Problem 7 - APEX Calculus

Question

In Exercises \(7-10,\) 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.) $$ m_{1}=4 \text { at } x=1 ; \quad m_{2}=3 \text { at } x=3 ; \quad m_{3}=5 \text { at } x=10 $$

Step-by-Step Solution

Answer

The center of mass \( \bar{x} \) is 5.25 cm.

Step 1: Understand the Problem

We have three point masses located on a line at different positions. We need to find the center of mass of these point masses. The point masses are given as \( m_1 = 4 \) at \( x_1 = 1 \), \( m_2 = 3 \) at \( x_2 = 3 \), and \( m_3 = 5 \) at \( x_3 = 10 \). All distances are in centimeters, and masses in grams.

Step 2: Use the Center of Mass Formula

The formula for the center of mass \( \bar{x} \) of point masses along a line is \( \bar{x} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} \). We will apply this formula using the given values of the masses and their respective positions.

Step 3: Calculate the Numerator

First, calculate the weighted sum of the positions: \( m_1 x_1 + m_2 x_2 + m_3 x_3 = (4)(1) + (3)(3) + (5)(10) \). Thus, \( = 4 + 9 + 50 = 63 \).

Step 4: Calculate the Denominator

Calculate the total mass: \( m_1 + m_2 + m_3 = 4 + 3 + 5 = 12 \).

Step 5: Compute the Center of Mass

Use the results from Steps 3 and 4 to find \( \bar{x} \). So, \( \bar{x} = \frac{63}{12} \), which simplifies to \( \bar{x} = 5.25 \).

Key Concepts

Point Masses

A point mass is a concept used to simplify the representation of objects in physics problems. When you think about point masses, imagine a small, concentrated mass that acts as if all its weight is at a single point in space. This makes calculations easier, especially in problems involving the center of mass. In the given exercise, we have three point masses:

• Mass 1 (\(m_1 = 4 \, \text{g}\)) located at position \(x_1 = 1 \, \text{cm}\)
• Mass 2 (\(m_2 = 3 \, \text{g}\)) located at position \(x_2 = 3 \, \text{cm}\)
• Mass 3 (\(m_3 = 5 \, \text{g}\)) located at position \(x_3 = 10 \, \text{cm}\)

These masses are considered point masses because they allow us to focus solely on their mass and position without worrying about their exact size and shape.
This simplifies our task as we focus on their interaction in terms of distance and mass.

Weighted Average

In the context of finding the center of mass, we deal with what's known as a weighted average. A weighted average lets us determine an average value by considering both the quantity and importance of each data point. Here, the importance is determined by the mass of each point mass.

• Each position of the point mass is multiplied by its respective mass. This places greater importance on masses that are larger.
• We then sum these products to get a total weighted sum.
• The result is then divided by the total mass to balance the consideration of all involved masses to find the average position, the center of mass.

The weighted average thus helps manage the influence that the mass of each object has on their overall combined position.
This ensures that larger masses contribute more significantly to the final position than smaller masses.

Center of Mass Formula

The center of mass is a critical concept in understanding the balance point of a system of masses. When dealing with multiple point masses situated on a line, we calculate their center of mass using the following formula:\[\bar{x} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3}\]This formula is straightforward:

• Multiply each mass by its position to get the weighted position for each mass.
• Sum up all these weighted positions to form the numerator.
• Add up all the mass values to determine the total mass, forming the denominator.

By dividing the summed weighted positions by the total mass, the formula computes the average position where the system's mass can be considered to be concentrated.
This is what we call the center of mass, and it tells us where the point of balance for the combined system is situated.

Linear Position

Linear position refers to the place along a line where a point mass is located. In problems involving linear positions, each point mass is placed at a specific spot along a one-dimensional line. This simplicity allows us to use basic arithmetic operations for calculating the center of mass.

• In our exercise, each mass has a linear position (e.g., \(x_1 = 1\, \text{cm}\)).
• The linear assumption avoids complex calculations and visual geometries of multidimensional spaces.
• All calculations happen along this one-dimensional line, making it easy to follow the logic.

Linear position helps in simplifying the context, as all movements and interactions are restricted to a single line. This confines calculations into straightforward steps that inherently align with the concept of balance along a straightforward path.