Deriving Equation of Reduced Mass W/ Angular Momentum

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SUMMARY

The discussion focuses on deriving the equation for reduced mass (m*) using angular momentum principles. The key equation presented is m* = mM/(m+M), where m is the mass of the electron and M is the mass of the nucleus. Participants emphasize the importance of expressing distances re and rn in terms of the total separation r and the respective masses. The correct approach involves balancing moments about the center of mass and ensuring that the sum of distances equals r.

PREREQUISITES
  • Understanding of angular momentum concepts, specifically Iω = m*r²ω
  • Knowledge of center of mass calculations for two-body systems
  • Familiarity with algebraic manipulation and solving equations
  • Basic principles of gravitational forces and equilibrium
NEXT STEPS
  • Learn how to derive expressions for center of mass in two-body systems
  • Study the principles of angular momentum conservation in physics
  • Explore the concept of reduced mass and its applications in quantum mechanics
  • Practice solving equilibrium problems involving moments and torques
USEFUL FOR

Students in physics, particularly those studying mechanics and quantum mechanics, as well as educators looking to enhance their understanding of angular momentum and reduced mass concepts.

Dgray101
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Homework Statement



Derive m*= mM/(m+M) *hint* total angular momentum Iω=m*r2ω equals the sum of the individual angular momenta, where r = re + rn; re and rn are the distances of the electron and nucleus respectively from the center of mass.

Homework Equations



Angular Momentum = mvr
Conservation of Angular momenta
m=mass of electron
M=mass of nucleus

The Attempt at a Solution



m*r2ω = m(re)2ω + M(rn)2ω

I write out this equation but algebraically I can't seem to solve for m* properly so I must be missing something :/ sorry about the notation in the equation I am still learning how to properly use the tools in the message figure.
 
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Write expressions for re and rn in terms of r, given that they are determined by the location of the center of mass.
 
So like m*r2ω = m(r-rn)^2 ω + M(r-re)^2 ω ?
 
Forget the angular velocity for the moment. The distances re and rn from the center of mass can be determined entirely by the masses and the total separation r.

Given m and M and r, what are those distances?
 
I have no idea :/ my guess is it has something to do with the gravitation equations. But I never took anything with this in first year because I was in engineering :/ not the normal physics stream.
 
I think that if you've done engineering you can find the center of mass given two objects and their separation. What's the distance of each from that center of mass?
 
I didn't do that well in engineering mechanics because it was my first year, and wasn't expecting it to be so brutal :/ but if I recall

Center of Mass = (1/M+m) *( (mre) + (Mrn) ) ?
 
Dgray101 said:
I didn't do that well in engineering mechanics because it was my first year, and wasn't expecting it to be so brutal :/ but if I recall

Center of Mass = (1/M+m) *( (mre) + (Mrn) ) ?

You've got the right idea, but there needs to be some reference point from which re and rn are measured for the expression to make sense. Since it's the center of mass itself that we want as the reference point, things would need adjusting.

I'll suggest another way to approach the problem. You want to get expressions for re and rn in terms of r and the masses m and M. So look at it as if it were two objects of mass M and m on a massless rod in a uniform gravitational field. You want to find the lengths re and rn from the pivot point where the system will be balanced:

attachment.php?attachmentid=63907&stc=1&d=1384365198.gif


Write an expression for the moments about the pivot point. That's one equation. For the other, use the fact that r is the sum of re and rn. Solve for re and rn.
 

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so... I had said that the sum of the moments about the center of mass would be...

remg - rnmg = r(m+M)g (I think anyways)

which becomes r(m+M) = Mrn + mre

Would it be helpful to perhaps take out a factor of mM in this case on the right hand side?
 
  • #10
Dgray101 said:
so... I had said that the sum of the moments about the center of mass would be...

remg - rnmg = r(m+M)g (I think anyways)
No, the moments need to balance for equilibrium. The right hand side should be zero (no net torque).

Then use the fact that the sum of the distances is r to write a second equation. Solve for re and rn.
 
  • #11
Oh okay so remg - rnMg = 0

re = (rnM) / m

Then pluging that into the sum of re and rn --> r = (M+m)rn / m

So now could we sub r into the original equation in the problem statement and solve for m*?
 
  • #12
Dgray101 said:
Oh okay so remg - rnMg = 0

re = (rnM) / m

Then pluging that into the sum of re and rn --> r = (M+m)rn / m

So now could we sub r into the original equation in the problem statement and solve for m*?

You don't want to replace r in the angular momentum equation, you want to replace both ##r_e## and ##r_n## in that expression.

So rearrange your equation above to isolate ##r_n##. Then solve for ##r_e## in a similar fashion. You should end up with two expressions:

rn = <something involving r,M,n only>

re = <something involving r,M,n only>

Those expressions will replace re and rn in your angular momentum equation.
 

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