Best youtube physics teacher,by far!!

WOW! That is some tricky cool math. It all fits like a puzzle and you have to relate one thing to another. Great explanation!

you are the first teacher who explained the I of sphere by a 2-D graph.

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.

@ECETNSAVIJAYAWADA

2 жыл бұрын

What are you doing now bro 😁

Sir ur techng skill is pretty good..... Thanks alot!!

Man I can't remember this is the how many th time you've saved my life. Thx so much

@MichelvanBiezen

Жыл бұрын

Glad you found our videos helpful! 🙂

Best youtube lesson I have taken....many thanks

very well explained - excellent quality

This was literally beautiful.

Your amazing! keep up the great work! :)

you're a great teacher!

Thanks you're great teacher .

Excellent Explanation..appreciated..thank you..!

How did you find dv?

it's brilliant...thanks a lot...

very well explained thank you , sir

this is so great thank you for the explanation

THANKS, YOU ARE THE best

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?

@MichelvanBiezen

4 жыл бұрын

We are still expanding all the various topics with our videos. It just takes time (and a lot of work) to do so.

Is it possible to find the I of the sphere by taking dm paralel to the axis z ?

@MichelvanBiezen

5 жыл бұрын

It would be much more difficult.

You 'r amazing ! love from India

@MichelvanBiezen

6 жыл бұрын

Welcome to the channel.

thank you,sir!!

You've helped me so much, Thank you🙏

@MichelvanBiezen

2 жыл бұрын

Glad to hear it. Thanks for sharing.

@frankdimeglio8216

Жыл бұрын

​@@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

Thank you sir from india.

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

@MichelvanBiezen

6 жыл бұрын

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

@andresrebata1958

6 жыл бұрын

Michel van Biezen Thanks!!! I guess the equation of moment of inertia of a rotating disk has in it of itself its own derivation??

@MichelvanBiezen

6 жыл бұрын

We have playlists on the moment of inertia. Take a look.

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.

@joao_pedro_c

5 жыл бұрын

How? If we use mx^2, the final result is 4/5*mR^2. Or did I make a mistake?

Q1: Where would you know where to slice? Q2: How would you know what limits you're setting for the integration?

@MichelvanBiezen

6 жыл бұрын

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.

@ashutoshkumar3921

4 жыл бұрын

@@MichelvanBiezen sir isnt it easy to do it by hollow sphere by taking limits 0 to R

if I calculates the x^4 equal R^4 (times) cos^4(theta), so how to I can solve this assignment?

Amazing!

Thanks a lot Sir!

good explanation.

Incredible thank you

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

@MichelvanBiezen

7 жыл бұрын

When you integrate you let dz approach 0, therefore in the limit, those 2 equations are identical.

@umerwaqas3916

7 жыл бұрын

got it. thanks...

If it is rotating along z-axis (vertically), why is he integrating the z-slices? shouldn't he be doing x-slices?

@MichelvanBiezen

6 жыл бұрын

Try using slices in a different direction and see if you can get the moment of inertia that way. (best way to find out).

Sir, why dI= 1/2dm x r(square) ?? Can you plz

@MichelvanBiezen

7 жыл бұрын

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

@abhishekkumaranand837

7 жыл бұрын

Actually I=MK^2, And K=r/root2, Which further Became (MR^2)/2

@Peter_1986

7 жыл бұрын

+Michel van Biezen Can this kind of method be used for a spherical shell as well?

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?

@MichelvanBiezen

5 ай бұрын

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.

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?

@MichelvanBiezen

5 жыл бұрын

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.

thank you sir this helped me a lot👍

@MichelvanBiezen

8 ай бұрын

Glad you found our videos. 🙂

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?

@MichelvanBiezen

9 жыл бұрын

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.

@baydood510

7 жыл бұрын

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.

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.

@MichelvanBiezen

4 жыл бұрын

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)

@David-ur7og

4 жыл бұрын

Michel van Biezen thank you very much.

You are great sir love from India

@MichelvanBiezen

6 жыл бұрын

Welcome to the channel!

Well explained

Thanks a lot for the explaination

@MichelvanBiezen

2 жыл бұрын

Glad it was helpful!

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

@MichelvanBiezen

7 жыл бұрын

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.

@baydood510

7 жыл бұрын

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?

@MichelvanBiezen

7 жыл бұрын

Then I didn't understand your original question. What specifically are you not sure about?

@baydood510

7 жыл бұрын

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?

@MichelvanBiezen

7 жыл бұрын

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).

Delicious Maths!

thank you sir

liked and subscribed. Thanks.

@MichelvanBiezen

2 жыл бұрын

Glad you found our videos. Welcome to the channel!

Sir do you have any video on product of inertia?

@MichelvanBiezen

3 жыл бұрын

Yes, look in this playlist: MECHANICAL ENGINEERING 12 MOMENT OF INERTIA in the mechanical engineering videos on this channel

@md.touhidulislam4668

3 жыл бұрын

@@MichelvanBiezen Thank you sir

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?

@MichelvanBiezen

6 жыл бұрын

The limits of integration are chosen as needed. There is no set rule.

@jarreddiaz839

6 жыл бұрын

Michel van Biezen so depending on the section chosen, the limits are different? Thank you

@MichelvanBiezen

6 жыл бұрын

Yes. the limits will be unique to the approach used.

awesome expln. love u sir

@MichelvanBiezen

4 жыл бұрын

So nice of you

why it is 1/2 of the whole thing ?

@MichelvanBiezen

6 ай бұрын

The moment of inertia of a solid disk is (1/2) MR^2

Thank u sir

Nice presentation 👍

@MichelvanBiezen

3 жыл бұрын

Thanks!

How can we derive a formula for an irregular solid?

@MichelvanBiezen

7 жыл бұрын

You will need to know the equations that define the solid (and it will be much more difficult).

@izzatredza854

7 жыл бұрын

Thank you sir

I Love You Sir!💞

@MichelvanBiezen

3 жыл бұрын

Thanks!

thank you sir nice explanation ❤

@MichelvanBiezen

Жыл бұрын

Thank you. 🙂

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!)

@MichelvanBiezen

4 жыл бұрын

The rule of thumb is that you want to use the technique that makes the integral the easiest.

Thankyou for ur video 😭

@MichelvanBiezen

Жыл бұрын

You are welcome.

is it polar moment of inertia?

@MichelvanBiezen

Жыл бұрын

The moment of inertia is not associated with "polar", but with the distance from the axis of rotation.

What are the units of mass and radius ?

@MichelvanBiezen

7 жыл бұрын

kg and m

@ilkertalatcankutlucan3257

7 жыл бұрын

thank you for instant answer:X

Excellent explanation.

why is the moment of inertia of a rotating disc 1/2*m*r^2 why is it not = mr^2

@MichelvanBiezen

9 жыл бұрын

shahid ahmed Look at the moment of inertia video (4 of 6). It is explained there.

@shahidahmed7511

9 жыл бұрын

Thnks

@MichelvanBiezen

9 жыл бұрын

alltherestaretaken Are you referring to a different video? On this video, dV = pi * x^2 dz

@MichelvanBiezen

9 жыл бұрын

alltherestaretaken The moment of inertia of a disk is: 1/2 m R^2

@shahidahmed7511

8 жыл бұрын

Thanks brother

thank you!

@MichelvanBiezen

5 ай бұрын

You're welcome!

It can also be solved by taking hollow sphere as elemental mass

@MichelvanBiezen

2 жыл бұрын

Yes, that will work. Did you work it out that way?

@sakibkhursheed1354

2 жыл бұрын

@@MichelvanBiezen yes, I got the same result

@sakibkhursheed1354

2 жыл бұрын

This disc method is little complex but hollow sphere method is more easy..

@MichelvanBiezen

2 жыл бұрын

It can indeed be done both ways. This is an illustration of how it would be done using disks.

Tq 👍

Amazing

@MichelvanBiezen

2 жыл бұрын

Thanks

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)

nice proof

Just confused. When u integrated from 0 to R shouldn't it be half the original integration? Why double the integration?

@jayeffiong6340

8 жыл бұрын

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.

@MichelvanBiezen

8 жыл бұрын

Exactly. You figured it out.

Sir, where did that half went sir

@rakshithsooriya3620

Жыл бұрын

In that 4th step, that is I = rho × phi integration of x ^4 dz limits 0 to R

@rakshithsooriya3620

Жыл бұрын

I got confused in that part.....

@MichelvanBiezen

Жыл бұрын

I changed the limits from -R to R to 0 to R because of the symmetry. That is why the 1/2 was dropped.

why does di = 1/2dmx^2 rather than x^2dm ?? plz help me .....

@MichelvanBiezen

5 жыл бұрын

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)

@Lynch-rp1fu

5 жыл бұрын

@@MichelvanBiezen thank you very much , I've realized that right after i calculate the I of the flat disk again by mu self . tks !!

Why are we moment of inertia =1/2x^2dm instead of =x^2dm

@MichelvanBiezen

5 жыл бұрын

that is the moment of inertia of a solid disk

@dust.7625

5 жыл бұрын

Thank you

Thanks sir

@MichelvanBiezen

6 ай бұрын

You are welcome.

I love you

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.

@MichelvanBiezen

4 жыл бұрын

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

@bakingenbraden853

4 жыл бұрын

@@MichelvanBiezen I see, thanks very much

great

@MichelvanBiezen

Жыл бұрын

Thank you

thank you sir . your work helps millions

@MichelvanBiezen

6 ай бұрын

You are welcome. Glad our videos are helpful.

why dm=1/2 dm x^2 is taken

@MichelvanBiezen

7 жыл бұрын

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

@nagaraju-pq6nv

7 жыл бұрын

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??

@MichelvanBiezen

7 жыл бұрын

The small dm is a solid cylinder, so you must use the same equation.

@nagaraju-pq6nv

7 жыл бұрын

how will be the small element will be solid cylinder it must be circular plate

@MichelvanBiezen

7 жыл бұрын

They both have the same formula for the moment of inertia.

how come dI = 1/2dm x^2?

@MichelvanBiezen

4 жыл бұрын

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)

@connorfitzgerald640

4 жыл бұрын

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.

the best explanation

You forgot to change the bounds of integration

@MichelvanBiezen

6 жыл бұрын

No, it was done correctly. Thanks for checking.

ı am dead

Mercy buckets

@MichelvanBiezen

Жыл бұрын

You are welcome.

Thank you sir

@MichelvanBiezen

Жыл бұрын

You are welcome. Glad you found it helpful. 🙂