A man stands at the centre of a circular platform holding his arms extended horizontally with 4kg block in each hand. He is set rotating about a vertical axis at 0.5 rev/s.The moment of inertia of the man plus platform is 1.6kg-m^2, assumed constant. The blocks are 90 cm from the axis of rotation. He now pulls the blocks in toward his body until they are 15 cm from the axis of rotation. Find the initial and final kinetic energy of the man and the platform

moment of inertia of blocks = 2Mr² =2*4*(0.9)2 (r=0.9m)

                                       =6.48kg-m²

moment of inertia of platform + man = 1.6                       (given)

total moment of inertia (I)= 6.48+1.6=8kg-m²

w = 2pi(f) = 2pi(0.5)=(pi) radsec^-1

kinetic energy(initial) = Iw²/2 = 8(pi)² /2=4pi² =39.4J          (approx)           (ans1)

finall blocks are at 15cm from axis of rotation ....angular velocity of blocks is changed now

applying conservation of angular momentam

IW = I_f W_f  .................1

kinetic energy final = I_fW_f²/2

                           =I_f (IW/I_f)² /2                       (from eq 1)

                           =(I/I_f)(IW²/2)

                           =(I/I_f) (KE)_i=(I/I_f) (4pi²)

I_f = 1.6 + 2Mr² = 1.6+.18=1.8                             (r =0.15m)

I/I_f = 4.44 so

final kinetic energy = 4.44*4pi² =175J     (approx)                 (ans2)

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