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A child stands at the centre of a turn table with his two arms out stretched. The turntable is set rotating with an angular speed of 40 rev/min. Now, the child folds his hands back and thereby reduces his moment of inertia to (2)/(5) times the initial value. The new kinetic energy of rotation is x times the initial kinetic energy of rotation. The value of x is : |
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Answer» 2.5 `therefore I_(1)omega_(1)=I_(2)omega_(2)` Now it is given that `I_(2)=(2)/(5)I_(1)` `therefore omega_(1)=5//2omega_(1)` Also `K.E.=(1)/(2)((2)/(5)I_(1))(5//2omega_(1))^(2)` `E_(k)=(1)/(2)xx5//2xxI_(1)omega_(1)^(2)` `=2.5xx(1)/(2)I_(1)omega_(1)^(2)` `therefore` value of `x=2.5` |
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