Angular Momentum of a system of particles

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SUMMARY

The total angular momentum of a system of two particles about the z-axis relative to point A at coordinates (0, -17 m) is calculated using the formula Lsystem = L1 + L2. Particle 1, with a mass of 26 kg and a speed of 43 m/s, contributes L1 = -19006 kg·m²/s, while Particle 2, with a mass of 63 kg and a speed of -37 m/s, contributes L2 = 39627 kg·m²/s. The final total angular momentum is Lsystem = 20621 kg·m²/s, indicating a net counterclockwise direction.

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Homework Statement



Two particles move in opposite directions along a straight line. Prticle 1 of mass m1 = 26kg at x1 = 23 m moves with a speed v1 = 43 m/s(to the right), while the particle 2 of mass m2 = 63 kg at x2= -22 m moves with a speed of -37 m/s(to the left).
Given: Counter clockwise is the positive angular diretion.

what is the total angular momentum of the system about the z-axis relative to point A along y-axis if A is at (0,-17m)?


Homework Equations



Lsystem=L1+L2
L=rXp
p=mv




The Attempt at a Solution



the angular momentum about A for particle 1 would be 17X(26)(43) which would be clockwise making a negative L so L1=-19006kgm^2/s
for particle2 it would be 17X(63)(37) counter clockwise spin about A so L2= 39627
L1+L2= 20621, but this does not seem to be correct
 
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the 2 particles are located on the x-axis if that helps at all
 

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