(a) A light, rigid rod of length e = 1.00 m joins two particles, with masses m, = 4.00 kg and m, = 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 theangular momentum of the system about the origin when the speed of each particle is 6.20 m/s. (Enter the magnitude to at least two decimal places in kg • m2/s.)magnitude|kg · m2/sdirection|---Select--- v(b) What If? What would be the new angular momentum of the system (in kg · m/s) if each of the masses were instead a solid sphere 13.0 cm in diameter? (Round your answer to at least two decimal places.)kg · m2/s

We know that, Angular momenteum, L for point masses= m×v×r Angular momenteum, L for larger masses=I×wGiven that, Both particles are moving at speed of 6.20 m/secAs per given conditions in question we can say both particles are point massesTotal angular momenteum, L=L1+L2then, L=m1×v×r1+m2×v×r2 equation 1where, m1=4.0 kg, m2=3.0 kg, v=6.20 m/secr1=r2=L2=1.002=0.50 msubstitute all known values in equation 1⇒L=4.00×6.20×0.50+3.00×6.20×0.50=21.70 kg m2 sec-1Answer: Angular momenteum will be 21.70 kg m2 sec-1, positive magnitude shows it in y-direction Part- BTotal angular momenteum will beL=L1+L2w= angular velocitty=V/r=6.20 m/sec/0.50 m=12.40 rad/secR= radius of sphere=13 cm/2 =6.5 cm=0.065 mIn that case , Net angular momenteum will beL=25m1R2+m1r2+25m2R2+m2r2=25×4.00kg×0.065 m2+4.00 kg×0.50m2+25×3.00kg×0.065 m2+3.00 kg×0.50m=0.00676+1+0.00507+0.75 ×12.8 in kgm2sec2 kg m2 value is multipled by 12.8 for conversion in SI unit=22.5514 kg m2 sec-1Answer: ANgular momenteum of the system will be 22.5514 kg m2 sec-1

Answered: (a) A light, rigid rod of length e =…