Physics 12 Moment of Inertia (2 of 7) Moment of Inertia of a Solid Sphere
Вставка
- Опубліковано 10 лют 2025
- Visit ilectureonline.com for more math and science lectures!
In this video I will find the moment of inertia of a solid sphere.
Next video in the moment of inertia series:
• Physics 12 Moment of ...
Best youtube physics teacher,by far!!
you are the first teacher who explained the I of sphere by a 2-D graph.
Man I can't remember this is the how many th time you've saved my life. Thx so much
Glad you found our videos helpful! 🙂
Wow... I know I keep saying this, but your derivations of formulas are amazing. I have a final coming up, and more than knowing how to do problems, I'm even more fascinated by how those equations on my formula sheet got there. My professor kind of skipped over how they were found (which led to me believing that moment of inertia was the same for ANY object), but you clearly demonstrated a way of deriving it that wasn't obvious to me, using techniques I already knew. Once again, thank you so much for doing what you do.
What are you doing now bro 😁
WOW! That is some tricky cool math. It all fits like a puzzle and you have to relate one thing to another. Great explanation!
Best youtube lesson I have taken....many thanks
thank you sir this helped me a lot👍
Glad you found our videos. 🙂
Sir ur techng skill is pretty good..... Thanks alot!!
You've helped me so much,
Thank you🙏
Glad to hear it. Thanks for sharing.
@@MichelvanBiezen Inertia/INERTIAL RESISTANCE is proportional to (or BALANCED with/as) GRAVITATIONAL force/ENERGY, AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE). This CLEARLY explains what is E=MC2 AND F=MA ON BALANCE. What is E=MC2 IS dimensionally consistent. What is GRAVITY IS, ON BALANCE, an INTERACTION that cannot be shielded or blocked. Consider TIME AND time dilation ON BALANCE, as the stars AND PLANETS are POINTS in the night sky. Accordingly, ON BALANCE, the rotation of WHAT IS THE MOON matches the revolution. A given PLANET (including what is THE EARTH) sweeps out equal area in equal TIME, AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE); AS TIME IS NECESSARILY possible/potential AND actual ON/IN BALANCE. Great.
By Frank Martin DiMeglio
You 'r amazing !
love from India
Welcome to the channel.
very well explained - excellent quality
Nice presentation 👍
Thanks!
You are great sir love from India
Welcome to the channel!
Q1: Where would you know where to slice?
Q2: How would you know what limits you're setting for the integration?
Q1: You want your slice such that it is easy to find the moment of inertia of the slice. If you took a vertical slice you would not be able to find the moment of inertia as easily. Q2: Due to symmetry, you can have limits from the center to the top and then double the result to account for the bottom part.
@@MichelvanBiezen sir isnt it easy to do it by hollow sphere by taking limits 0 to
R
This was literally beautiful.
awesome expln. love u sir
So nice of you
thank you sir nice explanation ❤
Thank you. 🙂
Thank you sir from india.
Thanks you're great teacher .
liked and subscribed. Thanks.
Glad you found our videos. Welcome to the channel!
Thanks a lot for the explaination
Glad it was helpful!
Thankyou for ur video 😭
You are welcome.
Very good video. Thanks. I have a question, how did you come up with the 1/2(m • x^2) before integrating? where did that x squared come from? In fact, all that formula, it kind of looks like the kinetic energy formula. Thanks
mx^2 is the moment of inertia of a point mass rotating about an axis with a radius of x. If it is a dist that is rotating, the equation for the moment of inertia becomes (1/2) mx^2
Michel van Biezen Thanks!!! I guess the equation of moment of inertia of a rotating disk has in it of itself its own derivation??
We have playlists on the moment of inertia. Take a look.
It can also be solved by taking hollow sphere as elemental mass
Yes, that will work. Did you work it out that way?
@@MichelvanBiezen yes, I got the same result
This disc method is little complex but hollow sphere method is more easy..
It can indeed be done both ways. This is an illustration of how it would be done using disks.
very well explained
thank you , sir
hey professor. your videos were and still are amazing. i just wanna ask you, have you ever thought to solve this problem in the Spherical Coordinates?
We are still expanding all the various topics with our videos. It just takes time (and a lot of work) to do so.
Your amazing! keep up the great work! :)
easy & simple derivation........but i have a question
sir u have written x^2+z^2=R^2...
but why it shouldn't be x^2+(z+dz)^2=R^2
When you integrate you let dz approach 0, therefore in the limit, those 2 equations are identical.
got it.
thanks...
why it is 1/2 of the whole thing ?
The moment of inertia of a solid disk is (1/2) MR^2
I Love You Sir!💞
Thanks!
you're a great teacher!
this is so great thank you for the explanation
How did you get the Z to use in the Pythagorean theorem? X is obvious because you related the thin disk slice as a distance X, and R was already stated so you just placed a mirror image. But you never mentioned the vertical distance 'Z', and I don't know what you are relating that with? Z is the vertical axis, and z is also the vertical side of the triangle?? I'm sorry please explain
If we replace z with y then the equation would be x^2 + y^2 = r^2 (instead of x^2 + z^2 = r^2) they are both the exact same triangle.
Thank you for replying professor, but now I don't see and understand the "y" comparison. Where does it say in this video about a "y"? This is what I don't get. How did you find z, and where is y if that is where you found or related z from?
Then I didn't understand your original question. What specifically are you not sure about?
at 4:00 you are trying to "find a way to convert from x to z". How did you get the z in the triangle you created and then used in your Pythagorean theorem?
We have a circle defined as x^2 + z^2 = r^2. 1) Solve that equation for x. 2) Then find x^4. 3) Then substitute x^4 in the integral by the expression you found in the previous step (2).
Excellent Explanation..appreciated..thank you..!
THANKS, YOU ARE THE best
Delicious Maths!
If we have taken the moment of inertia as mx² instead of ½mx²(I know this is moment of inertia of spherical disk) and detect the bounds as 0 to R instead of -R to R, we would find same result. Is it coinsidence or a thing that need to think deep.
How? If we use mx^2, the final result is 4/5*mR^2. Or did I make a mistake?
Incredible thank you
it's brilliant...thanks a lot...
So dv is the volume of the thin disk (px^2dz), and in the density function v is the whole volume of the sphere (4pir^3)
Why did you use two different Vs and what's the difference between these two?
Li Ping Hsu I don't understand the question
dV is the volume of the thin disk and approached zero in the limit as dz approaches zero, this is the concept of calculus
V is the volume of the sphere.
There is no second V
Unless you think that dV and V are two different volumes?
dV is the differential of V, which is a concept in calculus.
Hi Professor,
I think he did mean that dV and V are two different volumes. One of them is the volume of the thin disk (dv) and the other is the volume of the sphere (V). You solved for Rho = M/V in the beginning and plugged that in as the final value for Rho (after integration was over). He wants to know why couldn't you just use the other Rho (Rho = dm/dv) instead?
I think this is because the density (Rho) is Uniform which is why we have the same Rho with two different equations/formulas (Rho = dm/dv and Rho = M/V). So Dm is infinitesimal (and we already took the integration) so the Rho we plug into the final answer is (M/V). Sorry if I didn't explain it well but I was wondering the same thing until I thought of it this way.
if I calculates the x^4 equal R^4 (times) cos^4(theta), so how to I can solve this assignment?
Is it possible to find the I of the sphere by taking dm paralel to the axis z ?
It would be much more difficult.
Sir do you have any video on product of inertia?
Yes, look in this playlist: MECHANICAL ENGINEERING 12 MOMENT OF INERTIA in the mechanical engineering videos on this channel
@@MichelvanBiezen Thank you sir
For the infinitesimal disc isn’t the limit taken from 0 to R for it to be 1/2 x^2 dm? However in the video the same formula is for -R to R, this part confused me and I would be grateful for some help. Thank you for reading this.
You must integrate from - R to + R because you want to sum up the whole mass of the sphere. (the slices are summed up from the bottom of the sphere to the top of the sphere)
Michel van Biezen thank you very much.
In your cylinder video, you integrated from 0 to R treating 0 as the axis of rotation. Why isnt it the same in this one?
The limits of integration are chosen as needed. There is no set rule.
Michel van Biezen so depending on the section chosen, the limits are different? Thank you
Yes. the limits will be unique to the approach used.
why do i get a different answer when i take x=R*cosθ and dz=r*dθ (R being the radius of the sphere)
Is there something wrong with this substitution?
Why not use the method shown? I would need to see the drawings you use to come up with thos substitutions to determine if they are correct.
Sir, why dI= 1/2dm x r(square) ?? Can you plz
The moment of inertia of a solid disk is: I = (1/2) mR^2, so the moment of inertia of a this slice is: dI = (1/2) dm R^2
Actually I=MK^2, And K=r/root2, Which further Became (MR^2)/2
+Michel van Biezen
Can this kind of method be used for a spherical shell as well?
thank you sir . your work helps millions
You are welcome. Glad our videos are helpful.
Excellent explanation.
If it is rotating along z-axis (vertically), why is he integrating the z-slices? shouldn't he be doing x-slices?
Try using slices in a different direction and see if you can get the moment of inertia that way. (best way to find out).
good explanation.
Tq 👍
Well explained
Thank u sir
thank you!
You're welcome!
Thanks a lot Sir!
thank you,sir!!
How did you find dv?
thank you sir
Amazing
Thanks
why is the moment of inertia of a rotating disc 1/2*m*r^2
why is it not = mr^2
shahid ahmed Look at the moment of inertia video (4 of 6).
It is explained there.
Thnks
alltherestaretaken
Are you referring to a different video? On this video, dV = pi * x^2 dz
alltherestaretaken
The moment of inertia of a disk is: 1/2 m R^2
Thanks brother
Amazing!
Sir, where did that half went sir
In that 4th step, that is I = rho × phi integration of x ^4 dz limits 0 to R
I got confused in that part.....
I changed the limits from -R to R to 0 to R because of the symmetry. That is why the 1/2 was dropped.
Thanks sir
You are welcome.
How can we derive a formula for an irregular solid?
You will need to know the equations that define the solid (and it will be much more difficult).
Thank you sir
nice proof
i tried changing x into Rcosθ and dz into Rdθ since it can be viewed as the arc of a circle with radius R. So then i did the integral of sin^4θdθ from 0 to π/2. But it doesnt give the right answer and i cant find the mistake. can you help me?
If you want to use spherical coordinates, you have to use the proper dV = R^2 sin(phi) dr d(theta) d(phi) and integrate over those variables.
why we consider the height dz for a solid sphere and r.d(theta) for hollow sphere?
why can't we use r.d(theta) for both? (for solid sphere, it doesn't work!)
The rule of thumb is that you want to use the technique that makes the integral the easiest.
I love you
Just confused. When u integrated from 0 to R shouldn't it be half the original integration? Why double the integration?
I think I get it now. Integrating the expression from 0 to R is half the sphere. So from -R to R is double that.
Exactly. You figured it out.
great
Thank you
the best explanation
Why are we moment of inertia =1/2x^2dm instead of =x^2dm
that is the moment of inertia of a solid disk
Thank you
What are the units of mass and radius ?
kg and m
thank you for instant answer:X
You forgot to change the bounds of integration
No, it was done correctly. Thanks for checking.
why dm=1/2 dm x^2 is taken
it is dI = dm x^2 and the reason is that the equation of the moment of inertia for a solid disk is: I = (1/2) mR^2
yes but when we are going to find I from dI by integrating we should take integration of dm x^2 only no but 1/2 dm x^2 u have taken why??
The small dm is a solid cylinder, so you must use the same equation.
how will be the small element will be solid cylinder it must be circular plate
They both have the same formula for the moment of inertia.
why does di = 1/2dmx^2 rather than x^2dm ?? plz help me .....
The moment of inertial of a flat disk is: I = (1/2) MR^2 (See the other videos in the playlist: PHYSICS 12 MOMENT OF INERTIA)
@@MichelvanBiezen thank you very much , I've realized that right after i calculate the I of the flat disk again by mu self . tks !!
how come dI = 1/2dm x^2?
Since the moment of inertia is algebraically additive, we can calculate the moment of inertia of each slice and add them up to give the moment of inertia of the whole object. (This is an integration technique)
That is the formula definition of the moment of inertia of a thin disk along the z-axis:
en.wikipedia.org/wiki/List_of_moments_of_inertia#:~:text=Thin%2C%20solid%20disk%20of%20radius%20r%20and%20mass%20m.&text=Also%2C%20a%20point%20mass%20m,%2C%20with%20r1%20%3D%200.
You can look up the moment of inertia derivation of a thin disk if you want to know why.
When i take x = R cos (theta) and dz=R x d(theta) and integrate from theta = o to pi/2 , why don't we get the desired results ? (I mean the moment of inertia value comes different)
is it polar moment of inertia?
The moment of inertia is not associated with "polar", but with the distance from the axis of rotation.
Is it necessary to use di? I tried it by just filling in for r^2 and dm, and then integrating from -R to R, but this gives an answer that is twice what it should be. But this method works for other shapes like a rod for example. Yet for this one the di method seems necessary, i don't quite understand why there seem to be different methods for different shapes.
Your method assumes that you have a hollow object instead of a solid object. The moment of inertia of a solid disk = (1/2) MR^2
@@MichelvanBiezen I see, thanks very much
Mercy buckets
You are welcome.
ı am dead
Thank you sir
You are welcome. Glad you found it helpful. 🙂