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When a 12000 joule of work is done on a flywheel, its frequency of rotation increases from 10 Hz to 20 Hz. The moment of inertia of flywheel about its axis of rotation is `(pi^(2) = 10)`A. `1 kgm^(2)`B. `2 kgm ^(2)`C. `1.688 kgm^(2)`D. `1.5 kgm^(2)` |
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Answer» Correct Answer - B Given , work done W = 12000J , Initial frequnecy , `f_(1) = 10 Hz` and final frequnecy `f_(2) = 20 Hz` Angular velcocity for rotational motion is given kinetic energy . ltbr ` omega = 2pif` `therefore " " omega_(1) = 2pif_(1) = 2pi xx 10 = 2o pi rad//s` and `omega_(2) = 2pif_(2) = 2pi xx 20 = 40 pi rads//s` According to work -energy theorem, work done in rotation = change in rotational kinetic energy `rArr " "W = (1)/(2) /Omega_(2)^(2) - (1)/(2)lomega_(1)^(2) " "[because pi^(2) = 10]` `= (1)/(2)l(omega_(2)^(2) - omega_(1)^(2))" "......(i)` where , I = moment of inertia of the flywheel. Substituing given value of Eq.(i) ew get `12000 =(1)/(2)l(1600pi^(2) - 400pi^(2))` `rArr = (1)/(2)l(1200 xx 10)" "[because pi^(2) = 10]` `rArr" "l = (12000xx2)/(12000) = 2 kgm^(2)` |
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