Lect 23 - Angular Momentum

Lect 23 - Angular Momentum - AngularMomentum Chapter11...

Info iconThis preview shows pages 1–15. Sign up to view the full content.

View Full Document Right Arrow Icon
    Angular Momentum Chapter 11
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
    The Cross Product Given two vectors,  A  and  B The vector (cross) product of  A  and  B   is defined as a third vector,  A  x  B C  is read as “ A  cross  B The magnitude of  C  is  AB  sin  θ  is the angle between  A  and  B
Background image of page 2
    More About Cross Products The quantity  AB  sin  θ  is  equal to the area of the  parallelogram formed by  A  and  B The direction of  C  is  perpendicular to the  plane formed by  A  and  B The best way to  determine this direction is  to use the right-hand rule
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
    Properties of the Cross Product The vector product is  not  commutative.  The  order in which the vectors are multiplied is  important To account for order, remember A  x  B  = -  B  x  A If  A  is parallel to  B  ( θ  = 0 o  or 180 o ), then  A  x  B  = 0 Therefore  A  x  A  = 0
Background image of page 4
    Properties of the Cross  Product If  A  is perpendicular to  B , then | A  x  B | =  AB The vector product obeys the  distributive law A  x ( B  +  C ) =  A  x  B  +  A  x  C
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
    Properties of the Cross  Product The derivative of the cross product with  respect to some variable such as  t  is where it is important to preserve the  multiplicative order of  A  and  B d dt A × Β ( 29 = δΑ δτ × Β + Α × δΒ
Background image of page 6
    Cross Products of Unit  Vectors
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
    Cross Products of Unit  Vectors Signs are interchangeable in cross  products A  x (- B ) = -  A  x  B   ö i × - ϕ ( 29 = - ι × ϕ
Background image of page 8
    Using Determinants The cross product can be expressed as Expanding the determinants gives A × Β = ι ϕ κ Α ξ ψ ζ Β = ι - ϕ + κ A × Β = - Α ( 29 ι - Α - Α ( 29 ϕ + - Α ( 29 κ
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
    CPS Question
Background image of page 10
    The Cross Product and Torque The torque vector lies in  a direction  perpendicular to the  plane formed by the  position vector and the  force vector τ  =  r  x  F The torque is the cross  product of the position  vector and the force  vector
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
    CPS Question
Background image of page 12
    Torque Vector Example Given the force τ  = ? m ) ö 00 . 5 ö 00 . 4 ( N ) ö 00 . 3 ö 00 . 2 ( j i r j i F + = + = τ= r ´ F = ö i ö j ö k 4 5 0 2 3 0 = 0 ö i - 0 ö j + 4 ( ) 3 ( ) - 2 ( ) 5 ( ) é ë ù û ö k = 2 Ngm ö k
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
    Angular Momentum Consider a particle of mass  m  located  at the vector position  r  and moving with  linear momentum  p r × Φ= τ=ρ× δπ δτ Αδδινγ τηε τερμ δρ × π = 0 τ = δρ×π ( 29
Background image of page 14
Image of page 15
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 41

Lect 23 - Angular Momentum - AngularMomentum Chapter11...

This preview shows document pages 1 - 15. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online