Figuring the torque required for acceleration is the easy part. The formula is:
T= I*α
Where T= torque (N-m), I is the mass moment of inertia (kg-m^2), and α it the angular acceleration (radians/sec^2). For a disc with evenly distributed mass I = m*r^2/4. For the 10m, 4 tonne platform this gives an I = 25,000 kg-m^2. (But see the additional I elements noted below.)
You will need to develop the required maximum angular acceleration from either the operational criteria or the control system's error correction requirements. The precision you noted (0.01 deg) is the starting point for developing the maximum angular acceleration needed, but you will need to look at how fast you want the platform to be able to respond to commands. Start-up may be the worse case and have the need for the greatest amount of acceleration, however too high or too low acceleration rates can make the control system unstable or unable to correct errors.
You've also given a maximum speed of 1.03 m/min (0.0172 m/s), which is an angular velocity of 0.00343 rad/sec. You might want to start by arbitrarily selecting (guessing) at a time needed to reach full speed, say 10 sec. Using 10 seconds gives an angular acceleration rate of 0.0343 rad/sec^2. Feeding that into T=I*α gives a torque for acceleration of 857.5 N-m. A point of caution here, this arbitrary 10 second time has important consequences to the control system's capability. I would guess you will want a time more in the 1 second range (i.e. 10 times the above torque).
Okay, as I said that is the easy part. Estimating the friction and drag will be less straight forward. For this you will need to work from a conceptual (or better) design. How many wheels will be used (6, more?), what type of wheel (pneumatic, solid urethane...), rolling resistance of the wheels (including their bearings, which may not have significant drag), the drive system (gear units, flex couplings, brakes, etc.). Each of these elements will contribute loses that the motor will need to over come. Some rolling resistance coefficient values for pneumatic tires from an old reference use values of 0.01 – 0.03 of the load. Gear units add resistance as a result of internal friction and losses, think of its efficiency. There will probably be a need for a brake (unless a four quadrant controller is used) and brakes often contribute resistance, even when they are released.
Each of these elements also contributes some additional rotating mass, that should be added to the effective total I for calculating the torque required for acceleration. Adding this additional rotating mass isn't simply adding their individual I's to the total I. The wheels, motor, etc. will rotate at a greater angular velocity than the main platform. To reflect their contribution to the main platform their I's need to be increased by the square of their respective angular velocities – (ω(wheel) / ω(platform))^2. The motor, if running through a gear reduction, may have a surprisingly large contribution to the total I due to the combined wheel and gear ratio adjustment.
The drive wheel needs to have enough load and friction to avoid, or at least minimize, slippage. Pneumatic tires on dry rough concrete can have high friction of traction coefficients. However, when running on polished concrete or metal, that may be dirty, oily or wet, can end up with coefficients below 0.05. Using that value for the traction coefficient and the above torque leads to a drive tire load of approximately 1,800 N. Assuming that all of the above additional effects were to double the required torque, the value is still within the required vertical support load. However, you will probably want to include some form of suspension on the drive wheel to assure it maintains an adequate force against the platform should there be any unevenness to the platforms running surface.
In the above I referred to pneumatic tires, that is because I'm quite familiar with them. For this application I would be more likely to consider urethane tires.
I hope this helps and that I wasn't too long.
