A small ball with mass 1.70 kg is mounted on one end of a rod 0.750 m long and of negligible mass. The system rotates in a horizontal circle about the other end of the rod at 5030 rev/min.
(a) Calculate the rotational inertia of the system about the axis of rotation.
(b) There is an air drag of 2.20 10-2 N on the ball, directed opposite its motion. What torque must be applied to the system to keep it rotating at constant speed?
The Attempt at a Solution
So I got (a) which is simply I=mr^2, or I=(1.70)(.750)^2, I=.95625 kgm^2
I thought I had (b) but I was wrong. First I broke down the 5,030 rev/min to an angular acceleration which came out to 526 rad/s^2. Then I plugged that, along with the I into T=Ialpha which came out to, T=(.9562)(526), T=502.9Nm. I then added in the calculated torque from the air drag which was T=(.022)(.75), =.0165Nm to come out with a torque of 503Nm, which was wrong. I have absolutely no idea what detail(s) I'm missing because I was sure I had this one. Any help is greatly appreciated, thank you.