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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?

Short Answer

Expert verified
The centrifugal force causes outward radial acceleration; Coriolis has no effect on the rod.

Step by step solution

01

Understand the Setup

The problem involves a bead on a frictionless rod rotating about an origin at a constant angular velocity \( \Omega \). This setup implies that the only forces acting on the bead are the centrifugal and Coriolis forces because the rod is frictionless.
02

Define the Rotating Frame of Reference

We need to examine the motion in a rotating frame where \( x \) is along the rod and \( y \) is perpendicular to the rod. The rod rotates with a constant angular velocity \( \Omega \).
03

Identify the Forces in the Rotating Frame

In the rotating frame, two fictitious forces act on the bead: the centrifugal force directed radially outward and dependent on \( x \), and the Coriolis force, which is proportional to the velocity of the bead. The centrifugal force is \( F_{centrifugal} = m \Omega^2 x \) and the Coriolis force is \( F_{Coriolis} = -2m\Omega\dot{y} \).
04

Write Down the Equation of Motion

The net force along \( x \) from these fictitious forces can be set equal to \( m\ddot{x} \), the mass times the acceleration of the bead. Applying Newton's second law, we have \[ m \ddot{x} = m \Omega^2 x - 2m\Omega \dot{y} \].
05

Simplify and Solve the Equation

Assume that \( y = 0 \), since the bead is constrained to move on the rod (the \( x \)-axis). Hence \( \dot{y} = 0 \), simplifying our equation to \[ m \ddot{x} = m \Omega^2 x \]. This is a second-order differential equation: \( \ddot{x} = \Omega^2 x \).
06

Solve the Differential Equation

The general solution to \( \ddot{x} = \Omega^2 x \) is \( x(t) = Ae^{\Omega t} + Be^{-\Omega t} \), where \( A \) and \( B \) are constants determined by initial conditions of the problem.
07

Interpret the Roles of Forces

The centrifugal force provides the central term in the differential equation, indicating it causes the bead to experience an acceleration outward. The Coriolis force terms disappear due to the assumption \( y = 0 \), meaning it has no effect along the rod axis in this specific setup.

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Key Concepts

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

Centrifugal Force
In a rotating reference frame, the centrifugal force appears to act on objects moving in a circular path around a central point. This force isn't a real force but rather a perceived effect due to the rotation of the frame. It acts radially outward, away from the axis of rotation. In the context of our exercise, the centrifugal force is expressed as \( F_{\text{centrifugal}} = m \Omega^2 x \), where \( m \) is the mass of the bead, \( \Omega \) is the angular velocity of the rod, and \( x \) is the radial distance of the bead from the center of rotation.
The role of centrifugal force is crucial in determining how the bead behaves on the rod. Because it acts outward, it influences the bead's potential motion. In the solution, centrifugal force forms the core of the differential equation that describes the bead's trajectory. It causes an acceleration which is factored into the bead's equation of motion, effectively pulling the bead away from the pivot as the rod rotates.
Coriolis Force
The Coriolis force is another fictitious force that comes into play within a rotating reference frame. Just like the centrifugal force, it doesn't exist in inertial frames and arises due to the rotation of the reference frame. The Coriolis force is perpendicular to the object's velocity and the axis of rotation, mathematically represented as \( F_{\text{Coriolis}} = -2m\Omega \dot{y} \), where \( \dot{y} \) is the velocity component perpendicular to the radial direction.
In many practical scenarios, it causes moving objects to deflect from their path. However, in our exercise's equation of motion, the term \( \dot{y} \) is zero since we assume the bead moves along the rod without any perpendicular velocity component. This removes the Coriolis effect from the equation. It highlights the importance of considering initial conditions and motion constraints to understand the influence of forces in rotating frames.
Angular Velocity
Angular velocity \( \Omega \) describes how fast an object rotates around a central point. It is a vector quantity, with both magnitude (the rotational speed) and direction (the axis of rotation). In our exercise, the rod spins at a constant angular velocity, impacting both centrifugal and Coriolis forces present in the rotating frame.
Angular velocity is pivotal in analyzing the bead's motion. It directly influences the magnitude of the centrifugal force as it squares in the expression \( m\Omega^2 x \). This means that as the rod rotates faster, the outward force on the bead increases, leading to a potentially larger deflection if not stopped by some constraint. Angular velocity also features in the Coriolis force term, but its effect vanishes due to the bead's constrained motion along the rod, demonstrated by \( \dot{y} = 0 \).
  • Angular velocity affects how strongly forces act in rotating systems.
  • It is crucial in calculating fictitious forces in non-inertial frames.
  • Understanding it helps in predicting motion within rotating systems.

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Most popular questions from this chapter

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.

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 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\%.

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?]

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