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v9.34 \star\star\star At a point \(P\) on the earth's surface, an enormous perfectly flat and frictionless platform is built. The platform is exactly horizontal \(-\) that is, perpendicular to the local free-fall acceleration \(\mathbf{g}_{P}\). Find the equation of motion for a puck sliding on the platform and show that it has the same form as (9.61) for the Foucault pendulum except that the pendulum's length \(L\) is replaced by the earth's radius R. What is the frequency of the puck's oscillations and what is that of its Foucault precession? [Hints: Write the puck's position vector, relative to the earth's center \(O\) as \(\mathbf{R}+\mathbf{r},\) where \(\mathbf{R}\) is the position of the point \(P\) and \(\mathbf{r}=(x, y, 0)\) is the puck's position relative to \(P .\) The contribution to the centrifugal force involving \(\mathbf{R}\) can be absorbed into \(\mathbf{g}_{P}\) and the contribution involving \(\mathbf{r}\) is negligible. The restoring force comes from the variation of \(g\) as the puck moves. \(J\) To check the validity of your approximations, compare the approximate size of the gravitational restoring force, the Coriolis force, and the neglected term \(m(\boldsymbol{\Omega} \times \mathbf{r}) \times \boldsymbol{\Omega}\) in the centrifugal force.

Short Answer

Expert verified
The equation of motion matches the Foucault pendulum form, oscillation frequency is \( \omega = \sqrt{g/R} \), and precession frequency is \( \Omega \sin(\lambda) \).

Step by step solution

01

Setup the Problem

The problem involves analyzing the motion of a puck on a frictionless platform. The platform is horizontal with respect to the local gravity vector \( \mathbf{g}_P \). We begin by defining the puck's position vector \( \mathbf{R} + \mathbf{r} \), where \( \mathbf{R} \) is the earth's radius vector to the point \( P \) and \( \mathbf{r} = (x, y, 0) \) is the puck's position relative to \( P \).
02

Understand Gravitational Force Changes

The gravitational restoring force can be derived from changes in \( g \) as the puck moves. Since the platform is horizontal, we consider only the component of gravity for the dynamics in \( xy \)-plane, causing a restoring force aligned with these axes.
03

Apply Gravitational Force Approximation

Given the small displacement \( \mathbf{r} \), the variation of gravity \( g \) can be approximated assuming small horizontal shifts treat \( \mathbf{g}_P \) as uniform. We use Hooke's Law-like restoring force: \( \mathbf{F} = -m \mathbf{g}_P \frac{\mathbf{r}}{R} \).
04

Incorporate Coriolis Force

Coriolis force arises due to Earth's rotation. The force is \( -2m(\boldsymbol{\Omega} \times \dot{\mathbf{r}}) \), where \( \boldsymbol{\Omega} \) is Earth's angular velocity. With \( \mathbf{r} \) being horizontal, the force affects the puck's horizontal motion.
05

Final Equation of Motion

The equation of motion is derived by setting the sum of the forces (gravitational, Coriolis) equal to mass times acceleration: \[ m \ddot{\mathbf{r}} = -m \mathbf{g}_P \frac{\mathbf{r}}{R} - 2m(\boldsymbol{\Omega} \times \dot{\mathbf{r}}) \] This equation is of the same form as the Foucault pendulum's equation of motion with \( R \) replacing \( L \).
06

Determine Frequencies

The frequency of the puck's oscillations, \( \omega = \sqrt{g/R} \), comes from the restoring force term. The Foucault precession frequency is given by \( \Omega \sin(\lambda) \), where \( \lambda \) is the latitude of point \( P \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Foucault pendulum
The Foucault pendulum is a classic demonstration of Earth's rotation and illustrates the Coriolis effect. It consists of a heavy pendulum freely swinging in any vertical plane. Due to Earth's rotation, the plane in which the pendulum swings appears to rotate over time. This apparent rotation is due to the Coriolis force acting on the pendulum.

The motion of a Foucault pendulum at any point on Earth can be described by an equation similar to one that governs the puck on a horizontal platform. Just as the puck experiences a combination of gravitational and Coriolis forces, so does the Foucault pendulum. The pendulum's path appears to precess at a rate dependent on the latitude of its location.

For the Foucault pendulum, the precession frequency can be observed at any point other than the equator. At the poles, the pendulum completes a full rotation in one day, whereas, at other latitudes, the precession rate is slower, given by the formula:
  • Precession frequency = Earth's angular velocity \(\Omega\) times the sine of the latitude.
The phenomenon is not just a captivating demonstration; it is a precise proof of Earth's rotation that can be calculated and predicted.
Coriolis force
The Coriolis force acts on objects in motion within a rotating reference frame, such as Earth. It is not a real force in the traditional sense, but rather an apparent force that results from Earth's rotation.

When an object moves in the horizontal plane—the same plane in which the Foucault pendulum swings or the puck slides on the platform—Earth's rotation causes it to accelerate slightly to the right or left, depending on the hemisphere. This effect is what causes the Coriolis force.

The Coriolis force is defined as \(-2m(\boldsymbol{\Omega} \times \dot{\mathbf{r}})\), where \(m\) is the mass, \(\boldsymbol{\Omega}\) is Earth's angular velocity, and \(\dot{\mathbf{r}}\) is the velocity of the moving object.

In the context of the puck on a frictionless platform, this force acts sideways to its motion and causes the apparent motion to spiral, contributing to the puck's oscillatory pattern. Understanding this force is crucial in classical mechanics and helps explain not only pendulum movements but also large-scale motions like ocean currents and atmospheric winds on Earth.
Gravitational force
Gravitational force is a fundamental force responsible for the attraction between masses. On Earth, it manifests as the force pulling objects towards the ground, described by the acceleration due to gravity \(g\).

In this exercise, the gravitational force plays a primary role by providing the restoring force that brings the puck back to its initial position. When the puck is displaced, the change in gravity, despite being small, is enough to generate a restoring force that can be described by Hooke's Law for small displacements:
  • Restoring force = \(-m \mathbf{g}_P \frac{\mathbf{r}}{R}\)
Here, \(R\) is Earth's radius, and \(\mathbf{r}\) is the puck's horizontal displacement, approximated as small disturbances do not significantly change \(g\).

This approximation highlights how gravitational force ties into the oscillatory behavior of objects on Earth when they are subject to small displacements, contributing to the harmonic motion observed in both pendulums and the puck model on the hypothetical platform.

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Most popular questions from this chapter

Be sure you understand why a pendulum in equilibrium hanging in a car that is accelerating forward tilts backward, and then consider the following: A helium balloon is anchored by a massless string to the floor of a car that is accelerating forward with acceleration \(A\). Explain clearly why the balloon tends to tilt forward and find its angle of tilt in equilibrium. [Hint: Helium balloons float because of the buoyant Archimedean force, which results from a pressure gradient in the air. What is the relation between the directions of the gravitational field and the buoyant force?]

I am standing (wearing crampons) on a perfectly frictionless flat merry-go- round, which is rotating counterclockwise with angular velocity \(\Omega\) about its vertical axis. (a) I am holding a puck at rest just above the floor (of the merry-go-round) and release it. Describe the puck's path as seen from above by an observer who is looking down from a nearby tower (fixed to the ground) and also as seen by me on the merry-go-round. In the latter case explain what I see in terms of the centrifugal and Coriolis forces. (b) Answer the same questions for a puck which is released from rest by a long-armed spectator who is standing on the ground leaning over the merry-go-round.

A donut-shaped space station (outer radius \(R\) ) arranges for artificial gravity by spinning on the axis of the donut with angular velocity \(\omega .\) Sketch the forces on, and accelerations of, an astronaut standing in the station (a) as seen from an inertial frame outside the station and (b) as seen in the astronaut's personal rest frame (which has a centripetal acceleration \(A=\omega^{2} R\) as seen in the inertial frame). What angular velocity is needed if \(R=40\) meters and the apparent gravity is to equal the usual value of about \(10 \mathrm{m} / \mathrm{s}^{2} ?\) (c) What is the percentage difference between the perceived \(g\) at a six-foot astronaut's feet \((R=40 \mathrm{m})\) and at his head \((R=38 \mathrm{m}) ?\)

The center of a long frictionless rod is pivoted at the origin and the rod is forced to rotate at a constant angular velocity \(\Omega\) in a horizontal plane. Write down the equation of motion for a bead that is threaded on the rod, using the coordinates \(x\) and \(y\) of a frame that rotates with the rod (with \(x\) along the rod and \(y\) perpendicular to it). Solve for \(x(t)\). What is the role of the centrifugal force? What of the Coriolis force?

The Compton generator is a beautiful demonstration of the Coriolis force due to the earth's rotation, invented by the American physicist A. H. Compton (1892-1962, best known as author of the Compton effect) while he was still an undergraduate. A narrow glass tube in the shape of a torus or ring (radius \(R\) of the ring \(\gg\) radius of the tube) is filled with water, plus some dust particles to let one see any motion of the water. The ring and water are initially stationary and horizontal, but the ring is then spun through \(180^{\circ}\) about its east-west diameter. Explain why this should cause the water to move around the tube. Show that the speed of the water just after the \(180^{\circ}\) turn should be \(2 \Omega R \cos \theta,\) where \(\Omega\) is the earth's angular velocity, and \(\theta\) is the colatitude of the experiment. What would this speed be if \(R \approx 1 \mathrm{m}\) and \(\theta=40^{\circ} ?\) Compton measured this speed with a microscope and got agreement within 3\%.

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