One of the several important advantages of the three-phase, a.c. induction motor is the grace with which it delivers a variable torque. You did not mention the motor's number of poles or whether your power was of the 50-Hz, Old World kind or of the 60-Hz kind used in the New World. In the Old World, the maximum (not minimum) speed of a typically configured induction motor is 3000 rpm, and this only if it is a two-pole motor. Four-pole motors are more common: their maximum speed in the Old World is 1500 rpm.
For the sake of answering, let us suppose that you are in the Old World and that your motor has two poles: maximum speed, 3000 rpm. (If in the New World, you can scale this answer's speeds by [60 Hz]/[50 Hz] = 1.20: maximum speed, 3600 rpm.)
For moderate values of torque—usually up to about 115 percent of the motor's full rated load—the motor's speed will decrease slightly and approximately linearly as the mechanical load to which the motor delivers power demands torque. For example, if the motor turns at 2800 rpm at full rated load, then it will turn at about 2850 rpm at 75 percent of full rated load, 2900 rpm and 50 percent, 2950 rpm at 25 percent, and (almost) 3000 rpm when unloaded. At loads greater than 115 percent, the motor's speed will decrease more than the linear model predicts, until the motor's breakdown torque is reached (the exact value of breakdown torque depends on the motor, the ambient temperature and other factors, but 200 to 230 percent of full rated load might be a typical figure). At breakdown torque, the motor will spin down to a stop, thenceforth acting as an inductive winding and delivering the motor's rated locked rotor or starting torque to the load.
I give all this detail for three reasons. First, because it may contain the answer you seek. Second, to explain why there is no single torque, but are many torques involved. Third, to convey the crucial concept that, within the motor's operational domain, it is the mechanical load, not the motor, that determines the torque.
If you really need to measure the actual torque accurately under specific conditions, then use a dynamometer. However, if an approximate indication of torque is all you need, and if you are operating in the motor's normal, spun-up state, at no more than 115 percent of full load, then you can approximate the torque pretty accurately from the motor's measured speed and its nameplate data, using the linear relationship described. The approximation is so much easier than the dynamometer that I would recommend the approximation in most cases.
Note: A few, unusual three-phase a.c. induction motors do not have any breakdown torque. In other words, for a few induction motors, the locked-rotor torque is the maximum torque the motor can deliver. In the United States, such motors usually read "Design: D" on their nameplates. This is probably not the case for you, but I thought that I should at least mention the matter.
If it doesn't matter how fast the weight is raised, then any size motor will do.
The reason?
Lifting a certain weight requires tension in the cable, torque through the rest of the geartrain, and current in the motor.
Lifting at a certain speed requires m/s at the cable, radians/s through the geartrain, and voltage at the motor.
Lifting the weight at a certain speed requires power from end to end, always measured in watts. At the output, the power to lift the weight is speed*force in m/s and N. Through the geartrain the power is angular speed * torque, in radians/s and Nm. At the motor, the power is volts * amps. Losses in the geartrain and motor reduce the power available at the output, at best 50% through the motor and another 50% through the worm gear.
The ratios between torque, speed, force, current, voltage etc are determined by your gear sizes, gear ratios, motor voltage constants etc. The power is always power, no matter what the gear configuration and the motor type.
To put real numbers on it, start by putting your units into SI, appropriate for physics. A 100lb weight is about 50kg, is about 500N. If you want to lift that at 1m/s, you will need 500 W, or maybe 2kW electrical input. Not possible in your 50mm dia motor. If you want to lift at 1mm/s, you will need 0.5W, perhaps 2W electrical input. Much more possible.
Once you have a power budget, you look at a gear catalogue to see what ratios are available. Then convert forces through the gearing to get torque and speed at the motor. Then look at a motor catalogue to find a motor that will deliver those. Iterate gear choice as required. Rinse and repeat.
/edit/ Thanks to respawnedfluff for pointing out the static friction aspect. Any bearing will have a static, or 'breakaway', friction that is larger than its running friction. For a wormgear, that's true in spades. Do all the sums for dynamic operation. Then make sure your motor and power supply can deliver 2 to 3 times the operating torque, transiently, at the start to get things moving. Most DC PM motors will be able to do this, as will most batteries, and simple power supplies. A current-limited supply will have to be adequate for the starting torque however, not just the running torque.
Aristotle, or was it Archimedes, said 'give me a lever long enough and I will move the world'. Similarly, give me a worm gear with a large enough gear ratio, and I can shift a large weight with a tiny motor, albeit very slowly.
It is a very specialised worm gear that will run backwards, low ratio, nice materials, properly lubricated. For 'most' worm gears, their efficiency is so low that they cannot be backdriven by the load. The OP is correct that the worm will hold the load with the motor undriven. /edit/
As in most things in life, I find it's easier to do an analysis of a straw-man system, and then tweak that, rather than a synthesis from specifications. So... let's say you can get a 150:1 worm drive, and a 10 watt 3000 rpm motor, and ignore mechanical losses initially. 3krpm is about 300 rad/s, so the motor torque is 10/300 = 0.033Nm. After 150:1 gearing, we have 5Nm at the worm. To lift 500N, we need to wind the cable around an axle of radius 0.01m. 3krpm /150 = 20 rpm at the worm, 1 rev in 3 seconds, or 20mm/second.
Allowing for worm efficiency < 50%, we need to increase motor power, and to allow for starting torque, increase the power supply transient current above that. Is 20 watt motor and 20mm per second adequate?
If not, figure out where the biggest wrong assumption is, and rework.
You have drawn a 1:1 bevel. You have the opportunity to reduce motor speed and increase torque by using an N:1 bevel. Look at your gear catalogue for available ratios.
Best Answer
Your calculations assume the whole weight is applied at the end of the lever. In fact, it will be roughly halfway there, at 0.25m
Also, since it will be horizontal at the highest point you don't need any trigonometry here. Your minimal torque is simply the weight of a door applied at half the door height, or 5 kg at 0.25m. Which is 1.25 kg-m or 125 kg-cm.
Of course you need more than that for reliability, say 150-200 kg-cm. While it is possible to find geared motors like that, they will be either a) slow, b) high current, c) expensive or d) all of the above.
Now it is good time to recall that garage openers have tiny motors compared to the weight of the doors. That is because majority of the lifting is done by a counterweight or a spring.
So, my suggestion would be to use good spring-loaded door and relatively small motor with worm-gear, picking up the small portion of the weight. Automotive 12V power-window motor mentioned here seems to be good fit for applications like this. You can also find one of the possible simple schematics for controlling the motor at that link, no MCU necessary.
Note that when designing spring assistance you do need take into account the angular position of the door. This is simple mechanical problem and you should be able to find examples or calculate it yourself. The basic idea is that the point of application of relatively constant spring/weight force should move further from the pivot, exactly following the door's center of gravity moving further from vertical as door is rising.
Also note that regardless of the small torque of the motor required, the gearbox and all connectors should be strong enough to keep the door in fixed position should something happen with counterweight system.
Finally, make sure you have fuses in your electrical wiring (separate for each direction!) rated close enough to normal working current so that if a limb accidentally gets in the way the motor will stop rather than causing injury.
Note that 3PDT switch should be rated for both mains voltage and motor current. End switches should be rated for mains voltage and current and should be snap-action type to prevent extensive arcing.