gravity) is therefore only determined by the external forces.
If v1v2v3··· are the velocities of masses m1, m2, m3, ··· in any direction and V is the velocity of the centre of mass in the same direction, then V = S mv/S m , and, if the total mass S m = M, then VM = S mv. Hence the impulse of a system (body) is computed as if its mass is at its centre of mass. If j1, j2, j3, ··· are the accelerations of the point masses m1, m2, m3, ··· in any directions and F is the acceleration of the centre of mass in the same direction, we obtain the acceleration of the centre of mass in one direction by summing all force components in the same direction and dividing by the total mass. The centre of mass moves as if the total mass is located at it and the forces were applied to it. Only under the action of an external force does the centre of mass accelerate.
As an example, the theorem of conservation of momentum provides the basis for the study of impact (collision). Impact involves two events: During the first, the centres of gravity of the colliding bodies approach each other (so that they flatten out during their contact and deform), during the second, they move away from each other (because the deformation recedes). The first stage is called compression, the second restitution. If during the second stage the entire deformation recedes and during the impact none of the mechanical energy is changed into another form (for example, heat), the impact if called elastic, if there is no restitution, it is called inelastic. The real situations (the half-elastic cases) lie in between.
There does not yet exist a strict theory of impact and one is satisfied with an analysis of certain idealized limiting cases. We will speak here only of the central, straight impact of two spheres A and B, the centres of which move along the same straight line. An impact is called central or excentral, depending on whether the line joining the centres of gravity (the centres of the spheres) coincide with the impact normals or not ( that is, with the normals at the point of contact of the two surfaces).The impact is called straight, if ahead of it the bodies do not rotate and move relative to each other along the impact normal; otherwise it is called oblique.
The two spheres in Fig. 44a represent the system, the motion of which we will study. They move in the same direction, but B moves faster than A. At impact, B transfers some of its velocity to A: This transfer occurs during the compression period (that is, not instantaneously, as will be discussed below); A and B flatten each other. They do so until their velocities are equal. Then the compression stage ends and different motions will occur, depending on whether the impact is elastic or inelastic.
Let m1 and m2 be the two masses, v1 and v2 their velocities ahead of the collision, w1 and w2 their velocities afterwards (reckoned positive in the sense of the positive x-coordinate). If u1 is the velocity of the system, that is, of the common centre of mass, then ahead of the collision m1v1 + m2v2 = (m1+ m2)u1. There act no external forces, whence the motion of the centre of mass does not change during impact. If the collision is inelastic, the masses m1 and m2 remain in contact and proceed with the common velocity, that is with the velocity u1. The velocity, common to both masses, is therefore u1 = (m1v1 + m2v2)/(m1 + m2); this is in the case of soft (completely inelastic bodies) their common velocity after collision. Their velocity is zero, that is, they come to rest, if m1v1 + m2v2 = 0, that is, if their momenta are equal, but have opposite signs.
During elastic impact, the situation changes. As soon as the compression period ends, their common velocity is again u1 = (m1v1 + m2v2)/(m1 + m2). The mass m1, coming from C, has given away some of its velocity (action) and reduced it to u1, that is by (v1 - u1). During the restitution period, m1 receives an impulse towards C (reaction). Its velocity u1 loses thereby once again the same amount of velocity which it already had given away voluntarily. Its velocity sinks thereby to u1- (v1 - u1) = 2 u1- v1, its velocity (w1) after impact. Similarly, after collision, the velocity of m2 is w2 = 2 u1- v2. After substitution of the value of u1 in the formulae for w1 and w2, you find
w1 = (m1v1
+ m2(2v2 - v1)/(m1
+ m2),
w2 = (m2v2
+ m1(2v1 - v2)/(m1
+ m2).
If m1 = m2, then w1 = v2 and w2 = v1, that is, after impact, the two bodies have interchanged their velocities. If m1 was prior to the impact at rest, then after the impact m2 stays at rest and m1 has the initial velocity of m2. You can demonstrate this by means of two spheres made out of ivory, suspended side by side, by letting one fall against the other at rest (Fig. 44b).
The transfer of the motion takes place during the compression period, that is, not instantaneously. You can also show this: Let the first sphere in (Fig. 44c) drop from a certain height against the second, then the last of the spheres, which was at rest, rises to the initial height of the first, while all the others do not move. If at the impact of the first sphere against the second the motion were transferred instantaneously to the entire mass, then motion would occur as if the first sphere were impacting another one of six times its mass - there are six equal spheres! However, every sphere passes on the velocity originating from the first sphere to the next sphere.
You can also see that the motion from one body does not pass on instantaneously to another body, for example, by shooting a bullet at a glas pane, which will not splinter. - The ballistic pendulum is an application of the laws of inelastic impact; it is employed to measure the final velocity of small bullets.
a) A body at rest, which can rotate about a fixed axis
If the points A, B,D in Fig. 39 are fixed, no point of the body can move. If only A and B are held fixed, D can move, but only in such a manner that it moves out of the plane of the drawing (forwards or backwards). Since its distances from A and B cannot change, and therefore also not its distance CD from the straight line AB, it can describe a circle with the radius CD, which is perpendicular to AB, but no other motion (Fig. 45).
What is true for D, is true for every other point away from the straight line AB. Since points can only move in circles the planes of which are perpendicular to AB, only forces can cause motion which lie in planes perpendicular to AB. In other words, these forces can be visualized by straight lines in these planes. The direction of a motion is identical with that of the moving force, the direction of motion on a circle at each point with that of the tangent at that point, whence the forces must act along tangents to the circles, which are possible during the motion. Hence the lines of action of the forces cannot be radial, that is, it cannot intersect the axis.
Other forces can induce pressure or tension, but are balanced by the body's rigidity or the bearings at the points A and C. - The line AB is called an axis, the points A and B its poles, the motion rotation about the axis and the circles such described parallel circles ( they are parallel, because all of them are perpendicular to the same line AB).
Let the body be fixed at the points A and B by two immovable pins (Fig. 46) and be able to rotate about the vertical axis AB, so that gravity acts parallel to the axis and has no influence on the motion. Let a force act on the body at D. Place a plane (Fig. 47) through D at right angle to the axis of the body (and in the plane of the page ). If C represents its intersection with the axis AB, then B is vertically above, A vertically below C in the plane of the figure (Fig. 47). Let P be the force component which falls into the cross-section through D, perpendicular to the axis.
It generates motion because it acts in a plane which is perpendicular to the axis and does not intersect it. Due to its rigidity, the motion of the body does not change - as will be proved below - whether we place P at D or move it along its direction, so that its point of attack is somewhere along this line. We will consider both cases, in order to emphasize the significance of the static moment, which is important for the rotation of a body.
Displacement of the point of attack of a force
If a body is rigid, it does not matter where the force attacks, provided that this point lies on the line of attack of the force. It is immaterial, whether (Fig.48) the force P acts at A or at B on the line PC. In fact, you can let two forces P1 and P2 act at B, which have the same magnitude and opposite directions, so that they will balance each other! If you now let them be equal to P and act along its direction, then you can remove P and P2 without changing the state of the body, for they balance each other, because A and B are rigidly linked, that is, they tend to move equally strongly the rigid line AB in opposite directions. Hence you are left with the force P1 at B, which has the same direction and magnitude as P, since you have made P1 = P. Hence the point of attack of P has been shifted along its line of action from A to B.
You should imagine the rigid body to be like a frame on wheels (Fig. 49) and the forces P, P1, P2 three equally strong horses: It does not matter whether we place the horse P at A or at B.
