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(Solved): A man stands on the center of a rotating turntable with his arms outstretched. He holds a 5kg mass i ...



A man stands on the center of a rotating turntable with his arms outstretched. He holds a 5kg mass in each hand. He is set in motion about a vertical axis at 1 rev every 2.0 seconds. Assume the moment of the inertia of the man is constant 6kg-m^2. The original distance of the weights from the rotational axis is 1m, and their final distance is 15cm. Neglect ALL Friction. 

a) Find his new angular momentum after he drops his hands to his sides.
b) Calculate the change inkinetic energy 
c) Discuss where the extra energy came from 

3. Find his new digulor momedtim aAte he drops hits hands to tis sioch.
b. Cakeutate the change n kincec energy
C. Dexculss w
3. Find his new digulor momedtim aAte he drops hits hands to tis sioch. b. Cakeutate the change n kincec energy C. Dexculss where the isea knesc energy came from


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Moment of inertia of the man, I1=6kgm2 Now initial moment of Inertia of the system , Ii=I1+2mr2 Where, m=5kg r=1m So, Ii=6+2×(5×12)=16kgm2
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