How Does Angular Momentum Conservation Apply to Rotational Systems?

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cupid.callin
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Homework Statement


Textbook solution figure: a horizontal rod with points O, A and B left to right, with A at the midpoint, and arrows marking velocity directions above A and B. Below, the angular momentum calculation L = mvr = mω(d/2)(d/2) = (1/4)mωd^2.

The Attempt at a Solution



I did it like this:

[tex]\vec{L}_{B \ wrt \ Ground} = \vec{L}_{B \ wrt \ A} + \vec{L}_{A \ wrt \ Ground}[/tex]

[tex]\vec{L}_{B \ wrt \ A} = \vec{L}_{B \ wrt \ Ground} - \vec{L}_{A \ wrt \ Ground}[/tex]

so

[tex]\vec{L}_{B \ wrt \ A} = mwd^2 \ - \ \frac{1}{4}mwd^2[/tex]

[tex]\vec{L}_{B \ wrt \ A} = \frac{3}{4}mwd^2[/tex]

why is my method wrong?
 
Attachments
  • Textbook solution page: a diagram of a rod with points O, A and B, A at its midpoint, followed by the calculation of the angular momentum of B with respect to A as L = mvr = mω(d/2)(d/2) = (1/4)mωd², along the axis perpendicular to the plane of rotation.
    11b.jpg
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hi cupid.callin! :smile:
cupid.callin said:
[tex]\vec{L}_{B \ wrt \ A} = \vec{L}_{B \ wrt \ Ground} - \vec{L}_{A \ wrt \ Ground}[/tex]

sorry, but this doesn't work at all :redface:

LA(B) = rAB x (vB - vA)

= (rOB - rOA) x ((vB - v0) - (vA - vA)) …

bits of that are the RHS of your equation, but that's all
 
Why is does not work,:cry:

Is it because the eqn i used relates velocity or acceleration of particle wrt to a non inertial particle or frame ? :confused:
 
why should it work? :confused:

it obviously doesn't work …

the RHS of your equation is rOB x vB - rOA x vA