Considering the fact that dr is very close to zero, dr^2 can be approximated to 0 because it is way, way smaller by the same factor dr, which is already extremely small.
The mass of the total disc is constant however when we consider smaller mass elements the only thing constant is the mass 'distribution'/'density'. So you can't take dm out of the integral!
If you want to compute that I think you're gonna use how you compute for the area of a shape in polar coordinates, so you start from θ=0 to θ=2π and then do the rest
i'm kinda late but 1/4 MR^2 are the momments of inertia tha correspond to the x and z axis that are parallel to the disc while 1/2 MR^2 is the momment of inertia of the y axis which is perpendicular to it
U taught this topic better than most of the teachers in Indian coaching who who prepare us for iit adv and stuff.
Thank u😗
Bruh I’m here for the same purpose
can't concentrate on the physics cause his backwards writing is too impressive
He is writing normally but the video is flipped. You are now free to concentrate. ;)
OMG.....all this time I thought they were writing backwards lol!!!! Mind blown
Mind blown for the 2nd time
@@mitocw still impressive tho
@@MurasakiBubble it's russian technique
clean explanation and straight to the point. thanks
MIT: making worlds best tech people
Also MIT microphone: we use raw audio with noise
Thanks a Lot Sir, You made is so easy, as I saw a lot of derivations which has no explanation for dm
Thank you so much Sir
thank you! you explained it way better than my teacher did
woah very nice explanation
2:51 How can you say that the 2nd order differential dr is zero as dr approaches zero but the first order differential remains unchanged?
for the ease of calculation he did that
Considering the fact that dr is very close to zero, dr^2 can be approximated to 0 because it is way, way smaller by the same factor dr, which is already extremely small.
Use a calculator. If r = 0.001 then r^2 = .000001 and so on and on. You can see that dr^2 will approach 0 within that limit.
Why should r be the integration variable, why can't m be the integration variable?
Isn't the mass constant? Maybe that's why.
Since total mass is constant why couldnt we have integrated like this m ∫ r^2 dr from the definition of inertia
The mass of the total disc is constant however when we consider smaller mass elements the only thing constant is the mass 'distribution'/'density'. So you can't take dm out of the integral!
thanks
How are you writing this ...I'm confused! Are you writing mirrored or flipping the image?
Rotate 90 degree to left /right
Why we have taken dm x r^2 only as difinetion of MI of rotational disc
How do we calculate if the disc is inclined at an angle to the vertical?
Sir can we slove moment of interia of a circle by considering triangle strip. If can please tell me how to slove
If you want to compute that I think you're gonna use how you compute for the area of a shape in polar coordinates, so you start from θ=0 to θ=2π and then do the rest
The DM is the tricky part
This is wrong,,,Icm of disc is=MR^ 2/4
no, its 1/2 mr^2
i'm kinda late but 1/4 MR^2 are the momments of inertia tha correspond to the x and z axis that are parallel to the disc while 1/2 MR^2 is the momment of inertia of the y axis which is perpendicular to it
Thanks
Thanks
thanks