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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.

Short Answer

Expert verified
From the ground, the puck moves straight; from the merry-go-round, it curves due to fictitious forces.

Step by step solution

01

Define the Problem

The problem involves two scenarios where a puck is released from rest: (a) by a person standing on a rotating merry-go-round and (b) by a stationary observer standing on the ground. We need to analyze the puck's motion from two reference frames: a stationary reference frame fixed to the ground and a rotating frame attached to the merry-go-round.
02

Analyze from the Stationary Frame (Ground Observer)

In scenario (a), when the puck is released by the person on the merry-go-round, it moves along a straight line path in the inertial frame fixed to the ground. This is because once released, there are no forces acting on the puck other than gravity, which does not affect the horizontal motion. Therefore, the puck will continue in motion due to its inertia, starting from the point of release along a straight line, with no curves, in the direction it was given initially. For scenario (b), releasing the puck from the ground follows the same inertial logic - it just moves in a straight line along its initial direction of release.
03

Analyze from the Rotating Frame (Merry-go-round Observer)

In scenario (a), viewed by the person on the merry-go-round, the puck appears to follow a curved path due to the presence of fictitious forces. This is because, in a rotating reference frame, the puck is subjected to apparent forces: the centrifugal force, which seems to push it radially outward, and the Coriolis force, which deflects the path to the right in the direction of rotation. The observer sees the puck move in a path that spirals outward in a curved trajectory. In scenario (b), once released by a ground observer, the puck still appears to move in a curved path due to the Coriolis and centrifugal forces. However, since the initial release involves no rotation relative to the ground, the centrifugal component would dominate, making it seem to spiral outward from the merry-go-round observer's perspective.
04

Summarize Both Scenarios

In both scenarios, the puck appears to move in straight lines from the ground observer's frame, maintaining its initial velocity vector after release. From the perspective of the rotating observer on the merry-go-round, the puck appears to follow a curved path due to fictitious forces that act to deflect and push the puck radially outward due to the merry-go-round’s rotation.

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Key Concepts

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

Inertial Reference Frame
An inertial reference frame is a viewpoint where Newton's laws of motion are valid. This means objects in this frame either stay at rest or move at a constant speed in a straight line unless acted upon by an external force.
In the merry-go-round exercise, an observer on the ground, who is not rotating, represents an inertial reference frame.
When the puck is released from a standstill on the merry-go-round, from this observer's perspective, it should continue moving in a straight line.
This straight-line motion occurs because the puck's initial movement meets no additional forces that alter its uniform inertia-driven path.
  • This frame adheres strictly to Newton's first law.
  • Gravity affects only vertical motion, which is not part of the horizontal analysis here.
Non-Inertial Reference Frame
In a non-inertial reference frame, things get a little more complicated. Here, Newton's laws seem to break because this frame itself is accelerating or rotating. To explain motion in such frames, fictitious forces must be introduced.
The merry-go-round serves as a non-inertial reference frame. For an observer riding the merry-go-round, everything seems different.
As the puck is released, it appears to take a curved path. Despite no physical force acting on it, this happens due to the rotating motion of the frame itself.
  • Fictitious forces, like the Coriolis force and centrifugal force, help explain these observations.
  • The frame is not stationary; hence perceived movements differ from reality.
Coriolis Force
The Coriolis force is an apparent force experienced in rotating frames, influencing the trajectory of moving objects. It acts at right angles to the direction of motion and the axis of rotation.
On the merry-go-round, as the puck moves, the Coriolis force causes it to seem as though it's deflected to the right (for a counterclockwise rotation).
This creates a curved path.
  • It's not a real force, but a perceived one in rotating frames.
  • Works in tandem with the centrifugal force to alter the path of motion.
  • This force becomes noticeable when viewing from the merry-go-round, affecting how and where the puck 'appears' to go.
Centrifugal Force
Centrifugal force acts outwardly in a rotating system and is also a fictitious force. When you’re on a merry-go-round, it feels like you're being pushed outward away from the center.
The puck, when seen from this vantage point, seems to spiral outwards due to this force. However, it's essential to understand that this force arises solely from the rotating perspective.
  • From a stationary frame, no outward force on the puck exists.
  • This force gives an intuitive feel of what's happening in a rotating frame.
  • Though it helps explain motion, centrifugal force doesn't appear in the inertial frame analysis.

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Most popular questions from this chapter

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.)

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 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 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.

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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