when finding the moment of inertia with respect to the x-axis we considered the segment along x to have a base of (1-x) and a height of dy. However when we found the moment of inertia with respect to the y-axis we considered the segment to have a base of dx and a height of y. Why didn’t we consider the segment to have a height of (1-y) like we did for the base of segment one which was (1-x).
This question is very hard to explain without a visualization. But think of a line that has an overall length of 1 m. Then cut an amount x of that bar from the left corner and the remainder of the bar that was not cut is the (1-x)... try to visualize it and see if that helps with your question.
I also have the same doubt ,when finding centroid of shaded area,dA was taken as product of xdy not (1-x)dy in the previous video,why 1-x is only used here ,is the same logic sir said want to applied in the previous problem too,is nt ?
Hello, because when we were doing 10-5 with respect to the x-axis the area was on the right side of the curve, if the area was on the left side of the curve it would be just "x*dy" for the dA. In 10-6 when we are doing with respect to y-axis you can visualize it as under the curve area, so dA = y*dx only. IF the area was above the curve then it would be dA = (1-y)dx... I hope that helps. The key is to see the height of the "square"/section we are integrating. Is the height of that square equal to the height of the curve? or what else?
There where y^2 twice. The first one is given by the formula or the moment of inertia around Ix. The second Y^2 is explained in the video, when we solve for x becomes y^2. :) Hope it helps to understand it a bit better.
Ur explanations are spot-on, crisp clean, and right to the point.
I can actually pass my class now, thank you.
Hey man! Just wanted to say that you explain the problem and topic very well! Thank you
Hey thank you for the words. Happy that the video(s) helped you!
great job tmr is finally and i can say that u really helped me thanks bro
you're a legend sir.. helps me very much in understanding this chapter.. you made it look easy
Glad to hear! Keep it up.
Thanks for explaining the problem really well!
when finding the moment of inertia with respect to the x-axis we considered the segment along x to have a base of (1-x) and a height of dy. However when we found the moment of inertia with respect to the y-axis we considered the segment to have a base of dx and a height of y. Why didn’t we consider the segment to have a height of (1-y) like we did for the base of segment one which was (1-x).
This question is very hard to explain without a visualization. But think of a line that has an overall length of 1 m. Then cut an amount x of that bar from the left corner and the remainder of the bar that was not cut is the (1-x)... try to visualize it and see if that helps with your question.
@@LearningbyTeaching But why this relation was not used for the calculation of inertia with respect to y
I also have the same doubt ,when finding centroid of shaded area,dA was taken as product of xdy not (1-x)dy in the previous video,why 1-x is only used here ,is the same logic sir said want to applied in the previous problem too,is nt ?
Thank you, Very helpful video for the FE
your great man… i’m understanding these example thanks from iraq♥️.
Thank you for the amazing help again. 😀
Thank you this was really helpful
i finally know dx cannot in some case. I should imagine the point shooting from the axis and see what rectangle is form
Thanks a lot
Lifesaver
Thanks a lot man
Helpful , Thank you
Hi, can you elaborate on how you get x = y² from the given y= x^1/2
Also how you calculate the limit, I can't figure out how you'd get the answer, pls elaborate huhu, my final exam is tomorrow morning.
Shouldn't dA be the same? Meaning if we can subsitute dy by dx and subsitute y in terms of x. For Ix, I obtained 0.4
Hii, from Philippines, why in respect to y-axis, you did not subtract the y from 1 to get the dA of the formula?
Hello, because when we were doing 10-5 with respect to the x-axis the area was on the right side of the curve, if the area was on the left side of the curve it would be just "x*dy" for the dA.
In 10-6 when we are doing with respect to y-axis you can visualize it as under the curve area, so dA = y*dx only. IF the area was above the curve then it would be dA = (1-y)dx...
I hope that helps. The key is to see the height of the "square"/section we are integrating. Is the height of that square equal to the height of the curve? or what else?
we get diffrent answer for 10.5 if we use vertical strip .
Check it out again, I got my answers matching with the ones in the back of the book!! Keep me updated.
goat saved my ass
Solve it by parallel axis theorm
why was it y^2 ?
There where y^2 twice. The first one is given by the formula or the moment of inertia around Ix. The second Y^2 is explained in the video, when we solve for x becomes y^2. :) Hope it helps to understand it a bit better.
Product please