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`        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-m2, 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`
7 years ago

510 Points
```										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 = If Wf  .................1
kinetic energy final = IfWf2/2
=If (IW/If)2 /2                       (from eq 1)
=(I/If)(IW2/2)
=(I/If) (KE)i=(I/If) (4pi2)
If = 1.6 + 2Mr2 = 1.6+.18=1.8                             (r =0.15m)
I/If = 4.44 so
final kinetic energy = 4.44*4pi2 =175J     (approx)                 (ans2)
approve my ans if u like
```
7 years ago
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