Two Ways to Find the Moment of Inertia Tensor for a Solid Cube

แชร์
ฝัง
  • เผยแพร่เมื่อ 18 ธ.ค. 2024

ความคิดเห็น • 11

  • @ariessunfeld
    @ariessunfeld 3 ปีที่แล้ว +8

    5:21 - "I shouldn't even have to write it out, but I am writing it out, because I care about you, and I want you to be happy." Thank you. New subscriber.

  • @fizixx
    @fizixx 3 ปีที่แล้ว

    Loved it. I always like adding details like the color, thank you.

  • @alexanderlomashvili8501
    @alexanderlomashvili8501 3 ปีที่แล้ว +3

    I like your videos.

  • @anawaraliyyi899
    @anawaraliyyi899 2 ปีที่แล้ว +1

    Thank you

  • @mohammadsadeghashrafpoor6737
    @mohammadsadeghashrafpoor6737 7 หลายเดือนก่อน +1

    It was best explanation

  • @yadukrishnanp1013
    @yadukrishnanp1013 2 หลายเดือนก่อน

    Why the euatuons to find the components of the tensor is different for different axes. I mean why do you say that xx component is calculated like this and all. It may seem silly but iam not getting the point.

  • @atakan716
    @atakan716 2 ปีที่แล้ว +1

    thanks!

  • @enchantularity
    @enchantularity 3 ปีที่แล้ว

    A quick doubt. Why xy and yx components get a negative sign. Sorry if the question is too obvious for the answer.

    • @DotPhysics
      @DotPhysics  3 ปีที่แล้ว

      That's a great question - without an easy answer. Basically, you are taking individual points and finding the angular momentum (r x p). When you factor and gather terms, you get stuff that looks like the elements in the inertia tensor and they just happen to have a negative sign.
      The key is to realize where that tensor comes from - r x p for a bunch of particles.

  • @andyrundquist7672
    @andyrundquist7672 3 ปีที่แล้ว +1

    First, do you always draw capital sigma by drawing a 2 and then adding a flag? Second, nice Monte Carlo approach (sounds familiar :)

    • @DotPhysics
      @DotPhysics  3 ปีที่แล้ว +1

      am I drawing sigma wrong? If so, I blame my parents.