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Chapter 9: Problem 14

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.)

Short Answer

Expert verified
The surface of the water forms a parabola described by \\(z = \frac{V_0}{mg} - \frac{1}{2g}\Omega^2r^2\\).

Step by step solution

01

Understanding the Forces

We need to consider two main forces acting on the water in the bucket: the gravitational force and the centrifugal force due to rotation. Gravitational force acts downward, while the centrifugal force acts radially outward in the rotating frame.
02

Choosing the Coordinate System

We use cylindrical polar coordinates \(r, \theta, z\) where \(r\) is the radial distance from the rotation axis, \(\theta\) is the angular coordinate, and \(z\) is the height above a reference plane. The bucket rotates about the vertical axis (z-axis).
03

Expressing the Potential Energy

The potential energy involves both gravitational and centrifugal components. Gravitation potential energy is \(V_g = mgz\) and centrifugal potential energy is given by \(V_c = \frac{1}{2}m\Omega^2r^2\), where \(m\) is the mass of the water element.
04

Setting the Total Potential Energy

Since the surface of the water is an equipotential, the total potential \(V = V_g + V_c = mgz + \frac{1}{2}m\Omega^2r^2\) must be constant for any point \(r, z\) on the surface.
05

Equating the Potential Energy Terms

Set the potential energy to a constant value, \(V_0\). This gives: \(mgz + \frac{1}{2}m\Omega^2r^2 = V_0\). Solving for \(z\), we have \(z = \frac{V_0}{mg} - \frac{1}{2g}\Omega^2r^2\).
06

Identifying the Parabolic Surface

The equation \(z = \text{constant} - \frac{1}{2g}\Omega^2r^2\) represents a parabola. The water's surface, described by this equation, is thus a parabola opening downward along the \(z\)-axis due to the \(-r^2\) term.

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Key Concepts

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

Angular Velocity
Angular velocity is an important concept when dealing with rotation, such as spinning a bucket of water. It describes how fast an object is rotating around an axis. In our case, the rotation occurs around the vertical axis. Angular velocity is denoted by \( \Omega \) and is measured in radians per second. Here are some key points to remember about angular velocity:
  • It is a vector quantity, meaning it has both a magnitude and a direction.
  • The direction is determined by the right-hand rule, pointing along the axis of rotation.
  • It describes the rate at which an angle is swept out by a rotating object.
In cylindrical coordinates, angular velocity becomes extremely useful. It affects how forces, such as gravitational and centrifugal forces, interact to shape the water's surface. When the bucket rotates steadily, the angular velocity causes the centrifugal force to push water outwards. This interplay results in the water forming a parabolic shape in equilibrium.
Centrifugal Force
The centrifugal force comes into play when an object is in a rotating frame, like our bucket of water. It acts perpendicular to the axis of rotation, pointing outward. This force can be thought of as the inertia of an object moving in a curve, trying to "escape" the curve.Consider the following about centrifugal force:
  • It is proportional to the square of the angular velocity: \( F_c = m \Omega^2 r \), where \( r \) is the radial distance from the center.
  • It is a fictitious force experienced in a non-inertial frame鈥攁ppears due to rotation rather than any physical interaction.
  • In our rotating bucket scenario, this force pushes the water outwards, affecting the balance between the gravitational pull and the centrifugal push.
When water in the bucket spins, the centrifugal force increases with the square of the distance from the center. In equilibrium, this force acts alongside gravity to help shape the surface of the water, forming a parabola in cylindrical coordinates. The interplay of centrifugal force and angular velocity reveals how dynamic and fascinating rotations can be.
Gravitational Potential Energy
Gravitational potential energy is fundamental to understanding how systems like our spinning bucket interact with the surrounding forces. It is the energy an object has due to its position in a gravitational field, which in most cases, is Earth鈥檚 field.Gravitational potential energy is described by:
  • Formula: \( V_g = mgh \), where \( m \) is the mass, \( g \) is the gravitational acceleration, and \( h \) is the height.
  • It acts downward and affects how the shape of the liquid surface aligns with gravity.
  • In a gravitational field, this energy seeks to minimize itself, contributing to equilibrium conditions.
In the context of the rotating bucket, gravitational potential energy combines with centrifugal energy to determine the surface of the water. As water settles, the balance between these two types of potential energy dictates its final shape鈥攁 downward-opening parabola resting along its equipotential. By understanding how gravitational potential energy interacts with forces in rotation, one can appreciate how equilibrium is achieved in a rotating system.

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Most popular questions from this chapter

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 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?

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?

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.
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