A grindstone of moment of inertia 5 kg m ^2 is rotating at 600 rpm . A constant retarding torque of 10 N m is applied to it. The number of complete revolutions the grindstone makes before coming to rest is

Step-by-step solution

Given, initial angular velocity _i = 600 rpm = 600 2 60 rad/s = 20 rad/s Final angular velocity _f = 0 Moment of inertia I = 5 kg m ^2 Retarding torque = 10 N m The angular deceleration is: = I = 10 5 = 2 rad/s ^2 Using the third equation of rotational kinematics: _f^2 = _i^2 - 2 0 = (20 )^2 - 2(2 ) 4 = 400 ^2 = 100 rad The number of complete revolutions n is: n = 2 = 100 2 = 50 Answer: 50