A disc of radius 0.4 m and mass 1 kg rotates about an axis passing through its centre and perpendicular to its plane. The angular acceleration is 10 rad/s ^2 . The tangential force applied to the rim of the disc is

Step-by-step solution

The moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is given by I = 1 2 MR^2 The torque acting on the disc is related to the angular acceleration by = I Since the torque is provided by a tangential force F applied at the rim, we have = F R Equating the two expressions for torque: F R = 1 2 MR^2 F = 1 2 MR Substituting the given values ( M = 1 kg, R = 0.4 m, = 10 rad/s ^2 ): F = 1 2 1 0.4 10 F = 2 N Answer: 2 N