- #1

stunner5000pt

- 1,447

- 2

**the motion of a system of n particles of masses Mi and coordinates Ri (i=1,2,3,...,n) is described by netwon's equations [itex] \dot{\vec{p_{i}}} = \vec{F_{i}}. [/itex] Prove taht if the system is isolated (ie total force acting on it is zero) [itex] \vec{F} = \sum_{i=1}^{n} \vec{F_{i}} = 0 [/itex], then total angular momentum of the sytem [itex] \vec{P} = \sum_{i=1}^{n} \vec{p_{i}} [/itex] is a constant of motion and express P in terms of the Mi and the centre of mass coordinate of the system**

well for the forces being zero

[tex] F = F_{1} + F_{2} + ... F_{N} = 0 [/tex]

[tex] \dot{P} = \dot{P_{1}} + \dot{P_{2}} + ... + \dot{P_{N}} [/tex]

integrate both sides by dt and it gives

[tex] P = P_{1} + P_{2} + ... P_{N} = 0 [/tex]

and that gives us what we wanted

for the center of mass coordinate system. For two masses Mi and Mi+1, let the center of mass vector M, have position vector R and let

[tex] r_{i,i+1} = r_{i+1} - r_{i} [/tex]

Let [tex] \sum_{i} = M_{i} = M [/tex]

[tex] M_{i} + M_{i+1} = M_{i,i+1} [/tex]

then [tex] \dot{r_{i}} = \dot{R} + \frac{m_{i+1}}{M_{i,i+1}} \dot{r_{i+1}} [/tex]

and [tex] \dot{r_{i+1}} = \dot{R} + \frac{m_{i}}{M_{i,i+1}} \dot{r_{i}} [/tex]

then the momentum is given by

[tex] \sum_{i=1}^{n} m_{i} \left( \dot{R} + \frac{m_{i+1}}{M_{i,i+1}} \dot{r_{i+1}} \right)= \sum_{i=1}^{n} m_{i} \dot{R} + \sum_{i=1}^{n}\frac{m_{i+1}}{M_{i,i+1}} \dot{r_{i+1}} = M \dot{R} + \sum_{i=1}^{n}\frac{m_{i+1}}{M_{i,i+1}} \dot{r_{i+1}} [/tex]

im wondering about that last equality... R is supposed to be center of mass of the whole system... but wouldnt two masses have their own center of mass Ri,i+1 rather than the common R? So my question is is it possible to replace the sums by simple terms? I doubt it...

ANy help offered is greatly appreciated!