# Homework Help: Deriving Equation of Reduced Mass W/ Angular Momentum

1. Nov 12, 2013

### Dgray101

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

Angular Momentum = mvr
Conservation of Angular momenta
m=mass of electron
M=mass of nucleus
3. 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.

2. Nov 12, 2013

### Staff: Mentor

Write expressions for re and rn in terms of r, given that they are determined by the location of the center of mass.

3. Nov 12, 2013

### Dgray101

So like m*r2ω = m(r-rn)^2 ω + M(r-re)^2 ω ?

4. Nov 12, 2013

### Staff: Mentor

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. Nov 13, 2013

### Dgray101

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. Nov 13, 2013

### Staff: Mentor

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. Nov 13, 2013

### Dgray101

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. Nov 13, 2013

### Staff: Mentor

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:

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. Nov 13, 2013

### Dgray101

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. Nov 13, 2013

### Staff: Mentor

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. Nov 13, 2013

### Dgray101

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. Nov 13, 2013

### Staff: Mentor

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.