Meeting times, textbook, homework, exam, grading policies

Written solutions, study groups, and plagarism.

Plan for what we'll cover this semester, and what will be learned.

Review 5200.

In Taylor, read sections 9.1 and 9.2 for today, and 9.3 to 9.5 for Friday.

Newton's laws are valid in inertial reference frames only. Usually we can choose and inertial frame in which to apply Newton's laws, but sometimes it's convenient to work in a non-inertial frame. For instance, this happens for Earth-based problems covering relatively large areas, such as weather systems, or the firing of a rocket; it also happens in the general treatment of rotation of rigid bodies, as we'll encounter in chapter 10. It is also interesting to learn some of the concepts of working with non-inertial frames as a prelude to general relativity. In this chapter we will develop equations for applying Newton's laws in non-inertial frames.

There is an interesting quote from Albert Einstein, commonly known as "the happiest thought of my life",

I was sitting in a chair at the patent office in Bern, when all of a sudden a thought occurred to me: If a person falls freely, he will not feel his own weight. I was startled. This simple thought made a deep impression on me. It impelled me toward a theory of gravitation.

It is often more 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 the section on Newton's First Law, 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.

m**r**[ddot] = **F**_{real} - m**A**

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 **R**_{0} is the displacement *OO'* of the moving coordinate system, then

Taking derivatives, we find the relations for velocity and acceleration:

If the moving system is not accelerating with repect to the fixed system, then **A**_{0} = 0, and **a** = **a'**.
Then, since **F** = m**a** in the fixed coordinate system, in the moving system we also find that **F** = m**a'**.
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 **A**_{0} is no longer zero, and Newton's second law in the primed system
becomes **F** = m**A**_{0} + m**a'** or **F** - m**A**_{0} = m**a'**.
We can write this equation in the somewhat suggestive form **F'** = m**a'**, 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 -m**A**_{0}.
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** = -g**k**.
This will be the acceleration of the falling person's coordinate system: **A**_{0} = **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 mass m rests on a surface (a table) that is accelerating (horizontally).
If the coefficient of (static) friction between the block and the suface is μ_{s}, what is the maximum acceleration of the surface for which the block won't slip?

The block is always subject to gravity, the normal force, and the force of friction, independent of the frame of the observer.
Gravity and the normal force will balance in the vertical direction.
We are interested in the horizontal direction where friction is the only "real" force, F_{real} = F_{friction} ≤ μ_{s}mg.
In the frame of the surface we should include the intertial force -mA, yielding the equation of motion

This problem can be solved without using a noninertial reference frame, but let's try applying the idea to see how it works.
Let μ_{s} be the coefficient of static friction between the block and the
table top.
Then the force of friction **F** has a maximum value of μ_{s}mg.
The condition for slipping, from the reference frame of the block, is that the fictitous force -mA_{0} exceeds the frictional force, where A_{0} is the acceleration of the table.
Hence |-mA_{0}| > μ_{s}mg or A_{0} > μ_{s}g.

mx'[ddot] = F_{friction} - mA

If the block doesn't slip then

A simple pendulum of mass m and length L is mounted in a rail car that is accelerating to the right with constant acceleration **A**.
Find the angle φ_{eq} at which the pendulum will hang at rest inside the accelerating car, and find the frequency of small oscillations about this equilibrium angle.
Contrast working this problem in an inertial and non-inertial frame.

First let's solve this problem in the non-inertial frame of the rail car.
Begin by identifying the forces present in an inertial frame (the real forces) on the pendulum mass.
The mass feels the tension in the string and gravity, therefore, **F**_{real} = **T** + m**g**.
In the non-inertial frame of the rail care we add the fictitious force due to the acceleration of the car, **F**_{inertial} = -m**A**, so the equation of motion reads:

m**r**[ddot] = **T** + m**g** - m**A** = **T** + m(**g** - **A**) = **T** + m**g**_{eff}

Both the fictitious force and gravitational force are proportional to the mass, so they can be added, and the sum interpreted as an effective gravitational force.
The resulting equation of motion looks idential to that for a simple pendulum, but with a different gravitational acceleration, **g**_{eff} = **g** - **A**.

The pendulum will be in equilibrium when it hangs parallel to **g**_{eff}, which is at an angle φ_{eff} = arctan(A/g) to the vertical, opposite in direction from the acceleration **A**.
The frequency of small oscillations is ω = sqrt(g_{eff}/L) = sqrt(sqrt(g² + A²)/L)

This solution was rather simple and direct, reusing the result already determined for a simple pendulum in a stationary car. Contrast this with the solution determined from an inertial frame.

In an inertial frame, string tension and gravity still act on the mass.
To be in equilibrium with respect to the rail car, the pendulum must accelerate at the same rate as the rail car, therefore **T** + m**g** = m**A**.
Since **g** and **A** are known, we can solve for **T** = m(**A** - **g**).
The equilibrium position is the direction of the tension, which gives us the same and φ as above.
The frequency of small oscillations is found by considering how the mass will move when slightly displaced from this equilibrium position.
This is not trivial in this frame.