The torque required to start or stop the shaft is a function of it's rotational inertia. Imagine a car and you have to push it. The lighter the car is, the easier it is to accelerate it from a dead stop by pushing it. The torsional analogue of that is what you're interested in. Just as the car has some mass which resists accelerating, the shaft with the attached cylinder has rotational resistance due to its rotational inertia. You can calculate the rotational moment of inertia using standard formulas for simple shapes as shown here: https://webspace.utexas.edu/cokerwr/www/index.html/RI.htm
or if you can do the math, try a more advanced calculation following the equations provided by Wikipedia here:
http://en.wikipedia.org/wiki/Moment_of_inertia#Example_calculation_of_moment_of_inertia
Once you know the rotational inertia, and you want to find the torque required to change the rotational speed, use the formula for angular acceleration,
http://en.wikipedia.org/wiki/Angular_acceleration
http://hyperphysics.phy-astr.gsu.edu/hbase/n2r.html
http://theory.uwinnipeg.ca/physics/rot/node5.html
I've provided a few different sites but they all say the same thing. Note that angular acceleration is simply the change in angular rotational rate divided by the time taken to change, ie: [PLAIN]http://upload.wikimedia.org/math/f/d/9/fd97cb711276815954e9824fabee8baf.png. So for example, if you want to accelerate the cylinder from 0 to 8 RPM in 1 second, that's an acceleration rate of 8/60*2*pi*radians/s
2. If it takes longer or shorter than 1 second, divide by the number of seconds.
For the torque exerted by the cylinder being off center by 2", just add an additional torque of the cylinder weight times this 2" moment arm (ie: 6500 lb x 2" = 13,000 lb in). The amplitude of that torque is obviously going to vary sinusoidally but for your purposes (trying to determine the peak torque on the brakes or gearbox) you really don't care about that.