Problem 20 - Calculus

Question

A wire of constant density \(\rho\) lies on the semicircle \(x^{2}+y^{2}=a^{2}, y \geqslant 0 .\) Find its mass \(M\) and also its moment \(M_{x}=\int \rho y d s .\) Where is its center of mass \(\bar{x}=M_{y} / M, \bar{y}=M_{x} /\) \(M ?\)

Step-by-Step Solution

Answer

Mass: \(\rho a \pi\), Moment: \(2 \rho a^2\), Center of Mass: \((0, \frac{2a}{\pi})\)."

Step 1: Parameterize the Semicircle

The semicircle can be parameterized using polar coordinates. Suppose the semicircles lies above the x-axis, which means we consider the parameterization for the upper half of the circle:\[ x = a \cos\theta, \quad y = a \sin\theta \]where \(\theta\) ranges from \(0\) to \(\pi\).

Step 2: Determine the Differential Arc Length

The differential arc length \(d s\) for a parametric curve \((x(\theta), y(\theta))\) is given by:\[ d s = \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} \, d\theta \]Calculating the derivatives: \(\frac{dx}{d\theta} = -a\sin\theta\), \(\frac{dy}{d\theta} = a\cos\theta\).Thus,\[ d s = \sqrt{(-a\sin\theta)^2 + (a\cos\theta)^2} \, d\theta = a \, d\theta \]

Step 3: Calculate the Mass of the Wire

The mass \(M\) of the wire is given by integrating the density over the semicircle:\[ M = \int \rho \, d s \]Substitute \(d s\) from Step 2:\[ M = \int_0^\pi \rho \, a \, d\theta = \rho \, a \int_0^\pi d\theta = \rho \, a \pi \]

Step 4: Calculate the Moment About the x-axis

The moment \(M_x\) with respect to the x-axis is:\[ M_x = \int \rho \, y \, d s \]Substituting \(y = a \sin\theta\) and \(d s = a \, d\theta\):\[ M_x = \int_0^\pi \rho \, a \sin\theta \, a \, d\theta = \rho \, a^2 \int_0^\pi \sin\theta \, d\theta \]Evaluating the integral:\[ \int_0^\pi \sin\theta \, d\theta = [-\cos\theta]_0^\pi = (1 - (-1)) = 2 \]Thus:\[ M_x = 2 \rho \, a^2 \]

Step 5: Calculate the Center of Mass

The coordinates of the center of mass \((\bar{x}, \bar{y})\) are given by:\[ \bar{x} = \frac{M_y}{M} \quad \text{and} \quad \bar{y} = \frac{M_x}{M} \]Since the semicircle is symmetric about the y-axis, \(M_y = 0\) and thus \(\bar{x} = 0\).Using the mass and moment calculated:\[ \bar{y} = \frac{2 \rho \, a^2}{\rho \, a \pi} = \frac{2 a}{\pi} \]

Step 6: Conclusion

The mass of the wire is \(M = \rho a \pi\). The moment about the x-axis is \(M_x = 2 \rho a^2\). The center of mass is located at \((\bar{x}, \bar{y}) = \left(0, \frac{2a}{\pi}\right)\).

Key Concepts

Mass of Wire

The mass of a wire refers to the measure of its total amount due to its density and length. In this exercise, we're dealing with a semicircular wire with constant density \[ \rho \text{ (rho)} \].To find the mass \( M \) of the wire, we integrate its density over the length of the semicircle.
The semicircle is parameterized, and we express its length in terms of the polar coordinate \( \theta \). The integration involves calculating the arc length differential \( d s \). Due to the setup, the arc length \[ d s = a \, d \theta \],leading to the mass calculation:

• Integrate the density \( \rho \) along the semicircle.
• Use the integral formula: \( M = \int \rho \, d s \).
• Substitute \( d s = a \, d \theta \) into this formula.

This provides us with the mass \( M = \rho a \pi \), which shows how density and geometry affect the total mass of the wire.

Moment of a Semicircle

Moments help us describe the distribution of mass or density about a specific axis. For a semicircle, this exercise calculates the moment about the x-axis, denoted as \( M_x \).
The formula applied here is:\[ M_x = \int \rho \, y \, ds \]which uses the y-coordinate in its calculation to show the distribution vertically.

• Use the relation \( y = a \sin \theta \) for the semicircle's height.
• Incorporate \( d s = a \, d \theta \), from the arc length differential.

With these substitutions, the expression simplifies and the integral evaluates to:\[ M_x = 2\rho a^2 \].This calculation underscores the importance of how distributed mass is represented by moments, which balances around an axis, and provides insights into the wire's stability and balance.

Parametric Equations

Parametric equations are instrumental for expressing curves and shapes in a convenient form for integration and analysis. In this context, the semicircle's curve is expressed using polar parametric equations:\[x = a \cos \theta, \, y = a \sin \theta\]These equations exhibit how changing the parameter \( \theta \) allows us to trace out each point on the semicircle.

• \( x = a \cos \theta \) provides horizontal positioning.
• \( y = a \sin \theta \) gives vertical height.

By using a parametric form, we can simplify calculations, such as derivatives or integrals involving this curved path. The parameters highlight flexibility, allowing the expressions to adapt and provide clarity on geometric properties.

Arc Length Integration

Arc length integration is a clever tool in calculus, used here for finding the total stretch or distance along a curve, like the semicircular wire. The goal is to express the arc length differential \( ds \) for precise integration.
The general form for arc length \[ ds = \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} \, d\theta \]looks complex, but it simplifies significantly for circular paths.

• Compute derivatives for parametric expressions: \( \frac{dx}{d\theta} = -a \sin\theta \; \text{and} \; \frac{dy}{d\theta} = a \cos\theta \).
• The distance expression shrinks to \( ds = a \, d\theta \), rooted in the semicircle's constant radius.

This straightforward result occurs due to constant radial symmetry, easing our integration process and allowing us to proceed to further calculations like mass or moment integrations with ease.