Tensor Version of the Parallel Axis Theorem (Steiner's Theorem) | Classical Mechanics

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  • เผยแพร่เมื่อ 9 ม.ค. 2025

ความคิดเห็น • 7

  • @zanbutala7223
    @zanbutala7223 4 ปีที่แล้ว +5

    red bracket in last row should be multiplied by mass

  • @martinmikkelsen2766
    @martinmikkelsen2766 5 ปีที่แล้ว +2

    Beautiful!

  • @amarj9909
    @amarj9909 5 ปีที่แล้ว +1

    Hi sir, I looking for answer how r^2 came in moment of inertia I=Mr^2 ,could you explain why we are multiply r^2 distance from the axis of rotation , from the position of point particle.
    IF particle is from distance r
    Then it could be I= Mr why it will I=M r^2
    Thanks for video explaining moment of inertia in tensor

    • @PrettyMuchPhysics
      @PrettyMuchPhysics  5 ปีที่แล้ว +3

      *Short answer*: that‘s the definition of the moment of inertia.
      *Long answer*: the moment of inertia is defined as L/w, i.e. angular momentum divided by angular velocity. For a point particle, you have
      L = r p = r m v
      w = v / r
      Therefore,
      I = L / w = r m v / ( v / r ) = r² m

  • @JohnDoe-ce1yp
    @JohnDoe-ce1yp 3 ปีที่แล้ว +1

    I am in love with your method of teaching, is it possible for you to teach me over zoom, in return for payment?

    • @PrettyMuchPhysics
      @PrettyMuchPhysics  3 ปีที่แล้ว

      Thank you very much! :) We cannot do tutoring unfortunately since we both work full time :/