Problem 262 - Calculus Volume 1

Question

For the following exercises, compute the center of mass \(\bar{x}\). \(\rho=1\) for \(x \in(0,1)\) and \(\rho=2\) for \(x \in(1,2)\).

Step-by-Step Solution

Answer

The center of mass \(\bar{x}\) is \(\frac{7}{6}\).

Step 1: Define the Problem

We are tasked to compute the center of mass, \(\bar{x}\), for a density function \(\rho(x)\) that varies over the interval \(x \in (0, 2)\). The density is given as 1 for \(x \in (0, 1)\) and 2 for \(x \in (1, 2)\).

Step 2: Find the Total Mass

The total mass \(M\) is found by integrating \(\rho(x)\) over the interval \(x \in (0, 2)\). So we need to calculate: \[M = \int_{0}^{1} 1 \, dx + \int_{1}^{2} 2 \, dx\] This can be solved as: \[M = \left[x\right]_{0}^{1} + \left[2x\right]_{1}^{2}\] \[M = (1 - 0) + (4 - 2) = 1 + 2 = 3\]

Step 3: Set Up the Integral for the First Moment

The first moment about the origin is given by: \[M_x = \int_{0}^{1} x \cdot 1 \, dx + \int_{1}^{2} x \cdot 2 \, dx\]Let's compute these integrals separately.

Step 4: Compute the First Moment Integrals

First, solve \[\int_{0}^{1} x \, dx = \left[\frac{x^2}{2}\right]_{0}^{1} = \frac{1}{2}\]. Next, solve \[\int_{1}^{2} 2x \, dx = 2\left[\frac{x^2}{2}\right]_{1}^{2} = \left[x^2\right]_{1}^{2} = (4 - 1) = 3\].Thus, \[M_x = \frac{1}{2} + 3 = \frac{1}{2} + \frac{6}{2} = \frac{7}{2}\].

Step 5: Find the Center of Mass

The center of mass, \(\bar{x}\), is calculated by dividing the first moment by the total mass: \[\bar{x} = \frac{M_x}{M} = \frac{\frac{7}{2}}{3} = \frac{7}{6}.\]

Step 6: Conclusion

The computed center of mass \(\bar{x}\) is \(\frac{7}{6}\). This represents the point on the x-axis where the average mass is located, given the varying density of the object.

Key Concepts

Density Function

The density function, denoted \(\rho(x)\), is a crucial mathematical concept used in physics and engineering to represent how mass is distributed across an object or a region. In our exercise, \ \rho(x) \ varies over the interval from \(0\) to \(2\) and is defined piecewise as:

• \ \rho(x) = 1 \ for \(0 < x < 1\)
• \ \rho(x) = 2 \ for \(1 < x < 2\)

This density function tells us that for \(x\) values between \(0\) and \(1\), the density is uniform and equals \(1\). For \(x\) values between \(1\) and \(2\), the density increases and becomes \(2\). Understanding this change in density is key to determining the overall mass and the center of mass of the system. It informs us how mass is distributed over different regions of \(x\), which in turn affects the calculations of mass and moment integrals.

Integration

Integration is a mathematical process used to add up infinitely small values over a range to find quantities like total mass or moments. In this exercise, we apply integration to calculate these quantities using our density function.

For the total mass \(M\), we integrate the density function over its domain: \[ M = \int_{0}^{1} 1 \, dx + \int_{1}^{2} 2 \, dx \]

• The first integral \(\int_{0}^{1} 1 \, dx\) accounts for the range where density is \(1\).
• The second integral \(\int_{1}^{2} 2 \, dx\) encompasses the range where density increases to \(2\).

These integrals tell us the cumulative effect of density over each interval, leading to a total mass calculation.

Similarly, when computing the first moment, integration helps sum the product of the position \(x\) and the density over the intervals, providing insight into how position affects the overall distribution of mass.

Mass Calculation

Mass calculation is about determining the total mass \(M\) of an object using a given density function across a specified interval. We achieve this by integrating the density function over the desired range. In our problem, this entails two separate calculations:

• For \(x \) from \(0\) to \(1\), with a constant density of \(1\).
• For \(x \) from \(1\) to \(2\), with an increased density of \(2\).

The integral for each segment is solved separately and then summed, simplifying the calculation process.

Performing the integration gives us:

• \( \left[x\right]_{0}^{1} = 1 \) for the first interval, and
• \( \left[2x\right]_{1}^{2} = 2 \) for the second interval.

Adding these results provides a total mass \(M = 3\). Accurately calculating mass is essential for determining other properties like the center of mass, as it represents the cumulative amount of matter present in the object.

First Moment

The first moment about the origin, denoted as \(M_x\), measures the distribution of mass at various distances from the origin. This concept provides insight into the balancing point or center of mass for an object.

In our exercise, the first moment is calculated by integrating the product of the position \(x\) and the density \(\rho(x)\) across the specified intervals: \[ M_x = \int_{0}^{1} x \cdot 1 \, dx + \int_{1}^{2} x \cdot 2 \, dx \]

• \( \int_{0}^{1} x \, dx = \frac{1}{2} \)
• \( \int_{1}^{2} 2x \, dx = 3 \)

These integrals reflect how the mass is distributed along the length of the object.

Summation of these contributions gives the total first moment: \(M_x = \frac{7}{2}\).

Dividing this by the total mass yields the center of mass

• \(\bar{x} = \frac{7}{6}\)

This value ensures that the calculated point equates to the average position of mass distribution, where the entire weight "balances" on this point of the x-axis.