Problem 9 - Calculus of a Single Variable

Question

Find the center of mass of the given system of point masses. \(\begin{array}{|c|c|c|c|}\hline m_{i} & {5} & {1} & {3} \\ \hline\left(x_{i}, y_{i}\right) & {(2,2)} & {(-3,1)} & {(1,-4)} \\ \hline\end{array}\)

Step-by-Step Solution

Answer

The coordinates of the center of mass are \( (1.11m, -0.11m) \).

Step 1: Calculate the value of \(M\)

The total mass \(M\) can be found by adding up all the individual masses. \(M = m_{1} + m_{2} + m_{3} = 5kg + 1kg + 3kg = 9kg\).

Step 2: Calculate the x-coordinate of the center of mass

The x-coordinate of the center of mass \(x_{cm}\) can be calculated by summing the products of each mass and its x-coordinate, and then dividing by the total mass. \(x_{cm} = \frac{1}{M} \sum m_{i}x_{i} = \frac{1}{9kg} \times [(5kg \times 2m) + (1kg \times -3m) + (3kg \times 1m)] = \frac{1}{9kg} \times [10kgm + -3kgm + 3kgm] = \frac{10kgm}{9kg} = 1.11m\).

Step 3: Calculate the y-coordinate of the center of mass

The y-coordinate of the center of mass \(y_{cm}\) can be calculated by summing the products of each mass and its y-coordinate, and then dividing by the total mass. \(y_{cm} = \frac{1}{M} \sum m_{i}y_{i} = \frac{1}{9kg} \times [(5kg \times 2m) + (1kg \times 1m) + (3kg \times -4m)] = \frac{1}{9kg} \times [10kgm + 1kgm + -12kgm] = \frac{-1kgm}{9kg} = -0.11m\).

Key Concepts

Point Masses

In physics, a 'point mass' is a simplified model where an object is assumed to occupy a single point in space. This means it has a certain mass but no volume or shape. It's a useful way to simplify problems, especially when dealing with systems of multiple objects. For our exercise:

• The point masses are given as: 5 kg, 1 kg, and 3 kg.
• They are located at coordinates (2,2), (-3,1), and (1,-4) respectively.

Understanding that each mass is concentrated at a single point helps in calculations of the center of mass, as we focus only on these defined points.

Coordinate Calculation

To find the center of mass of a system of point masses, coordinate calculation is essential. This involves determining the average position of all the point masses in the system. The center of mass acts as a balance point for the object where it can be freely rotated without an offset. In calculations, we deal separately with the x-coordinates and y-coordinates. We compute them one by one to identify the system's center of mass in a two-dimensional space.

Total Mass

The total mass of a system is the sum of all its point masses. In simple terms, it's like adding up the weight of all involved objects to see how much it all weighs together. For the example given:

• We have three masses: 5 kg, 1 kg, and 3 kg.
• Thus, the total mass, denoted by \( M \), is calculated as:

\[M = 5 \, \text{kg} + 1 \, \text{kg} + 3 \, \text{kg} = 9 \, \text{kg}\]This total is crucial as it's used to average out the positions of the point masses when calculating the exact location of the center of mass.

X-coordinate

The x-coordinate of the center of mass represents its position along the horizontal axis. To calculate it, we sum up the products of each point mass and its respective x-coordinate, then divide by the total mass.For the problem at hand:

• We computed the x-coordinate as:

\[x_{cm} = \frac{1}{M} \left( \sum m_{i}x_{i} \right) = \frac{1}{9 \, \text{kg}} \left[ (5 \, \text{kg} \times 2 \text{m}) + (1 \, \text{kg} \times -3 \text{m}) + (3 \, \text{kg} \times 1 \text{m}) \right]\]Which simplifies to:\[x_{cm} = \frac{10 \, \text{kg} \cdot \text{m}}{9 \, \text{kg}} \approx 1.11 \, \text{m}\]Here, \( x_{cm} \) indicates the system's average horizontal position based on the point mass distribution.

Y-coordinate

Much like the x-coordinate, the y-coordinate of the center of mass identifies the vertical position of the center. We calculate it similarly by summing the products of each point mass's y-coordinate with its mass, then dividing by the total mass.In this case:

• The calculation for the y-coordinate is:

\[y_{cm} = \frac{1}{M} \left( \sum m_{i}y_{i} \right) = \frac{1}{9 \, \text{kg}} \left[ (5 \, \text{kg} \times 2 \text{m}) + (1 \, \text{kg} \times 1 \text{m}) + (3 \, \text{kg} \times -4 \text{m}) \right]\]Leads to:\[y_{cm} = \frac{-1 \, \text{kg} \cdot \text{m}}{9 \, \text{kg}} \approx -0.11 \, \text{m}\]This result \( y_{cm} \) signifies where the center of mass lies in terms of height when considering all the weighted mass points.