Converting Potential Energy into a Long Rotation of heavy object

I think your question is really about converting gravitational potential energy to rotational kinetic energy. Once you have that you can change direction and rotational axis using common components like bevel gears.

Converting gravitational potential energy to rotational kinetic energy is very common in hydroelectric energy. An option would be to let the weight hit a paddle or some other feature connected to a gear to add torque to the system. Alternatively you could drill a hole through the shaft where the gear is mounted, tie some rope/twine to the shaft and tie the other end to your mass. Wind up the system and then let the mass go. You'd want twine set up so that it releases from the shaft when the mass gets to the bottom of travel.

I'm not sure what is meant by slow for a long period of time. You can reduce the rpm and increase torque with a simple spur gears. The larger gear will have higher torque and slower speed. How slow it goes is dependent on the speed of the smaller gear and ration of diameters for the two gears. If you want it to go for a long time you need to fight frictional losses. Meaning you'll need good bearings and lubrication between the components. Best case scenario is operating in a vacuum with magnetic bearings.

Let's say you want to roll a shaft made of wood with a diameter of 10 cm weighing 1 kg on a long, smooth track with, say, a rolling friction of $0.05%, which is a plausible number.

As a basic mechanism, you can tie a cord around a handle on the shaft and pull the cord around a pulley hanging the 2 kg weight. This mechanism will start first by rolling your shaft and accelerating it to a speed that will reach the maximum speed of air resistance. The forces acting on the shaft are:

$$2 -1kg.0.05=1.995kg force, F $$

This $F$ is doing two tasks. Let's call them F1 and F2.

This force causes a torque in the shaft and also pushes the shaft forward. $$T=F_1*r$$ Assuming $g=10$

$$T=F_1*r$$ $$I_{shft}=\frac{1}{2}mr^2$$ Calling angular acceleration of shaft $\alpha$

$$T= I_{shaft} \cdot\alpha$$ At the same time, $F_2$ is pushing the shaft with $$F_2= m \alpha_{2}$$, causing it to accelerate.

The F division between F1 and F2 is not addressed here, but it is not relevant to our point.

The shaft will accelerate continuously until it reaches a speed V that will receive a head wind resistance, which will stop the acceleration. From that point on, the shaft will roll with a steady speed.

That force is called drag force $$F_D = \frac{1}{2} C_d \rho A V^2 $$

• $F_D$ drag force
• $c_d$ is the drag coefficient (75-80)% for a cylinder
• $A$ the area of shaft facing the wind
• $v$ is the speed of shaft.
• $\rho\ $ is the density of the air.

Depending on the situation, the wheel may reach the ground before the cylinder reaches its final speed.