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S E C T I O N 9 . 5 • The Center of Mass 273 2 0 2 1 1 3 y(m) x(m) 3 m 1 m 2 m 3 (a) r CM m 3 r 3 Mr CM m 1 r 1 m 2 r 2 (b) r CM Figure 9.21 (Example 9.13) (a) Two 1.0-kg particles are located on the x axis and a single 2.0-kg particle is located on the y axis as shown. The vector indicates the location of the system’s center of mass. (b) The vector sum of m i r i and the resulting vector for r CM . The position vector to the center of mass measured from We can verify this result graphically by adding together 1 r 1 # m 2 r 2 # m 3 r 3 and dividing the vector sum by M, the total mass. This is shown in Figure 9.21b. (0.75 iˆ # 1.0jˆ) m r CM
! x CM iˆ # y CM jˆ ! ! 4.0 kg)m 4.0 kg ! 1.0 m ! (1.0 kg)(0) # (1.0 kg)(0) # (2.0 kg)(2.0 m) 4.0 kg y CM ! " i m i y i M ! m 1 y 1 # m 2 y 2 # m 3 y 3 m 1 # m 2 # m 3 Example 9.14 The Center of Mass of a Rod - ! . x, where . is a constant. Find the x coordinate of the center of mass as a fraction of L. Solution In this case, we replace dm by - dx, where - is not CM is ! . M
% L 0
x 2 dx ! . L 3 3M x CM ! 1 M
%
x
dm ! 1 M
% L 0 x-
dx ! 1 M
% L 0
x.
x dx (A) Show that the center of mass of a rod of mass M and length L lies midway between its ends, assuming the rod has Solution The rod is shown aligned along the x axis in Fig- CM ! z CM ! 0. Furthermore, if we call the mass per unit length - (this quantity is called the linear Because - ! M/L, this reduces to One can also use symmetry arguments to obtain the same (B) Suppose a rod is nonuniform such that its mass per unit length varies linearly with x according to the expression L 2 x CM ! L 2 2M
# M L $ ! x CM ! 1 M
%
x
dm ! 1 M
% L 0 x-
dx ! - M
x 2 2
) L 0 ! - L 2 2M L x dm = ldx y dx O x Figure 9.22 (Example 9.14) The geometry used to find the center of mass of a uniform rod. |