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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

Consider a frictionless puck on a horizontal turntable that is rotating counterclockwise with angular velocity \(\Omega\). (a) Write down Newton's second law for the coordinates \(x\) and \(y\) of the puck as seen by me standing on the turntable. (Be sure to include the centrifugal and Coriolis forces, but ignore the earth's rotation.) (b) Solve the two equations by the trick of writing \(\eta=x+\) iy and guessing a solution of the form \(\eta=e^{-i \alpha t} .\) [In this case \(-\) as in the case of critically damped SHM discussed in Section \(5.4-\) you get only one solution this way. The other has the same form (5.43) we found for the second solution in damped SHM.] Write down the general solution. (c) At time \(t=0,\) I push the puck from position \(\mathbf{r}_{\mathrm{o}}=\left(x_{\mathrm{o}}, 0\right)\) with velocity \(\mathbf{v}_{\mathrm{o}}=\left(v_{\mathrm{xo}}, v_{\mathrm{yo}}\right)(\) all as measured by me on the turntable). Show that $$\left.\begin{array}{l} x(t)=\left(x_{0}+v_{x} t\right) \cos \Omega t+\left(v_{y o}+\Omega x_{o}\right) t \sin \Omega t \\ y(t)=-\left(x_{0}+v_{x 0} t\right) \sin \Omega t+\left(v_{y o}+\Omega x_{o}\right) t \cos \Omega t \end{array}\right\\}.$$ (d) Describe and sketch the behavior of the puck for large values of \(t .\) [Hint: When \(t\) is large the terms proportional to \(t\) dominate (except in the case that both their coefficients are zero). With \(t\) large, write (9.72) in the form \(x(t)=t\left(B_{1} \cos \Omega t+B_{2} \sin \Omega t\right),\) with a similar expression for \(y(t),\) and use the trick of (5.11) to combine the sine and cosine into a single cosine \(-\) or sine, in the case of \(y(t) .\) By now you can recognize that the path is the same kind of spiral, whatever the initial conditions (with the one exception mentioned).]

The Coriolis force can produce a torque on a spinning object. To illustrate this, consider a horizontal hoop of mass \(m\) and radius \(r\) spinning with angular velocity \(\omega\) about its vertical axis at colatitude \(\theta\). Show that the Coriolis force due to the earth's rotation produces a torque of magnitude \(m \omega \Omega r^{2} \sin \theta\) directed to the west, where \(\Omega\) is the earth's angular velocity. This torque is the basis of the gyrocompass.

I am spinning a bucket of water about its vertical axis with angular velocity \(\Omega\). Show that, once the water has settled in equilibrium (relative to the bucket), its surface will be a parabola. (Use cylindrical polar coordinates and remember that the surface is an equipotential under the combined effects of the gravitational and centrifugal forces.)

A bullet of mass \(m\) is fired with muzzle speed \(v_{\mathrm{o}}\) horizontally and due north from a position at colatitude \(\theta\). Find the direction and magnitude of the Coriolis force in terms of \(m, v_{\mathrm{o}}, \theta,\) and the earth's angular velocity \(\Omega .\) How does the Coriolis force compare with the bullet's weight if \(v_{\mathrm{o}}=1000 \mathrm{m} / \mathrm{s}\) and \(\theta=40\) deg?

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 \((R=38 \mathrm{m}) ?\)

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