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Basic Physics of Mechanical Drive Systems MDF 03 23 08-2p

Basic Physics of Mechanical Drive Systems MDF 03 23 08-2p -...

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Basic Physics of Mechanical Drive Systems MDF 03 23 08 Introduction Electric drives have constraints on torque and speed set by the basic physics of the electromechanical system. Linear Motion Consider a load of constant mass M acted on by a constant force f. (1) f M = d dt H Mu L = M du dt = Ma where a is the acceleration (2) a = du dt = f M M . In MKS, a net force of 1 Newton, acting on a constant mass of 1 kg, results in an acceleration of 1 Meter ë Second 2 . Integrating the acceleration with respect to time we obtain, (3) u H t L = u H 0 L + 0 t a H t L „t Integrating the speed with respect to time, position can be determined by (4) x H t L = x H 0 L + 0 t u H t L „t . Example 1. Assume constant force, f M . Also assume zero initial velocity and position, u(0)=0, x(0)= 0. Then u(t) will be f M M t and x(t) will be f M M t 2 2 . Work The differential of work is

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(5) dW e = f e dx Fig. 1. Motion of mass M due to action of forces. The external force f e , is opposed by the load force f L , resulting in a net mechanical force, f M . Power is the time rate at which work is done. Differenting both sides of Eq. 5 with respect to time, we have (6) p e H t L = d W e dt = f e d x dt = f e u . also f M = f e - f L and (7) p M H t L = dW M dt = f M dx dt = f M u . Note that it takes a certain amount of energy to bring a mass from rest to a terminal velocity, resulting in kinetic energy of motion that can be recovered. Recognizing that f M = M du dt , we have, (8) p M H t L = Mu du dt , and the energy input can be calculated by integrating both sides of 8 as follows: (9) W M = 0 t p M H t L „t = 0 t Mu du d t „t = 0 U Mu u = 1 2 MU 2 , where U is the final velocity. 2 Basic Physics of Mechanical Drive Systems MDF 03 23 08-2p.nb
Rotating Systems θ M g β Fig. 2. Geometry for calculation of holding torque Most electric motors are the rotating type. When a force f is applied perpendicular to the direction of the radius r to the pivot point, (10) T = fr Now assume that a mass M is hung from the tip of a lever, and calculate the holding torque required to just keep the lever from turning. (11) f = Mg Cos @ b D , where b is the angle between the normal to the pivor and the direction of gravitation force on the mass, and g = 9.8 Meter ë Second 2 is the gravitational acceleration. Since b = q [see Fig. 2], we have for the required holding torque: (12) T h = Mgr Cos @ q D . Tcm ω M T L Motor Torque in electric motor MDF 03 24 08 Fig. 3. Mechanical aspects of an electric motor.

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Basic Physics of Mechanical Drive Systems MDF 03 23 08-2p -...

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