/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 A system consists of the followi... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 10

A system consists of the following masses in the \(x y\) -plane: \(4.0 \mathrm{~kg}\) at coordinates \((x=0, y=5.0 \mathrm{~m}), 7.0 \mathrm{~kg}\) at \((3.0 \mathrm{~m}, 8.0 \mathrm{~m})\), and \(5.0 \mathrm{~kg}\) at \((-3.0 \mathrm{~m},-6.0 \mathrm{~m})\). Find the position of its center of mass. $$ \begin{array}{l} x_{\mathrm{cm}}=\frac{\sum x_{i} m_{i}}{\sum m_{i}}=\frac{(0)(4.0 \mathrm{~kg})+(3.0 \mathrm{~m})(7.0 \mathrm{~kg})+(-3.0 \mathrm{~m})(5.0 \mathrm{~kg})}{(4.0+7.0+5.0) \mathrm{kg}}=0.38 \mathrm{~m} \\ y_{\mathrm{cm}}=\frac{\sum y_{i} m_{i}}{\Sigma m_{i}}=\frac{(5.0 \mathrm{~m})(4.0 \mathrm{~kg})+(8.0 \mathrm{~m})(7.0 \mathrm{~kg})+(-6.0 \mathrm{~m})(5.0 \mathrm{~kg})}{16 \mathrm{~kg}}=2.9 \mathrm{~m} \end{array} $$

Short Answer

Expert verified
The center of mass is at (0.38 m, 2.9 m).

Step by step solution

01

Write down the formula for center of mass

The formula to find the center of mass of a system of particles in the plane is given by:\[ x_{\text{cm}} = \frac{\sum x_i m_i}{\sum m_i} \] and \[ y_{\text{cm}} = \frac{\sum y_i m_i}{\sum m_i} \]where \(x_i\) and \(y_i\) are the coordinates of each mass and \(m_i\) are the masses.
02

List the masses and coordinates

Here we have three masses: 1. \(4.0\, \text{kg}\) at \((x = 0, y = 5.0\, \text{m})\)2. \(7.0\, \text{kg}\) at \((x = 3.0\, \text{m}, y = 8.0\, \text{m})\)3. \(5.0\, \text{kg}\) at \((x = -3.0\, \text{m}, y = -6.0\, \text{m})\)

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Key Concepts

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

Mass Distribution
In physics, mass distribution refers to how mass is spread out in a given system or configuration. It is crucial when calculating the center of mass because it directly influences the center's position. If the mass is evenly distributed, the center of mass will be at the geometrical center. However, if the distribution is uneven, the center of mass will shift towards where more mass is concentrated.

For instance, consider the system in the exercise: different masses at different coordinates. These masses are:
  • 4.0 kg at (0, 5.0 m)
  • 7.0 kg at (3.0 m, 8.0 m)
  • 5.0 kg at (-3.0 m, -6.0 m)
The differing values and locations of these masses imply that their mass distribution affects the overall center of mass in the system.

To determine how the distribution influences the center of mass, you would calculate a weighted average of the masses' positions, highlighting the importance of each mass's location and value.
Coordinate System
A coordinate system helps precisely define the position of points, objects, or in this case, masses within a plane. It provides a systematic framework to locate these positions using coordinates like (x, y). In this exercise, the coordinate system is the 2D plane where each mass has x and y values describing its position.

The choice of coordinate system affects how you perceive and solve problems related to mass distribution. A common approach to simplify calculations and understanding is to initially assume that the coordinate system's origin (0,0) is at a convenient point. In our exercise, each mass is placed at specific coordinates:
  • The first mass is at (0, 5.0 m)
  • The second mass is at (3.0 m, 8.0 m)
  • The third mass is at (-3.0 m, -6.0 m)
By being aware of these positions, you utilize the coordinate system to calculate where the system's center of mass lies. This refers to finding the weighted center of these coordinates.
Particle System
A particle system in physics refers to a group of discrete masses that can interact or behave as a single entity. It is a fundamental concept where systems with multiple bodies/small particles exhibit collective properties.

In this exercise, dealing with three separate masses, the task is to understand how these masses interact to form a single center of mass for the entire system. Each mass within the system is treated as a particle, and these are:
  • 4.0 kg at (x = 0, y = 5.0 m)
  • 7.0 kg at (x = 3.0 m, y = 8.0 m)
  • 5.0 kg at (x = -3.0 m, y = -6.0 m)
Despite being distinct, the particle system concept allows one to consider these masses' overall effect by calculating their collective center of mass, as determined by the formulas:\[ x_{\text{cm}} = \frac{\sum x_i m_i}{\sum m_i} \] and \[ y_{\text{cm}} = \frac{\sum y_i m_i}{\sum m_i} \]These formulas enable a holistic understanding of the system's mass distribution across the plane by providing a mathematical framework to find the center of mass of the entire group.

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Most popular questions from this chapter

A force of \(1000 \mathrm{~N}\) is applied to a small space satellite for a time of \(10.0\) minutes. If the craft has a mass of \(200 \mathrm{~kg}\), what will be its final speed? [Hint: Be careful with those exponents when using a calculator.]

Three masses are placed on the \(y\) -axis: \(2 \mathrm{~kg}\) at \(y=300 \mathrm{~cm}, 6 \mathrm{~kg}\) at \(y\) \(=150 \mathrm{~cm}\), and \(4 \mathrm{~kg}\) at \(y=-75 \mathrm{~cm}\). Find their center of mass.

A ball is dropped from a height \(h\) above a tile floor and rebounds to a height of \(0.65 h\). Find the coefficient of restitution between ball and floor. Assign floor quantities the subscript 1 , and ball quantities the subscript \(2 .\) The initial and final velocities of the floor, \(u_{1}\) and \(v_{1}\), are zero. Therefore, $$ e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}}=-\frac{v_{2}}{u_{2}} $$ Since we know both the drop and rebound heights ( \(h\) and \(0.65 h\) ), we can write equations for the interchange of \(\mathrm{PE}_{\mathrm{G}}\) and \(\mathrm{KE}\) before and after the impact \(\sqrt{665}=0.8\) Notice that the coefficient of restitution equals the square root of the final rebound height over the initial drop height.

Typically, a tennis ball hit during a serve travels away at about 51 \(\mathrm{m} / \mathrm{s}\). If the ball is at rest mid-air when struck, and it has a mass of \(0.058 \mathrm{~kg}\), what is the change in its momentum on leaving the racket?

A \(2.0\) -kg brick is moving at a speed of \(6.0 \mathrm{~m} / \mathrm{s}\). How large a force \(F\) is needed to stop the brick in a time of \(7.0 \times 10^{-4} \mathrm{~s}\) ? Since we have a force and the time over which it acts, that suggests using the impulse equation (i.e., Newton's Second Law): $$ \begin{array}{l} \text { Impulse on brick }=\text { Change in momentum of brick }\\\ \begin{aligned} F \Delta t &=m v_{f}-m v_{i} \\ F\left(7.0 \times 10^{-4} \mathrm{~s}\right) &=0-(2.0 \mathrm{~kg})(6.0 \mathrm{~m} / \mathrm{s}) \end{aligned} \end{array} $$ from which \(F=-1.7 \times 10^{4} \mathrm{~N}\). The minus sign indicates that the force opposes the motion.
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