Utilizing the top triangle as a reference, y/x = h/L so y = x(h/L) for the top portion. For the y limits, all you really had to do is integrate from 0 to x(h/L) and then multiply that by 2. This is nice because it matches polar coordinates. But you did it in rectilinear form which is just fine too because you avoided using trigonometric manipulations. This is pretty straight forward based upon the fact that your density in this case is mass/area where the summed area cancels and you have the mr^2 over the surface of the triangle. Do you agree.
you can invert the whole video in post-processing to make it look natural. or shoot into a mirror to double the inversion and eventually cancel them out
Sinister悪魔, Not writing backwards (I'm not that talented). The board is called Learning Glass. You can check it out at www.learning.glass Cheers, Dr. A
This was a fun problem. Any chance you have an example where the density of the plate isn't uniform? I guess if the density were a function of position then you would just keep it inside the integral and slog through it?
It doesn't have to be. For non-isosceles triangles just re-orient your bounds appropriately with trigonometric rules. Your centroid will not be easy to find, though...
This method isn't really that much harder once you get used to it, and I think that it might be slightly more general. But yes, you could probably use a thin strip element and integrate that thing as well.
Interesting, what takes a lot of work is using spherical coordinates to find the moment of a pyramid. ha ha. That's triple integration and very tricky to do. ha ha
Arbaaz khan phymath tales, Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
An absolutely shit method, he says that the figure is not very symmetric but proceeds to assume that the triangle has length 'h' above and below the axis. Also using similarity of triangles and the height of it mitigates the problem and also makes it more general and makes the calculation easier.
Excellent points! We could have exploited the symmetry much more. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
Excellent good vibes physics lectures!! Making it easy to comprehend.
Are you writing backwards.
Mirror edit bro
This is glorious. Love from Canada
Utilizing the top triangle as a reference, y/x = h/L so y = x(h/L) for the top portion. For the y limits, all you really had to do is integrate from 0 to x(h/L) and then multiply that by 2. This is nice because it matches polar coordinates. But you did it in rectilinear form which is just fine too because you avoided using trigonometric manipulations. This is pretty straight forward based upon the fact that your density in this case is mass/area where the summed area cancels and you have the mr^2 over the surface of the triangle. Do you agree.
Happy to have u professor😄😄
Would the Moment of Inertia still come down to a simple general solution if it was an oblique triangle?
I can't focus on the topic as i am wondering whether you are writing mirror images of all letters??..means how are you doing it??
And the legend is back
Very good! Thanks so much! Hugs from Brazil
Bruno Albuquerque,
You're very welcome. Hello to Brazil!
You might also like my new site: www.universityphysics.education
Cheers,
Dr. A
Any Idea about what technology is he using for board. He seems to write on a glass between him and us , yet we do not see inverted/mirrored writing.
you can invert the whole video in post-processing to make it look natural. or shoot into a mirror to double the inversion and eventually cancel them out
Professor Anderson I did not get why did you put a 2 infront of the areal density at the time 7:20 in the video .....??????
because 1 - (-1) = 2. And 1something - (-1something) = 2something
How do you write it flipped!? That's some serious dedication!
Sinister悪魔,
Not writing backwards (I'm not that talented). The board is called Learning Glass. You can check it out at www.learning.glass
Cheers,
Dr. A
Sir I want your help in finding moment of inertia of triangular loop.please help me
Cheers,
Dr. A
This was a fun problem. Any chance you have an example where the density of the plate isn't uniform? I guess if the density were a function of position then you would just keep it inside the integral and slog through it?
I know how to do it.
Oh I caught what he did. It's rotating about the apex and not the base.
That changes the problem.
Sir,is the axis of rotation perpendicular to the plane of triangle or is it parallel?
perpendicular
Are u still there
Thank you, I'm now even more confused than before I started watching the video.
Sir but it's only useful for a isosceles triangle right as you took altitude on side 2h as midpoint of side2h
It doesn't have to be. For non-isosceles triangles just re-orient your bounds appropriately with trigonometric rules. Your centroid will not be easy to find, though...
What if I want to rotate on the y axis?
I think you have to change r but use the same dm
sir when a unit circle complete a revolution how much distance covered by it
Circumference = 2πr. If your r=1 (in meters, say), then the distance = 2π (meters).
Cheers,
Dr. A
Great it's very helpful sir
How is he writing?
Oh rotating about the apex.
opps. Yeah!
why dont you integrate strip of length x and width dx it might be simple.. will it be?
This method isn't really that much harder once you get used to it, and I think that it might be slightly more general.
But yes, you could probably use a thin strip element and integrate that thing as well.
Thank you so much.
Doing good job
Fascinating.
Thanks sir Benedict Cumberbatch 😃 if you know what I mean
I know what you mean.
Or don't I?
Cheers,
Dr. A
Yes Dr. Strange for MCU fans
Interesting, what takes a lot of work is using spherical coordinates to find the moment of a pyramid. ha ha.
That's triple integration and very tricky to do. ha ha
Thx sir
thx!
You're welcome.
Cheers,
Dr. A
very nice
Arbaaz khan phymath tales,
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
what he is showing is pretty hard. u can actually find moi by taking an elemental strip and then integrating it
An absolutely shit method, he says that the figure is not very symmetric but proceeds to assume that the triangle has length 'h' above and below the axis. Also using similarity of triangles and the height of it mitigates the problem and also makes it more general and makes the calculation easier.
Excellent points! We could have exploited the symmetry much more.
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A