A wheel having a moment of inertia of 2 kg m ^2 is rotating at 300 rpm. It is brought to rest by a constant frictional torque in 25 revolutions. The magnitude of this frictional torque is :

Step-by-step solution

Initial angular speed, ₀ = 300 2 60 = 10 rad/s Final angular speed, = 0 Angular displacement, = 25 2 = 50 rad Using the equation of rotational kinematics, ^2 = ₀^2 + 2 0 = (10 )^2 + 2 (50 ) = - 100 ^2 100 = - rad/s ^2 Magnitude of frictional torque, = I| | = 2 = 2 N m Answer: 2 N m