Problem 540
Question
Three particles of the same mass lie in the \((\mathrm{X}, \mathrm{Y})\) plane, The \((X, Y)\) coordinates of their positions are \((1,1),(2,2)\) and \((3,3)\) respectively. The \((X, Y)\) coordinates of the centre of mass are \(\\{\mathrm{A}\\}(1,2)\) \(\\{\mathrm{B}\\}(2,2)\) \(\\{\mathrm{C}\\}(1.5,2)\) \(\\{\mathrm{D}\\}(2,1.5)\)
Step-by-Step Solution
Verified Answer
The center of mass is given by the weighted average position of the particles. We calculated the X- and Y-coordinates of the center of mass to be \((x_{cm}, y_{cm}) = (2, 2)\). Thus, the correct answer is \(\mathrm{B}(2,2)\).
1Step 1: Calculate the X-coordinate of the center of mass
The X-coordinate of the center of mass, \(x_{cm}\), is given by the weighted average of the X-coordinates of the three particles:
\[x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3}\]
Because the three particles have the same mass, we can write:
\[x_{cm} = \frac{1}{3}(x_1 + x_2 + x_3)\]
Substitute the X-coordinates of the particles:
\[x_{cm} = \frac{1}{3}(1 + 2 + 3) = \frac{1}{3}(6) = 2\]
2Step 2: Calculate the Y-coordinate of the center of mass
Similarly, the Y-coordinate of the center of mass, \(y_{cm}\), is given by the weighted average of the Y-coordinates of the three particles:
\[y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3}\]
Because the three particles have the same mass, we can write:
\[y_{cm} = \frac{1}{3}(y_1 + y_2 + y_3)\]
Substitute the Y-coordinates of the particles:
\[y_{cm} = \frac{1}{3}(1 + 2 + 3) = \frac{1}{3}(6) = 2\]
3Step 3: Identify the correct answer
We found that the \((X, Y)\) coordinates of the center of mass are \((2, 2)\). Therefore, the correct answer is \(\mathrm{B}(2,2)\).
Key Concepts
Coordinate PlaneWeighted AverageParticles with Uniform MassCoordinate Geometry
Coordinate Plane
The coordinate plane, often referred to as the Cartesian plane, is a two-dimensional surface where each point is uniquely specified by a pair of numerical coordinates. These coordinates are typically denoted as
- X-coordinate (x-axis): This represents the horizontal position of a point.
- Y-coordinate (y-axis): This represents the vertical position of a point.
Weighted Average
A weighted average is a mathematical tool used to calculate the average of a set of values, taking into account the importance, or weight, of each value. In finding the center of mass, the concept of a weighted average is pivotal.
The formula given for the center of mass’s X-coordinate, rackets{ x_{cm} }, is: \[ x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} \]With equal masses, this simplifies to an unweighted average: \[ x_{cm} = \frac{1}{3}(x_1 + x_2 + x_3) \]
This logic also applies to finding rackets{ y_{cm} }, as a weighted average affords a balanced view of all inputs, giving us the collective position of distributed particles.
The formula given for the center of mass’s X-coordinate, rackets{ x_{cm} }, is: \[ x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} \]With equal masses, this simplifies to an unweighted average: \[ x_{cm} = \frac{1}{3}(x_1 + x_2 + x_3) \]
This logic also applies to finding rackets{ y_{cm} }, as a weighted average affords a balanced view of all inputs, giving us the collective position of distributed particles.
Particles with Uniform Mass
In physics, when dealing with particles with uniform mass, each particle contributes equally to the calculation of physical properties like the center of mass. This uniformity simplifies calculations because we only need to consider the geometric configuration rather than differing weights.
In this problem, each particle has the same mass, rackets{ m }, which allows us to simplify the weighted average formula for determining the center of mass of several particles. Consequently, the masses cancel out, leading to simpler arithmetic where \[ x_{cm} = \frac{1}{3}(x_1 + x_2 + x_3) \] and correspondingly for the y-coordinate. Understanding uniform mass particles is crucial, as this uniformity is often applicable in exercises involving symmetrical distributions or identical modular components.
In this problem, each particle has the same mass, rackets{ m }, which allows us to simplify the weighted average formula for determining the center of mass of several particles. Consequently, the masses cancel out, leading to simpler arithmetic where \[ x_{cm} = \frac{1}{3}(x_1 + x_2 + x_3) \] and correspondingly for the y-coordinate. Understanding uniform mass particles is crucial, as this uniformity is often applicable in exercises involving symmetrical distributions or identical modular components.
Coordinate Geometry
Coordinate geometry, sometimes known as analytic geometry, is the study of geometry using a coordinate system. This field bridges algebra and geometry by using algebraic equations to describe geometric shapes and their properties in the coordinate plane.
When finding the center of mass in this particular exercise, coordinate geometry is employed to identify the regions and positional attributes of the objects involved. Each particle’s placement in the rackets{ (X, Y) } coordinate system is used to craft equations that reveal broader geometric truths.
The systematic approach of coordinate geometry allows for solving spatial problems with precision. It ensures that each point’s relationship and resulting calculations, like the weighted average leading to a center of mass, are coherent and mathematically accurate.
When finding the center of mass in this particular exercise, coordinate geometry is employed to identify the regions and positional attributes of the objects involved. Each particle’s placement in the rackets{ (X, Y) } coordinate system is used to craft equations that reveal broader geometric truths.
The systematic approach of coordinate geometry allows for solving spatial problems with precision. It ensures that each point’s relationship and resulting calculations, like the weighted average leading to a center of mass, are coherent and mathematically accurate.
Other exercises in this chapter
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