Question

Three odd-shaped blocks of chocolate have the following masses and center-of- mass coordinates: (1) 0.300 kg, (0.200 m, 0.300 m); (2) 0.400 kg, (0.100 m, \(-\)0.400 m); (3) 0.200 kg, (\(-\)0.300 m, 0.600 m). Find the coordinates of the center of mass of the system of three chocolate blocks.

Short answer

The center of mass is at (0.044 m, 0.056 m).

Step-by-step solution

Identify given data

We are given the masses and center-of-mass coordinates for three blocks of chocolate: - Block 1: Mass = 0.300 kg, Coordinates = (0.200 m, 0.300 m) - Block 2: Mass = 0.400 kg, Coordinates = (0.100 m, -0.400 m) - Block 3: Mass = 0.200 kg, Coordinates = (-0.300 m, 0.600 m)

Use the center of mass formula

The center of mass (CM) of a system is given by the formula: \[ x_{CM} = \frac{\sum{m_i x_i}}{\sum{m_i}} \] and \[ y_{CM} = \frac{\sum{m_i y_i}}{\sum{m_i}} \] where \(m_i\) are the masses and \((x_i, y_i)\) are the coordinates of each mass.

Calculate the total mass

Calculate the total mass of the blocks:\[ m_{\text{total}} = 0.300 \, \text{kg} + 0.400 \, \text{kg} + 0.200 \, \text{kg} = 0.900 \, \text{kg} \]

Calculate x-coordinate of CM

Calculate the x-coordinate of the center of mass:\[ x_{CM} = \frac{(0.300 \times 0.200) + (0.400 \times 0.100) + (0.200 \times (-0.300))}{0.900} \]\[ = \frac{0.060 + 0.040 - 0.060}{0.900} \]\[ = \frac{0.040}{0.900} \approx 0.044 \text{ m} \]

Calculate y-coordinate of CM

Calculate the y-coordinate of the center of mass:\[ y_{CM} = \frac{(0.300 \times 0.300) + (0.400 \times (-0.400)) + (0.200 \times 0.600)}{0.900} \]\[ = \frac{0.090 - 0.160 + 0.120}{0.900} \]\[ = \frac{0.050}{0.900} \approx 0.056 \text{ m} \]

Compile the CM coordinates

The coordinates for the center of mass of the system are approximately (0.044 m, 0.056 m).

Key concepts

Understanding Mass Distribution

When solving problems related to the center of mass, understanding mass distribution is crucial. Mass distribution refers to how mass is spread out in space. In the exercise, each chocolate block has a different mass and position, which contributes differently to the system's center of mass.

• Each block's mass acts as a weight at its respective coordinates.
• Heavier blocks have a larger impact on the overall center of mass location.

The way mass is distributed affects the balance of the system. Think of it like balancing a seesaw; where the mass is located determines the balance point. Properly calculating this distribution allows us to pinpoint the exact center. In our exercise, calculating the mass distribution involves multiplying each block's mass by its respective coordinates. This provides insight into how each block contributes to the total system's center of mass.

Exploring Coordinate Systems

Coordinate systems are essential in problems involving positions, like center of mass calculations. In our exercise, each block's position is given in a Cartesian coordinate system, typically using an (x, y) format. This allows you to visualize each block's location in space.

• The coordinates indicate where on the plane each block is positioned.
• An (x, y) value of (0, 0) would be the origin or the central point of the plane.
• Positive and negative values determine directions along the axes.

With these coordinates, we calculate an average or weighted balance of the system components through the formula for the center of mass. Understanding the coordinate system is imperative because errors in interpreting these coordinates can lead to incorrect conclusions about the center of mass.

Introduction to Composite System Analysis

Composite system analysis involves understanding how multiple objects, or blocks in our exercise, interact to form a larger system.

• A composite system combines individual systems or objects into one whole.
• The behavior and properties of the system depend on the interaction between its components.

In our example, analyzing the composite system means considering each block's mass and position to find the collective center of mass. The center of mass represents the "average" location of all the mass distribution. This analysis allows us to understand how changing one component might affect the overall system. This can be particularly useful in engineering, physics, and design where balancing components precisely is crucial to functionality and safety.