Problem 78 - College Physics

Question

A \(3.0-\mathrm{kg}\) rod of length \(5.0 \mathrm{~m}\) has at opposite ends point masses of \(4.0 \mathrm{~kg}\) and \(6.0 \mathrm{~kg}\). (a) Will the center of mass of this system be (1) nearer to the \(4.0-\mathrm{kg}\) mass, (2) nearer to the 6.0 -kg mass, or (3) at the center of the rod? Why? (b) Where is the center of mass of the system?

Step-by-Step Solution

Answer

(a) Nearer to the 6.0 kg mass. (b) Center of mass is at 2.88 m from the 4.0 kg mass.

Step 1: Identify the System and Components

The system consists of a rod with point masses attached at its ends. Masses are \(4.0\, \text{kg}\) at one end and \(6.0\, \text{kg}\) at the other end. The rod itself has a mass of \(3.0\, \text{kg}\) and length \(5.0\, \text{m}\).

Step 2: Understand the Concept of Center of Mass

The center of mass of a system is the point where the total mass of the system can be considered concentrated. For a system of particles, the center of mass is calculated based on their relative positions and masses.

Step 3: Derive the Formula for Center of Mass of the System

For a linear system, the center of mass \(x_{cm}\) can be calculated using the formula: \[ x_{cm} = \frac{m_1 \cdot x_1 + m_2 \cdot x_2 + m_3 \cdot x_3}{m_1 + m_2 + m_3} \] where \(m_1, m_2, m_3\) are the masses and \(x_1, x_2, x_3\) are their positions.

Step 4: Assign Coordinates to Masses and Rod

Let's set the position of the \(4.0\, \text{kg}\) mass at \(x_1 = 0\), the \(6.0\, \text{kg}\) mass at \(x_3 = 5.0\, \text{m}\) (since they are at opposite ends of the rod). Assume the rod is uniform, so its mass' center is at its midpoint \(x_2 = 2.5\, \text{m}\).

Step 5: Calculate Center of Mass

Plug the masses and positions into the center of mass equation: \[ x_{cm} = \frac{4.0 \cdot 0 + 3.0 \cdot 2.5 + 6.0 \cdot 5.0}{4.0 + 3.0 + 6.0} = \frac{0 + 7.5 + 30}{13.0} = \frac{37.5}{13} \approx 2.88 \text{ m} \] The center of mass is approximately \(2.88\, \text{m}\) from the \(4.0\, \text{kg}\) mass.

Step 6: Analyze the Result

The calculated center of mass position \(x_{cm} \approx 2.88\, \text{m}\) is closer to the \(6.0\, \text{kg}\) mass because it is more massive compared to the \(4.0\, \text{kg}\) mass at the opposite end.

Key Concepts

Understanding Linear Systems in Mass Calculation

When dealing with a physics problem involving multiple objects affecting each other, you often consider it a linear system. This means that the components of the system can be laid out in a straight line, like in the case of a rod with several masses attached to it. The solution involves considering each part of the system and how it contributes to the overall behavior, particularly when finding the center of mass.
In our original exercise, we have a rod with point masses at each end and the rod’s own mass. These are pictured as a linear element, where each mass has a specific position along a straight line. To find the center of mass, each object's position and mass is taken into account. Understanding how to align the system linearly allows us to use formulas designed for such systems to derive meaningful results. This simplified setup helps in minimizing complex three-dimensional considerations, making the problem easier to solve.

Role of Mass Distribution in Center of Mass

The center of mass is significantly influenced by how mass is distributed along an object or a system of objects.
To visualize, imagine balancing a seesaw. If one side is heavier, that side naturally tips, illustrating that the center of mass is drawn towards the heavier section.
In the given problem, we have three key masses: a point mass of 4.0 kg, a rod with a mass of 3.0 kg, and another point mass of 6.0 kg. The leader in determining the center of mass is the object with the greatest mass – in this case, the 6.0 kg mass.

• The position of each mass affects the center of mass location.
• The distribution is not only about weight but also about where each mass is placed.

In our scenario, although the rod contributes to the total weight, it is the 6.0 kg mass that has a larger influence, causing the center of mass to shift closer to it. This shows that the mass spread is vital in determining the physical influence over the object's balance point.

Practical Physics Problem Solving

Physics often poses real-world problems requiring calculation, like finding the center of mass. The process involves identifying known variables, applying suitable equations, and carrying out calculations. Starting with clarity in understanding the problem is key. To solve the center of mass problem, we:

• Identified key masses and their positions on the rod.
• Used the center of mass formula, fitting the problem into a linear equation setup.
• Carried out the computations step-by-step, ensuring accuracy at each stage.

Using known physics steps simplifies problem solving, such as placing coordinates for each mass position and substituting into the formula. This breaks down a seemingly complex task into smaller, manageable steps. Calculations like these in physics enable accurate and informed predictions about object behavior in real-world scenarios.