29.6 Deep Dive - Derivation of the Parallel Axis Theorem

MIT 8.01 Classical Mechanics, Fall 2016
View the complete course: ocw.mit.edu/8-0...
Instructor: Dr. Peter Dourmashkin
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
More courses at ocw.mit.edu

[@pantheraonca3635]

how was the expression
(sigma)(mj)(rcmj)=0 obtained? where did Dr. Dourmashkin get this?

[@serden8804]

I did not understand, too

[@KerbalLauncher]

Definition of center of mass.

[@comebackata2]

I think i got it!
Imagine you have a sphere with the center of mass marked at the middle.
To the right of the CM you get +r, to the left of the CM you get -r.
Sum them up you get zero.

[@shadon_official2510]

comebackata2 doesn’t matter cause you square it anyway. I still don’t see why it’s zero.

[@leoliu7492]

@@comebackata2 This is helpful, thanks!

[@jackoboyle3195]

How do they display the professor's writing not backward? Is he in of a green screen and writing on some board that electronically stores what he writes then they add it the writing to the picture?

[@krishnadhulipala9861]

He's likely writing on a large glass pane. However, the camera's view of the writing would be mirror-imaged, so they likely flip the video, which produces the desired effect.

[@markbrightwell3536]

This proof assumes the rigid object is essentially flat, i.e., thin in the dimension perpendicular to the illustrated view.

[@Wild4lon]

Hi does anyone know what about points that lie between the COM and the new point through which we define the axis? We must have a vector sum of displacements taking the COM as the origin going on here. So I suppose in a way we could say we are transforming the coordinate system by a translation to the left? And then RCM is a variable vector while d is a vector to the right, and when squaring, we simply take the modulus squared?

[@nebpoma]

The proof is not comprehensive

[@markkennedy9767]

Hi you might be able to help me with this:
If I apply the parallel axis theorem to a body of mass m in uniform circular motion around a point a distance r away, then I get the angular momentum of the body about the point is
(I + mr^2) omega where I is the body's moment of inertia around its COM and omega is the body's angular velocity around the point.
However if we compute the angular momentum by just getting the orbital angular momentum about the point (where the body isn't spinning around its COM), we get r × mv = (mr^2 omega) which is clearly different to what we got above.
Can you point out where I might be making a mistake here.
Your help is greatly appreciated.
Thanks.

[@sabith-p.ksachu8358]

L = I omega
This relation does not always hold true.
It only holds true for bodies that are symmetrical with respect to the axis of rotation

[@sabith-p.ksachu8358]

And secondly both mass are not the same as you have given,
In parallel axis theorem we use the total mass of the body and get the total angular momentum of the system we consider
Whereas in the 2nd relation u typed, we use that particle's mass and it only gives L of that particle

[@markkennedy9767]

Hi I'm not sure if those actually address the question but thanks anyway. No I figured out that my concept of spin angular momentum was off. So If I accept that the first scenario has spin angular momentum (like the moon has spin even though it has a half that never faces the earth) then this problem Is resolved since both orbital and spin angular momentum are omega.
Thanks.

[@saisathvik3587]

so parallel axis theorm is applicable only for Icm and and any other axis

[@porte_bosho]

At 4:24 what is that term called? I mean integral of dm*rcom

[@5gallonsofwater495]

1 year late reply... but in an another video (ruclips.net/video/ol1COj0LACs/видео.html), it was written in summation form. its apparently the "definition of the center of mass."

[@samyakroul8087]

At 04:46, does that mean that the integral for the y-component also equals zero?

[@balkabag250]

yes, both x and y components equal zero by the definition of center of mass

[@yangyue20]

well, I think the diagram is not accurate, just the axis does not pass the centre of mass, it should pass the centre of block?

[@juliakim8226]

Does the tilt of an axis also influence the body's rotational inertia?

[@MJ-ou9ff]

Yes it does

[@MathwithMunaza]

Very helpful.Thanks

[@MJ-ou9ff]

Thanks a ton:)

[@haslan4885]

great video

[@zakariachahiedine6761]

Why does that middle term go to zero?

[@glokta1]

It's the definition of the center of mass

[@comebackata2]

I think i got it!
Imagine you have a sphere with the center of mass marked at the middle.
To the right of the CM you get +r, to the left of the CM you get -r.
Sum them up you get zero.

[@shadon_official2510]

meep elaborate more

[@Fosgen]

Sir, so does it mean, Earth is in fact, flat?

[@user-wn3bo9vo7f]

Precisely

[@joseps637]

What does that per thing mean

[@rehtronicles879]

its actually 'perp' he is just referring that the its the perpendicular distance from the axis

[@premjangir326]

ok

[@roy_narten]

Most impressive part is how he can write backwards so quickly. lol

[@swaree]

bruh, the image is mirrored

[@premjangir326]

Wonderfull dear

[@skaegi8867]

Wow he writes everything
mirrored...

[@astrosertao]

he writes normally and then mirror the video