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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.

Short Answer

Expert verified
The torque due to the Coriolis force is \( m \omega \Omega r^2 \sin \theta \) westward.

Step by step solution

01

Understanding the Coriolis Force

The Coriolis force is an inertial force that acts on objects moving within a rotating reference frame. It is defined by the expression: \( F_c = -2m (\vec{\Omega} \times \vec{v}) \), where \(\vec{\Omega}\) is the angular velocity of the Earth's rotation and \(\vec{v}\) is the velocity of the moving object.
02

Calculate Velocity Vector

As the hoop spins with an angular velocity \( \omega \), a point on the hoop at a distance \( r \) from the center moves with velocity \( \vec{v} = \vec{r} \times \vec{\omega} \). In simplified form, due to circular motion in a horizontal plane, the velocity of any point on the hoop is given by \( v = r\omega \).
03

Identify the Coriolis Force

Substitute the velocity into the Coriolis force equation. We have \( F_c = -2m (\vec{\Omega} \times \vec{v}) \). The cross product yields a force component perpendicular to both the velocity \( \vec{v} \) and the Earth's rotation vector \( \vec{\Omega} \).
04

Simplify the Cross Product

Given that \( \vec{\Omega} \) is the Earth's rotational angular velocity vector pointing north and \( \vec{v} \) is directed tangentially due to the hoop's rotation, their cross product in magnitude becomes \( |\vec{\Omega} \times \vec{v}| = \Omega v \sin \theta = \Omega r \omega \sin \theta \).
05

Determine Torque Direction and Magnitude

The Coriolis force acts in a direction causing a torque around the vertical axis of the hoop. The magnitude of this torque due to a tangential force is given by \( \tau = rF_c = m \omega \Omega r^2 \sin \theta \). The direction of the torque is such that it is directed toward the west due to the right-hand rule and the direction of rotation.

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Key Concepts

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

Coriolis effect
The Coriolis effect is a fascinating phenomenon that occurs in rotating systems, like our very own Earth. It arises because of the Coriolis force, which is an inertial force acting on objects in motion within a rotating reference frame. Imagine you are on a merry-go-round and you toss a ball to a friend sitting on the opposite side. Even though you aim straight, your friend sees the ball curve. This is due to the Coriolis effect.

On Earth, as the planet rotates, this effect influences the path of moving objects like air currents, ocean currents, and even tiny particles. This force doesn't change the speed of the object but alters its direction of motion. The Coriolis force is mathematically described by the equation: \[F_c = -2m (\vec{\Omega} \times \vec{v})\], where \(\vec{\Omega}\) is the angular velocity vector of Earth, and \(\vec{v}\) is the velocity of the moving object.

In our exercise, this effect helps create a torque on a spinning hoop, illustrating how the Coriolis force causes rotation itself to produce additional rotational effects.
torque
Torque is a measure of the force that causes an object to rotate. You can think of it as the rotational equivalent of linear force. Imagine turning a wrench: when you apply force at a handle's end, you exert torque, causing the bolt to turn. The torque's magnitude depends on the force applied, the distance from the axis of rotation, and the angle at which the force is applied.

In mathematical terms, torque \( \tau \) is calculated by: \[\tau = rF \sin \theta\], where \(r\) is the distance from the pivot point, \(F\) is the applied force, and \(\theta\) is the angle the force makes with the line from the pivot.

In the context of our spinning hoop, the Coriolis force creates a torque that is proportional to the mass \(m\), the hoop's angular velocity \(\omega\), Earth's angular velocity \(\Omega\), and the radius squared \(r^2\), specifically along the hoop's rotation direction. This is the basis of the gyrocompass, guiding vessels by aligning them with Earth's axis rotation.
angular velocity
Angular velocity is a vector quantity that describes how fast something rotates and in which direction. It provides both the rate of rotation and the axis around which the object rotates. Imagine watching a fan spin: the blades rotate around a fixed point at a specific speed, which is quantified by the angular velocity.

Mathematically, angular velocity \( \omega \) is expressed in radians per second. It can be derived using \( \omega = \frac{d\theta}{dt}\), where \(\theta\) is the angle the object has rotated, and \(t\) is time.

In our spinning hoop problem, the hoop rotates with angular velocity \(\omega\) about a vertical axis. This spin is subject to another angular velocity, \(\Omega\), from Earth's rotation. Thus, multiple angular velocities interact, contributing to complex motions due to the Coriolis effect. Here, the combined rotations result in measurable torques applied to the hoop.
rotating reference frame
A rotating reference frame is a perspective that rotates with the object or system being considered. It's like being on a carousel: you see everything relative to the carousel's rotation, leading to phenomena not present in a non-rotating frame.

Objects in rotating reference frames experience fictitious forces, like the Coriolis force, which appear due to the rotation. These forces are not observed in an inertial, non-rotating frame. It's essential to understand rotating frames for analyzing systems affected by Earth's spin, like weather patterns or gyroscopic devices.

In our exercise, the hoop is in a rotating reference frame because it spins as Earth rotates. The Coriolis force acts in this frame, producing a torque. This frame enables us to understand behaviors that wouldn't be apparent purely by observations made from a static perspective, and are vital for applications spanning from meteorology to navigation and beyond.

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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?]

What are the directions of the centrifugal and Coriolis forces on a person moving (a) south near the North Pole, (b) east on the equator, and (c) south across the equator?

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.

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\%.

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.

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