Timeline for The moment of inertia of two rods joined at the center forming a cross about the $x$-axis
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 6, 2021 at 15:05 | vote | accept | Blue Green | ||
| Nov 2, 2021 at 19:17 | comment | added | Kshitij Kumar | Welcome again! Actually moment of inertia for cylinder perpendicular to axis wouldn't be same as for rod $1/12mr^2$. Again because of cylinder is 3d and rod is 1d, both have different kind of mass distribution. Here you need calculus to find moment of inertia. Other cylinders moment of inertia is of course $1/2mr^2$. | |
| Nov 2, 2021 at 19:04 | comment | added | Blue Green | aaaa that makes so sense! Thank you so much! So if both of them were cylinders, then the inertia of the whole thing would be 1/12 mr^2 + 1/2 mr^2, right? | |
| Nov 2, 2021 at 19:01 | comment | added | Kshitij Kumar | Yeah ok so I got to know your confusion. Actually in image about $I$ for rod about axis is actually cylinder. You see rod is in idealistic sense is just a 1 dimensional line. So $I$ about axis along our rod is meaningless. Think of it. You can't rotate line in same way you are rotating cylinder about that axis. Cylinder is a 3d object so it can rotate meaningfully and provides inertia. So in this case, rod along x axis don't provide any moment of inertia. | |
| Nov 2, 2021 at 18:55 | comment | added | Blue Green | Wait sorry to bother you but I am talking about the rods, not cylinders. Both formulas in the image are for the same rod. So it gives the inertia of the rod rotated in the x-axis when it is parallel to it, and the other rod rotated in the x-axis when it is perpendicular to the axis. Therefore, the inertia when the rods are joined to make a cross would be 1/12 mr^2 + 1/2 mr^2, right? | |
| Nov 2, 2021 at 18:45 | comment | added | Kshitij Kumar | Wait, you are confusing. If you meant rod is to be treated like rod only then there is no meaning of $I$, about the axis. And you will treat other rod, like rod only. So $I_f=1/12mr^2$ On other hand treating rod like cylinder, you will get $I_1=1/2mr^2$ for cylinder across the x axis. For other cylinder you would have to find $I_2$ manually. $I_{\text{final}}=I_1+I_2$ | |
| Nov 2, 2021 at 18:41 | comment | added | Blue Green | Exactly! :) And then also the other rod which has I = 1/12 mr^2 | |
| Nov 2, 2021 at 18:40 | comment | added | Kshitij Kumar | You mean about axis which has $I=\frac{1}{2}mr^2$? | |
| Nov 2, 2021 at 18:39 | comment | added | Blue Green | So around the x-axis. Please check out the image that I linked :) | |
| Nov 2, 2021 at 18:38 | comment | added | Kshitij Kumar | What do you mean by 'around axis'? | |
| Nov 2, 2021 at 18:37 | comment | added | Blue Green | So I know that the inertia of a rod about center is simply 1/12 ml^2. Also, the inertia of a rod around axis is 1/2 ml^2. So does this mean that the inertia of the whole thing would be the inertias added? Check this out for what I mean: cdn.kastatic.org/ka-perseus-images/… | |
| Nov 2, 2021 at 18:35 | comment | added | Kshitij Kumar | Welcome! If you are new, you can upvote answers/questions which you find useful, and can select answers which you think are best by clicking tick button. | |
| Nov 2, 2021 at 18:31 | comment | added | Blue Green | Hi! Thank you so much!! | |
| Nov 2, 2021 at 18:27 | history | answered | Kshitij Kumar | CC BY-SA 4.0 |
