10.5 Center of Mass Consider two point-like particles with masses m1 and m2 . Choose a coordinate system   with a choice of origin such that body 1 has position r1 and body 2 has position r2 (Figure 10.5). Figure 10.5 Center of mass coordinate system.  The center of mass vector, R cm , of the two-body system is defined as    m1 r1 + m2 r2 . R cm = m1 + m2 (10.5.1) We shall now extend the concept of the center of mass to more general systems. Suppose we have a system of N particles labeled by the index i = 1, 2, 3, , N . Choose a  coordinate system and denote the position of the i th particle as ri . The mass of the system is given by the sum 10-1 i= N msys = ∑ mi (10.5.2) i=1 and the position of the center of mass of the system of particles is given by  1 i= N  R cm = ∑m r . msys i=1 i i (10.5.3) (For a continuous rigid body, each point-like particle has mass dm and is located at the  position r′ . The center of mass is then defined as an integral over the body,  R cm = ∫  dm r′ body ∫ dm . (10.5.4) body Example 10.2 Center of Mass of the Earth-Moon System The mean distance from the center of the earth to the center of the moon is rem = 3.84 ×108 m . The mass of the earth is me = 5.98 × 1024 kg and the mass of the moon is mm = 7.34 ×1022 kg . The mean radius of the earth is re = 6.37 × 106 m . The mean radius of the moon is rm = 1.74 ×106 m . Where is the location of the center of mass of the earthmoon system? Is it inside the earth’s radius or outside? Solution: The center of mass of the earth-moon system is defined to be  1 i= N  1   R cm = (mere + mmrm ) . ∑ miri = msys i=1 me + mm (10.5.5) Choose an origin at the center of the earth and a unit vector î pointing towards the moon,   then re = 0 . The center of mass of the earth-moon system is then  R cm =    mm rem m r 1 (me re + mm rm ) = = m em ˆi me + mm me + mm me + mm  (7.34 ×1022 kg)(3.84 ×108 m) ˆ R cm = i = 4.66 ×106 m ˆi 24 22 (5.98 ×10 kg + 7.34 ×10 kg) (10.5.6) (10.5.7) The earth’s mean radius is re = 6.37 × 106 m so the center of mass of the earth-moon system lies within the earth. 10-2 Example 10.3 Center of Mass of a Rod A thin rod has length L and mass M . =M/L (x) = M 0 2 L x2 M L L a) uniform rod b) non-uniform rod Figure 10.6 a) Uniform rod and b) non-uniform rod (a) Suppose the rod is uniform (Figure 10.6a). Find the position of the center of mass with respect to the left end of the rod. (b) Now suppose the rod is not uniform (Figure 10.6b) with a linear mass density that varies with the distance x from the left end according to λ λ (x) = 20 x 2 (10.5.8) L where λ0 is a constant and has SI units [kg ⋅ m -1 ] . Find λ0 and the position of the center of mass with respect to the left end of the rod. Solution: (a) Choose a coordinate system with the rod aligned along the x -axis and the origin located at the left end of the rod. The center of mass of the rod can be found using the definition given in Eq. (10.5.4). In that expression dm is an infinitesimal mass element and r is the vector from the origin to the mass element dm (Figure 10.6c). =M/L x x=0 +x dm = dx x=L Figure 10.6c Infinitesimal mass element for rod Choose an infinitesimal mass element dm located a distance x′ from the origin. In this problem x′ will be the integration variable. Let the length of the mass element be dx′ . Then (10.5.9) dm = λ dx′  The vector r = x′ î . The center of mass is found by integration 10-3   1 1 x 1 2 R cm = x′ ∫ r dm = ∫ x′dx′ ˆi = M body L x′=0 2L x′=L x′=0 ˆi = 1 (L2 − 0) ˆi = L ˆi . (10.5.10) 2L 2 (b) For a non-uniform rod (Figure 10.6d), (x ) = x x=0 0 2 L 2 x +x dm = (x ) dx x=L Figure 10.6d Non-uniform rod the mass element is found using Eq. (10.5.8) λ0 2 x′ dx′ . (10.5.11) L2  The vector r = x′ î . The mass is found by integrating the mass element over the length of the rod dm = λ ( x′)dx′ = λ = x=L M = ∫ dm = ∫ λ ( x′)dx′ = body x′ =0 x′=L λ0 x=L 2 λ λ λ = 02 (L3 − 0) = 0 L . (10.5.12) x′ dx′ = 02 x′3 2 ∫ x′=0 L x′=0 3L 3L 3 Therefore λ0 = 3M L (10.5.13) The center of mass is again found by integration  1 3 x 3 x 3  R cm = ∫ r dm = ∫ λ (x′)x′ dx′ î = 3 ∫ x′ dx′ î M body λ0 L x′=0 L x′=0  x′= L 3 3 3 R cm = 3 x′ 4 î = 3 (L4 − 0) î = L î. x′=0 4 4L 4L (10.5.14) . 10-4