Question

two happy astronaunts, each having a mass of 75 kg are connected by a rod length 10.0 and a mass 140 kg. The system is isolated in space, orbiting the center of the rod at an angular speed of 0.50 rad/s counterclockwise. Treat the astronaunts as point masses. The moment of inertia of a rod is I=(1/12)ML^2

what is the initial moment of intertia of the total rod-and-astronaunts system?

By pullinh on the rod, the astronaunts shorten the distance between them to 5.0 m. What is the final angular velocity of the rod-and-astronauts system in rad/s?

Answer 1

Mass of astronaunt m = 75 kg

Mass of rod M = 140 kg

Length fo the rod L = 10 m

The moment of inertia of a rod is I =(1/12)ML²

                                                 = (1/12)x140 x10 ²

                                                 = 1166.66 kg m ²

The initial moment of intertia of the total rod-and-astronaunts system is

      I ' = I + 2(mr ² )   where r = L/2 = 5 m

         = 1166.66 +[2x75 x5 ² ]

         = 1166.66 +3750

         = 4916.66 kg m ²

Angular speed = 0.5 rad/s

The astronaunts shorten the distance between them after pulling is 5.0 m

i.e., the distance between center of the rod to the astronaunts r ' = 5 m/ 2 = 2.5 m

So, new moment of inertia of rod-and-astronaunts system I " = I + 2(mr' ²)

                                                                                        = 1166.66 +[2x75x2.5 ²]

                                                                                         = 1166.66 + 937.5

                                                                                         = 2104.16 kg m ²

From law of conservation angular momentum, I ' = I " '

From this ' = I ' / I "

                    = (4916.66 x 0.5 ) / 2104.16

                    = 1.168 rad/s