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A fly wheel having a moment of inertia of 10^(7) g cm^(2) and revolving at the rate of 120 rpm is stopped by a brake in course of 5 seconds. Find the controlling torque. |
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Answer» `2.51 xx 10^(7)` DYNE-cm Initial angular speed of the fly wheel, `omega_(1)=(2pir)/60=(2pixx120)/60=4PI"rad"//"sec"` Final angular speed = `omega_(2) = 0`, time= t = 5 sec. Now, applying the FORMULA `omega_(2) = omega_(1) - alphat`, we get `0=4pi-alphaxx5` or `alpha=(4pi)/5` rad/se`c^(2)` We know, torque = moment of Inertia (I) `xx alpha` `therefore` Torque, `tau=10^(7)xx(4pi)/5=2.51xx10^(7)` dyne cm |
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