Mathematics of Physics and Engineering

Non-Rigid Systems of Points 67

with position vector

1 n

We will see that some information about the motion of S can be obtained
by considering a single point mass M with position vector TCM-
EXERCISE 2.2.1. c Verify that a change of the reference point O does
not change either the location in space of the center of mass or formula
(2.2.1) for determining the location: if O' is any other frame and fj is
the position vector of rtij in O', then the position vector of CM in O' is
n > ^
rcM = (1/-W) 5Z mj Tj. Hint: fj = Vj + O'O and so TCM = TCM + O'O,
which is the same point in space.
EXERCISE 2.2.2. (a)B Show that the center of mass for three equal
masses not on the same line is at the intersection of the medians of the
corresponding triangle. (b)A Four equal masses are at the vertices of a
regular tetrahedron. Locate the center of mass.
The velocity and acceleration of the center of mass are TCM and fcM,
respectively. Differentiating (2.2.1) with respect to t, we obtain the relations

1 n

rcM = jjYlmii'i> (^2 -^2 -^2 )

J=I

1 "

fCM = jf^Zmjfj. (2.2.3)

3 = 1

To study the motion of the center of mass, suppose that the reference

frame O is inertial and denote by Fj, 1 < j < n, the force acting on the

point mass rrij. Then Fj = rrij fj, and, multiplying (2.2.3) by M, we get

the relation

n m

MfCM = Yl m 3 *3 =J2Fi = F (^2 -^2 -^4 )

3=1 3=1

Equality (2.2.4) suggests that the total force F can be assumed to act on

a point mass M at the position TCM- As a next step, we will study the

structure of the force F.