but it also needs to be able to lift about 5lb/ft (roughly 900 oz/in).
What are pounds per foot and ounces per inch? Not torque I'm afraid, I think you mean "foot pounds-force" (ft-lbf)
and "inch ounce-force" (in-ozf)
Use SI units of newton metres (Nm) to make things easy to calculate. 5 ft-lbf is 6.7 Nm
power = torque x angular velocity (no conversion factors needed - torque in Nm, angular velocity
in radians/second and power in watts.
30rpm is pi radians/s, so power needed is = 6.7 x 3.14 = 21W. Of course we need a safety margin and to allow
for gear losses, so call it 30 or 40W. If your motor / power-supply cannot handle that sort of power level you'll
need to find ones that can (is this continuous or intermittant duty BTW?)
Smallish DC motors are not greatly efficient, so you'll have to allow for perhaps 10 to 15W heat dissipation in the
motor as well, and make sure there is enough air-flow (these motors will have built-in fans).