Find the equations of motion for a particle in a frame rotating with variable angular velocity \(\boldsymbol{\omega}\), and show that there is another apparent force of the form \(-m \dot{\boldsymbol{\omega}} \wedge \boldsymbol{r}\). Discuss the physical origin of this force.

Short Answer

Expert verified
Based on the steps provided, provide a short answer: To find the equations of motion for a particle in a frame rotating with variable angular velocity, we consider the kinematics of a rotating frame and differentiate the position vector of a point in the rotating frame to find velocity and acceleration. We then calculate the centrifugal force term, the Coriolis force term, and the additional force connected with variable angular velocity. Combining all these forces, we obtain the equations of motion for a particle in a rotating frame. The additional force arising from the variation of angular velocity, similar to the Euler force, arises due to the time-dependent rotation of the frame and plays an important role in understanding the motion of objects in rotating frames of reference.

Step by step solution

01

Review the kinematics of a rotating frame

We know that if a particle is moving in an inertial frame of reference, it will obey Newton's second law: \(\boldsymbol{F}=m\boldsymbol{a}\). To describe the motion of a particle in a rotating frame of reference, we need to find the relationship between the linear acceleration in the inertial frame \(d^2\boldsymbol{r}/dt^2\) and the acceleration in the rotating frame \(\boldsymbol{a}\).
02

Find the total derivative of the position vector in the rotating frame

Consider a point P with position vector \(\boldsymbol{r}\). Its position vector in the rotating frame is given as \(\boldsymbol{r'}=\boldsymbol{r}-\boldsymbol{R}(t)\). We can differentiate \(\boldsymbol{r'}\) to find the velocity and acceleration in the rotating frame: $$ \frac{d\boldsymbol{r'}}{dt} = \frac{d\boldsymbol{r}}{dt} - \frac{d\boldsymbol{R}}{dt} = \boldsymbol{v} - \dot{\boldsymbol{R}}, $$ $$\boldsymbol{a'} = \frac{d^2\boldsymbol{r'}}{dt^2} = \boldsymbol{a} - \ddot{\boldsymbol{R}}. $$
03

Calculate the centrifugal force term and the Coriolis force term

Now we can calculate the centrifugal force term which is given by the cross product \(-m(2 \boldsymbol{\omega} \times \boldsymbol{v'}+\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\boldsymbol{r'}))\). Next, we need to calculate the Coriolis force term which is given by the cross product \(-2m(\boldsymbol{\omega}\times\boldsymbol{v'})\).
04

Calculate the additional force connected with variable angular velocity

We are now ready to calculate the additional force connected with variable angular velocity, as given by the term \(-m\dot{\boldsymbol{\omega}} \wedge \boldsymbol{r}\). To do this, we simply need to compute the cross product: $$ -m \dot{\boldsymbol{\omega}} \wedge \boldsymbol{r'} = -m \frac{d \boldsymbol{\omega}}{dt} \wedge (\boldsymbol{r} - \boldsymbol{R}(t)) $$
05

Combine all forces to get the final equations of motion

Adding all the relevant forces components (actual force, centrifugal force, Coriolis force, and additional force), we have a total apparent force in the rotating frame: $$ m\boldsymbol{a'} = \boldsymbol{F} - m(2 \boldsymbol{\omega} \times \boldsymbol{v'} + \boldsymbol{\omega}\times(\boldsymbol{\omega}\times\boldsymbol{r'})) - 2m(\boldsymbol{\omega} \times \boldsymbol{v'}) - m \frac{d\boldsymbol{\omega}}{dt} \wedge (\boldsymbol{r} - \boldsymbol{R}(t)) $$ Dividing by m gives us the equations of motion for a particle in a rotating frame of reference: $$ \boldsymbol{a'} = \frac{\boldsymbol{F}}{m} - (2 \boldsymbol{\omega} \times \boldsymbol{v'} + \boldsymbol{\omega}\times(\boldsymbol{\omega}\times\boldsymbol{r'})) - 2(\boldsymbol{\omega} \times \boldsymbol{v'}) - \frac{d\boldsymbol{\omega}}{dt} \wedge (\boldsymbol{r} - \boldsymbol{R}(t)) $$
06

Discuss the physical origin of the additional force

The additional force \(-m\dot{\boldsymbol{\omega}}\wedge\boldsymbol{r}\) is connected with the variation of the angular velocity, and it is similar to the Euler force found in classical mechanics. This force can be described as a pseudo force that arises in the rotating frame due to the time-dependent rotation of the frame itself. A common manifestation of this force can be seen in the case of spinning objects, where the Euler force leads to complex movement like precession and nutation. This force, along with centrifugal and Coriolis forces, helps us to understand the motion of objects in rotating frames of reference.

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Key Concepts

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

Centrifugal Force
If you've ever been on a merry-go-round, you might have felt like you were being pushed outward as it spun. That sensation is due to what we refer to as the centrifugal force. It's important to note that centrifugal force is not a 'real' force in physics, but rather an apparent force that is felt by an object in a rotating frame of reference. It arises because of inertia, the property that makes objects resist changes in their state of motion. In a rotating reference frame, this force appears as a reaction to the centripetal force that is necessary to keep an object moving in a circle. The formula for this force is \( -m\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\boldsymbol{r'}) \) where \( \boldsymbol{r'} \) is the position in the rotating frame and \( \boldsymbol{\omega} \) is the frame's angular velocity.

In our exercise, the centrifugal force is one of the components that need to be considered to understand the equations of motion for a particle in a rotating frame.
Coriolis Force
Now let's move on to another fascinating apparent force: the Coriolis force. This effect is observed in rotating systems and can be quite counterintuitive. It is responsible for deflections in trajectories of objects in motion within a rotating frame. For instance, Earth's rotation causes wind patterns to curve, which is a significant factor in meteorology. The Coriolis force is given by \( -2m(\boldsymbol{\omega}\times\boldsymbol{v'}) \) where \( \boldsymbol{v'} \) is the velocity relative to the rotating frame.

In our problem, the Coriolis force contributes to the total apparent force on a particle in a rotating frame when considering its motion equations. The particle's path appears curved as this force acts perpendicular to its velocity.
Euler Force
The Euler force, often less discussed than the Coriolis or centrifugal forces, is nonetheless an essential part of the dynamics in non-uniformly rotating systems. It appears in response to changes in the rotation rate of the frame, specifically when the angular velocity of the frame is not constant. In the context of our exercise, we find the additional force term \( -m \dot{\boldsymbol{\omega}} \wedge \boldsymbol{r'} \) that stems from a varying angular velocity, \( \dot{\boldsymbol{\omega}} \). This force, akin to the Euler force in classical mechanics, manifests during time-dependent rotational movements, leading to effects like precession or nutation in spinning objects.
Angular Velocity
Angular velocity, denoted by \( \boldsymbol{\omega} \), is a vector quantity that represents how fast an object rotates or revolves relative to another point, usually the center of a circle or another axis. In a sense, it's the rotational equivalent of linear velocity. In our exercise, we deal with a frame that is rotating with a variable angular velocity, meaning \( \boldsymbol{\omega} \) changes with time. This change contributes to the Euler force and must be included in the equations of motion for the particle. Understanding angular velocity is crucial because it directly influences the magnitude and direction of the centrifugal, Coriolis, and Euler forces in a rotating system.

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Most popular questions from this chapter

Another way of deriving the equation of motion (5.16) is to use Lagrange's equations. Express the kinetic energy \(\frac{1}{2} m(\mathrm{~d} \boldsymbol{r} / \mathrm{d} t)^{2}\) in terms of \((x, y, z)\), and show that Lagrange's equations (3.44) reproduce (5.16) for the case where the force is conservative.

An aircraft is flying at \(800 \mathrm{~km} \mathrm{~h}^{-1}\) in latitude \(55^{\circ} \mathrm{N}\). Find the angle through which it must tilt its wings to compensate for the horizontal component of the Coriolis force.

Find the velocity relative to an inertial frame (in which the centre of the Earth is at rest) of a point on the Earth's equator. An aircraft is flying above the equator at \(1000 \mathrm{~km} \mathrm{~h}^{-1}\). Assuming that it flies straight and level (i.e., at a constant altitude above the surface) what is its velocity relative to the inertial frame (a) if it flies north, (b) if it flies west, and (c) if it flies east?

Find the centrifugal acceleration at the equator of the planet Jupiter and of the Sun. In each case, express your answer also as a fraction of the surface gravity. (The rotation periods are 10 hours and 27 days, respectively, the radii \(7.1 \times 10^{4} \mathrm{~km}\) and \(7.0 \times 10^{5} \mathrm{~km}\), and the masses \(1.9 \times 10^{27} \mathrm{~kg}\) and \(\left.2.0 \times 10^{30} \mathrm{~kg} .\right)\)

The angular velocity of the electron in the lowest Bohr orbit of the hydrogen atom is approximately \(4 \times 10^{16} \mathrm{~s}^{-1}\). What is the largest magnetic field which may be regarded as small in this case, in the sense of \(\S 5.5 ?\) Determine the Larmor frequency in a field of \(2 \mathrm{~T}\).

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