Attached to each end of a thin steel rod of length 1.00 m and mass 6.80 kg is a small ball of mass 1.15 kg. The rod is constrained to rotate in a horizontal plane about a vertical axis through its midpoint. At a certain instant, it is rotating at 43.0 rev/s. Because of friction, it slows to a stop in 25.0 s. Assuming a constant retarding torque due to friction, compute (a) the angular acceleration, (b) the retarding torque, (c) the total energy transferred from mechanical energy to thermal energy by friction, and (d) the number of revolutions rotated during the 25.0 s.
K = Kf - Ki = (1/2)Iwf^2 - (1/2)IWi^2 = W
T = Ia (a = angular acceleration)
The Attempt at a Solution
I found the angular acceleration to be -10.816 rad / s^2
and the number of revolutions to be 537.42 rev
For torque i used the equation T=Ia where I = 1/12ML^2 where M is the total mass of the balls attached to each end of the rod plus the mass of the rod itself which is a total of 9.10 kg. I got T = -8.202 [N m] and for the work I used the work equation but got a really high solution which was -27723 J.
Both are wrong according to my homework's website