Angular Momentum of Two Particles System

Therefore, the total angular momentum of the system is constant and conserved, which is significant in understanding the dynamics of the two body system. In summary, the total angular momentum of a two body system is the sum of the angular momenta of the center of mass and the relative motion. This total angular momentum is conserved when there is no external torque acting on the system, resulting in both Lcm and Lrel being constant of motion.
  • #1
stunner5000pt
1,461
2
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!
 
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  • #2
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
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
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
 
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What is angular momentum?

Angular momentum is a physical quantity that describes the rotational motion of an object. It is a vector quantity and is defined as the product of an object's mass, its velocity, and the distance from the axis of rotation.

How is angular momentum of a two-particle system calculated?

The angular momentum of a two-particle system is calculated by multiplying the individual particle's angular momentum by their respective distances from the axis of rotation, and then adding these values together.

What is the conservation of angular momentum?

The conservation of angular momentum states that the total angular momentum of a system remains constant as long as there are no external torques acting on the system. This means that the initial angular momentum of a system will be equal to the final angular momentum of the system, even if the system undergoes changes in its rotational motion.

What happens to the angular momentum of a two-particle system if the particles move closer to each other?

If the particles in a two-particle system move closer to each other, their distances from the axis of rotation will decrease. This will result in an increase in their individual angular momenta, and therefore, an increase in the total angular momentum of the system.

Can the angular momentum of a two-particle system be negative?

Yes, the angular momentum of a two-particle system can be negative. This would occur if the particles have opposite directions of angular momentum, canceling each other out. In this case, the total angular momentum of the system would be zero.

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