this format of "students" discussing and figuring it out, ESPECIALLY about that last part of the moment of inertia of a single particle being THE SAME as a thin hoop. I view it now as just a large particle where the distance to the axis stays the same, except that it's just more massive... The students observe and ask the right questions. Thanks!!!!
thank you so much for these videos! i hope you keep on making them because ive only recently gotten into physics, and your videos are really helping me understand these concepts. :)
Thank you Sir your videos are the best for a student. I have a question, should the axis of rotation be at the center of mass of that rigid body or it could be anywhere since the rigid body is in static equilibrium ?
Interesting. I do not recall ever seeing limits on an integral of 1 with respect to a variable. In this case, because it is dM, the limits would be 0 to M. That gives you M.
@@FlippingPhysics I have been thinking about this question hard all week. I came up with an interesting alternative derivation that definitely isnt necessary for this course, but is fun. I'll send it to you. Great video and explanation, and thanks for taking the time to respond!
@@Reallycoolguy1369 To put limits on this integral, I would switch the variable of integration to theta, and use 0 to 2*pi. This gives us path of integration we can easily visualize. Each mass element, dm, would therefore equal M/(2*pi) * dtheta. Pull out M/(2*pi) as a constant, and we integrate dtheta from 0 to 2*pi. We end up getting (2*pi)/(2*pi) as part of our result, which cancels out, and are left with M*R^2.
There's a much longer derivation where you integrate around the loop from 0 to 2 pi radians, using a value of dm gotten by comparing the linear density of the whole hoop to a segment. I used to do that one in class but I like this shorter version better. My students caught onto the shorter version and would ask why bother going through the longer version- from now on, I'll do the shorter derivation in class and post the longer version if they want to see it (most won't, lol).
this format of "students" discussing and figuring it out, ESPECIALLY about that last part of the moment of inertia of a single particle being THE SAME as a thin hoop. I view it now as just a large particle where the distance to the axis stays the same, except that it's just more massive... The students observe and ask the right questions. Thanks!!!!
So glad I could help!
thank you so much for these videos! i hope you keep on making them because ive only recently gotten into physics, and your videos are really helping me understand these concepts. :)
You are welcome!
Thank you Sir your videos are the best for a student. I have a question, should the axis of rotation be at the center of mass of that rigid body or it could be anywhere since the rigid body is in static equilibrium ?
sir please keep them coming
i am in love with these videos ❤
I will do my best. I am glad you enjoy them!
Looks great!
Thanks!
Very nice episode
Thanks!
sir do you make jee content
All of my videos are for JEE content. They are organized into JEE playlists here: flippingphysics.com/playlists.html
👍👍👍👍
How come there are not limits of integration for this one?
Interesting. I do not recall ever seeing limits on an integral of 1 with respect to a variable. In this case, because it is dM, the limits would be 0 to M. That gives you M.
@@FlippingPhysics I have been thinking about this question hard all week. I came up with an interesting alternative derivation that definitely isnt necessary for this course, but is fun. I'll send it to you. Great video and explanation, and thanks for taking the time to respond!
@@Reallycoolguy1369 To put limits on this integral, I would switch the variable of integration to theta, and use 0 to 2*pi. This gives us path of integration we can easily visualize. Each mass element, dm, would therefore equal M/(2*pi) * dtheta. Pull out M/(2*pi) as a constant, and we integrate dtheta from 0 to 2*pi. We end up getting (2*pi)/(2*pi) as part of our result, which cancels out, and are left with M*R^2.
@@carultch that's very interesting, thanks
There's a much longer derivation where you integrate around the loop from 0 to 2 pi radians, using a value of dm gotten by comparing the linear density of the whole hoop to a segment. I used to do that one in class but I like this shorter version better. My students caught onto the shorter version and would ask why bother going through the longer version- from now on, I'll do the shorter derivation in class and post the longer version if they want to see it (most won't, lol).