Question

A sanding disk with rotational inertia 8.6 x 10^-3 kg·m² is attached to an electric drill whose motor delivers a torque of magnitude 15 N·m about the central axis of the disk. About that axis and with torque applied for 86 ms, what is the magnitude of the (a) angular momentum and (b) angular velocity of the disk?

Answer 1

Given
  
   disk rotational inertia I = 8.6*10^-3 kg m2

   torque T = 15 N, time period t = 86*10^-3 s

a) angular momentum

   we know the relation between torque and angular momentum is
  
           T = dL/dt

           L = T*t = 15*86*10^-3
           L = 1.29 Js

b) angular speed
   from angular momentum L = I*W

               W = L/I

               W = (1.29)/(8.6*10^-3) rad/s
               W = 150 rad/s