Answer to Question #1906 in Mechanics | Relativity for nirmala

Question

Consider a system of three equal mass particles moving in space; their positions are given by .
For particle 1, a1=3t^2+4,b1=0,c1=0
For particle 2, a2=7t+5,b2=0,c2=4
For particle 3, a3=2t,b3=3t+4,c3=t^2
Determine the position and velocity of the centre of mass as functions of time.

Answer 1

The position of the centre of mass is determined as
Sum(miri) / Sum(mi) = {m(r_1a +r_2a + r_3a) + m(r_1b +r_2b + r_3b) + m(r_1c +r_2c + r_3c) }/3m =
= (3t²+4 + 7t+5+2t)_a/3 + (0+0+3t+4)_b/3 + (0+4+t2)_c/3 = (t²+3t+3)_a +(3t+4)_b/3 + (t²+4)c/3
|r(t)|² = ra²+ rb² +rc² = (t²+3t+3)² +(3t+4)²/9 + (t²+4)²/9 = 10t⁴/9 + 6t³ + 152t²/9 + 186t/9 + 113/9

Components of the velocity for each particle are
N Va Vb Vc
1. 6t 0 0
2. 7 0 0
3. 2 3 2t
The velocity of the centre of mass is
Sum(mivi) / Sum(mi) = (m (6t +7+2)a + 3mb +2tmc) / 3m = (2t+3)a + 1b + (2t/3)c
|v(t)|² = 4t² + 12t + 9 +1 +4t²/9 = 40t²/9 + 12t + 10
v(t) = sqrt(40t²/9 + 12t + 10)