Moment of Inertia: Ignoring Small Bond Axis Molecule

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SUMMARY

The discussion centers on the moment of inertia of diatomic molecules about their bond axis, emphasizing that this value is significantly smaller than the moments of inertia about axes perpendicular to the bond axis. Participants clarify that the bond axis is defined as the line connecting the nuclei of the two atoms, while the moment of inertia is calculated using the formula I = Σ(m_i * r_i²), where m_i is the mass and r_i is the distance from the axis of rotation. It is established that the moment of inertia about the bond axis is on the order of 10-10 times that of the perpendicular axes, due to the relative sizes of bond lengths and nuclear dimensions.

PREREQUISITES
  • Understanding of molecular structure and diatomic molecules
  • Familiarity with the concept of moment of inertia
  • Basic knowledge of atomic mass and nuclear dimensions
  • Ability to perform calculations involving mass and distance
NEXT STEPS
  • Study the calculation of moment of inertia for various molecular geometries
  • Learn about the relationship between bond lengths and molecular rotational dynamics
  • Explore the concept of center of mass in diatomic molecules
  • Investigate the effects of molecular symmetry on rotational degrees of freedom
USEFUL FOR

Chemists, physicists, and students studying molecular dynamics, particularly those interested in rotational motion and molecular structure analysis.

  • #61
Bystander said:
Yup.
Thanks a lot.You helped me a lot.I must say You are truly homework helper and science advisor but not a Bystander,you totally got involved in my problem.Thanks again.
 
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  • #62
gracy said:
Thanks a lot.You helped me a lot.I must say You are truly homework helper and science advisor but not a Bystander,you totally got involved in my problem.Thanks again.
Actually i got the answer which i have asked in my original post but i am stuck on a point.The video which i have posted in my post 42 from time 3:15 to 3:30 and in this video
;from time 2:05 t0 3:00 ,why they haven't included rotation about x-axis in yz plane ?According to your 17th and vagn 18th and my 59th post they should include. As x-axis is not same as bond axis.Please reply.It's my humble request.
 
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  • #63
Note: Some people have trouble thinking in terms of "order of magnitude" ...

@gracy - the picture you drew for the diatomic molecule is basically a dumbell shape.
Do you not know how to find the moment of inertia of two spheres attached by a rod?

For the sake of the argument, let's say that this particular diatomic molecule has the same element at each end ... say it's H2 or something.
So we have two spheres diameter d (approx 10-15m) whose centers are separated by distance L (~10-10m)

The moment of inertia of a sphere is given by: ... (look it up)
The parallel axis theorem is: ... (look it up)

So: for the sake of notation, let the x-axis be the bond axis, and have the origin at the center of mass.
What is the moment of inertia about the x axis? Ix = ...
What is the moment of inertia about the (say) y axis? Iy = ...

What is Ix/Iy ?

See why Ix is negligible compared with Iy ?

[edit]Oh you seem to have sorted it out...
 
  • #64
they should include. As x-axis is not same as bond axis
... consider: there are an infinite number of axes that are not the bond axes.
 
  • #65
Simon Bridge said:
let the x-axis be the bond axis
But how can i assume this?this is what my question is difference between bond axis and x axis.
 
  • #66
If you choose x as the bond axis, you have the other 2 axes where a lot of physics happens and your x-axis is where you just twist the peas. I don't even think spinning the toothpick is truly physically relevant, does the bond itself have angular momentum?
 
  • #67
I am totally lost.Please bystander reply to my 62 post.Because you are the only one who has been in conversation from start and i am not getting your 17th post. Thanks to jerromyjon and simon bridge also.
 
  • #68
gracy said:
As x-axis is not same as bond axis.Please reply.It's my humble request.
It is the same. We agreed it's the same. It's not considered in the videos because the angular moment of inertia about it is insignificantly small due to the dimensions of the nuclei --- remember? Square 10-15m and compare it to the square of 10-10m? It's one 10 billionth the magnitude of the other two moments.
 
  • #69
Bystander said:
It is the same.
That's what i wanted to listen.Now i am having no doubt.It completes this thread.
Thanks again Bystander .You are a great teacher.
 
  • #70
Remember, you get to pick the axes; and, life is easier when you pick them to minimize the work you have to do. Minimum work for diatomic molecules is to pick the bond axis as one of the three, and pick the other two at right angles to that, and to each other, through the center of mass which is located on the bond axis.
 
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  • #71
gracy said:
But how can i assume this?this is what my question is difference between bond axis and x axis.
... you are free to call the bond axis anything you like. If you don't like "x" how about "y" or "z" (z is a very popular choice because: cylindrical coordinates)... or "bob" or "kate" or anything you like. It's just a bit of shorthand.

Earlier on - you drew a diagram with a big arrow on it labelled "x-axis", and you immediately ran against a convention in maths drawings that an arrow with an axis label written next to it is the axis in question but you intended it to indicate the axis. That's just you learning how to talk in maths, and it seems to have got you sidetracked a bit.

One of the things that can happen to people as they learn maths is they start thinking of x-y-z as somehow being fixed in space and everything else is positioned against these fixed axes. But this is not the case. You can choose any direction and call it anything you like... it's just that Cartesian axes are handy labels. As soon as we pick something else, we then have to write down a bunch of stuff to define what we mean and that's already been done for x-y-z 400-odd ago(?) by Rene Decartes. So we all use his by default.

In physics, our axes are chosen to correspond to some physical object or process ... in your case it makes sense to put one cartesian axis along the bond-axis. It does not matter where the others point. I chose x because you seemed to be used to using the x axis. But you can pick another one if you like. You don't even have to line up the cartesian axes to the bond if you don't want to - it makes only cosmetic differences to the final result while making the calculation itself a lot harder.
 
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