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 4.40 m/s. magnitude kg
Solution
L=J?, where ?=v/R and J=J1+J2=m1*R²+m2*R²=(m1+m2)R² ==>
L=(m1+m2)R²v/R=(m1+m2)Rv= (4+3)*0.5*5.4=18.9 kg·m²/s
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 4.40 m/s. magnitude kg
Solution
L=J?, where ?=v/R and J=J1+J2=m1*R²+m2*R²=(m1+m2)R² ==>
L=(m1+m2)R²v/R=(m1+m2)Rv= (4+3)*0.5*5.4=18.9 kg·m²/s
