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Troublshooting your pc for dummies - Bourg D.M.

Bourg D.M. Troublshooting your pc for dummies - Wiley publishing , 2002. - 350 p.
ISBN 0-596-00006-5 Previous << 1 .. 5 6 7 8 9 10 < 11 > 12 13 14 15 16 17 .. 128 >> Next d2ar = (xcgcar Xcg) T (y^gcar leg) d2r = (100 ft - 99.7 ft)2 + (100 ft - 100.1 ft)2 = 0.1 ft2
^driver = (xcg driver — Xcg) + (ycg driver — leg)
driver = <103 ft - 59.7 ft)2 + (105 ft - 100.1 ft)2 = 34.9 ft2
= (^Cgtuel Xcg) 4“ (yCg fuel Xcg)
d2el = (93 ft - 99.7 ft)2 + (100 ft - 100.1 ft)2 = 44.9 ft2 Now we can apply the parallel axis theorem as follows:
^cgcar — "I" TTl?
Icgcar = 2800 lb-s2-ft + (39B lb/32.174 ft/s2) (0.1 ft2) = 2812lb-s2-ft
fcg driver — lu 4“ md
Icgdriver = 5.5 lb-s2-ft + (190 lb/32.174 ft/s2) (34.9 ft2) = 211.6 lb-sz-ft fegfuel “ lo 4“ ttld leg fuel = 6.1 lb-s2-ft + (210 lb/32.174 ft/s2)(44.9 ft2) = 299.2 lb-s2-ft
Notice the obvious relatively large contribution to Icg for the driver and the fuel due to the md2 terms. In this example, the local inertia of the driver and fuel are only 2.7% and 2.1%', respectively of their corresponding md2 terms.
14 | Chapter 1: Basit Concepts
Finally, we can obtain the total moment of inertia of the body about its own neutral axis by summing the Icg contributions of each component as follows:
fcgcocai ~ -leg car T -leg driver H” -^cgfuel *
-leg total = 2812 lb-s2-ft + 211.6 lb-s2-ft + 299.2 lb-s2-ft = 3322.8 lb-s2-ft
In summary, the mass properties of the body, that is, the combination of the car, driver and full tank of fuel, are shown in Table 1-3.
Table 1-3. Example Summary of Mass Properties
Property I -I;.. ? Computed Value
Total Mass (weight) 134.2 slugs (4317 lb)
Combined Center of Mass Location foy) = (99.7 ft, 100.1 ft)
Mass Moment of Inertia 3322.8 lb-sJ-ft '
It is important that the concepts illustrated in this example are well understood because as we move on to more complicated systems and especially to general motion in 3D, these calculations are only going to get more complicated. Moreover, the motion of the bodies to be simulated are functions of these mass properties, in that mass will determine how these bodies are affected by forces, center of mass will be used to track, position, and mass moment of inertia will determine how these bodies rotate under the action of noncentroidal forces.
So far, we have looked at moments of inertia about the three coordinate axes in 3D space. However, in general 3D rigid body dynamics, the body may rotate about any axis, not necessarily one of the coordinate axes, even if the local coordinate axes pass through the body center of mass. This complication implies that we must add a few more terms to our set of 1 ’s for a body to handle this generalized rotation. I will address this topic further in the last section of this chapter, but before I do that, I need to go over Newton’s second law of motion in detail.
Newton's Second Law of Motion
As I stated in the first section of this chapter, Newton’s second law of motion is of particular interest in the study 6f mechanics. Recall that the equation form of Newton’s second law is
F = ma
where F is the resultant force acting on the body, m is the mass of the body, and a is the linear acceleration of the body center of gravity.
If you rearrange this equation as
F fm = a
you can see bow the mass of a body acts as measure of resistance to motion. Observe here that as mass increases in the denominator for a constant applied force, the resulting
Newton's Second Law of Motion 15
acceleration of the body will decrease. It can be said that the body of greater mass offers greater resistance to motion. Similarly, as the mass decreases for a constant applied force, the resulting acceleration of the body will increase, and it can be -s^rid that the body of smaller mass offers lower resistance to motion.
Newton’s second law also states that the resulting acceleration is in the same direction as the resultant force on the body; therefore, force and acceleration must be treated as vector quantities. In general, there may be more than one force acting on the body at a given time, which means that the resultant force is the vector sum of all forces acting on the body. Thus, you can now write
y^F = ma where a represents the acceleration vector.
In 3D, the force and acceleration vectors will have x,y, and z components in the Car tesian reference system. In this case, the component equations of motion are written as follows:
Fx = max
J2Fy= may .
Fz = maz
An alternative way to interpret Newton’s second law is that the sum of all forces acting on a body is equal to the rate of change of the body’s momentum over time, which is the derivative of momentum with respect to time. Momentum equals mass times velocity, and since velocity is a vector quantity, so is momentum. Thus,
G ' mv
where G is linear momentum of the body, m is the body’s mass, and v is velocity of the center of gravity of the body. The time rate of change of momentum is the derivative of momentum with respect to time:
dG/dt = d/dt(mv) -
Assuming that the body mass is constant (for now), you can write ' dG/dt —mdv/dt
Observing that the time rate of change of velocity, dv/dt, is acceleration, we arrive at Previous << 1 .. 5 6 7 8 9 10 < 11 > 12 13 14 15 16 17 .. 128 >> Next 