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Angular momentum cross product  Please help! 
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#1
Dec112, 02:28 PM

P: 4

1. The problem statement, all variables and given/known data
A 1.47kg particle moves in the xy plane with a velocity of v = (4.59i  3.28j)m/s. Determine the magnitude of the particle's angular momentum when its position vector is r = (1.35i + 2.57j)m. 2. Relevant equations p = mv L = r x p (the x is supposed to be a cross product and not a variable) L = r x mv 3. The attempt at a solution First I scaled the velocity vector: v = (4.59i  3.28j)m/s by the mass, 1.47 kg, to get a new momentum vector (6.75i  4.82j)kg*m/s. Then I took the cross product of the r vector with the new momentum vector: (1.35i + 2.57j)m x (6.75i  4.82j)kg*m/s (I let a=1.35, b=2.57, c=0, d=6.75, e=4.82, and f=0, the got the k vector cross product by doing k=aebd) The answer I got was 23.9 kg*m^2/s, which wasn't right. What did I do wrong? Am I even anywhere near the correct solution/answer? 


#2
Dec112, 03:16 PM

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#3
Dec112, 03:33 PM

P: 4

I didn't, but I just went back and tried it and it picked up the k as being part of the unit. So, it's not that, I don't think. The question also just asks for magnitude.



#4
Dec112, 03:36 PM

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Angular momentum cross product  Please help!



#5
Dec112, 03:38 PM

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23.9 kg*m^2/s



#6
Dec112, 03:39 PM

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Are magnitudes ever negative?



#7
Dec112, 03:41 PM

P: 4

Nope. Wow, I feel a little stupid now. Thanks!



#8
Dec112, 03:44 PM

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