Classical Mechanics

  • #1
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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!
 

Answers and Replies

  • #2
quasar987
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stunner5000pt said:
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
[itex] \vec{P} = \sum_{i=1}^{n} \vec{p_{i}} [/itex] is the total linear momentum of the system, not angular.

stunner5000pt said:
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
You got the right ideas and you might get all allowed points for this because it is a physics homework (=sloppy math fest), but your math is wrong. Personally, I would phrase the solution as:

"We want to show that P(t) = k, or equilavently, that dP/dt = 0, provided that [itex] F = F_{1} + F_{2} + ... F_{N} = 0 [/itex]. But [itex] \dot{P} = \dot{P_{1}} + \dot{P_{2}} + ... + \dot{P_{N}} = F_{1} + F_{2} + ... F_{N} = F \Rightarrow \dot{P} = 0[/itex] QED"

But if you want to keep your way, just replace [itex] P = P_{1} + P_{2} + ... P_{N} = 0 [/itex] by [itex] P = P_{1} + P_{2} + ... P_{N} = k \in \mathbb{R} [/itex]

stunner5000pt said:
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
So my question is is it possible to replace the sums by simple terms? I doubt it...

ANy help offered is greatly appreciated!

Dude, that is immensely complicated for nothing. The proof of this is as direct as the first one. I regret that I cannot answer your question directly, but I can hardly follow what you do in your weird notation. Let me show you a simpler path instead...

"We want to express P in terms of the Mi and the center of mass coordinate.

By definition, the center of mass coordinate is

[tex]\vec{R}_{CM} \sum_{i} M_i = \sum_{i}M_i \vec{R}_i[/tex]

And

[tex]\vec{P} = \sum_{i} \vec{P}_i = \sum_{i} M_i \vec{v}_i = \sum_{i} M_i \frac{d\vec{R}_i}{dt} = ...[/tex]"

You finish.
 
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  • #3
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so then all i needed to do was
[tex]\vec{P} = \sum_{i} \vec{P}_i = \sum_{i} M_i \vec{v}_i = \sum_{i} M_i \frac{d\vec{R}_i}{dt}[/tex] and i would be done?
 
  • #4
quasar987
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Of course not! I wrote "..." and added "You finish". Do you know what "express P in terms of the Mi and the centre of mass coordinate" means?

It means you have to find a way to write P = "something involving Mi a R_CM". I see Mi but I don't see R_CM in that last equality. Hence, you're not done.
 
  • #5
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couldnt you just do this
[tex] \frac{\vec{R_{CM}}}{dt} \sum_{i} M_{i} = \sum_{i} M_{i} \frac{dR_{i}}{dt} [/tex]

Rcm is certainly not a constant... so its deriavtive is not zero.
 
  • #6
quasar987
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That is indeed the missing step!

[tex]\vec{P} = M\dot{\vec{R}}_{CM}[/tex]

And pushing this equation a step further, you get that

[tex]\dot{\vec{P}} = \vec{F} = M\ddot{\vec{R}}_{CM}[/tex]

This is my favorite theorem of classical mechanics. It says that you're not wasting your time calculating the effect of forces on pointlike particles that do not even exist in nature. That is, because for any body (i.e. system of particle) whatesoever, however eratic the motion of the particles composing it, you know that the total force applied on it will have the effect of inducing a motion of its center of mass identical to that of a point particle of mass M located at said center of mass.
 
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  • #7
Galileo
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Ehm, weren't you supposed to show the conservation of angular momentum?

Anyway, I think the proof is incomplete. You said: Since the total force is zero: [itex] F = F_{1} + F_{2} + ... F_{N} = 0 [/itex]
What are the [itex]F_i[/itex]? From the descrition of the problem it is clear that it is the external force on the i-th particle. But from the next step you took it to mean the total force on the i-th particle. But there is no a priori reason to think the sum of the internal forces is zero without invoking Newton's third law. It is given that the total external force on the system is zero, so you only need to deal with the internal forces. These cancel only because Newton's 3rd law says so, not from the premise of the problem. This reasoning isn't shown in your solution.
 
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  • #8
quasar987
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Mmh, yes.

A system is isolated if the sum of the external forces on it is 0. The way the problem is worded is missleading. He says: "Consider an isolated system, i.e. such that [itex]\vec{F}=\sum \vec{F}_i = 0[/itex]", which can easily be interpreted, if you overlooked the key word "isolated", as "such that the sum of the total force is 0".

Since this is not the case, we must first and foremost prove that the sum of the total force IS in fact 0. This boils down to proving that the sum of the internal forces is 0. You got some more job to do stunner! :tongue2:
 
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  • #9
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prove the sum of the forces is zero? Do we really need to do that ? Isnt that basic assumption?

And dont we want the angular momentum of the system on the whole and not hte particles within?

i guess the question mislead because it uses P for the angular momentum rather the usual L

the L of the entire system
[tex] L = \sum_{i} m_{i} (r_{i} \times \dot{r_{i}}) [/tex]
isnt it ?
 
  • #10
quasar987
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I'm reaaally confused here. I thought that when you wrote "angular momentum", that was just a typo and that you meant "linear momentum". You're telling me that Angular momentum was not a typo? Who uses P for angular momentum?!
 
  • #11
Galileo
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If you consider [itex]P_i[/itex] to be angular momentum, then why did you use [itex]F_i=\dot P_i[/itex]?

The question has an error. Either they asked for the total linear momentum or they made a typo and meant [itex]\vec L = \sum_i =\vec r_i \times \vec p_i[/itex].
If you can do one you can easily do the other, since the proofs are very similar.
 

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