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Chapter 13: Problem 9

Point masses are given along a line or in the plane. Find the center of \(\operatorname{mass} \bar{x}\) or \((\bar{x}, \bar{y}),\) as appropriate. (All masses are in grams and distances are in cm.) $$ \begin{array}{l} m_{1}=2 \text { at }(-2,-2) ; \quad m_{2}=2 \text { at }(2,-2) ; \\ m_{3}=20 \text { at }(0,4) \end{array} $$

Short Answer

Expert verified
The center of mass is at (0, 3).

Step by step solution

01

Identify the Masses and Coordinates

We have three point masses given with their respective locations. Mass 1, \(m_1 = 2\) grams, is located at \((-2, -2)\). Mass 2, \(m_2 = 2\) grams, is at position \((2, -2)\). Mass 3, \(m_3 = 20\) grams, is located at \((0, 4)\).
02

Use the Formula for Center of Mass

The center of mass \((\bar{x}, \bar{y})\) can be found using the formulas: \[\bar{x} = \frac{\sum_{i=1}^{n} m_i x_i}{\sum_{i=1}^{n} m_i} \quad \text{and} \quad \bar{y} = \frac{\sum_{i=1}^{n} m_i y_i}{\sum_{i=1}^{n} m_i}\]where \(m_i\) is the mass at coordinate \((x_i, y_i)\).
03

Calculate Total Mass

Find the total mass: \[M = m_1 + m_2 + m_3 = 2 + 2 + 20 = 24 \, \text{grams}\]
04

Calculate \(\bar{x}\) Component

Substitute the values into the formula for \(\bar{x}\):\[\bar{x} = \frac{(m_1 \cdot x_1) + (m_2 \cdot x_2) + (m_3 \cdot x_3)}{M} = \frac{(2 \times -2) + (2 \times 2) + (20 \times 0)}{24} \]This simplifies to:\[\bar{x} = \frac{-4 + 4 + 0}{24} = \frac{0}{24} = 0 \]
05

Calculate \(\bar{y}\) Component

Substitute the values into the formula for \(\bar{y}\):\[\bar{y} = \frac{(m_1 \cdot y_1) + (m_2 \cdot y_2) + (m_3 \cdot y_3)}{M} = \frac{(2 \times -2) + (2 \times -2) + (20 \times 4)}{24}\]This simplifies to:\[\bar{y} = \frac{-4 - 4 + 80}{24} = \frac{72}{24} = 3\]
06

Interpret the Results

The center of mass for the given system of point masses is located at \((\bar{x}, \bar{y}) = (0, 3)\).

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Key Concepts

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

Point Masses
In the realm of physics and mathematics, point masses represent objects that have mass but occupy very minimal or no space. These hypothetical masses make calculations involving large systems more manageable. Even though objects in real life have dimensions and occupy space, they can often be approximated as point masses for the sake of simplicity.
When dealing with point masses, we look at their individual contributions to the overall system. For our exercise, we have three point masses located at different coordinates. Each mass has a specific weight, such as 2 grams or 20 grams, making it essential to account for both
  • the magnitude of each mass, and
  • their positions.
Using these attributes, one can calculate the center of mass for the entire system, providing insights into how the masses are distributed across a plane or along a line.
Coordinate Geometry
Coordinate geometry is a powerful tool that allows us to explore points, lines, and surfaces using their geometric relationships. In the case of calculating the center of mass, we locate point masses in a two-dimensional space using a coordinate system with an
  • x-axis (horizontal) and
  • y-axis (vertical)
.

In our given problem, point masses are situated at coordinates
  • -2 on both x and y for the first mass,
  • 2 on x and -2 on y for the second mass, and
  • 0 on x and 4 on y for the third mass.

Using the known positions, we apply mathematical formulas to figure out where the overall center of mass for this system sits in the coordinate plane. This is achieved by addressing each axis separately, resulting in
  • a horizontal (x) component and
  • a vertical (y) component
as part of the center of mass calculation.
Mass Distribution
Mass distribution describes how mass is spread over a particular space or volume. In this context, it refers to how individual point masses contribute to the total mass system. Understanding mass distribution is vital when calculating the center of mass as it's indicative of how the mass "pushes" the balance point toward more densely packed areas.

When calculating the center of mass, it鈥檚 essential to consider
  • both the mass of each point and
  • its position
in the overall coordinate plane. Larger masses apply more weight (or influence) in pulling the center of mass towards their location.
For instance, the mass at (0, 4) in our exercise weighs significantly more (20 grams) than those at other positions (2 grams each). Consequently, it has a greater effect on where the center of mass is located, ultimately resulting in a point at (0, 3). This balance between masses is crucial in understanding how the assignment of individual masses impacts the calculation.

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Most popular questions from this chapter

A region is space is described. Set up the triple integrals that find the volume of this region using rectangular, cylindrical and spherical coordinates, then comment on which of the three appears easiest to evaluate. The region enclosed by the cone \(z=\sqrt{x^{2}+y^{2}}\) and plane \(z=1 .\)

Set up the triple integrals that give the volume of \(D\) in all 6 orders of integration, and find the volume of \(D\) by evaluating the indicated triple integral. \(D\) is bounded by the planes \(y=0, y=2, x=1, z=0\) and \(z=(3-x) / 2\) Evaluate the triple integral with order \(d x d y d z\).

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In Exercises \(11-14,\) an iterated integral in rectangular coordinates is given. Rewrite the integral using polar coordinates and evaluate the new double integral. $$ \int_{0}^{5} \int_{-\sqrt{25-x^{2}}}^{\sqrt{25-x^{2}}} \sqrt{x^{2}+y^{2}} d y d x $$
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