sizing DC motor needs
If all the motor needs to do is spin the wheel around 3- 5 RPM then I suggest finding a simple Brushed DC motor with an integrated gear box that spins at that RPM when the rated voltage is applied.  When looking for a motor you should make sure the torque the motor outputs can defeat the friction of your drive train and have enough torque left over to accelerate the wheel up to speed in a reasonable amount of time.

Working through your problem:
Lets say we want to spin up to 5 RPM in five seconds.

We know that rotational acceleration can be described by 

$$ \mathrm d \omega = {T \over J} $$

Where:


\$\mathrm d \omega\$ is rotational acceleration,
\$T\$ is torque applied, and
\$J\$ is inertia of load (Your wheel in this case, we can ignore the motor's inertia for it will be very small compared to the load)


We know dw because we want to reach 5 RPM (0.514 Rad/s) from 0 RPM in at most 5 seconds.

$$ \mathrm d \omega
 = {0.514 \over 5}
 = 0.105\:\mathrm{Rad/(s^2)}
 = 1\:\mathrm{RPM/s} $$

We can approximate the wheel as a disk of uniform density:

$$ J = {Mr^2 \over 2} $$

Where \$M\$ = Mass (Kg) and \$r\$ = radius (meters)

$$ J = 0.585\:\mathrm{kg\:m^2} $$

The needed torque to accelerate the load then can be calculated to be:

$$ T = J \mathrm d \omega = 0.0613\:\mathrm{Nm} = 6.13\:\mathrm{Ncm} $$

But this is only one part of the load the motor will experience.  The other part is friction.  The friction torque is something you can determine experimentally by seeing how much torque it takes to move the wheel at standstill.  Add the frictional torque to the torque needed to accelerate your load and you have the minimum torque your motor needs to be rated for.

Hope this helps.