Problem 4 - Single Variable Calculus

Question

Find the center of mass of the system comprising masses \(m_{k}\) located at the points \(x_{k}\) on a coordinate line. Assume that mass is measured in kilograms and distance is measured in meters. $$ \begin{array}{l} m_{1}=6, \quad m_{2}=4, \quad m_{3}=5, \quad m_{4}=8, \quad m_{5}=4 \\ x_{1}=-4, \quad x_{2}=-2, \quad x_{3}=0, \quad x_{4}=3, \quad x_{5}=6 \end{array} $$

Step-by-Step Solution

Answer

The center of mass of the given system is at a position \( x_{cm} = \frac{16}{27} \) meters.

Step 1: Calculate the weighted sum of positions

Multiply each mass with its corresponding position and add all the products: $$ \sum_{k} m_k x_k = m_1 x_1 + m_2 x_2 + m_3 x_3 + m_4 x_4 + m_5 x_5 $$ $$ = 6(-4) + 4(-2) + 5(0) + 8(3) + 4(6) = -24 - 8 + 0 + 24 + 24 = 16 $$

Step 2: Calculate the total mass of the system

Add all individual masses: $$ \sum_{k} m_k = m_1 + m_2 + m_3 + m_4 + m_5 = 6 + 4 + 5 + 8 + 4 = 27 $$

Step 3: Calculate the center of mass

Divide the weighted sum calculated in step 1 by the total mass calculated in step 2: $$ x_{cm} = \frac{\sum_{k} m_k x_k}{\sum_{k} m_k} = \frac{16}{27} $$ So, the center of mass of the system is at a position $$ x_{cm} = \frac{16}{27} $$ meters.

Key Concepts

Weighted Sum

The concept of a weighted sum is essential when calculating the center of mass. Imagine you have different objects, each with its own mass and position on a line. To find a common point or center of these masses, you combine their positions, but not just by simply adding them. Their mass influences the contribution of each position. This is where the idea of a weighted sum comes in: it involves multiplying each mass with its corresponding position and adding them all together.

So, how does it work in practice? Let's break it down:

• Multiply each mass by its position.
• Add all these products together to get the weighted sum.

In the example given, we multiply each mass like this:

• \( m_1x_1 = 6(-4) = -24 \)
• \( m_2x_2 = 4(-2) = -8 \)
• \( m_3x_3 = 5(0) = 0 \)
• \( m_4x_4 = 8(3) = 24 \)
• \( m_5x_5 = 4(6) = 24 \)

Adding these results: \(-24 - 8 + 0 + 24 + 24 = 16\). This is your weighted sum.

Coordinate Line

A coordinate line is a simple number line that helps us visualize and solve problems involving positions and distances. In this exercise, it's used to locate the positions of different masses.

Think of a coordinate line as a ruler laid out in front of you. The numbers on this line represent distances (in meters in our scenario) where masses are placed. Each mass has a unique position, determined by a number on the line.

For our problem:

• Mass \( m_1 \) is at position \( x_1 = -4 \).
• Mass \( m_2 \) is at position \( x_2 = -2 \).
• Mass \( m_3 \) is at position \( x_3 = 0 \).
• Mass \( m_4 \) is at position \( x_4 = 3 \).
• Mass \( m_5 \) is at position \( x_5 = 6 \).

These positions, along with the respective masses, are vital to determining the center of mass on the coordinate line.

Masses and Positions

Understanding the relationship between masses and their positions is crucial when determining the center of mass. Each mass is not only distinct due to its weight but also carries significance because of where it is placed on the coordinate line.

Consider the following: a heavier mass will "pull" the center of mass closer to its own position than a lighter mass would if everything else is equal. Therefore, examination of both the size of the mass and its position is vital.

For our task:

• The mass of \(6\) kg is at \(-4\) meters, substantially far from the zero point.
• Another mass of \(4\) kg is at \(-2\) meters, bringing its influence closer.
• A mass of \(5\) kg at \(0\) does not contribute to the directional shift, since it is on the zero point.
• A mass of \(8\) kg is at \(3\) meters, potentially shifting the center further right.
• Finally, \(4\) kg at \(6\) meters pulls even more to the right.

These combined influences from mass and position guide us to find the equilibrium point, the center of mass.

Total Mass Calculation

To find the center of mass, it’s essential to consider the total mass of all objects combined. This is done by summing up all individual masses in the system. Calculating this total mass provides the denominator when we find the center of mass, refining the balance of influence from all contributing masses.

In our case, we have:

• Mass \( m_1 = 6 \),
• Mass \( m_2 = 4 \),
• Mass \( m_3 = 5 \),
• Mass \( m_4 = 8 \),
• Mass \( m_5 = 4 \).

Adding these gives a total mass of \( 6 + 4 + 5 + 8 + 4 = 27 \) kilograms.

This total mass is crucial as it balances the weighted sum: it tells us how much the combined mass pulls or pushes the center to its final position on the coordinate line. Dividing the weighted sum by the total mass, we achieve the center of mass: \( x_{cm} = \frac{16}{27} \) meters.