A bicycle wheel, of radius 0.330 m and mass 2.00 kg (concentrated on the rim), is rotating at 4.12 rev/s. After 48.0 s the wheel comes to a stop because of friction. What is the magnitude of the average torque due to frictional forces? First I did I = (1/2)(mr^2) = (1/2)(2kg)(0.330^2 m) = 0.1089 kg m^2 Then i used the equation Torque = I*alpha = (I)(omega)/(time) =((0.1089 kg m^2)(4.12 rev/s x 2pi rad/rev))/(48 s) =0.0587 N m however i am getting this question wrong. what am i doing wrong?