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Generalized Newton’s second law Hiroki Okubo P The term F becomes the vector sum of all forces acting on all particles of the isolated system from sources external to the system, and f becomes the vector sum of all forces on all particles produced by the internal actions and reactions between particles. This last sum is identically zero since all internal forces occur in pairs of equal and opposite actions and reactions. By using Eq. (1), we have 00 ρi m11 i 00 11 G 00 11 00 11 00 11 ’ ρ ri i ρ r O rP P X Figure 1: Relative motion F =m d2 r dt2 (4) Equation (4) is the generalized Newton’s second law of motion for a mass system and is called the 1 Generalized Newton’s sec- equation of motion of m. The equation states that the resultant of the external forces on any system ond law of masses equals the total mass of the system times We extend Newton’s second law of motion to cover the acceleration of the center of mass. This law a general mass system which we model by consid- expresses the so-called principle of motion of the ering n mass particles bounded by a closed surface mass center. Equation (4) may be expressed in component in space. Figure 1 shows a representative particle of mass mi of the system isolated with forces form using x-y-z coordinates. Thus, F 1 , F 2 , F 3 ... acting on mi from sources external X d2 y X d2 z d2 x X to the system boundary, and forces f 1 , f 2 , f 3 ... Fy = m 2 , Fz = m 2 Fx = m 2 , dt dt dt acting on mi from sources internal to the bound(5) ary. The internal forces are forces of reaction with other mass particles within the boundary. The particle of mass mi is located by its position vector r i 2 Work-energy measured from the nonaccelerationg origin O. The center of mass G of the isolated system of particles 2.1 Work-energy relation is located by the position vector r which is given For the entire system, the sum of the work-energy by P X equations written for all particles is (U1−2 ) = mr = mi r i (1) P ∆Ti , which may be represented by Newton’s second law gives T1 + U1−2 = T2 (6) F 1 + F 2 + · · · + f 1 + f 2 + · · · = mi r̈ i (2) P where U1−2 = (U1−2 ), the work done by all forces, and ∆T is the change in the total kinetic Similar equation may be written for each particle. P energy T = T of the system. If these equations written for all particles of the i For a nonrigid mechanical system, a part of the system are added together, the result is work done by the external forces goes into changX X X F+ f= mi r̈ i (3) ing the internal elastic potential energy Ve . If the 1 P The time derivative P of H O is Ḣ O = P (r i × mi v̇ i ). The summation is (r i × mi v̇ i ) = (r i × F i ), which is the vector sum of the moments about O of P all forces acting on all particles of the system, M O. work done by the gravity forces is excluded from the work term and is accounted for instead by the changes in gravitational potential energy Vg , then ′ done on the system we may equate the work U1−2 during an interval of motion to the change ∆E in the mechanical energy of the system ′ T1 + Vg1 + Ve1 + U1−2 = T2 + Vg2 + Ve2 3.2.2 (7) About the mass center G The angular momentum of the mass system about the mass center G is X HG = ρi × mi ṙ i (13) We may write the velocity of the representative particle as v i = v + ρ̇i (8) By using Eq. (8) we obtain X where v is the velocity of the mass center G and ρ̇i HG = ρi × mi ρ̇i (14) is the velocity of mi with respect to a translating reference frame moving with the mass center G. We The expression of Eq. (13) is called the absolute write the kinetic energy of the system as angular momentum, and the expression of Eq. (14) is relative angular momentum. Differentiating Eq. X1 X 1 1 mi v i · v i = mv 2 + mi | ρ̇i |2 (9) (13) with respect to time gives T = 2 2 2 X P because mi ρi = 0. Equation (9) expresses the Ḣ G = ρi × mi r̈ i (15) fact that the total kinetic energy of a mass system P P ρi × mi r̈ i = ρi × F i . Thus equals the kinetic energy of mass-center translation The summation is of the system as a whole plus the kinetic energy due X M G = Ḣ G (16) to motion of all particles relative to the mass center. 2.2 3 3.1 Kinetic energy expression 3.2.3 Impulse-momentum About an arbitrary point P The angular momentum about an arbitrary point P as shown in Fig. 1 can be expressed as X X HP = ρ′i × mi ṙ i = (ρ + ρi ) × mi ṙ i (17) Linear momentum The linear momentum of the system is defined as the vector sum of the linear momenta of all of its The rearranging gives particles. By substituting Eq. (8), we obtain X H P = H G + ρ × mv (18) (10) G= mi (v + ρ̇i ) = mv P M P , we make use of the The linear momentum of any system of constant In order to calculate mass is the product of the mass and the velocity of principle of moment. Thus X its center of mass. The time derivative of G is the M P = Ḣ G + ρ × ma (19) resultant external force acting on the system. X F = Ġ (11) Equation (19) enables us to write the moment equation about any convenient moment center P . 3.2 3.2.1 Angular momentum References About a fixed point O [1] J. L. Meriam and L. G. Kraige, (2001), Engineering Mechanics, Volume 2, Dynamics, 5th edition, Wiley The angular momentum of the mass system about the fixed point O is defined as X HO = (r i × mi v i ) (12) 2