Differentialgleichung Drehmoment?
Hallo ich habe ein Problem mit folgender Aufgabe:
Die dementsprechende Lösung sieht wie folgt aus:
Mein Problem:
Ich habe keinerlei Ahnung wie man auf die Formel für M_a gekommen ist, noch was M_a in Beziehung zu der Aufgabe darstellen soll.
1/w kann ich umformulieren zu M_m/M_l, dies multipliziert mit M_l ergibt M_m.
Der Anteil davor ergibt aber auch M_m, demnach ist M_a=2*M_m. Ich habe keinen blassen schimmer, wie ich dies plausibel zuordnen soll.
Das sonstige Vorgehen mit immer weiter auflösen und dann rückgängig einsetzen ist mir soweit klar.
Vielleicht kann mir da ja jemand weiterhelfen, schonmal danke dafür.
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You don’t make us answer your question easily. The formula mark M_m does not appear, for example, in the calculation. And with 1/w you probably mean But also this formula character does not appear. There are only two different speeds in this whole configuration either omega_m (motor speed) or omega_l (speed of the rope winch).
And the ratio of two moments does not yield a rotational speed, but at most a transmission ratio.
That’s why I can only explain at the moment how the formula for M_a was created.
M_a is composed of two parts. The first summand describes the moment of inertia of the motor itself. The second term describes the moment of inertia of all attached loads, which are all summarized in the term M_l. Between the motor and the cable winch there is a gear that reduces the speed for the cable winch by a factor of 15. This the Torque (non) the Torque) simultaneously raised by the factor. From the point of view of the motor drive, however, the load torque is reduced by the factor 15. Therefore, M_l is divided by the factor ü (ü=15).
The remaining consideration is only about further clarifying the load torque M_l. The cable winch itself has a moment of inertia, the deflection roller also. Finally, even translational loads are converted to torques. This is the function of the deflection roller with its lever arm r.
I’m sorry, I meant 1/o upstairs.
And with M_m I mean the torque resulting from J_m*(d/dt*w_m), i.e. the torque of the machine.
In addition to solving, we have agreed that ü=w_m/w_l=M_l/M_m is and thus 1/ü=M_m/M_l.
Yes, exactly. ü is a transmission ratio with effect on torque and speed.
But caution J_m*(d/dt*w_m) is not the torque of the engine, but only the moment of inertia of the engine which acts during the acceleration phase.
At the end, an M_a is accounted for here, i.e. the sum of all torques to be applied at least by the machine.
Thank you, I think I understand.