1.

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

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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