If Moment of Inertia units are mm^4, how do you use it in calculations?

[Femme_physics]

So, we learned that the units of Moment of Inertia are mm^4. My question is how do they use moment of inertia in static calculation (where we use Newtons), if Moment of Inertia is in mm^4?

 

[maimonides]

Please doblecheck your notes or your textbook.
The unit of moment of inertia is mass times length squared, in SI kgm^2.
In statics you often use a related quantity called radius of gyration.
If you really need to use mm^4: You know how to convert mm to m, square mm to sqare m, cubic mm to cubic m, don´t you?.

 

[xxChrisxx]

EDIT: WAit.

He's not talkign about angular mass, he's talking about second moment of area or polar moment of inertia.
They are all so similarlally named it gets really confusing.

OP could you clarify which moment of inertia you would like to know about :D

 

[Femme_physics]

Please doblecheck your notes or your textbook.
The unit of moment of inertia is mass times length squared, in SI kgm^2.
In statics you often use a related quantity called radius of gyration.
If you really need to use mm^4: You know how to convert mm to m, square mm to sqare m, cubic mm to cubic m, don´t you?.

1) Double checking my notes the units for the result of Moment of Inertia are [mm^4]. Perhaps it's because we're working in 2D, or haven't started calculus yet?

2) So radius of gyration is simply the Inertia divided by the area, and that's the quality you're normally given in a statics equation in order to solve a problem whether something moves or not?

3) To answer your last question, yes-- unit conversion is easy peasy.

 

[maimonides]

Sorry about the confusion with the moment(s).
In Newton, you have m as length unit. So you better convert all lengths to m. (But I think you´ve got it.)
Edit: I didn´t notice the "they". You´re probably dealing with small (mm- or centimeter sized) rods (or whatever); so they use mm´s to avoid zeros or having to write 10^-12 all the time. Might be different for a bridge.
Second afterthought: There are "engineering formulae" , which only work for specific (not always consistent) units. (Opposed to "physics formulae", which require consistent units) They have the conversion factors built in somewhere. Sometimes somebody does not state clearly enough which kind of formula is used.

 

[Femme_physics]

np, I did scan my class notes if it helps (they're in hebrew but the math is there). I'm not so worried about conversion, just trying to understand how it works.

Let's say I applied on a circular object a 50 Newton force at an ideal location to cause rotation (i.e. away from the COM). I need to relate that 50 Newton force to cubic m [or let's say to mm^4], yes?


(Scanned class notes attached below)

 

[maimonides]

I´m a bit confused.
On the first page, you have I = mr^2 and then something that looks like Steiners theorem\parallel axis theorem. (Moment of inertia for a body rotating around an axis not through its CoM). ("A" ought to be a mass, maybe there was some density factor, that got left out. The formula also applies to sMoA calculations, then A is correct, but the index CoM is doubtful)
Thne you switch to I ~ r^4, and on the second page you have I = h*b^3/12. Those I´s are, as Chris pointed out, second moments of area (sMoA). They have to do with bending of bars, not with rotation.
You can´t mix I(rot.) and I(sMoA).
Is your problem about bending a bar/rod or about rotating a body?

These links might be helpful
http://en.wikipedia.org/wiki/Moment_of_inertia
http://en.wikipedia.org/wiki/Second_moment_of_area
http://en.wikipedia.org/wiki/Steiner's_theorem

 

[maimonides]

Application of I in statics:
http://en.wikipedia.org/wiki/Euler–Bernoulli_beam_equation#Examples
or here
http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/flexure-formula
http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/chapter-6-beam-deflections

 

[Femme_physics]

Well, that's what I copied from the board, maim.

My problem is about rotating a body, not bending bar/rod.

Let me gather my thoughts and make a proper reply. I want to draw a couple of my own diagrams to show you exactly what I mean but I lack time right now...but I'll definitely post back here when I'm freed.

 

[Gokul43201]

Let's say I applied on a circular object a 50 Newton force at an ideal location to cause rotation (i.e. away from the COM). I need to relate that 50 Newton force to cubic m [or let's say to mm^4], yes?

If you're given the moment of inertia in mm^4, you need to know the thickness and density of the circular disc, so you can figure out the moment of inertia in units of M*L^2

 

[Femme_physics]

I think I fully understand now. My only issue is that we were doing static calculations with pulley we never considered thickness an density, but I guess that's because we were doing simple problems. Anyway, I got it worked out now :) Thanks.

 

[marius10ster]

Thank you so much Gokul... and Femme ofcourse thank u too