Deriving Equation of Reduced Mass W/ Angular Momentum

In summary, the electron has more angular momentum than the nucleus due to their greater distance from the center of mass.
  • #1
Dgray101
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0

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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  • #2
Write expressions for re and rn in terms of r, given that they are determined by the location of the center of mass.
 
  • #3
So like m*r2ω = m(r-rn)^2 ω + M(r-re)^2 ω ?
 
  • #4
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?
 
  • #5
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.
 
  • #6
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?
 
  • #7
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) ) ?
 
  • #8
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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  • #9
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.
 

1. What is the equation for reduced mass with angular momentum?

The equation for reduced mass with angular momentum is given by μ = m1m2 / (m1 + m2) * r^2, where μ is the reduced mass, m1 and m2 are the masses of the two particles, and r is the distance between them.

2. How is the reduced mass related to the angular momentum?

The reduced mass is directly related to the angular momentum through the equation L = μvr, where L is the angular momentum, μ is the reduced mass, v is the velocity of the system, and r is the distance between the particles.

3. What is the significance of the reduced mass in angular momentum calculations?

The reduced mass is significant because it takes into account the relative masses of the two particles in a system. It allows for a more accurate calculation of angular momentum, taking into account the individual masses and their distance from each other.

4. How is the reduced mass calculated for a system with three or more particles?

For a system with three or more particles, the reduced mass is calculated by taking the reciprocal of the sum of the reciprocals of each individual mass. In other words, μ = 1/ (1/m1 + 1/m2 + 1/m3 + ...).

5. Can the reduced mass ever be larger than the mass of either individual particle?

No, the reduced mass can never be larger than the mass of either individual particle. This is because it is a calculated value that takes into account the masses of both particles and is always smaller than the smallest individual mass.

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