Problem 51 - University Physics with Modern Physics

Question

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

Step 1: Understanding the Formula for Center of Mass

To find the center of mass of a system of particles, use the formula: \( \bar{x} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} \) and \( \bar{y} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} \), where \(m_i\) are the masses and \((x_i, y_i)\) are the coordinates of each object.

Step 2: Plug in the Values for the X-Coordinate

Using the formula for \( \bar{x} \), substitute the given values: \( m_1 = 0.300 \) kg, \( x_1 = 0.200 \) m; \( m_2 = 0.400 \) kg, \( x_2 = 0.100 \) m; \( m_3 = 0.200 \) kg, \( x_3 = -0.300 \) m.\[ \bar{x} = \frac{(0.300)(0.200) + (0.400)(0.100) + (0.200)(-0.300)}{0.300 + 0.400 + 0.200} \]

Step 3: Calculating the Numerator for \(\bar{x}\)

Calculate each component:- \((0.300)(0.200) = 0.060 \) - \((0.400)(0.100) = 0.040 \)- \((0.200)(-0.300) = -0.060 \)Add these values to find the numerator: \[ 0.060 + 0.040 - 0.060 = 0.040 \]

Step 4: Calculating the Denominator for \(\bar{x}\)

Find the sum of the masses: \[0.300 + 0.400 + 0.200 = 0.900 \text{ kg}\]

Step 5: Find the X-Coordinate of the Center of Mass

Plug the results from Steps 3 and 4 into the formula for \(\bar{x}\):\[ \bar{x} = \frac{0.040}{0.900} \approx 0.044 \text{ m} \]

Step 6: Plug in the Values for the Y-Coordinate

Using the formula for \( \bar{y} \), substitute the given values: \( m_1 = 0.300 \) kg, \( y_1 = 0.300 \) m; \( m_2 = 0.400 \) kg, \( y_2 = -0.400 \) m; \( m_3 = 0.200 \) kg, \( y_3 = 0.600 \) m.\[ \bar{y} = \frac{(0.300)(0.300) + (0.400)(-0.400) + (0.200)(0.600)}{0.300 + 0.400 + 0.200} \]

Step 7: Calculating the Numerator for \(\bar{y}\)

Calculate each component:- \((0.300)(0.300) = 0.090 \)- \((0.400)(-0.400) = -0.160 \)- \((0.200)(0.600) = 0.120 \)Add these values to find the numerator:\[ 0.090 - 0.160 + 0.120 = 0.050 \]

Step 8: Calculate the Denominator for \(\bar{y}\)

The denominator is already calculated as the sum of the masses: \[0.300 + 0.400 + 0.200 = 0.900 \text{ kg}\]

Step 9: Find the Y-Coordinate of the Center of Mass

Plug the results from Steps 7 and 8 into the formula for \(\bar{y}\):\[ \bar{y} = \frac{0.050}{0.900} \approx 0.056 \text{ m} \]

Step 10: Conclusion - Coordinates of the Center of Mass

The coordinates of the center of mass for the system of three chocolate blocks are approximately \( (0.044 \, \text{m}, 0.056 \, \text{m}) \).

Key Concepts

Mass Distribution

Mass distribution refers to how mass is spread out in a system. Imagine you have several chocolate blocks, all with different shapes and masses. The way their mass is distributed affects where the center of mass, or balance point, of the entire system is located.
Understanding mass distribution is crucial in identifying the center of mass in physics. This concept helps in determining how objects will behave when forces are applied. For instance, in our chocolate blocks problem, each block has a specific mass and precise location in space, all of which contribute to the system's mass distribution.
When finding the center of mass, we look at each object's mass and position in the coordinate system. This helps ensure that our calculations accurately represent the full picture of where the collective "center" of these masses lies. Consider it like finding a balancing point; the heavier or further an object is from this point, the more it affects the overall balance.

Coordinate System

In physics, a coordinate system allows us to locate objects precisely on a grid. Typically it involves an X (horizontal) and Y (vertical) axes to help pinpoint the exact position of an object in a two-dimensional space. When working with complex systems made of several objects, using a coordinate system can simplify the problem by providing a clear framework.
In the case of the chocolate blocks, each has specific coordinates, such as \((0.200 \text{ m}, 0.300 \text{ m})\), which tell you exactly where it lies on this grid. This helps everyone involved in the calculation to speak the same "language" and make the process straightforward.
Understanding how coordinates work is essential for correctly applying the formulas needed to calculate the center of mass. By placing all objects in a consistent coordinate system, it becomes easier to perform and understand the required calculations.

Physics Problem-Solving

Problem-solving in physics often involves several clear steps to ensure accuracy and understanding. It usually begins with understanding the parameters and what is being asked. For the example given with the chocolate blocks, the ultimate goal is to find the center of mass of these three blocks.
**Step-by-Step Approach:**

• Specification: Clearly define each object involved—know its mass and position.
• Formula Application: Use the correct formulas to integrate these known values into a solution. Here, the center of mass formula helps compile different components into a singular average point.
• Calculation: Perform the math carefully, ensuring each step is laid out, like calculating numerators and denominators separately before combining into a final solution.
• Interpret Results: Understanding the significance of the numbers obtained, such as \((0.044 \text{ m}, 0.056 \text{ m})\), provides a solid grasp of the object's balance point.

By following a methodical approach, physics problem-solving becomes a logical progression from given information to conclusive results, making complex problems more approachable and easier to tackle.