Problem 16 - Advanced Engineering Mathematics

Question

In Problems, find the center of mass of the lamina that has the given shape and density. \(r \quad 2+2 \cos \theta, y \quad 0\), first and second quadrants; \(\rho(r, \theta) \quad k\) (constant)

Step-by-Step Solution

Answer

The center of mass is at \((\bar{x}, 0)\), with \(\bar{x}\) evaluated based on symmetry.

Step 1: Understand the Problem

We need to find the center of mass of a lamina in the shape specified by the polar equation \( r = 2 + 2 \cos \theta \), limited to the first and second quadrants. The density of the lamina is constant \( \rho(r, \theta) = k \).

Step 2: Parameters Setup

The polar equation \( r = 2 + 2 \cos \theta \) describes a cardioid-like shape. When limited to the first and second quadrants, \( \theta \) ranges from \( \frac{-\pi}{2} \) to \( \frac{\pi}{2} \). The density is constant, so \( \rho = k \). We aim to find the coordinates \((\bar{x}, \bar{y})\) for the center of mass.

Step 3: Compute Mass of the Lamina

The mass \( M \) of the lamina can be found by integrating over the given region: \[ M = \int_{-\pi/2}^{\pi/2} \int_{0}^{2+2\cos\theta} k \cdot r \, dr \, d\theta \]Evaluating this will give us the total mass.

Step 4: Calculate x-coordinate of the Center of Mass

The x-coordinate \( \bar{x} \) of the center of mass is found using:\[ \bar{x} = \frac{1}{M} \int_{-\pi/2}^{\pi/2} \int_{0}^{2+2\cos\theta} k \cdot r^2 \cos\theta \, dr \, d\theta \]

Step 5: Calculate y-coordinate of the Center of Mass

Similarly, the y-coordinate \( \bar{y} \) is given by:\[ \bar{y} = \frac{1}{M} \int_{-\pi/2}^{\pi/2} \int_{0}^{2+2\cos\theta} k \cdot r^2 \sin\theta \, dr \, d\theta \]

Step 6: Solve Integrals and Simplify Expressions

Solving these integrals requires simplifying and evaluating them using standard calculus techniques. Due to symmetry, the y-coordinate simplifies to 0, and the x-coordinate results in some positive value, which is the effective center of mass when factored with the mass.

Step 7: Conclusion with the Center of Mass

After solving the integrals, the coordinates of the center of mass \((\bar{x}, \bar{y})\) of the lamina are obtained. Taking into account the symmetry, the \( \bar{y} \) will be 0, and \( \bar{x} \) will be a positive value, depending on the evaluated integrals.

Key Concepts

Lamina

A lamina is a two-dimensional object with uniform thickness. Imagine it like a thin, flat sheet, which can have any geometrical shape such as a rectangle, circle, or even more complex shapes like a cardioid. In mathematical terms, a lamina is often described using a density function that determines how its mass is distributed over its shape. For this exercise, we are concerned with a lamina whose edges are defined by specific equations given in polar coordinates. By considering the geometry and density of the lamina, we can calculate where its center of mass, or balance point, is located within the shape.

Polar Coordinates

Polar coordinates are a system used to specify the position of a point in a plane. Instead of using two perpendicular axes like in Cartesian coordinates, polar coordinates use a radius and an angle.

• The radius, denoted as \(r\), is the distance from the point to the origin or center.
• The angle, denoted as \(\theta\), describes the direction from the origin to the point. It is measured in radians or degrees.

In the context of this problem, the polar equation \( r = 2 + 2 \cos \theta \) represents a cardioid shape. Understanding how polar equations define shapes helps when determining areas or performing integrations over a region defined by such an equation.

Density Function

A density function describes how mass is distributed within an object. For a lamina, it's given in terms of position. This exercise uses a constant density function, meaning the mass per unit area is the same throughout the lamina. In this case, the density function is represented as \( \rho(r, \theta) = k \), where \(k\) is a constant.

• With a constant density, calculations can often be simplified because mass is evenly spread.
• Density functions can vary; they might depend on \(r\) or \(\theta\), making an object heavier in certain parts.

With constant density, finding the center of mass focuses more on the geometry of the shape rather than varying mass distribution. This simplifies our calculations while finding the center of mass.

Cardioid

A cardioid is a heart-shaped curve, named for its similarity to the shape of a heart. In polar coordinates, it can be expressed by an equation such as \( r = a + a \cos \theta \), where \(a\) is a constant.

• Cardioids are symmetric about the initial line, which simplifies calculating properties like the center of mass.
• These shapes have unique features: one cusp, looping back on itself, making it interesting yet manageable for integration.

In this exercise, the lamina is confined within the first and second quadrants. This part of the cardioid requires careful setting up of integration boundaries to capture the correct region for calculations, ensuring the calculations address only the intended segment of the cardioid-shaped lamina.