Problem 48 - College Physics

Question

\(\cdot\) Three odd-shaped blocks of chocolate have the following masses and center-of-mass coordinates: \((1) \quad 0.300\) kg, \((0.200 \mathrm{m}, 0.300 \mathrm{m}) ;\) (2) \(0.400 \mathrm{kg}, \quad(0.100 \mathrm{m},-0.400 \mathrm{m})\) (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 at approximately (0.044 m, 0.056 m).

Step 1: Calculate the Total Mass

To find the total mass of the system, add together the masses of all three blocks. \[ m_{\text{total}} = 0.300\, \text{kg} + 0.400\, \text{kg} + 0.200\, \text{kg} = 0.900\, \text{kg} \]

Step 2: Calculate the X-coordinate of the Center of Mass

The x-coordinate of the center of mass \((x_{\text{cm}})\) is calculated using the formula: \[ x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_{\text{total}}} \] Substitute the given values: \[ x_{\text{cm}} = \frac{(0.300)(0.200) + (0.400)(0.100) + (0.200)(-0.300)}{0.900} \] Calculate each term: \[ (0.300 \times 0.200) = 0.060, \ (0.400 \times 0.100) = 0.040, \ (0.200 \times -0.300) = -0.060 \] Sum these up: \[ 0.060 + 0.040 - 0.060 = 0.040 \] Divide by the total mass: \[ x_{\text{cm}} = \frac{0.040}{0.900} \approx 0.044 \text{ m} \]

Step 3: Calculate the Y-coordinate of the Center of Mass

The y-coordinate of the center of mass \((y_{\text{cm}})\) is calculated similarly: \[ y_{\text{cm}} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_{\text{total}}} \] Substitute the given values: \[ y_{\text{cm}} = \frac{(0.300)(0.300) + (0.400)(-0.400) + (0.200)(0.600)}{0.900} \] Calculate each term: \[ (0.300 \times 0.300) = 0.090, \ (0.400 \times -0.400) = -0.160, \ (0.200 \times 0.600) = 0.120 \] Sum these up: \[ 0.090 - 0.160 + 0.120 = 0.050 \] Divide by the total mass: \[ y_{\text{cm}} = \frac{0.050}{0.900} \approx 0.056 \text{ m} \]

Step 4: Conclusion

Combine the calculated x and y coordinates to obtain the center of mass of the system: \((0.044\, \text{m}, 0.056\, \text{m})\). This is the center of mass of all three chocolate blocks combined.

Key Concepts

Mass Distribution

Understanding mass distribution is crucial when calculating the center of mass. Imagine several objects, each having its own unique mass and position. These masses must be "distributed" across a certain space. In our chocolate block example, each block has a different mass and position. The center of mass is influenced by both the individual mass of each object and how these masses are placed or distributed in space.

Mass distribution involves knowing:

• Each object's mass
• The position of each mass within the chosen coordinate system

In short, by considering how the masses are spread out, we can determine the point that represents the average position — the center of mass. This helps in understanding the balance and movement of the combined system.

Coordinate System

To find the center of mass, it is essential to use a clear and defined coordinate system. A coordinate system helps in pinpointing the precise location of each mass point in space. Let's look at some essentials of using a coordinate system:

• The coordinate system includes axes, usually labeled as x, y, and sometimes z for three-dimensional problems.
• Each block of chocolate in our problem is positioned at a specific point on this system, denoted as \((x, y)\) coordinates.

By assigning each mass a coordinate position, you can easily plug these values into formulas to find the center of mass. Think of it like plotting points on a map — without a coordinate system, calculating the center of mass would be nearly impossible!

Calculation Steps

Calculating the center of mass involves straightforward math once you understand the distribution and coordinate system. Here's a streamlined view of the steps involved:

• First, determine the total mass of the system by summing up the masses of all individual objects.
• Next, calculate the weighted average for the x-coordinates. This involves multiplying each mass by its x-position and summing these products.
• Divide this sum by the total mass to find the x-coordinate of the center of mass.
• Repeat the process for the y-coordinate using the y-positions of each mass.

Remember, it’s like blending colors. Each mass contributes to the final position statement — the combined point that represents the entire system best: the center of mass.