geerky42 2 years ago Help Needed with Inertia...

1. geerky42

2. Mashy

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

3. geerky42

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

4. geerky42

Not sure what to do in next step...

5. Mashy

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???

6. Mashy

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7. geerky42

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

8. Mashy

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

9. Mashy

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

10. geerky42

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

11. Mashy

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

12. geerky42

You mean $$\large I = ml^2$$?

13. geerky42

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

14. Mashy

its $\sum_{ }^{ } mR ^{2}$

15. Mashy

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

16. geerky42

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

17. Mashy

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

18. Mashy

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

19. Mashy

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

20. geerky42

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

21. Mashy

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..

22. geerky42

I see. Thanks!

23. Mashy

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