Moment of Inertia: Ignoring Small Bond Axis Molecule

[gracy]

Draw a line from the point at one end through the toothpick to the point at the other; strictly speaking that is the "major" axis.

Am i right?

 

[Bystander]

Yes. Yes. Yes. Yes. Yes. Think also of the axes of the cartesian coordinate system, imaginary lines intersecting at right angles to one another.

 

[gracy]

No. No. No.
I will give you a toothpick. You will stick one baby green pea on one end and a second baby green pea on the other end. The bond length is the length of the toothpick, or the distance between the peas. The peas are the two atomic nuclei in the diatomic molecule. Lay it on the table in front of you. You can spin it on the table (horizontal plane) around the center of the toothpick (the point midway between the peas), or you can roll it as if the toothpick were an axle and the peas were little bitty tires, or you can rotate it in a vertical plane about the center of the toothpick. These are the three axes of rotation. The moments of inertia for rotation in either the horizontal or vertical planes are calculated by squaring the distances of the peas from the center of the toothpick, multiplying the squares by the masses of the peas and adding them. The moment of inertia for rolling the toothpick is calculated by finding how far the peas are from the axis of the toothpick, and that is at most only half the diameter of the peas (the toothpick is stuck through them), squaring that distance and multiplying by the masses of the peas.

You do NOT square one distance for one moment and multiply it by the square of another distance for a different moment.

The rolling of toothpick is similar to my the moment of inertia of a molecule about the bond axis and because we are multiplying square of half of diameter (in the order of 10-30m2)with masses of pea.so it's value reaches nearly zero.

 

[gracy]

The rolling of toothpick is similar to my the moment of inertia of a molecule about the bond axis and because we are multiplying square of half of diameter (in the order of 10-30m2)with masses of pea.so it's value reaches nearly zero.right.?That's what my original question was.

 

[Bystander]

Yes.

 

[gracy]

Yes.

Ok last question

Lay it on the table in front of you. You can spin it on the table (horizontal plane) around the center of the toothpick (the point midway between the peas)

This is same as motion in xz plane i.e perpendicular or about to y axis.

or you can rotate it in a vertical plane about the center of the toothpick.

This is same as motion in xy plane i.e perpendicular to or about z axis.
Right?

 

[Bystander]

You got it.

 

[gracy]

You got it.

Ok.Now it's finally my last question.Just a tiny doubt.What about rolling case?isn't it same as motion in yz plane i.e about or perpendicular to x axis. .According to your 17 post i don't think so.how x.axis differ from bond axis?just that bond axis passes from center of mass and x-axis doesn't?If we take toothpick example ,how would motion in yz plane look like?

 

[gracy]

Ok.Now it's finally my last question.Just a tiny doubt.What about rolling case?isn't it same as motion in yz plane i.e about or perpendicular to x axis. .According to your 17 post i don't think so.how x.axis differ from bond axis?just that bond axis passes from center of mass and x-axis doesn't?If we take toothpick example ,how would motion in yz plane look like?

Wait.I think i have figured it out.motion in YZ plane would be same as in

Lay it on the table in front of you. You can spin it on the table (horizontal plane) around the center of the toothpick (the point midway between the peas),

Just instead of horizontal plane it would be vertical plane.Right?

 

[Bystander]

Yup.

 

[gracy]

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.

 

[gracy]

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.

 

[Simon Bridge]

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? I_x = ...
What is the moment of inertia about the (say) y axis? I_y = ...

What is I_x/I_y ?

See why I_x is negligible compared with I_y ?

[edit]Oh you seem to have sorted it out...

 

[Simon Bridge]

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.

 

[gracy]

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.

 

[jerromyjon]

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?

 

[gracy]

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.

 

[Bystander]

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.

 

[gracy]

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.

 

[Bystander]

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.

 

[Simon Bridge]

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.