Question

Help Needed with Inertia...

Answer 1

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

Answer 2

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

Answer 3

Not sure what to do in next step...

Answer 4

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???

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

thats the general formula.. and in my question .. since only one point mass is present.. you can say its just

\[ml ^{2} \]

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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..

Answer 20

I see. Thanks!

Answer 21

your welcome :)