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.