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Angular Momentum & Torque

  1. Apr 1, 2009 #1
    1. The problem statement, all variables and given/known data
    At the instant of Fig. 11-42, a 6.60 kg particle P has a position vector [tex]\stackrel{\rightarrow}{r}[/tex] of magnitude 3.80 m and angle θ1 = 42.0° and a velocity vector [tex]\stackrel{\rightarrow}{v}[/tex] of magnitude 9.70 m/s and angle θ2 = 33.0°. Force [tex]\stackrel{\rightarrow}{F}[/tex], of magnitude 8.20 N and angle θ3 = 30.0° acts on P. All three vectors lie in the xy plane. About the origin, what are the magnitude of (a) the angular momentum of the particle and (b) the torque acting on the particle?

    3bvb7.gif

    2. Relevant equations
    [tex]\tau[/tex] = r x F
    l = r x p = m (r x v)

    3. The attempt at a solution
    I found the x and y components of r and v.
    I used those in the 2nd equation above to solve for l which is angular momentum.
    Then I used the 1st equation to solve for torque.
    I found the x and y components of F and solved for torque.
    I don't know what I'm doing wrong here.
    I used cross product as indicated. =/
     
  2. jcsd
  3. Apr 1, 2009 #2

    rl.bhat

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    Homework Helper

    Please show your calculations.
     
  4. Apr 3, 2009 #3
    rcosΘ = 3.80cos(42)
    rsinΘ = 3.80sin(42)

    vcosΘ = 9.7cos(33)
    vsinΘ = 9.7sin(33)

    L = M (R x V)
    Lx = (6.60) (3.80cos(42)) (9.7cos(33))
    Ly = (6.60) (3.80sin(42)) (9.7sin(33))
    L = [tex]\sqrt{Lx^2 + Ly^2}[/tex] = 223

    FcosΘ = 8.20cos(30)
    FsinΘ = 8.20sin(30)
    T = R x F
    Tx = (3.80cos(42)) (8.20cos(30))
    Ty = (3.80sin(42)) (8.20sin(30))
    T = [tex]\sqrt{Tx^2 + Ty^2}[/tex] = 28.3
     
  5. Apr 4, 2009 #4

    rl.bhat

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    Homework Helper

    Is it necessary to the components r?
    Take the components of V and F parallel and perpendicular to r.
    Only perpendicular components will contribute to angular moment and torque.
     
  6. Apr 6, 2009 #5
    Oh okay I see my mistake now.
    Thank you very much =]
     
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