I'm pretty sure I watched and commented on this video two or three years ago for my physics 1 class in my university. Now I'm about to take the final exam for my senior level classical mechanics course and I find myself reviewing with your videos again haha! Thank you so much for making these videos. People like you make UA-cam such a wonderful place. I hope your students appreciate what a great teacher they have.
So, right now I'm taking High School level physics (and calc), and I was annoyed that my textbook didn't show me how to derive these various moment-of-inertia formulae (they only supply you with the 3 or 4 formulae that they'll have you use...) AND THIS VIDEO WAS EXACTLY WHAT I WANTED, no joke! Thanks a ton!!
From Turkey. Very clear language and understandable speech. that's very important for non-native English people to understand the content. Thank you very much :-)
A clear understanding is pivotal in deriving every core riddle of Physics...REALLY INDEBTED TO THE EFFORTS OF A LEARNED MAN CONSOLIDATING THE BASICS OF PHYSICS.
You explained this really, really clearly. I'm a senior mechanical engineering student at the University of Houston and used this to brush up on some things for Solid Mechanics. Thanks!
The confusion totally finishes off when we understand the concept of integrals , Linear density and ro.................. Great Lecture helped me a lot by clearifying my little confusions... Thank you mr.Fullerton
Wow, that's mighty kind of you, and I sure appreciate the thought. How about instead of a donation you tell some other folks about the videos and the APlusPhysics site and kick butt on your AP exam for me? Best of luck!
After struggling with moment of inertia this video really helped me to finally understand it. The process of calculating moment of inertia of a 3D shape such as the cylinder was explained very clearly and I understood it better than any other explanation I have found. Great video!
For other people who were as confused as me: when he wants to find dm in terms of dr, he differentiated both the total mass with respect to r and the total volume with respect to r, as (dm/dr) / (dv/dr) = dm/dv = dm / ( (dv/dr) * dr ) = rho, so dm = rho * (dm/dr) * dr.
Great question!!!, and absolutely you can do this, but it becomes a bit of a trivial integral. The integral of a thin disk is 0.5 mR^2, so if you integrate from -L/2 to L/2 with respect to L, 0.5R^2 is a constant and comes out of the integral, leaving you with the integral from -L/2 to L/2 of m*dl (where m is the mass of the little slice of disc). Of course, the integral from -L/2 to L/2 of m*dl is just going to give you the total mass, M, therefore you end up with 1/2MR^2 again. :-)
@@DanFullerton Teachers like you who make their videos available are a HUGE reason why I can understand the things that I do and I appreciate it and hope to pass it forward.
Both of those are correct. Delta omega / t is the rate of change of angular velocity, which is angular acceleration. If you know linear acceleration, and the radius, you can also find angular acceleration alpha = a/R. Both will give you the same answer!
It is quite interesting that the moment of inertia of the cylinder is independent of its length: I = (1 / 2)MR^2, which is the same as the moment of inertia of a thin planar circular disk. Thank you for this resource.
Depends on what's given. If you're dealing with charge in a line, try a linear charge density (lambda). If it's charge on a surface, use sigma for surface charge density. And if it's charge distributed in a volume, use volume charge density (rho). Good luck!
Hello! Thank you for making these awesome videos! Seriously cannot help but thank you and all the other teachers that make these review videos for AP prep, our AP scores probably go from 2s and 3s to 4s and 5s because of you! Thank you and I will try to do my best tomorrow! :D
I have 3 or 4 problems that I do within the lecture itself as examples. Beyond that, it'll be a while before I get around to augmenting these lectures with more examples (my "to-do" list of videos and other items is pretty long at the beginning of the school year). There are a number of great physics for engineers - type books that have more sample problems and solutions, though, or feel free to check out the APlusPhysics site -- post any problems in the AP Physics C and see what you get! Tnx!
I love it when physics teachers get down and dirty with the math. The IB physics textbook gives an explanation on this same topic that allows people to assume that infinitesimal calculus is some black box whose processes can't be understood, ironically while IB math requires students to learn calculus.
Glad you like the video... you'll find most of the AP-C videos show the math in detail (as is the goal of the upcoming "AP Physics C Companion" book, in the works now! Thanks for the great feedback, and have a fantastic day.
That would be more exact. I'm simplifying the problem by making the assumption they are point masses, therefore the moment of inertia of the bowling ball thorugh which the axis of rotation is passing is zero (or practically negligible compared to the other).
Thank for your lesson! I didn`t find any information on the Internet about how to determine the moment of inertia about the axis of the rod that passes through an arbitrary point of the rod (not through the center and through the end)? May you explain it, please?
Hello sir, I love how the way you explained about the concept. Just one thing i was a bit unsure of, is why at 9:51, you set the limit from -L/2 to L/2? On what coordinate does it lies actually because looking at when rotating it at its tips, the limit looks like it lies on the x-axis?
You need to know how to derive the moments of inertia of objects, but it's probably a good idea to memorize some common moments of inertia (rod about its center, rod about end, hollow sphere, solid sphere, etc.)
Thanks, this was very helpful. My question is to find the moment of inertia for a soild cylinder: can we not make it the sum of multiple disks going from top to bottom or vice versa as we use to find the M.O.I for a sphere? We could make the sum for all of those disks with a thickness dz. If we can would we get to the same result? If I am not wrong may you help me with the calculation please? Thanks again.
Hi Aishwarya. The Perpendicular Axis Theorem is outside the scope of the AP Physics C course, so it will be quite some time before I'd catch up to do that one (I'm already behind on AP-2 videos), but Dr. Walter Lewin gives it some great coverage in this video: video.mit.edu/watch/19-rotating-rigid-bodies-moment-of-inertia-parallel-axis-and-perpendicular-axis-theorem-r-12554/
I really love your videos, they help me so much to prep for the upcoming AP Physics C tests! Is there any way I can donate to you to show my appreciation?
Mr. Fullerton, angular acceleration is [(delta w) / (t)] right? But how come that I saw a formula of angular acceleration (alpha=a/R)??? I'm confused Mr. Fullerton please help me.
why is it that we ignore the radius of the uniform rod? or are we considering it so thin that it basically doesn't contribute much? i'm not sure i'm able to integrate it at this point, as it seems to require multiple integrals (varying length-wise radius and also the radius of the cylinder width-wise). e: imagine we try out two rods: we know length doesn't matter, as long as the density is uniform. one is 1m long with a 5cm radius, one is 0.005m long with a 1m radius. i'd imagine they'd spin a bit differently, even in a frictionless vacuum. or do they not? i'll try doing the double integral later...
Sure... imagine a soda can. You cut it along the long edge and spread it out to make a rectangular box. The volume of that rectangle would be the circumference of the can (2*pi*r), the height of the can (L), and its thickness is dr. We multiple that volume by the mass density to get the total mass. Then, to get the solid cylinder, we start at a 0 radius, and then keep increasing r as we build fatter and fatter coke cans, then add the volume of all those cans together to get the entire volume
Our little bit of mass, dm, is a thin cylindrical shell which expands outward, so we need the mass of that infinitely thin hollow cylindrical shell. And we always begin by integrating with respect to m (that's our general formula for moment of inertia).
great video, but for future videos, can you define some of these variables like row and lambda and dx with more detail. i think the problem with most people trying to learn this is wrapping their heads entirely around the concepts and derivations, so being very clear with the variables would be extremely helpfull. thanks
Thanks, I appreciate the feedback. Please keep in mind that these are part of a series of videos covering the entire course, so some items such as rho and lambda and dx are covered in earlier videos in the series (which could certainly add to the confusion). Make it a great day!
Sir while we were calculating the rod rotating about its center we wrote +L/2(integral) -L/2 but after that we used R(integral)0 for solid cylinder rotating about its center too so why did we not write R/2(integral)-R/2 for it ?
+Afrasyab Alper Hi Afrasyab. When we're moving up the rod from -L/2 to L/2, we're adding up little pieces of rod dy. When we're doing the cylinder, we're adding up lots of hollow cylinders, with a minimum radius of 0, and a maximum of R. A negative radius doesn't have a physical meaning.
+Denzel Brown It's the same derivation as what you had on the left part of the screen in white -- since you'd already done most of that work, I didn't repeat it on the right. Integral of x^2 is x^3/3, so I pulled the 1/3 out of the parentheses to make it clearer.
Apralana -- while solving for I with the axis passing through the center, your r will vary from -L/2 to L/2. If you wanted to take the limits of integration as 0 to L, you would have to define and measure all your positions (r's) from L/2 -- which is doable, but complicated.
Integral of x^2 is x^3/3. Using the same work we did on the left (in white) to speed up the integration on the right. Notice that we did the same integration previously, just with different limits.
Sorry but one more thing...even though we arrive at the same answer: Would it still be considered as deriving the equation or is the other method more correct in a way to derive the equation. Why is it that one would choose one way or the other to derive the equation? Well I believe the way in the video is more important because the solid cylinder is the sum of infinite hollow cylinders but my way would leave a limited number of disks which defeats the entire purpose. Am I right?
I'm pretty sure I watched and commented on this video two or three years ago for my physics 1 class in my university. Now I'm about to take the final exam for my senior level classical mechanics course and I find myself reviewing with your videos again haha! Thank you so much for making these videos. People like you make UA-cam such a wonderful place. I hope your students appreciate what a great teacher they have.
Cody Kellogg Thanks Cody, and great to hear from you again! Best of luck to you with your final exam, and make it a great day!
learned more in 18 mins than in my 1 and a half hour lecture. thank you for your videos!
Glad to hear the video helped!
So, right now I'm taking High School level physics (and calc), and I was annoyed that my textbook didn't show me how to derive these various moment-of-inertia formulae (they only supply you with the 3 or 4 formulae that they'll have you use...) AND THIS VIDEO WAS EXACTLY WHAT I WANTED, no joke! Thanks a ton!!
Thrilled to hear it helped you out!
From Turkey. Very clear language and understandable speech. that's very important for non-native English people to understand the content. Thank you very much :-)
This is the first time that I can recall actually understanding the concept of inertia mathematically. Thank you!
if you want to learn physics the smooth clear way...watch this guy. My teacher on the hand teach us witchcraft and sorcery.
Thanks Danilo. Glad to hear this is helping your physics. Hope your witchcraft and sorcery studies are going just as well!
You just explained inertia to me better than ANYONE else! Thank you :)
I'm so glad to hear things are becoming clear! Make it a great day...
A clear understanding is pivotal in deriving every core riddle of Physics...REALLY INDEBTED TO THE EFFORTS OF A LEARNED MAN CONSOLIDATING THE BASICS OF PHYSICS.
11 years late to the party but you just saved a senior's whole semester. Thank you❤
I'm trying to self study ap physics c in 3 weeks. Your videos are making it possible!! Thank you so much!!
How’d it go?
@@washington0005 He died of depression because it was too much stress.
You explained this really, really clearly. I'm a senior mechanical engineering student at the University of Houston and used this to brush up on some things for Solid Mechanics. Thanks!
+Tony Richmond You're very welcome, and good luck on a fantastic senior year!
The confusion totally finishes off when we understand the concept of integrals , Linear density and ro..................
Great Lecture helped me a lot by clearifying my little confusions... Thank you mr.Fullerton
Wow, that's mighty kind of you, and I sure appreciate the thought. How about instead of a donation you tell some other folks about the videos and the APlusPhysics site and kick butt on your AP exam for me? Best of luck!
After struggling with moment of inertia this video really helped me to finally understand it. The process of calculating moment of inertia of a 3D shape such as the cylinder was explained very clearly and I understood it better than any other explanation I have found. Great video!
If only my professor was half as clear as this, would've saved me so much time.
Thank you so much! You have no idea how much your videos have helped me and my friends throughout the year!
Thanks boss. Very specific and clear. I had learned these before but the delivery has never been this decent. Cheers
For other people who were as confused as me: when he wants to find dm in terms of dr, he differentiated both the total mass with respect to r and the total volume with respect to r, as (dm/dr) / (dv/dr) = dm/dv = dm / ( (dv/dr) * dr ) = rho, so dm = rho * (dm/dr) * dr.
You are brilliant. The video kept me alert till the end, even made the time pass faster. Those 18 felt like 5 min.
kudos to you bro, you are saving my butt in college,.....kudos to you
Nice to hear the videos are helping you out!
thanks for your help you are a true physic teacher
+John Farfan You are very welcome!
Great question!!!, and absolutely you can do this, but it becomes a bit of a trivial integral. The integral of a thin disk is 0.5 mR^2, so if you integrate from -L/2 to L/2 with respect to L, 0.5R^2 is a constant and comes out of the integral, leaving you with the integral from -L/2 to L/2 of m*dl (where m is the mass of the little slice of disc). Of course, the integral from -L/2 to L/2 of m*dl is just going to give you the total mass, M, therefore you end up with 1/2MR^2 again. :-)
Your lectures are in sequence of our syllabi! I thank you a lot sir for these!
THANK YOU! Why Khan academy decided to skim over the calc is disappointing, but I knew it would be out there somewhere !
Glad this helped you out!
@@DanFullerton Teachers like you who make their videos available are a HUGE reason why I can understand the things that I do and I appreciate it and hope to pass it forward.
Glad to hear it helped, you're very welcome!
Both of those are correct. Delta omega / t is the rate of change of angular velocity, which is angular acceleration. If you know linear acceleration, and the radius, you can also find angular acceleration alpha = a/R. Both will give you the same answer!
sir you really understand the concept of this topic !
clear with full explanation...
so useful
thank you
This was very helpful. Made the parallel axis theorem easy.
Thanks for the help! It provided a fine intuition of how we got the MI's of common objects.
It is quite interesting that the moment of inertia of the cylinder is independent of its length: I = (1 / 2)MR^2, which is the same as the moment of inertia of a thin planar circular disk. Thank you for this resource.
Depends on what's given. If you're dealing with charge in a line, try a linear charge density (lambda). If it's charge on a surface, use sigma for surface charge density. And if it's charge distributed in a volume, use volume charge density (rho). Good luck!
Hello! Thank you for making these awesome videos! Seriously cannot help but thank you and all the other teachers that make these review videos for AP prep, our AP scores probably go from 2s and 3s to 4s and 5s because of you! Thank you and I will try to do my best tomorrow! :D
Umama Ahmed You're very welcome, and good luck!
Thanks sir.You and your videos are Gr8!!Cleared all my concepts
This Videos really helped me!
Greeting from the Technical University of Munich🙋♀️🙋♀️
Greetings right back at you! :-)
I have 3 or 4 problems that I do within the lecture itself as examples. Beyond that, it'll be a while before I get around to augmenting these lectures with more examples (my "to-do" list of videos and other items is pretty long at the beginning of the school year). There are a number of great physics for engineers - type books that have more sample problems and solutions, though, or feel free to check out the APlusPhysics site -- post any problems in the AP Physics C and see what you get! Tnx!
You saved my life i cant tell how blessed i am 🙏🏻
You made understanding M.O.I. really easy, thank you!!!
My pleasure!
I love it when physics teachers get down and dirty with the math. The IB physics textbook gives an explanation on this same topic that allows people to assume that infinitesimal calculus is some black box whose processes can't be understood, ironically while IB math requires students to learn calculus.
Glad you like the video... you'll find most of the AP-C videos show the math in detail (as is the goal of the upcoming "AP Physics C Companion" book, in the works now! Thanks for the great feedback, and have a fantastic day.
thnxx sir
Thanks from Clemson University! Makes a lot more sense now!
You're welcome, and go Tigers!
it's beautiful...I'm tearing up. sniff sniff. Well done
Don't cry, it'll be OK. Thanks!
Yes. On the parallel axis theorem slide you should be able to find it in the diagram. :-)
Thank you so much. You just saved me right before my exam!
Good luck with your test!
well done sir! you have made physics easy and quick to learn!
Great presentation, examples and graphics. thanks
That would be more exact. I'm simplifying the problem by making the assumption they are point masses, therefore the moment of inertia of the bowling ball thorugh which the axis of rotation is passing is zero (or practically negligible compared to the other).
Glad to hear it Nick, have a great year!
Brilliant video mate! Learnt so much!
Wow thank you very much! This was clearly explained. I am doing this for my Extended Essay (a requirement for the IB). Thank you sir.
This supplemented my knowledge of Physics B well :)
Thank for your lesson!
I didn`t find any information on the Internet about how to determine the moment of inertia about the axis of the rod that passes through an arbitrary point of the rod (not through the center and through the end)?
May you explain it, please?
Hi Dan!
Thank u. You kept it clear and simple.
My Greatest Gratitude Sir!!!!
thanks Mr Dan. very good explanation .
Hello sir, I love how the way you explained about the concept. Just one thing i was a bit unsure of, is why at 9:51, you set the limit from -L/2 to L/2? On what coordinate does it lies actually because looking at when rotating it at its tips, the limit looks like it lies on the x-axis?
+Shafique Abdul razor I'm integrating there from x=-L/2 to x=+L/2, so across the length of the rod, treating the center of the rod as x=0.
Very clear explanation.
You need to know how to derive the moments of inertia of objects, but it's probably a good idea to memorize some common moments of inertia (rod about its center, rod about end, hollow sphere, solid sphere, etc.)
Very good job seriously!
Thank you!
Glad you found the video helpful!
Could you also explain the Perpendicular Axis Theorem in any of your future videos?
for the example of solid cylinder can you elaborate more on how is dm=2(pi)rL(rho) and thanks a lot for posting this videos! :)
Thrilled you liked it!
Thanks, this was very helpful. My question is to find the moment of inertia for a soild cylinder: can we not make it the sum of multiple disks going from top to bottom or vice versa as we use to find the M.O.I for a sphere? We could make the sum for all of those disks with a thickness dz. If we can would we get to the same result? If I am not wrong may you help me with the calculation please? Thanks again.
Dan, you helped me out a lot. Thank you so much!
Hi Aishwarya. The Perpendicular Axis Theorem is outside the scope of the AP Physics C course, so it will be quite some time before I'd catch up to do that one (I'm already behind on AP-2 videos), but Dr. Walter Lewin gives it some great coverage in this video: video.mit.edu/watch/19-rotating-rigid-bodies-moment-of-inertia-parallel-axis-and-perpendicular-axis-theorem-r-12554/
Not anymore !!
@@Adam-dr9crNow available here: ua-cam.com/video/fDJeVR0o__w/v-deo.html
I really love your videos, they help me so much to prep for the upcoming AP Physics C tests! Is there any way I can donate to you to show my appreciation?
Dan! You're the man! THANK YOU!!!
Joshua Walters Thanks! Make it a great day!
Perfect explanation. Thanks alot!
Glad you like it, you're welcome!
Do we need to memorize that? Or it is just a proof of that? Btw thank you so much u have the best physic tutorial video in UA-cam.
thanks a lot .. it helped me on my semester
Thrilled to hear it!
I understood it thank you so much but i dont get from the problem of the rod. Why lamda = M/ L. And what does it represent?
You are very welcome!
nice video sir, really got my mind goin :)
Excellent. Make it a great day!
thanks, cleared up some confusion i had,
Mr. Fullerton, angular acceleration is [(delta w) / (t)] right? But how come that I saw a formula of angular acceleration (alpha=a/R)??? I'm confused Mr. Fullerton please help me.
when it comes to the parallel azis theorem, i'm assuming d is the perpendicular separation between l and l'?
Very helpful tutorial. Thanks a lot :)
Impressive. Embarrassing traditional public education with well-done and energetic lectures. For free.
Glad you found it useful! Thanks!
Hi. So I'm trying to understand why you have to use lamda=m/l to integrate this. What is it about density that we need?
Using the linear density function allows you to process the integral of r^2dm in a fairly straightforward fashion.
why is it that we ignore the radius of the uniform rod? or are we considering it so thin that it basically doesn't contribute much? i'm not sure i'm able to integrate it at this point, as it seems to require multiple integrals (varying length-wise radius and also the radius of the cylinder width-wise).
e: imagine we try out two rods: we know length doesn't matter, as long as the density is uniform. one is 1m long with a 5cm radius, one is 0.005m long with a 1m radius. i'd imagine they'd spin a bit differently, even in a frictionless vacuum. or do they not? i'll try doing the double integral later...
Yes, we're assuming it's infinitely thin to simplify our calculations.
Sure... imagine a soda can. You cut it along the long edge and spread it out to make a rectangular box. The volume of that rectangle would be the circumference of the can (2*pi*r), the height of the can (L), and its thickness is dr. We multiple that volume by the mass density to get the total mass. Then, to get the solid cylinder, we start at a 0 radius, and then keep increasing r as we build fatter and fatter coke cans, then add the volume of all those cans together to get the entire volume
Great explanation!
when calculatin I for the solid cylinder what's your reasoning for what dm is and why are you integrating with respect to m and not r?
Our little bit of mass, dm, is a thin cylindrical shell which expands outward, so we need the mass of that infinitely thin hollow cylindrical shell. And we always begin by integrating with respect to m (that's our general formula for moment of inertia).
your awesome, you have any lecture on forces moment?
Sure -- check out the APlusPhysics website!
wow......u are so amazing !!!!! thank u so much for this video u made
great video, but for future videos, can you define some of these variables like row and lambda and dx with more detail. i think the problem with most people trying to learn this is wrapping their heads entirely around the concepts and derivations, so being very clear with the variables would be extremely helpfull. thanks
Thanks, I appreciate the feedback. Please keep in mind that these are part of a series of videos covering the entire course, so some items such as rho and lambda and dx are covered in earlier videos in the series (which could certainly add to the confusion). Make it a great day!
why is the piece took
from the uniform rod dm? And why dm=入dx? Can anyone help me thx!
for the uniform cylinder, why is dm=2pi*r*L*(the rest) and not dm=r^2*pi*L?
Sir while we were calculating the rod rotating about its center we wrote +L/2(integral) -L/2 but after that we used R(integral)0 for solid cylinder rotating about its center too so why did we not write R/2(integral)-R/2 for it ?
+Afrasyab Alper Hi Afrasyab. When we're moving up the rod from -L/2 to L/2, we're adding up little pieces of rod dy. When we're doing the cylinder, we're adding up lots of hollow cylinders, with a minimum radius of 0, and a maximum of R. A negative radius doesn't have a physical meaning.
Dan Fullerton Hmm Thanks sir :)
Please someone help quick; why is he multiplying by lamda over 3 at 10:12?
+Denzel Brown It's the same derivation as what you had on the left part of the screen in white -- since you'd already done most of that work, I didn't repeat it on the right. Integral of x^2 is x^3/3, so I pulled the 1/3 out of the parentheses to make it clearer.
thank you !
thank you so much....... keep doing more videos
lambda = M/l is the linear mass density of the rod.
while taking out the I of rod with the axis of rotation passing through the centre, why can't we take the limits of integration as 0 to L?
Apralana -- while solving for I with the axis passing through the center, your r will vary from -L/2 to L/2. If you wanted to take the limits of integration as 0 to L, you would have to define and measure all your positions (r's) from L/2 -- which is doable, but complicated.
Great video! But how do we solve sphere problem?
What sphere problem?
At 10:18, why lambda over 3 instead of just lambda?
Integral of x^2 is x^3/3. Using the same work we did on the left (in white) to speed up the integration on the right. Notice that we did the same integration previously, just with different limits.
+Dan Fullerton (APlusPhysics) Ooh, now I see. I've got love for your videos, sir. Thanks!
thanks Dan that was very helpful
Sorry but one more thing...even though we arrive at the same answer: Would it still be considered as deriving the equation or is the other method more correct in a way to derive the equation. Why is it that one would choose one way or the other to derive the equation? Well I believe the way in the video is more important because the solid cylinder is the sum of infinite hollow cylinders but my way would leave a limited number of disks which defeats the entire purpose. Am I right?
This was really helpful! Thank you!
You're welcome!
Glad you liked it!