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A solid cylinder of mass 2 kg and radius 4 cm rotating about its axis at the rate of 3 rpm . The torque required to stop after 2pi revolutions is |
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Answer» `2 xx 10^(6) Nm` Initial ANGULAR speed `omega_(0) = 3 rpm = 3 xx (2pi)/(60)` rad/s = `(pi)/(10)` rad/s We know that , `omega_(2) = omega_(0)^(2) + 2 ALPHA theta` `implies 0 = ((pi)/(10))^(2) + 2 xx alpha xx 2pi xx 2pi implies alpha = (-1)/(800) "rad/s"^(2)` MOMENT of inertia of a solid cylinder , `I = (MR^(2))/(2) = (2 xx ((4)/(100))^(2))/(2) = (16)/(10^(4))` Torque `tau = I alpha = ((16)/(10^4)) xx (- (1)/(800)) = -2 xx 10^(-6) N m` |
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