A disc of mass 10 ~kg and radius 0 1 ~m is rotating at 120 r.p.m. A retarding torque brings it to rest in 10 ~s . If the same torque is due to force applied tangentially on the rim of the disc then magnitude of force is

Step-by-step solution

frequency f =120 rpm = 120 60 =2 rps =2 f =2 2=4 rad / s Angular acceleration = = ₂- ₁ t = 0-4 10 = -2 5 rad / s ² Moment of inertia I = MR ² 2 = 10 (0.1)² 2 =0.05 ~kg ~m ² Torque = I =0.05 2 5 =0.02 Nm . = Fr F = r = 0.02 0.1 =0.2 N