Angular momentum of particle in xy plane

  • Thread starter barz1
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  • #1
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Homework Statement


Okay so, particle of mass 4kg moves in positive x direction with v=8m/s along line y=4m. Find angular momentum relative to point (x,y) = (1m, 10m)


Homework Equations


Angular momentum L=mrxv

The Attempt at a Solution


Basically, I set my velocity vector = (8i+0j)m/s and my position vector = (i -6j)m. I determined this position vector from the point (1,10) to the position of the velocity vector(0,4) and then just used the equation 4[(8i+0j)x(i-6j)] and came up with the answer -192 kg*m^2/s.

So my question I guess is did I determine my position and velocity vectors correctly and did is this how the angular momentum equation should be done?
Thanks!
 

Answers and Replies

  • #2
tiny-tim
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Homework Helper
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welcome to pf!

hi barz1! welcome to pf! :smile:
Angular momentum L=mrxv

Basically, I set my velocity vector = (8i+0j)m/s and my position vector = (i -6j)m. I determined this position vector from the point (1,10) to the position of the velocity vector(0,4) and then just used the equation 4[(8i+0j)x(i-6j)] and came up with the answer -192 kg*m^2/s.

yes that's fine :smile:, except you used vxr instead of rxv :redface:

(personally, i'd have done an easy question like this much less formally … i'd say that the perpendicular from P to L is obviously -6j, so the angular momentum is 4 times -6j x 8i, = 192 i x j = 192 :wink:)
 
  • #3
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Hey thanks a lot tiny-tim ill remember your recommendation for test time.
 

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