2‐5
In Example 2-3, calculate the kinetic energy stored in the combined inertia at a speed of 1800 rpm.
Solution
From Eq. (2-29),
2‐4 FRICTION
Friction within the motor and the load acts to oppose rotation. Friction occurs in the bearings that support rotating structures. Moreover, moving objects in air encounter windage or drag. In vehicles, this drag is a major force that must be overcome. Therefore, friction and windage can be considered as opposing forces or torque that must be overcome. The frictional torque is generally nonlinear in nature. We are all familiar with the need for a higher force (or torque) in the beginning (from rest) to set an object in motion. This friction at zero speed is called stiction. Once in motion, the friction may consist of a component called Coulomb friction, which remains independent of speed magnitude (it always opposes rotation), as well as another component called viscous friction, which increases linearly with speed.
In general, the frictional torque Tf in a system consists of all of the aforementioned components. An example is shown in Fig. 2-10; this friction characteristic may be linearized for an approximate analysis by means of the dotted line. With this approximation, the characteristic is similar to that of viscous friction in which
(2-30)
where B is the coefficient of viscous friction or viscous damping.
The aerodynamic drag force in automobiles can be estimated as fL = 0.046 Cw A u 2, where the coefficient 0.046 has the appropriate units, the drag force is in N, Cw is the drag coefficient (a unit‐less quantity), A is the vehicle cross‐sectional area in m2, and u is the sum of the vehicle speed and headwind in km/h [1]. If A = 1.8 m2 for two vehicles with Cw = 0.3 and Cw = 0.5, respectively, calculate the drag force and the power required to overcome it at the speeds of 50 and 100 km/h.
Solution
The drag force is fL = 0.046 Cw Au 2 and the power required at the constant speed, from Eq. (2-6), is P = fL u where the speed is expressed in m/s. Table 2-1 lists the drag force and the power required at various speeds for the two vehicles. Since the drag force FL depends on the square of the speed, the power depends on the cube of the speed.
Traveling at 50 km/h, compared to 100 km/h, requires 1/8th of the power, but it takes twice as long to reach the destination. Therefore, the energy required at 50 km/h would be 1/4th that at 100 km/h.
Fig. 2-10 Actual and linearized friction characteristics.
TABLE 2-1 Drag Force and the Power Required
| Vehicle | u = 50 km/h | u = 100 km/h | ||
| Cw = 0.3 | fL = 62.06 N | P = 0.86 kW | fL = 248.2 N | P = 6.9 kW |
| Cw = 0.5 | fL = 103.4 N | P = 1.44 kW | fL = 413.7 N | P = 11.5 kW |
2‐5 TORSIONAL RESONANCES
In Fig. 2-7, the shaft connecting the motor with the load was assumed to be of infinite stiffness, that is, the two were rigidly connected. In reality, any shaft will twist (flex) as it transmits torque from one end to the other. In Fig. 2-11, the torque Tshaft available to be transmitted by the shaft is
This torque at the load‐end overcomes the load torque and accelerates it,
(2-32)
Fig. 2-11 Motor and load‐torque interaction with a rigid coupling.
The twisting or flexing of the shaft, in terms of the angles at the two ends, depends on the shaft torsional or the compliance coefficient K:
where θM and θL are the angular rotations at the two ends of the shaft. If K is infinite, θM = θL. For a shaft of finite compliance, these two angles are not equal, and the shaft acts as a spring. This compliance in the presence of energy stored in the masses, and inertias of the system, can lead to resonance conditions at certain frequencies. This phenomenon is often termed torsional resonance. Such resonances should be avoided or kept low; otherwise they can lead to fatigue and failure of the mechanical components.
2‐6 ELECTRICAL ANALOGY
An analogy with electrical circuits can be very useful when analyzing mechanical systems. A commonly used analogy, though not a unique one, is to relate mechanical and electrical quantities, as shown in Table 2-2.
