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Plan for what we'll cover this semester, and what will be learned.
You wre requested to read Taylor 9.1 to 9.3 for today.
In Taylor, read sections 9.4 to 9.5 for Wednesday.
Newton's laws are valid in inertial reference frames only. Usually we can choose an inertial frame in which to apply Newton's laws, but sometimes it's more 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.
The idea is that a person at rest in a frame accelerating uniformly in a direction opposite to gravity with magnitude equal to the local acceleration due to gravity will feel no net force. The acceleration of gravity will be canceled by the acceleration of the person.
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.
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
Taking derivatives, we find the relations for velocity and acceleration:
If the moving system is not accelerating with respect 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 with 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 in their noninertial reference system the force of gravity is exactly canceled by the fictitous force.
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 μs, 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, Freal = Ffriction ≤ μsmg.
This problem can be solved without using a noninertial reference frame, but let's try applying the idea to see how it works. Let μs be the coefficient of static friction between the block and the table top. Then the force of friction F has a maximum value of μsmg. The condition for slipping, from the reference frame of the block, is that the fictitous force -mA0 exceeds the maximum frictional force, where A0 is the acceleration of the table. Hence the no-slip requirement is |-mA0| < μsmg or |A0| < μsg.
If the block doesn't slip then it remains at rest in the accelerating frame, i.e. it accelerates with the surface. If it slips, then it can remain at rest in the inertial frame. This is the idea behind the trick of pulling a table cloth off a table, leaving the plates, glasses, and utensils on the table.
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 φeq 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, Freal = T + mg. In the non-inertial frame of the rail car we add the fictitious force due to the acceleration of the car, Finertial = -mA, so the equation of motion reads:
Both the fictitious force and gravitational force are proportional to the mass, so the sum can be 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, geff = g - A.
The pendulum will be in equilibrium when it hangs parallel to geff, which is at an angle φeff = arctan(A/g) to the vertical, in the direction away from the acceleration A. The frequency of small oscillations is ω = sqrt(geff/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 + mg = mA. 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 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 -- a difference in acceleration.
Tides aren't experienced in the midwest, but are commonplace along the coasts. They are of considerable importance to people and organisms living near the shores. The tides are a twice daily rising and falling of the level of the ocean and nearby connected water (rivers, bays, sounds, and so on). For a long time it's been understood that the tides are somehow connected to the moon, but a proper understanding eluded people until Newton.
To understand the origin of the tides, imagine the earth as a solid sphere covered by a layer of water to represent the oceans, with the moon in a fixed location. In the absence of the moon, the oceans would form into a uniform layer around the earth. The presence of the moon produces a distortion to the oceans so that the depth is not uniform all the way around. You might imagine that the ocean would be pulled by the moon so that it bulged out on the side of the earth facing the moon. That is incorrect.
The correct explanation requires that we evaluate the combined effect of the earth's and the moon's gravity.
Please read this section.
We denote the magnitude of the angular velocity of an object by ω. For the purposes of this chapter, we can assume that an object is rotating about a fixed axis, whose direction is given by a unit vector we can call û. To express both the magnitude of the angular velocity and the direction of the axis of rotation, we can combine these quantities into an angular velocity vector,
Assume we have a system with fixed point O, and an angular velocity ω. What is the velocity of a point located a distance r from O? The rotation of the system means that the point moves in a circle around the axis, and the radius of the circle is r⊥ = |r| sinθ where θ is the angle between r and ω. The velocity of an object moving around a circle of radius r⊥ with angular velocity ω is v = ωr⊥, and the direction is tangential to the circle (perpendicular to r and ω). Using the expression for r⊥, and the relation ω×r = |r| ω sinθ, we can write
Finally, since the velocity vector is perpendicular to r and ω, we can write
This relation is more general, and works for any vector and its derivative:
Relative angular velocity vectors add just like relative velocity vectors. That is, if we know that the velocity of object 2 with respect to object 1 is v21, and that the velocity of object 3 with respect to object 2 is v32, then the velocity of object 3 with respect to object 1 is
Now use the relation derived above to show how angular velocities add.
therefore
So relative angular velocities add just like relative velocities.
I'll denote the angular velocity of the coordinate system (its rotation about the origin) by Ω.