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Polar moment of inertia in a rod?

  • Thread starter smr101
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  • #1
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How is the polar moment of inertia in a rod calculated?

Thanks.
 

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  • #2
BvU
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No idea what it is. This link something ?
 
  • #3
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Yeah, I've had a look at that. Formula doesn't quite make sense to me. I know what moment of inertia is but the 'polar' bit has thrown me off. Thanks anyway.
 
  • #4
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Since there is no context whatsoever in the original posting, it's difficult to assist any further here.
 
  • #5
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Since there is no context whatsoever in the original posting, it's difficult to assist any further here.
Here's the problem., 4. (b) (i) is the one I'm looking at.

H7jim.jpg
 
  • #6
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Well, at least now I can understand your confusion. However, it doesn't literally ask for the polar moment of inertia (a.k.a area moment of inertia), but for the polar mass moment of inertia. I see that used for the rotational moment of inertia, so with the context of b (ii) and b(iii) that seems the most logical choice.
 
  • #7
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Well, at least now I can understand your confusion. However, it doesn't literally ask for the polar moment of inertia (a.k.a area moment of inertia), but for the polar mass moment of inertia. I see that used for the rotational moment of inertia, so with the context of b (ii) and b(iii) that seems the most logical choice.
What is the formula for rotational moment of inertia?
 
  • #8
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Can't imagine you haven't seen it before ! What did you use for 4a ?

## I=\int dI = {\displaystyle \int_0^M r^2 \; dm}##
 
  • #9
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Can't imagine you haven't seen it before ! What did you use for 4a ?

## I=\int dI = {\displaystyle \int_0^M r^2 \; dm}##
I'm on 4. (a) (ii) currently. Where should it be used in (a)?

Ok, thanks, I have seen it but haven't used it previously.
 
  • #10
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Can't imagine you haven't seen it before ! What did you use for 4a ?

## I=\int dI = {\displaystyle \int_0^M r^2 \; dm}##
I've found this... http://hyperphysics.phy-astr.gsu.edu/hbase/mi2.html#irod3

Looking at the bottom set of equations, so dm = (M/L) x dr...

So essentially, the equation is r^2 x (M/L) x dr...

Using these number that would be 0.075^2 x (20/2) x 0.15 x 0.075...

Is that correct?
 
  • #11
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Tempting indeed (*), but I think this time the rod is turning around its length axis. I.e. it's a solid cylinder !

And you will need a moment of inertia in 4a (ii) as well: you don't just have to accelerate the cart, but also the drum.
However, in 4a (ii) they give you a radius of gyration, so you use that to calculate the moment of inertia..


(*) well, not really. did you really think dr stands for diameter x radius (because I see a 0.15 x 0.075, and then a few dots, indicating justified doubt, I hope) ?
 
  • #12
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Ok, I am familiar with the equation for mass moment of inertia for a cylinder, which fits with this.

So are you saying the equation you gave me wouldn't work with this? I find it odd they say it's a rod and expect a cylinder equation to be used.

And yes I knew it wasn't the correct answer as I have the solutions and it didn't match...I see that dr is used in the integration equation now.

So what is the formula?
dm = elemental mass =
(M/L) x dr so the whole equation is r^2 x (M/L) x dr
And then?
 
  • #13
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It's a drilling rod. Look here
 

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