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Energy of 484 J is spent in increasing the speed of a flywheel 60 rpm to 360 rpm. Find the moment of inertia of thewheel.10.7 Kg m'l |
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Answer» SINCE 484J ENERGY IS SUPPLIED TO IT. THEN CHANGE IN ROTATIONAL KINETIC ENERGY IS EQUAL TO ENERGY SUPPLIED TO IT.SOLUTION: (1/2)IWF2 – (1/2)IWi2 = 484 (WHERE WF AND WI ARE INITIAL AND FINAL ANGULAR VELOCITIES)WI = 6 RAD/SWF = 37 RAD/SHENCE PUTTING THE VALUES(1/2)I(37*37)-(1/2)I(6*6)= 484684.5I-18I = 484HENCE MOMENT OF INERTIA= 0.72 |
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