# 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.

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(r

= (3t

|r(t)|

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)|

v(t) = sqrt(40t

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 == (3t

^{2}+4 + 7t+5+2t)_{a}/3 + (0+0+3t+4)_{b}/3 + (0+4+t2)_{c}/3 = (t^{2}+3t+3)_{a}+(3t+4)_{b}/3 + (t^{2}+4)c/3|r(t)|

^{2}= ra^{2}+ rb^{2}+rc^{2}= (t^{2}+3t+3)^{2}+(3t+4)^{2}/9 + (t^{2}+4)^{2}/9 = 10t^{4}/9 + 6t^{3}+ 152t^{2}/9 + 186t/9 + 113/9Components 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 }= 4t^{2}+ 12t + 9 +1 +4t^{2}/9 = 40t^{2}/9 + 12t + 10v(t) = sqrt(40t

^{2}/9 + 12t + 10)Need a fast expert's response?

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