Problem 43 - College Physics

Question

Three odd-shaped blocks of chocolate have the following masses and center-of- mass coordinates: \(\begin{array}{lll}\text { (1) } 0.300 \text { kg. }\end{array}\) \((0.200 \mathrm{~m}, 0.300 \mathrm{~m});$$\begin{array}{lll}\text { (2) } 0.400 \text { kg. }\end{array}$$(0.100 \mathrm{~m},-0.400 \mathrm{~m});\)(3) \(0.200 \mathrm{~kg},(-0.300 \mathrm{~m}, 0.600 \mathrm{~m})\). Find the coordinates of the center of mass of the system of three chocolate blocks.

Step-by-Step Solution

Answer

The center of mass is approximately (0.0444 m, 0.0556 m).

Step 1: Understanding the Problem

We have three chocolate blocks, each with a given mass and center-of-mass coordinates. We need to find the overall center of mass for these three blocks combined. The formula to find the center of mass for a system is given by \((x_{cm}, y_{cm}) = \left(\frac{\sum m_ix_i}{\sum m_i}, \frac{\sum m_iy_i}{\sum m_i}\right)\), where \(m_i\) are the masses and \((x_i, y_i)\) are their respective coordinates.

Step 2: Calculate Total Mass

First, sum up the masses of all the chocolate blocks to find the total mass, \(\sum m_i\). The masses are 0.300 kg, 0.400 kg, and 0.200 kg. Therefore, \(\sum m_i = 0.300 + 0.400 + 0.200 = 0.900\) kg.

Step 3: Calculate Weighted Sum for X-Coordinate

Calculate \(\sum m_ix_i\) by multiplying the mass of each block by its x-coordinate and summing them up:\(0.300 \times 0.200 + 0.400 \times 0.100 + 0.200 \times (-0.300) = 0.060 + 0.040 - 0.060 = 0.040\) m.

Step 4: Calculate Weighted Sum for Y-Coordinate

Calculate \(\sum m_iy_i\) by multiplying the mass of each block by its y-coordinate and summing them up:\(0.300 \times 0.300 + 0.400 \times (-0.400) + 0.200 \times 0.600 = 0.090 - 0.160 + 0.120 = 0.050\) m.

Step 5: Determine Center of Mass Coordinates

For the center of mass coordinates, divide each weighted sum by the total mass:\(x_{cm} = \frac{0.040}{0.900} \approx 0.0444\) m and \(y_{cm} = \frac{0.050}{0.900} \approx 0.0556\) m. Thus, the coordinates are approximately (0.0444 m, 0.0556 m).

Key Concepts

Understanding Mass

In physics, mass refers to the amount of matter contained in an object. It's a fundamental property and is commonly measured in kilograms. In the context of the center of mass calculations, each chocolate block's mass is crucial because it determines the influence of that block's position on the overall center of mass of the system.
This means that heavier objects exert a greater influence on the center of mass's position than lighter ones.
For our chocolate blocks, the masses given are:

• Block 1: 0.300 kg
• Block 2: 0.400 kg
• Block 3: 0.200 kg

These values are used to calculate the weighted averages necessary to find the center of mass. Hence, knowing the mass helps us understand how each block contributes to the overall balance of the system.

Using Coordinates Effectively

Coordinates in physics are numbers that define the position of an object in space. In our problem, every chocolate block has specific coordinates that describe where its center of mass is located. For each position, we have an x-coordinate and a y-coordinate.

• Block 1: (0.200 m, 0.300 m)
• Block 2: (0.100 m, -0.400 m)
• Block 3: (-0.300 m, 0.600 m)

These coordinates are essential in calculating the overall center of mass.
The coordinates tell us the exact position of each block, allowing us to determine how they are spatially arranged. By finding the weighted average of these coordinates, we can pinpoint where the entire system’s center of mass lies, considering each block’s specific position.

Calculating a Weighted Average

A weighted average gives more weight or importance to certain values in a set when calculating an average. For finding a system's center of mass, we apply this concept by considering each block's mass when averaging their coordinates.
The formula used is:

• For x-coordinate: \[ x_{cm} = \frac{\sum m_i x_i}{\sum m_i} \]
• For y-coordinate: \[ y_{cm} = \frac{\sum m_i y_i}{\sum m_i} \]

Here, \( m_i \) represents the mass of each block, while \( x_i \) and \( y_i \) are the respective coordinates.
Using weighted averages means the larger the mass, the more it influences the average. This process helps us find a point that represents the average position of all blocks, considering their different weights and spots.

Approach to Physics Problem Solving

Solving textbook problems in physics, such as finding a center of mass, involves a systematic approach. This method helps simplify complex ideas and ensures accurate results.
Key steps include:

• Understand the problem: Clearly identify what's given and what's needed. For instance, knowing masses and coordinates of blocks like in this example.
• Use appropriate formulas: Apply relevant physics equations. Here, the center of mass formula integrates weighted averages.
• Perform calculations carefully: Calculate totals, such as \( \sum m_i \) for total mass, and verify each step.
• Check final results: Ensure the answer logically fits within the context. Does the coordinate make sense with the positions and masses provided?

Following these steps, as showcased in the chocolate block exercise, makes problem-solving manageable and helps build a strong foundation for handling various physics topics.