A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of 25 Nm for 40 s, the speed increases to 2100 rpm. The diameter of the disk is ________ m.

Step-by-step solution

Given, m =1 ~kg aligned & _ i =1800 rpm =1800 2 60 =60 rad sec & _ f =2100 rpm =2100 2 60 =70 rad sec & _ e x t =25 Nm & t =40 sec aligned Using equation of motion aligned & _ f = _ i + t & 70 =60 + (40) & = 4 rad / sec ^2 aligned Also, = I = mR ^2 4 aligned & 25 = 1 R ^2 4 4 & R =20 ~m aligned Hence, diameter of disk =2 R =2 20=40 ~m