Help Needed with Inertia...

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- geerky42

Help Needed with Inertia...

- schrodinger

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- geerky42

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- anonymous

ah... so the first thing you need to do is calculate the location of the centre of mass.. can you do that?

- geerky42

Yeah, it's 1.95 m away from 1kg mass.

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- geerky42

Not sure what to do in next step...

- anonymous

ok.. i won't check that then..
fine.. now answer this.. if i know that there is a mass that is connected to a string.. which has a length 'l'.. and i swirled it.. what would be the moment of inertia???

- anonymous

|dw:1359715230549:dw|

- geerky42

\(\large I = \dfrac{1}{3}ml^2\), right?

- anonymous

noooOOO :O.. how did you get that? :O

- anonymous

its a point mass.. imagine its a point mass!!

- geerky42

Idk, it's a formula for end of rod, though this is what you asked for, sorry

- anonymous

no.. what is the general formula for moment of inertia?? i think the concept itself is not clear to you.. huh?

- geerky42

You mean \(\large I = ml^2\)?

- geerky42

it's a general formula for inertia, i think.

- anonymous

its \[\sum_{ }^{ } mR ^{2}\]

- anonymous

thats the general formula.. and in my question .. since only one point mass is present.. you can say its just
\[ml ^{2} \]

- geerky42

So for my question, I just use the sum of inertia?

- anonymous

so its basically summasation of all point mass times their distance from the axis of rotation squared.. so in your question.. you have 2 point masses.. so find each one's moment of inertia and sum it up

- anonymous

and don't call it inertia.. its called moment of inertia :P

- anonymous

inertia = resistance to change for linear motion
moment of inertia = resistance to change for rotational motion

- geerky42

ok I see. does this formula \(\displaystyle\int r^2 \text{d}m\) work too? I tired this way and I got lost.

- anonymous

that is only if you have a rigid object.(continuos distribution of mass). but in your case its not a rigit object... its just 2 point masses.. !! so you can indivisually find their moment's of inertia about the rotational axis and sum them up..

- geerky42

I see. Thanks!

- anonymous

your welcome :)

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