Turning and acceleration of flywheel masses
Support » FAQ » Stepper motors
Often inquiries are received where only a well mounted rotary disk (e.g. 5 kg, diameter 30 cm) must be rotated and only the measurements and the weight are known as well as the required dynamics.
First, the moment of inertia of the mass must be calculated. For a solid cylinder (formula J = m * r²) that rotates around the axis of symmetry, the following applies: (J [kg*m^2], m [kg], r [m]) = 0.056 kgm^2 = 562.500 gcm^2
The moment of inertia can then be calculated theoretically from ([rad/s^2]), e.g. the retainer should be repositioned by precisely 360° in 1 s. The average speed is then 1 rev/s, it is first accelerated in 0.5 s to f = 2 rev/s and then braked to 0 again in 0.5 s (triangular profile). The following then applies to the angular acceleration: 25 rad/s^2 => M = 1.4 Nm.
But be careful:
The problem here is the acceleration. Because the SM does not begin at 0 but with a starting speed (that often has to be selected quite high to avoid resonances), the true acceleration is appreciably higher than that calculated here. As this is difficult to calculate or measure, however, (the encoder signal would have to be evaluated) the motor selection is made using the following rule of thumb using a known external moment of inertia.
Thus, a motor must be selected whose rotor has at least 1/20th of the external moment of inertia, here 28125 gcm^2 (Jred = Jex / i2). In this case a gear must be used which reduces the external moment of inertia quadratic with the ratio: A gear with 10:1 gives a reduction factor of 100 for 281 gcm^2 here (Jm of the ST59.. is ..). This means the combination of ST5818M2008 + GPLE40-2S-12 is suitable.
Due to the gear the max. speed is naturally also limited which still needs to be taken into account in this case but presents no problem.
