previous lecture | next lecture | 5200 index
F&C sections 3.4 and 3.6
mx" + cx' + kx = 0
Three cases to consider, overdamped, critically damped, and underdamped.
| quantity | relations | comments |
| w0 | w02 = k/m | positive |
| wd | wd2 = w02 - g2 | positive |
| g | g = c/2m | positive |
We now have the solution to the homogeneous differential equation. If an additional driving force is applied, then we need to consider the equation:
mx" + cx' + kx = F
In general, the force can take on any form, but a particularly interesting case is if the force is time dependent, F = F(t). Then, the differential equation is a non-homogeneous, linear, ODE, and techniques exist to solve it. We will solve the case where the force varies cosinusoidally with time: F(t) = F0cos(wt).
The quote at the beginning of Chapter 5 is quite interesting. It is a quote from Albert Einstein, commonly known as "the happiest thought of my life", which introduces us to his starting point for the development of general relativity! The fact that it is printed at the beginning of this chapter may cause some pause, since general relativity is not in the course syllabus. No worry, we're not going to tackle that subject, but it is an indication that we will be skirting the boundaries between Newtonian mechanics and general, and special, relativity.
It is often convenient to solve a problem from the perspective of a moving coordinate system than from a fixed coordinate system. In fact, if you refer back to section 2.1, Newton's First Law, Inertial Reference Systems, you will recall that we discussed the fact that it is probably impossible to define a truly fixed coordinate system. Instead, we settle for coordinate systems that are "approximately" inertial. We will now investigate what happens when a non-inertial system is used.
First, let's consider the case where our coordinate system is translating relative to a "fixed" coordinate system. Let Oxyz represent the fixed, inertial coordinate system, and unprimed vectors represent vectors in this coordinate system, for instance, r will be the position of a point P in this coordinate system. Let O'x'y'z' represent a coordinate system translating with respect to Oxyz. If r' is the position vector of the same point P, and R0 is the displacement OO' of the moving coordinate system, then
r = R0 + r'
Taking derivatives, we find the relations for velocity and acceleration:
v = V0 + v' a = A0 + a'
If the moving system is not accelerating with repect to the fixed system, then A0 = 0, and a = a'. Then, since F = ma in the fixed coordinate system, in the moving system we also find that F = ma'. This is called a Galilean coordinate transformation. Coordinate systems in relative motion, but where Newton's law is valid, are called inertial coordinate systems.
What happens if the prime system is accelerating relative to the unprimed system? Then A0 is no longer zero, and Newton's second law in the primed system becomes F = mA0 + ma' or F - mA0 = ma'. We can write this equation in the somewhat suggestive form F' = ma', implying that the second law works in the accelerating frame, but with a modified force. The acceleration of the moving reference frame can be taken into account by modifying the force by the term -mA0. This term is called a (non)inertial force or fictitous force. Notice that there are no fictitous forces in an inertial reference system. This is the basic requirement for an inertial reference system -- no fictitous forces exist. Therefore, for instance, an object in motion, not acted on by any manifest force, should continue in constant, unchanging motion.
What is the force "felt" by a person in free fall?
The force that is felt by the person is the force F' in a coordinate system that moves along with the falling person. In a coordinate system fixed to the earth (this is an inertial coordinate system for the purposes of this problem), the person is accelerating like a = -gk. This will be the acceleration of the falling person's coordinate system: A0 = a. Therefore, F' = [-mg -(-mg)]k = 0, such that the person feels no force. (Of course we have neglected air drag, which would modify this result.)
This points out a misconception perpetuated by the common usage of the phrase "zero-gravity of space". Astronauts in orbit around the earth, or anywhere for that matter, are not in a region of zero-gravity. It is gravity that keeps them in orbit around the earth, and even away from earth, there is no region of space without gravity! But the astronauts are in free-fall, and they feel no force in their noninertial reference system.
A block of wood rests on a rough horizontal table. If the table is accelerated in a horizontal direction, under what conditions will the block slip?
This problem can be solved without using a noninertial reference frame, but let's try applying the idea to see how it works. Let ms be the coefficient of static friction between the block and the table top. Then the force of friction F has a maximum value of msmg. The condition for slipping, from the reference frame of the block, is that the fictitous force -mA0 exceeds the frictional force, where A0 is the acceleration of the table. Hence |-mA0| > msmg or A0 > msg.
A pendulum is suspended from the ceiling of a railroad car accelerating uniformly in the right (+x) direction. An observer in the car sees the pendulum hanging at an angle q to the left (-x) of vertical, due to a fictitous force. An observer outside the car also sees the pendulum hanging at the same angle, but sees no fictitous force. Explain these different views.
The observer outside the car sees the pendulum being pulled along by the train car. Since it is being accelerated, Newton's second law says
© 06 April 1999 R. Harr