Angular momentum

  • #1
1,444
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Consider a system of two particles of (fixed) masses m1 and m2 with position vectors r1 and r2 respectively (relative to a fixed origin of coordinates)
a) Expres the total angular momentum L = r1 x mv1 + r2 x mv2 of this two body system in terms of hte cnetre of mass vector R amd the relative coordinate vectpr r = r1 - r2 Interpret your reult

wouldnt it be something like this
[tex] r_{1} = R + \frac{m_{2}}{M} r [/tex]
[tex] r_{2} = R + \frac{m_{1}}{M} r [/tex]
then [tex] \dot{r_{1}} = \dot{R} + \frac{m_{2}}{M} \dot{r} [/tex]
[tex] \dot{r_{2}} = \dot{R} + \frac{m_{1}}{M} \dot{r} [/tex]

then [tex] L = (R + \frac{m_{2}}{M} r) \times m \dot{r_{1}} = \dot{R} + \frac{m_{2}}{M} \dot{r} + (R + \frac{m_{1}}{M} r) \times m \dot{r_{2}} = \dot{R} + \frac{m_{1}}{M} \dot{r} [/tex]

dont think i can go much past this point .... can i ?

please help!
 

Answers and Replies

  • #2
Galileo
Science Advisor
Homework Helper
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Check your expression for [itex]r_2[/itex]. As it is now, both [itex]r_1[/itex] and [itex]r_2[/itex] point in the same direction from the center of mass. Intuitively this can't be right, you missed a minus sign.
 
  • #3
1,444
2
so then r2 is
[tex] r_{2} = R - \frac{m_{1}}{M} r [/tex]

then
[tex] L = (R + \frac{m_{2}}{M} r) \times m_{1}(\dot{R} + \frac{m_{2}}{M} \dot{r}) + (R - \frac{m_{1}}{M} r) \times m_{2} (\dot{R} - \frac{m_{2}}{M} \dot{r}) [/tex]

can the cross product be computed from this? Or can it be left liek this?

The second part of this questions asks
Under wht conditions the agular momentum of the centre of mass Lcm nad the angular momentum of hte rlation motion L rel are constants of motions and what physical significance of this is.

this would happen when the frame from which the angularm ometnu mis viewed is non acclerating and the body which is being viewed is non accelerating as well. SO inertial frames from both perspectives?
 
  • #4
1,444
2
solving that cross product i get
[tex] L = (m_{1}+m_{2}) \vec{R} \times \dot{\vec{R}} + \left( \frac{m_{1}m_{2}^2}{M^2} + \frac{m_{1}^2m_{2}}{M^2} \right) \vec{r} \times \vec{\dot{r}} [/tex]
well m1 + m2 = M
also \frac{m_{1}m_{2}^2}{M^2} + \frac{m_{1}^2m_{2}}{M^2} = \frac{\mu^2}{m_{1}} + \frac{\mu^2}{m_{2}} = \mu^2 \left(\frac{1}{m_{1}} + \frac{1}{m_{2}} right) = \mu[/tex]

[tex] L = M(\vec{R} \times \dot{\vec{R}}) + \mu \left( \vec{r} \times \dot{\vec{r}} \right)= L_{CM} + L_{Rel} [/tex]

what is the interpretation of this result?
means the total angular momentum of a two body system is the sum of the momenta wrt thecenter of mass and the angular momentum of the center of mass itself? Is that a correct interpretation?

Explain under what conditions the angular momentum of the center of mass Lcm and the angular momentum of the relative motion Lrel are constnats of motion and what physical significance this possesses.

Since Torque is the relative of the angular momentum, if Lcm is zero, then the torque on the center of mass is zero. If L rel is zero, means the entire system , origin and all, have no net external torque being exerted on them
 
Last edited:

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