Angular Momentum Cross Product

[@rahuldubey7490]

Best physics teacher u r giving us free and quality education very much thanks for this sir and the hardwork you do behind each video is very much appreciable .

[@FlippingPhysics]

Thanks for the love!

[@johnunverzagt9387]

That fact about not needing the rigid body might be weird, but it does have a real world use. In orbital mechanics, using those three vectors: speed, velocity, and angular momentum one can calculate all of the properties of the given object’s orbit. My mind was blown when I found out that a satellite has angular momentum with respect to the orbit’s gravitational focus. (I.e whatever planet or body it is orbiting). That fact was not mentioned in my Newtonian physics classes, and I first encountered it as a sophomore in undergrad whilst taking Introduction to Astronautics. Thank you for your wonderfully clear explanation!

[@FlippingPhysics]

Thanks for the observations. Very nice to have that perspective.

[@andrewjustin256]

Mr. P, thank you so much for this video! I frankly love those teachers who care about derivation of the equations nit just hurling them at students! And I wish you the most joyful of the New Year as well!

[@FlippingPhysics]

Thanks for the love my friend!

[@andrewjustin256]

@FlippingPhysics Thank you! I hope to the utmost degree you celebrated the new year with incessant zeal and effervescence!

[@Mallikadinesh-h2q]

This same concept is being thought to us now in college . Thank youuuu .

[@flanker1497]

Dedication at its peak!! U are really an inspiration for many students Lots of love

[@FlippingPhysics]

Thanks for the love!

[@sahasras2647]

That's such an clear and wonderful explanation! Thanks for the efforts sir!

[@FlippingPhysics]

Glad it was helpful!

[@MohiniGiri704]

U helped me a lot for cleaning concept of physics ❤️love from india 💓 love for flipping physics

[@FlippingPhysics]

I am very glad to help you learn physics!

[@technotux7835]

Do I have to study 3D vectors in order to study dot and cross vector multiplication?

[@dingus42]

for cross product yes, because it is an operation defined for 3D vectors :)

[@carultch]

In the special case that both given vectors are in the x-y plane, the cross product will be in the z-direction. This simplifies the amount of complexity you could have with 3D vectors, because all inputs to the cross product would reduce to 2D vectors in the x-y plane, and the output would be limited to the dimension of the z-axis.

[@gowrissshanker9109]

Respected sir, IS A photon of violet light is brighter than A photon of red light...what does brightness depends on?? Amplitude or energy?? Thank you sir

[@lastmiles]

That is a quantum physics question and you asked about a "particle" of light while dragging in wavelength of light at the same time. Very messy or you are just being tricky.

[@carultch]

Brightness depends on the relative sensitivity of human eyes to each color of light. There is a bell curve across the visible spectrum that peaks at 540 THz, where it takes the fewest W/m^2 of light to generate 1 lux of brightness, and that bell curve is close to symmetric. In a hypothetical alternate reality where humans were uniformly sensitive to the entire spectrum, brightness would depend on intensity of light. I.e. the Watts/meter^2. This means that when you see red light and violet light that appear equally bright (assuming that you compare two spectral colors that humans are equally sensitive to seeing), that the red light would have a much larger population density of photons per second per square meter. As for amplitude (as in maximum Newtons/Coulomb of electric field, or maximum Teslas of magnetic field), the amplitude of an electromagnetic wave is the same for all EM waves of the same intensity, regardless of frequency. See the concept of Poynting vector, and the cross product relationship of wave speed, electric field and magnetic field. Intensity is proportional to amplitude squared, and the equation is independent of frequency.

[@malamudtania1053]

Your videos help me a lot, un abrazo desde Argentina😄

[@FlippingPhysics]

Gracias por el abrazo. Envío uno desde los Estados Unidos. 😁

[@Test-ph1vk]

Hi. I think there's something wrong. The vector formula Torque = Inertia x Angular Acceleration is *incorrect*. It's because the moment of inertia may be different for each axis when considering the x,y,z components of the formula.

[@Test-ph1vk]

Btw, this is mentioned on HRK's Physics

[@carultch]

@@Test-ph1vk It's not that it is incorrect, just simplified to the case of uniaxial torque & rotation. Indeed moment of inertia is different for the different axes, even given the same shape. As long as the moment of inertia used in this formula is consistent with the axis of rotation, it is a correct formula. The full concept of moment of inertia requires a 3x3 tensor called the inertia tensor, rather than a simple scalar, and there is a form of multiplication that enables the tensor to interact with the other two formulas. This is far beyond the scope of what a high school student is expected to understand.

[@Mallikadinesh-h2q]

The fact that you agreed in the end of the video that it is weard makes me feel good. 😁 . I'm not the only one . 🙄

[@FlippingPhysics]

You are not the only one

[@krishoberoi]

why cant we see sun's shadow?

[@flanker1497]

because we are blind

[@carultch]

Because the Sun is a light source, rather than a light sink. We can see the shadow that objects in the Sun cast, and we can see the shadows from the Moon and the Earth during special conditions called eclipses, but the Sun itself has no intrinsic shadow.

[@AyalaMrC]

Yup. That's pretty weird. Good episode. 👍

[@FlippingPhysics]

Agreed. Thanks for previewing!

[@julielangenbrunner9212]

Looks good!

[@FlippingPhysics]

Thanks!

[@xicad1533]

You should speak with mandlbaur... he's incorrectly using this

[@FlippingPhysics]

yeah...

[@xicad1533]

@@FlippingPhysics i see you've heard of him lol

[@carultch]

​@@FlippingPhysics I investigated Mandlbaur's claim that angular momentum isn't conserved when radius changes. The problem I made up to do so, uses a point mass that would be in uniform circular motion, but receives a constant force toward the origin that exceeds the initial m*v^2/r, in order to move closer to the origin with zero torque. From first principles and your lesson on Euler's method, it confirms conservation of angular momentum, and consistency of KE with work. Mandlbaur called it an "appeal to tradition fallacy" when I showed it to him. The shape of the path surprised me. I expected it to spiral in toward the origin, and continue to get closer. It appears to be an ellipse, centered on the origin, whose major axis is rotating about the origin as it is drawn. But it "bounces back" to its full original radius, shortly after 180 degrees of rotation. As expected, there is the special case where it reduces to uniform circular motion.

[@krishoberoi]

who named god "GOD"?

[@flanker1497]

God's father

[@carultch]

The Dutch did.