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Answer to Question #1906 in Mechanics | Relativity for nirmala

Question #1906
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.
Expert's answer
The position of the centre of mass is determined as
Sum(miri) / Sum(mi) = {m(r1a +r2a + r3a) + m(r1b +r2b + r3b) + m(r1c +r2c + r3c) }/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)

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