II.—THE CONSERVATION OF MOMENT OF MOMENTUM.
We give here an elementary investigation of the fundamental dynamical principle which has been of such importance throughout this volume.
§ 8. Case where there are no forces.
Newton’s first law of motion tells us that a particle in motion if unacted upon by force, will move continuously in a straight line without change of velocity.
Let A0, Fig. [60], be the position of the particle at any moment. Let A1 be its position after the time t; A2 be the position at the time 2t; A3 be the position at the time 3t, and so on.
Then the first law of motion tells us that the distances A0 A1, A1 A2, A2 A3, A3 A4, must form parts of the same straight line and must be all equal.
If lines O A0, O A1, O A2, etc., be drawn from any fixed point 0, then the areas of the triangles O A0 A1, O A1 A2, O A2 A3, 0 A3 A4, will be all equal. For each area is one-half the product of the base of the triangle into the perpendicular O T from O on A0 A1, and, as the bases of all the triangles are equal, it follows that their areas are equal.
Thus we learn that a particle moving without the action of force will describe around any fixed point O equal areas in equal times.
Fig. 60.—First Law of Motion exemplifies
Constant Moment of Momentum.
The product of the mass of the particle and its velocity is termed the momentum. If the momentum be multiplied by O T the product is termed the moment of momentum around O. We have in this case the simplest example of the important principle known as the conservation of moment of momentum.
The moment of momentum of a system of particles moving in a plane is defined to be the excess of the sum of the moments of momentum of those particles which tend round O in one direction, over the sum of the moments of momentum of those particles which tend round O in the opposite direction.
If we deem those moments in one direction round O as positive, and those in the other direction as negative, then we may say that the moment of momentum of a system of particles moving in a plane is the algebraical sum of the several moments of momentum of each of the particles.
§ 9. A Geometrical Proposition.
The following theorem in elementary geometry will be required:—
Fig. 61.—A Useful Geometrical Proposition.
Let A B and A C be adjacent sides of a parallelogram, Fig. [61], of which A D is the diagonal, and let O be any point in its plane. Then the area O A C is the difference of the areas O A D and O A B.
Draw D Q and C P parallel to O A. Then O A D = O A Q, whence O A D – O A B = O B Q = O A P = O A C.
§ 10. Relation Between the Change of Moment of Momentum and the Force Acting on the Particle.
Fig. 62.—Acceleration of Moment of Momentum equals Moment of Force.
Let A1 and A2, Fig. [62], be two adjacent points on the path of the particle, and let A1 Q and A2 R be the tangents at those points. Let S Q represent the velocity of the particle at A1, and SR the velocity of the particle at A2. Then Q R represents both in magnitude and direction the change in velocity due to the force F, which we suppose constant both in magnitude and direction, while the particle moves from A1 to A2 in the small time t; we have also Q R = F t ÷ m.
Complete the parallelogram S Q R U, and let fall O P1, O P2, O T perpendiculars from O on S Q, S R, S U respectively. Since S Q is the velocity of the particle when at A1 the moment of momentum is m O P1 × S Q; when the particle is at A2 the moment of momentum is m O P2 × S R. Whence the difference of the moments of momentum at A1 and A2 is m (O P2 × S R - O P1 × S Q) = 2 m (O S R - O S Q) = 2 m O S U = m O T × S U = m O T. Q R = F t × O T. But in the limit S coincides with A1 and A2, and we see that the gain in moment of momentum is t times the moment of the force around O. Hence we deduce the following fundamental theorem, in which, by the expression acceleration of moment of momentum, we mean the rate at which the moment of momentum increases:—
If a particle under the action of force describes a plane orbit, then the acceleration of the moment of momentum around any point in the plane is equal to the moment of the force around the point.
If the force is constantly directed to a fixed point, then the moment of the force about this point is always zero. Hence the acceleration of the moment of momentum around this point is zero, and the moment of momentum is constant. Thus we have Kepler’s law of the description of equal areas in equal times, and we learn that the velocity is inversely proportional to the perpendicular on the tangent.
§ 11. If Two or More Forces Act on a Point, then the Acceleration of the Moment of Momentum, due to the Resultant of these Forces, is Equal to the Algebraic Sum of the Moments of Momentum due to the Action of the Several Components.
Let A D, Fig. [61], be a force, and A C and A B its two components. Then, since O A D = O A B + O A C, we see that the moment of A D around O is equal to the sum of the moments of its components. Hence we easily infer that if a force be resolved into several components the moment of that force around a point is equal to the algebraical sum of the moments of its several components.
The acceleration of the moment of momentum around O, due to the resultant of a number of forces, is equal to the moment of that resultant around O. But, as we have just shown, this is equal to the sum of the moments of the separate forces, and hence the theorem is proved.
§ 12. If any Number of Particles be Moving in a Plane, and if they are not Subjected to any Forces save those which arise from their Mutual Actions, then the Algebraic Sum of their Moments of Momentum round any Point is Constant.
This important theorem is deduced from the fact stated in the third law of motion, that action and reaction are equal and opposite. Let us take any two particles; then, the acceleration of the moment of momentum of one of them, A, by the action of the other, B, will be the moment of the force between them. The acceleration of the moment of momentum of B by the action of A will be the same moment, but with an opposite sign. Hence the total acceleration of the moment of momentum of the system by the mutual action of A and B is zero. In like manner we dispose of every other pair of actions, and thus, as the total acceleration of the moment of momentum is zero, it follows that the moment of momentum of the system itself must be constant.
This fundamental principle is also known as the doctrine of the conservation of areas. It may be stated in the following manner:—
If a system of particles are moving in a plane under the influence of their mutual actions only, the algebraic sum of the areas swept out around a point, each multiplied by the mass of the particle, is directly proportional to the time.
§ 13. If a Particle of Mass m, is Moving in Space under the Action of any Force F, then the Projection of that Particle on any Fixed Plane will Move as if it were a Particle of Mass m Acted upon by that Component of F which is Parallel to the Plane.
This is evident from the consideration that the acceleration of the particle parallel to the plane must be proportional to this component of F.
Let us now suppose a system of particles moving in space under their mutual actions. The projections of these particles on a plane will move as if they were the particles themselves subjected to the action of forces which are the projections of the actual forces on the same plane, and as the reactions between any two particles are equal and opposite, the projections of those reactions on the plane are equal and opposite. Hence the proof already given of the constancy of the moments of momentum of a plane system, will apply equally to prove the constancy of the moments of momentum of the projections of the particles on the plane. Hence we have the following important theorem:—
Let a system of particles be moving in space under the action of forces internal to the system only. Let any plane be taken, and any point in that plane, and let the momentum of each particle be projected into the plane, then the algebraic sum of the moments of these projections around the point is constant.
§ 14. On the Principal Plane of a System.
Let us suppose a system of particles moving under the influence of their mutual actions. Let O be any point, and draw any plane L through O. Then the moment of momentum of the system around the point O and projected into the plane L is constant. Let us call it S. If another plane, L´, had been drawn through O, the similar moment with regard to L´ is S´. Thus for each plane through O there will be a corresponding value of S. We have now to show that one plane can be drawn through O, such that the value of S is greater than it is for any other plane. This is the principal plane of the system.
If v be the velocity of a particle, then in a small time t it moves over the distance v t. If p be the perpendicular from O on the tangent to the motion, then the area of the triangle swept round O in the time t is ½ p v t, and we see that the momentum is proportional to the mass of the particle multiplied into the area swept over in the time t. The quantity S will, therefore, be proportional to the sum of the projections of the areas in L, swept over in the time t, each increased in the proportion of the mass of the particle. It is easily seen that the projection of an area in one plane on another is obtained by multiplying the original area by the cosine of the angle between the two planes. For if the area be divided into thin strips by lines parallel to the line of intersection of the planes, then in the projection of these strips the lengths are unchanged, while the breadths are altered by being multiplied by the cosine of the angle between the two planes. If, therefore, we mark off on the normal to a plane L a length h proportional to any area in that plane, then the projection of this area on any other plane L´ may be measured by the projection of h on the normal to L´.
Fig. 63.—Moment of Momentum unaltered
by Collision.
To determine the moment of momentum resolved in any plane we therefore proceed as follows: Draw a plane through O, and the tangent to the path of one of the particles, and mark off on the normal drawn through O to this plane a length l proportional to the moment of momentum. Repeat the same process for each of the other particles with lengths l´, l″, etc., on their several normals. Suppose that l, l´, l″ represent forces acting at O, and determine their resultant R. Then R, resolved along any other direction, will give the component of moment of momentum in the plane to which that direction is normal. In any plane which passes through R the component of moment of momentum is zero. The plane perpendicular to R contains the maximum projection of moment of momentum. This is the principal plane of the system which we have seen to be of such importance in connection with the nebular theory.
§ 15. Collisions.
The conservation of moment of momentum remains true in a system, even though there may have been actual collisions between the several parts. This is included in the proof already given, for collisions are among the mutual actions referred to. It may, however, be instructive to give a direct proof of a particular case.
Let two particles collide when meeting in the directions A P and B P (Fig. [63]) respectively. Whether the particles be elastic or inelastic is quite immaterial, for in both cases the action and reaction must be equal and opposite, and take place along some line P Q. The action on the particle moving along A P will give to it an acceleration of moment of momentum which is equal to the moment of the action around O. The acceleration of the moment of momentum coming along B P will be equal and opposite. Thus the total acceleration of the moment of momentum is zero. Hence the collision has no effect on the total moment of momentum.
§ 16. Friction and Tides.
We have shown that such actions as collisions cannot affect the moment of momentum of the system, neither can it be affected by friction of one body on another. Here, as in the former case, the actions and reactions are equal and opposite, and consequently the accelerations of moment of momentum are zero. Nor is it possible for any tidal action to affect the total moment of momentum of the system. Every such action must be composed of the effects of one particle in the system on another, and as this must invariably produce an equal and opposite reaction the total moment of momentum is unaltered.