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.
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 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
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
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?
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).
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.
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
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.
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)
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!)
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)
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?
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.
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.
