Angular Velocity Dynamics (Easy?)

[Stingarov]

A 73.9 kg shaft with a radius of 8.9 cm is spinning at a rate of 364.5 rpm. If a board leans against the outside providing a frictional force of 59.92 N, how long will it take for the shaft to stop rotating?

Answer says 2.094 s.

Problem I have is that my Inertia table doesn't list shaft inertia or anything. Unsure how to go about starting this?

[Doc Al]

Treat the shaft as a cylinder--see if that works.

[Stingarov]

So Inertia = Solid cylinder = 1/2mr squared.
I = .5 * 73.9 * .089 sq
I = .293

Do I use the Net Torque = 0 from here? I'm still pretty lost for some reason from here.

[Stingarov]

Alright I figured it out surprisingly by conceptual thinking.

Force: I/r
1) 1/2mr = 3.28855
2) * angular velocity = 3.28855 * 38.17 radians per second

3) 125.52 / F2 = time
= 2.094