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A copper ball of mass `m = 1 kg` with a radius of `r = 10 cm` rotates with angular velocity `omega = 2 rad//s` about an axis passing through its centre. The work should be performed to increase the angular velocity of rotation of the ball two fold is. |
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Answer» `KE=1/2(2/5)mv^(2)implies KE=2/10 mv^(2)` If `omega` is to be doubled `v` is also doubled. Therefore `/_KE-KE_(2)-KE_(1)=2/10m[v_(2)^(2)-v_(1)^(2)]` `implies /_KE=2/10m[r^(2)omega_(2)^(2)-r^(2)omega_(1)^(2)]` `=2/10xx1[(1/10)^(2)][4^(2)-2^(2)]=2.4xx10^-(2)J` |
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