A child is standing with folded hands at the centre of a platform rotating about its central axis. The kinetic energy of the system is K. The child now stretches his arms so that the moment of inertia of the system doubles. The kinetic energy of the system now is

Asked by ayush.gupta3052 | 10th Jul, 2021, 07:01: PM

Expert Answer:

Let begin mathsize 14px style straight I end stylei be initial moment of inertia of system and ωi be the initial angular speed of rotation .
Let begin mathsize 14px style straight I end stylebe final moment of inertia of system and ωf be the final angular speed of rotation .
By Conservation of angular momentum , we have , 
 
begin mathsize 14px style straight I end styleωi = begin mathsize 14px style straight I end stylef ωf ..................... (1)
Final moment of Inertia begin mathsize 14px style straight I end stylef = 2 begin mathsize 14px style straight I end stylei ...................... (2)
Hence , we get from eqn.(1) ,  ωf = (1/2)ωi 
 
Final kinetic energy = (1/2) begin mathsize 14px style straight I end stylef ωf2 = (1/2) ( begin mathsize 14px style straight I end stylei ) [ (1/2)ωi ]2 = (1/4) begin mathsize 14px style straight I end styleωi2 
 
Hence Final kinetic energy becomes (1/4)th of initial kinetic energy

Answered by Thiyagarajan K | 10th Jul, 2021, 08:16: PM

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