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Verify that the positions of two particles can be written in terms of the CM and relative positions as \(\mathbf{r}_{1}=\mathbf{R}+m_{2} \mathbf{r} /\mathbf{M}\) and \(\mathbf{r}_{2}=\mathbf{R}-m_{1} \mathbf{r} / M .\) Hence confirm that the total \(\mathrm{KE}\) of the two particles can be expressed as \(T=\frac{1}{2} \mathbf{M} \dot{\mathbf{R}}^{2}+\frac{1}{2} \mu \dot{\mathbf{r}}^{2},\) where \(\mu\) denotes the reduced mass \(\mu=\mathbf{m}_{1} \mathbf{m}_{2} / \mathbf{M}.\)

Short Answer

Expert verified
The positions and kinetic energy relations are verified, showing that the total KE can be split into CM and relative parts.

Step by step solution

01

Define the Center of Mass and Relative Position

The center of mass (CM) position \( \mathbf{R} \) for two particles is given by the equation: \[ \mathbf{R} = \frac{m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2}{M} \] where \( M = m_1 + m_2 \) is the total mass of the system. The relative position \( \mathbf{r} \) between the two particles is defined as: \[ \mathbf{r} = \mathbf{r}_1 - \mathbf{r}_2 \]
02

Express Individual Positions in terms of CM and Relative Position

We need to express \( \mathbf{r}_1 \) and \( \mathbf{r}_2 \) in terms of \( \mathbf{R} \) and \( \mathbf{r} \). From the definitions in Step 1 and solving for \( \mathbf{r}_1 \) and \( \mathbf{r}_2 \): \[ m_1 \mathbf{r}_1 = m_1 \mathbf{R} + m_1 m_2 \mathbf{r}/M \]\[ m_2 \mathbf{r}_2 = m_2 \mathbf{R} - m_1 m_2 \mathbf{r}/M \]From which,\[ \mathbf{r}_1 = \mathbf{R} + \frac{m_2 \mathbf{r}}{M} \]\[ \mathbf{r}_2 = \mathbf{R} - \frac{m_1 \mathbf{r}}{M} \]
03

Derive Expressions for Velocities

Differentiate the expressions for \( \mathbf{r}_1 \) and \( \mathbf{r}_2 \) with respect to time to get the velocities:\[ \dot{\mathbf{r}}_1 = \dot{\mathbf{R}} + \frac{m_2 \dot{\mathbf{r}}}{M} \]\[ \dot{\mathbf{r}}_2 = \dot{\mathbf{R}} - \frac{m_1 \dot{\mathbf{r}}}{M} \]
04

Calculate Kinetic Energy of Each Particle

Use the velocity expressions to calculate the kinetic energy (KE) of the particles:\[ KE_1 = \frac{1}{2} m_1 \dot{\mathbf{r}}_1^2 = \frac{1}{2} m_1 \left( \dot{\mathbf{R}} + \frac{m_2 \dot{\mathbf{r}}}{M} \right)^2 \]\[ KE_2 = \frac{1}{2} m_2 \dot{\mathbf{r}}_2^2 = \frac{1}{2} m_2 \left( \dot{\mathbf{R}} - \frac{m_1 \dot{\mathbf{r}}}{M} \right)^2 \]
05

Simplify the Total Kinetic Energy Expression

The total kinetic energy (T) is the sum of \( KE_1 \) and \( KE_2 \).Substitute the expression \( M = m_1 + m_2 \) and simplify:\[ T = KE_1 + KE_2 = \frac{1}{2} m_1 \left( \dot{\mathbf{R}}^2 + 2\dot{\mathbf{R}} \frac{m_2 \dot{\mathbf{r}}}{M} + \left( \frac{m_2 \dot{\mathbf{r}}}{M} \right)^2 \right) + \frac{1}{2} m_2 \left( \dot{\mathbf{R}}^2 - 2\dot{\mathbf{R}} \frac{m_1 \dot{\mathbf{r}}}{M} + \left( \frac{m_1 \dot{\mathbf{r}}}{M} \right)^2 \right) \]
06

Combine and Recognize the Reduced Mass Term

Combine like terms and factor expressions:\[ T = \frac{1}{2} M \dot{\mathbf{R}}^2 + \frac{1}{2} \left( \frac{m_1 m_2}{M} \right) \dot{\mathbf{r}}^2 \]Recognize that \( \mu = \frac{m_1 m_2}{M} \), the reduced mass, thus:\[ T = \frac{1}{2} M \dot{\mathbf{R}}^2 + \frac{1}{2} \mu \dot{\mathbf{r}}^2 \]

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Key Concepts

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

Relative Position
The relative position of two particles in a system is a fundamental concept in physics. Imagine two particles moving in space, each with their respective positions \(\mathbf{r}_1\) and \(\mathbf{r}_2\). The relative position \(\mathbf{r}\) is simply defined as the difference between these two positions: \[ \mathbf{r} = \mathbf{r}_1 - \mathbf{r}_2 \] This vector tells us how far one particle is from the other and in which direction. It's a way to look at the system from the perspective of one particle with respect to the other.
  • Helps in understanding the internal configuration of a system of particles.
  • Provides the basis for defining other quantities like relative velocity and relative acceleration.
By using the relative position, we can express each particle's location as a combination of the center of mass (an average position of the entire system) and their separation. Recognizing and using \(\mathbf{r}\) allows for the simplification of complex equations involving multiple particle systems.
Reduced Mass
Reduced mass is an innovative concept that simplifies the study of two-body problems, such as the motion of electrons around a nucleus. When dealing with interactions between two particles with masses \(m_1\) and \(m_2\), the actual system behaves as if it were a single particle with a mass called the reduced mass \(\mu\). It is defined as: \[ \mu = \frac{m_1 m_2}{M} \] where \(M\) is the total mass \(M = m_1 + m_2\). This concept is particularly useful because it allows us to transform a complicated two-body problem into a simpler single-body problem.
  • The reduced mass appears in the equations of motion and energy conservation for two-body systems.
  • It provides insights into how forces and energy are distributed between interacting objects.
By using the reduced mass, physicists and engineers can solve problems like orbital mechanics and molecular dynamics more efficiently, gaining clear insights into the system's behavior without getting bogged down by complicated interactions between the two masses.
Kinetic Energy
Kinetic energy (KE) is a measure of how much energy an object has due to its motion. In the case of two particles, each one contributes to the total kinetic energy of the system. For each particle, the kinetic energy is given by: \[ KE = \frac{1}{2} m v^2 \] where \(m\) is the mass of the particle and \(v\) its velocity. When considering a system of two particles, we derive an expression for the total kinetic energy \(T\) by considering the center of mass and relative velocities: \[ T = \frac{1}{2} M \dot{\mathbf{R}}^2 + \frac{1}{2} \mu \dot{\mathbf{r}}^2 \]
  • The first term represents the kinetic energy due to the motion of the system's center of mass.
  • The second term captures the energy contribution from the relative motion between the particles, involving the reduced mass.
This division into two parts makes it easier to analyze systems with multiple particles by separating external movement from internal dynamics. Understanding these components helps in studying dynamics in mechanics, chemistry, and even quantum physics.

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Most popular questions from this chapter

An earth satellite is observed at perigee to be \(250 \mathrm{km}\) above the earth's surface and traveling at about \(8500 \mathrm{m} / \mathrm{s}\). Find the eccentricity of its orbit and its height above the earth at apogee. [Hint: The earth's radius is \(R_{e} \approx 6.4 \times 10^{6} \mathrm{m} .\) You will also need to know \(G M_{\mathrm{e}},\) but you can find this if you remember that \(\left.G M_{\mathrm{e}} / R_{\mathrm{e}}^{2}=g .\right]\)

What would become of the earth's orbit (which you may consider to be a circle) if half of the sun's mass were suddenly to disappear? Would the earth remain bound to the sun? [Hints: Consider what happens to the earth's KE and PE at the moment of the great disappearance. The virial theorem for the circular orbit (Problem 4.41) helps with this one.] Treat the sun (or what remains of it) as fixed.

Consider two particles of equal masses, \(m_{1}=m_{2},\) attached to each other by a light straight spring (force constant \(k\), natural length \(L\) ) and free to slide over a frictionless horizontal table. (a) Write down the Lagrangian in terms of the coordinates \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\), and rewrite it in terms of the \(\mathrm{CM}\) and relative positions, \(\mathbf{R}\) and \(\mathbf{r},\) using polar coordinates \((r, \phi)\) for \(\mathbf{r} .\) (b) Write down and solve the Lagrange equations for the CM coordinates \(X, Y\). (c) Write down the Lagrange equations for \(r\) and A. Solve these for the two special cases that \(r\) remains constant and that \(\phi\) remains constant. Describe (i) the contant the corresponding motions. In particular, show that the frequency of oscillations in the second case is \(\omega=\sqrt{2 k / m_{1}}.\)

Suppose that we decide to send a spacecraft to Neptune, using the simple transfer described in Example 8.6 (page 318). The craft starts in a circular orbit close to the earth (radius 1 AU or astronomical unit) and is to end up in a circular orbit near Neptune (radius about \(30 \mathrm{AU}\) ). Use Kepler's third law to show that the transfer will take about 31 years. (In practice we can do a lot better than this by arranging that the craft gets a gravitational boost as it passes Jupiter.)

Consider two particles interacting by a Hooke's law potential energy, \(U=\frac{1}{2} k r^{2},\) where \(\mathbf{r}\) is their relative position \(\mathbf{r}=\mathbf{r}_{1}-\mathbf{r}_{2},\) and subject to no external forces. Show that \(\mathbf{r}(t)\) describes an ellipse. Hence show that both particles move on similar ellipses around their common CM. [This is surprisingly awkward. Perhaps the simplest procedure is to choose the \(x y\) plane as the plane of the orbit and then solve the equation of motion (8.15) for \(x\) and \(y .\) Your solution will have the form \(x=A \cos \omega t+B \sin \omega t,\) with a similar expression for \(y .\) If you solve these for sin \(\omega t\) and \(\cos \omega t\) and remember that \(\sin ^{2}+\cos ^{2}=1,\) you can put the orbital equation in the form \(a x^{2}+2 b x y+c y^{2}=k\) where \(k\) is a positive constant. Now invoke the standard result that if \(a\) and \(c\) are positive and \(a c>b^{2}\) this equation defines an ellipse.]

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