Starting and stopping a heavy rotating disc

Question

I am currently working at a project involving a heavy disc rotating around its center. The disc weights around 2 kg and has a radius of 0,25 m. At every angle π/6, there exists smaller discs of masses from range 0,1 kg to 1 kg 0,15 m from the center. The radius for these are 0,05 m. A picture to illustrate:

I have roughly calculated the moment of inertia when all smaller discs weight 1 kg. Using the formula for a circular plate and Steiner's theorem, the result is:

$$ Inertia = \frac{0,25^2}{2} + 12(\frac{0,05^2}{4} + 0,15^2) = 0,34 kgm^2$$

Now I want the disc to be able to spin and stop at these specific angles. Say for instance I want the disc to rotate from 0 to π. This means I need a precise way to control my disc. My plan is to use a servo and some gears to drive this. I need the disc to turn 180° in at least 3 seconds (preferably less). With this angular velocity and inertia, I have realized it might not be the easiest thing to stop this spinning wheel, let alone accelerate it. Here is another image:

The motor does not need to be positioned like that, it would also be possible to drive the disc by positioning the motor on the edge of the disc for example.

What kind of motor should I be looking for, how would I handle stopping the disc? I am looking for general tips on how to accomplish this.

Answer 1

Some of the issues here are the same as in Rotate (and stop) a large disk in very tiny increments, and some are different. See the second answer there for drawings of one possible arrangement of the motor driving the plate.

I think it would make sense to use parts that now are widely used for 3D printer axis motion, such as Nema 17 stepper motors, encoders, timing-belt pullies / sprockets, and timing belts or timing belt material. If a timing belt wraps around the 25-cm disk and then is driven by a 1-cm timing-belt pulley flanked by two tension idlers, the 25:1 mechanical reduction drops the required motor torque into the range that Nema 17 stepper motors can handle while running at a speed high enough to meet your requirements.

Such motors can run at 6000 to 7000 half-steps per second when providing 0.03 Nm torque. (See, for example, the "Pull out Torque Curve" diagram in C140-A+datasheet.jpg at adafruit.com.) This is 900 to 1050 RPM, somewhat above the necessary 250 RPM stepper speed to rotate the large disk half a turn in 3 seconds, which is 10 RPM. (Ie, 10 RPM = (0.5/3 RPS)*60 S/M.) The required torque at the stepper appears to be just under 0.03 Nm, as follows: For the large disk, τ = Ia, where τ is torque, in Nm; I is moment of inertia, listed in the question as 0.34 kg m²; and a is angular acceleration. Here, the required angular acceleration is positive for 1.5 seconds and negative for 1.5 seconds, assuming the disk is at rest, moves, and is again at rest. Moving π radians in 3 seconds from rest to rest is π/3 radians per second net, via sustained accelerations of ±2π/3 radians per second². Thus τ = 0.34 kg m² * 2π/3 /s² = 0.712 Nm. Divided by 25, this is 0.028 Nm.

Note, running the stepper motor at a given angular-acceleration rate implies accelerating and decelerating it, vs immediately starting it out at 250 RPM. That is, you would start out sending a fairly slow train of pulses to the stepper; during 1.5 seconds increase to maximum rate of 500 RPM; then for 1.5 seconds decrease the pulse rate back to zero. Without stepper acceleration and deceleration, you risk losing pulses, which would lead to inaccuracy, due to exceeding stepper motor torque limits. _{^{[Note: edit 1 revised this paragraph]}}

Note, it would make sense to have an encoder attached to the large wheel, to track its position rather than depending on open-loop operation of the stepping system.