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Chapter 3: Problem 15

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

[Computer] A grenade is thrown with initial velocity \(\mathbf{v}_{\mathrm{o}}\) from the origin at the top of a high cliff, subject to negligible air resistance. (a) Using a suitable plotting program, plot the orbit, with the following parameters: \(\mathbf{v}_{\mathrm{o}}=(4,4), g=1,\) and \(0 \leq t \leq 4\) (and with \(x\) measured horizontally and \(y\) vertically up). Add to your plot suitable marks (dots or crosses, for example) to show the positions of the grenade at \(t=1,2,3,4 .\) (b) At \(t=4,\) when the grenade's velocity is \(\mathbf{v},\) it explodes into two equal pieces, one of which moves off with velocity \(\mathbf{v}+\Delta \mathbf{v} .\) What is the velocity of the other piece? (c) Assuming that \(\Delta \mathbf{v}=(1,3),\) add to your original plot the paths of the two pieces for \(4 \leq t \leq 9 .\) Insert marks to show their positions at \(t=5,6,7,8,9\). Find some way to show clearly that the CM of the two pieces continues to follow the original parabolic path.

Consider a system comprising two extended bodies, which have masses \(M_{1}\) and \(M_{2}\) and centers of mass at \(\mathbf{R}_{1}\) and \(\mathbf{R}_{2}\). Prove that the \(\mathrm{CM}\) of the whole system is at $$ \mathbf{R}=\frac{M_{1} \mathbf{R}_{1}+M_{2} \mathbf{R}_{2}}{M_{1}+M_{2}} $$ This beautiful result means that in finding the CM of a complicated system, you can treat its component parts just like point masses positioned at their separate centers of mass - even when the component parts are themselves extended bodies.

A juggler is juggling a uniform rod one end of which is coated in tar and burning. He is holding the rod by the opposite end and throws it up so that, at the moment of release, it is horizontal, its \(\mathrm{CM}\) is traveling vertically up at speed \(v_{\mathrm{o}}\) and it is rotating with angular velocity \(\omega_{\mathrm{o}} .\) To catch it, he wants to arrange that when it returns to his hand it will have made an integer number of complete rotations. What should \(v_{\mathrm{o}}\) be, if the rod is to have made exactly \(n\) rotations when it returns to his hand?

Use spherical polar coordinates \(r, \theta, \phi\) to find the CM of a uniform solid hemisphere of radius R, whose flat face lies in the \(x y\) plane with its center at the origin. Before you do this, you will need to convince yourself that the element of volume in spherical polars 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.)

In the early stages of the Saturn V rocket's launch, mass was ejected at about \(15,000 \mathrm{kg} / \mathrm{s}\), with a speed \(v_{\mathrm{ex}} \approx 2500 \mathrm{m} / \mathrm{s}\) relative to the rocket. What was the thrust on the rocket? Convert this to tons (1 ton \(\approx 9000\) newtons) and compare with the rocket's initial weight (about 3000 tons).
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