A4.2-m-diameter merry-go-round is rotating freely with an angularvelocity of 0.80 rad/s. its total moment of inertia is1760kgm2. four people standing on the ground, eachof mass 65 kg, suddenly step onto the edge of themerry-go-round. a. what is the angular velocity of themerry-go-round now? b. what if the people were on it initiallyand then jumped off in a radial direction (relative to themerry-go-round)?

Answer

given,

mass of people = 65 kg

number of people =

diameter of the  merry-go-round = 4.2 m

radius = 2.1 m

angular velocity = 0.8 rad/s

moment of inertia =  1760 kg m²

Using the law of conservation of angular momentum, we have

                             L₁=  L₂

                          I₁ω₁ =I₂ω₂

                            I₁ω₁= ( I₁+ 4 m r²)ω²

                         ( 1760 x 0.8) = ( 1760 + 4 x 65 x 2.12)ω²

                         1408 = 2311.2 ω²

                         ω² = 0.609

                         ω  = 0.781 rad/s

b) If the people jump of the merry-go-round radially, they exert no torque.

hence,  it will not change the angular momentum of the merry-go-round. It will continue to move with the same   ω .

i think its c if i'm wrong