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**How is the polar moment of inertia in a rod calculated?**

Thanks.

Thanks.

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- Thread starter smr101
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Thanks.

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massmoment 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?

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## I=\int dI = {\displaystyle \int_0^M r^2 \; dm}##

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## 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?

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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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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?

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