InterviewSolution
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InterviewSolution
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The earth has a mass of `6 xx 10^(24) kg` and a radius of `6.4 xx 10^(6)m`. Calcualte the amount of work that must be done to slow down its rotation so that duration of day becomes `30` hrs instead of 24 hours. Moment of inertia of earth `= (2)/(5)MR^(2)`. |
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Answer» Here, `M = 6 xx 10^(24) kg, R = 6.4 xx 10^(6)m, W =?` `T_(1) = 24 hrs, T_(2) = 30 hrs`. `I = (2)/(5)MR^(2) = (2)/(5) xx 6 xx 10^(24) xx (6.4 xx 10^(6))^(2)` `= 9.83 xx 10^(37)kg m^(2)` Work done = Decrease in rotational `KE` `W = (1)/(2)I (omega_(2)^(2) - omega_(1)^(2)) = (I)/(4) xx 4pi^(2) [(1)/(T_(2)^(2))-(1)/(T_(1)^(2))]` `=(2)/(5) MR^(2) xx 2pi^(2) ((T_(1)^(2) - T_(2)^(2))/(T_(1)^(2) T_(2)^(2)))` `= 9.83 xx 10^(37) xx 2 xx (22)/(7) xx (22)/(7)` `xx{((24 xx 60 xx 60)^(2) - (30 xx 60 xx 60)^(2))/((24 xx 60 xx 60)^(2) (30 xx 60 xx 60)^(2))}` `W =- 9.36 xx 10^(28) J` |
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