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Find the position of the center of mass of three particles lying in the \(x y\) plane at \(\mathbf{r}_{1}=(1,1,0)\) \(\mathbf{r}_{2}=(1,-1,0),\) and \(\mathbf{r}_{3}=(0,0,0),\) if \(m_{1}=m_{2}\) and \(m_{3}=10 m_{1} .\) Illustrate your answer with a sketch and comment.

Short Answer

Expert verified
The center of mass is at \( \left( \frac{1}{6}, 0, 0 \right) \). It lies on the x-axis, closer to the larger mass at the origin.

Step by step solution

01

Understand the Concept of Center of Mass

The center of mass of a system of particles is the point that moves as if the entire mass of the system were concentrated there and all external forces were applied at that point. For particles in a plane, it can be found using the formula: \( \mathbf{R}_{cm} = \frac{1}{M} \sum m_i \mathbf{r}_i \). Here, \( M \) is the total mass and \( \mathbf{r}_i \) are the position vectors of the particles.
02

Determine the Position Vectors and Masses

The position vectors of the particles are given as \( \mathbf{r}_1 = (1, 1, 0) \), \( \mathbf{r}_2 = (1, -1, 0) \), and \( \mathbf{r}_3 = (0, 0, 0) \). The masses are \( m_1 = m_2 \), and \( m_3 = 10m_1 \). To proceed, let's call \( m_1 = m \) so \( m_2 = m \) and \( m_3 = 10m \).
03

Calculate the Total Mass

Sum up all the masses: \[ M = m_1 + m_2 + m_3 = m + m + 10m = 12m \].
04

Calculate the Center of Mass Position

Using the formula for center of mass: \[ \mathbf{R}_{cm} = \frac{1}{12m} \left( m \begin{pmatrix} 1 \ 1 \ 0 \end{pmatrix} + m \begin{pmatrix} 1 \ -1 \ 0 \end{pmatrix} + 10m \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix} \right) \]Calculate each component separately:- For \( x \): \( x_{cm} = \frac{1}{12m} (m(1) + m(1) + 10m(0)) = \frac{2m}{12m} = \frac{1}{6} \).- For \( y \): \( y_{cm} = \frac{1}{12m} (m(1) + m(-1) + 10m(0)) = \frac{0}{12m} = 0 \).- For \( z \): Since the particles lie in the \( xy \)-plane, \( z_{cm} = 0 \).
05

State the Result

Therefore, the position of the center of mass is \( \mathbf{R}_{cm} = \left( \frac{1}{6}, 0, 0 \right) \), which shows that the center of mass is along the \( x \)-axis, slightly displaced from the origin.
06

Sketch the Arrangement and Center of Mass

Draw a coordinate system. Plot points at (1, 1), (1, -1), and (0, 0) representing the positions of the particles, marking the larger mass at (0, 0). Indicate the center of mass at \( (\frac{1}{6}, 0) \), showing its location along the \( x \)-axis due to the dominance of the mass at the origin.

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Key Concepts

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

Position Vectors
Position vectors play a crucial role in finding the center of mass, especially when dealing with particles in a plane. A position vector describes the location of a particle in space or on a plane in relation to an origin. In our exercise, we have three particles with distinct position vectors.
  • \( \mathbf{r}_1 = (1, 1, 0) \) which places the first particle at the point (1, 1) in the plane.
  • \( \mathbf{r}_2 = (1, -1, 0) \) which positions the second particle at (1, -1).
  • \( \mathbf{r}_3 = (0, 0, 0) \) locates the third particle at the origin.
This arrangement of particles on the plane allows us to apply the center of mass formula, which aggregates these vectors based on their respective masses. Understanding these vectors is the foundation for determining how their collective positioning influences the center of mass.
Particles in a Plane
When dealing with particles in a plane, as shown in our example, the objective is to find the location where the total mass seems to concentrate. This is known as the center of mass. Here, the three particles lie in the plane defined by coordinates (x, y) while their z-component is zero, meaning they do not extend above or below the plane.
  • The task is simplified to two dimensions, focusing primarily on the movement and distribution along the x and y axes.
  • In two-dimensional space, each particle contributes to the center of mass depending on its mass and coordinate position.
  • The formula for the center of mass for particles in a plane becomes:\[ \mathbf{R}_{cm} = \frac{1}{M} \sum m_i \mathbf{r}_i \] where \( M \) is the total mass and \( \mathbf{r}_i \) are the position vectors.
This simplifies the calculation of the center of mass and provides insight into how mass distribution affects its position.
Total Mass
Calculating the total mass is a fundamental step in determining the center of mass. The total mass is the sum of the masses of all particles in the system. In our exercise, the masses are structured as follows:
  • \( m_1 = m_2 = m \)
  • \( m_3 = 10m \)
To find the overall mass \( M \), we add the individual masses together:\[ M = m_1 + m_2 + m_3 = m + m + 10m = 12m \]
This total mass \( M \) is crucial as it is the denominator in the center of mass equation. It scales the contribution of each particle's position vector according to their respective mass. By calculating the total mass, you ensure that the center of mass reflects the system's entire distribution, not just the individual particles.

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Most popular questions from this chapter

Consider a gun of mass \(M\) (when unloaded) that fires a shell of mass \(m\) with muzzle speed \(v\). (That is, the shell's speed relative to the gun is \(v\).) Assuming that the gun is completely free to recoil (no external forces on gun or shell), use conservation of momentum to show that the shell's speed relative to the ground is \(v /(1+m / M).\)

Many applications of conservation of momentum involve conservation of energy as well, and we haven't yet begun our discussion of energy. Nevertheless, you know enough about energy from your introductory physics course to handle some problems of this type. Here is one elegant example: An elastic collision between two bodies is defined as a collision in which the total kinetic energy of the two bodies after the collision is the same as that before. (A familiar example is the collision between two billiard balls, which generally lose extremely little of their total kinetic energy.) Consider an elastic collision between two equal mass bodies, one of which is initially at rest. Let their velocities be \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}=0\) before the collision, and \(\mathbf{v}_{1}^{\prime}\) and \(\mathbf{v}_{2}^{\prime}\) after. Write down the vector equation representing conservation of momentum and the scalar equation which expresses that the collision is elastic. Use these to prove that the angle between \(\mathbf{v}_{1}^{\prime}\) and \(\mathbf{v}_{2}^{\prime}\) is \(90^{\circ} .\) This result was important in the history of atomic and nuclear physics: That two bodies emerged from a collision traveling on perpendicular paths was strongly suggestive that they had equal mass and had undergone an elastic collision.

The masses of the earth and moon are \(M_{\mathrm{e}} \approx 6.0 \times 10^{24}\) and \(M_{\mathrm{m}} \approx 7.4 \times 10^{22}\) (both in kg) and their center to center distance is \(3.8 \times 10^{5} \mathrm{km}\). Find the position of their \(\mathrm{CM}\) and comment. (The radius of the earth is \(\left.R_{e} \approx 6.4 \times 10^{3} \mathrm{km} .\right).\)

A shell traveling with speed \(v_{\mathrm{o}}\) exactly horizontally and due north explodes into two equal-mass fragments. It is observed that just after the explosion one fragment is traveling vertically up with speed \(v_{\mathrm{o}} .\) What is the velocity of the other fragment?

Show that the moment of inertia of a uniform solid sphere rotating about a diameter is \(\frac{2}{5} M R^{2}\). The sum (3.31) must be replaced by an integral, which is easiest in spherical polar coordinates, with the axis of rotation taken to be the \(z\) axis. The element of volume is \(d V=r^{2} d r \sin \theta d \theta d \phi\). (Spherical polar coordinates are defined in Section 4.8. If you are not already familiar with these coordinates, you should probably not try this problem yet.)

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