Question

A light, rigid rod of length l = 1.00 m joins two particles, with masses m₁ = 4.00 kg and m₂ = 3.00 kg, at its ends. The combination rotates in the xy plane about a pivot through the center of the rod (see figure below). Determine the angular momentum of the system about the origin when the speed of each particle is 3.20 m/s.

magnitude	kg · m2/s
direction	chose the right one ( +x , -x , +y , -y , +z , -z )

Answer 1

mass m1 = 4 kg

mass m2 = 3 kg

speed of each particle v = 3.2 m/s

length of the rod l = 1 m

angular momentum L = mvr

from this ,

     total angular momentum of the system is

          L = L₁ + L₂

             = {m₁vr + m₂vr } k^

             = {[m₁+ m₂]vr} k^

             = [4 kg + 3 kg](3.2 m/s)(1 m/2)}

             = {(7 kg)(3.2 m/s)(0.5 m)} k^

             = 11.2 kg.m² /s k^

direction : +z