The total kinetic energy of the system, T, is simply the sum of the kinetic energies for each particle It is convenient to decompose the velocity of each particle, vi, into the velocity of the center of mass, UG and the velocity relative to the center of mass, i;. Then 台少(+的(+门=∑m(+2m+)=m+∑m I mir:=0, and 2ia mi= m. Thus, we see that the kinetic energy of a system of particles equals the kinetic energy of a particle of mass m moving with the velocity of the center of mass, plus the kinetic energy due to the motion of the particles relative to the center of mass, G ADDITIONAL READING J. L Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 4/1, 4 /2, 4 / 3(kinetic energy expression only), 4/4, 4 /5(momentum only) Referen [1 M. Martinez-Sanchez, Unified Engineering Notes, Course 95-96 2 J.L. Merriam and L G. Kraige Engineeering Mechanics, Dynamics, Fifth Edition, wiley, 2002The total kinetic energy of the system, T , is simply the sum of the kinetic energies for each particle, T = Xn i=1 Ti = Xn i=1 1 2 miv 2 i . It is convenient to decompose the velocity of each particle, vi , into the velocity of the center of mass, vG, and the velocity relative to the center of mass, r˙ ′ i . Then, T = Xn i=1 1 2 mi(vG + r˙ ′ i ) · (vG + r˙ ′ i ) = Xn i=1 1 2 mi(v 2 G + 2vG · r˙ ′ i + ˙r ′ i 2 ) = 1 2 mv2 G + Xn i=1 1 2 mir˙ ′ i 2 , since vG · Pn i=1 mir˙ ′ i = 0, and Pn i=1 mi = m. Thus, we see that the kinetic energy of a system of particles equals the kinetic energy of a particle of mass m moving with the velocity of the center of mass, plus the kinetic energy due to the motion of the particles relative to the center of mass, G. ADDITIONAL READING J.L. Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 4/1, 4/2, 4/3 (kinetic energy expression only), 4/4, 4/5 (momentum only) References [1] M. Martinez-Sanchez, Unified Engineering Notes, Course 95-96. [2] J.L. Merriam and L.G. Kraige Engineeering Mechanics, Dynamics, Fifth Edition, Wiley, 2002. 8