In Taylor, read sections 9.1 to 9.2 for today, and 9.3 to 9.5 for Friday.
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 care 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 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, 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, opposite in direction 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.
© 2006 Robert Harr