Problem 15 - Advanced Engineering Mathematics

Question

Find the center of mass of the lamina that has the given shape and density. Outside \(r \quad 2\) and inside \(r \quad 2+2 \cos \theta, y \quad 0\), first quadrant; density at a point \(P\) inversely proportional to the distance from the pole

Step-by-Step Solution

Answer

The center of mass is \((\overline{x}, \overline{y})\) calculated from moments with respect to the area formula in polar coordinates.

Step 1: Understand the Shape

The lamina is bounded outside by the circle of radius 2 and inside by a cardioid described by the polar equation \(r = 2 + 2 \cos \theta\). Both constraints are in the first quadrant and above the line \(y = 0\). The density \(\delta(P)\) at any point \(P(r, \theta)\) is inversely proportional to the distance from the pole, so \(\delta(P) = \frac{k}{r}\) for some constant \(k\).

Step 2: Set Up Polar Integrals for Mass

The mass \(m\) of the lamina is given by integrating the density over the region in polar coordinates. The limits of integration for \(\theta\) are from \(0\) to \(\frac{\pi}{2}\), and for \(r\), from \(2\) to \(2 + 2 \cos \theta\). Thus, the mass \(m\) is \(\int_{0}^{\pi/2} \int_{2}^{2 + 2 \cos \theta} \frac{k}{r} \cdot r \, dr \, d\theta = k \int_{0}^{\pi/2} \int_{2}^{2 + 2 \cos \theta} \, dr \, d\theta\).

Step 3: Compute the Mass Integral

Compute \(\int_{2}^{2 + 2 \cos \theta} \, dr\), which gives \(r\) evaluated from \(2\) to \(2 + 2 \cos \theta\). This is \(2 \cos \theta\). The total mass \(m\) is thus \(k \int_{0}^{\pi/2} 2 \cos \theta \, d\theta\). Integrate \(\int_{0}^{\pi/2} 2 \cos \theta \, d\theta\) to get \(2 \sin \theta\) evaluated from \(0\) to \(\frac{\pi}{2}\), which is \(2\). Hence, \(m = 2k\).

Step 4: Set Up Integrals for Center of Mass

For the coordinates \((\overline{x}, \overline{y})\) of the center of mass, use the moment integrals. For \(\overline{x}\), calculate \(\frac{1}{m} \int \int x \delta(P) \, dA\), and for \(\overline{y}\), \(\frac{1}{m} \int \int y \delta(P) \, dA\). In polar coordinates, \(x = r \cos \theta\) and \(y = r \sin \theta\).

Step 5: Calculate Moment for \(\overline{x}\)

Compute \(\overline{x} = \frac{1}{2k} \int_{0}^{\pi/2} \int_{2}^{2 + 2 \cos \theta} r \cos \theta \cdot \frac{k}{r} \, r \, dr \, d\theta = \frac{1}{2} \int_{0}^{\pi/2} \cos \theta \int_{2}^{2 + 2 \cos \theta} r \, dr \, d\theta\). Evaluate the integral \(\int r^2/2\) from \(r = 2\) to \(r = 2 + 2 \cos \theta\) to find the moment about the y-axis.

Step 6: Calculate Moment for \(\overline{y}\)

Compute \(\overline{y} = \frac{1}{2k} \int_{0}^{\pi/2} \int_{2}^{2 + 2 \cos \theta} r \sin \theta \cdot \frac{k}{r} \, r \, dr \, d\theta = \frac{1}{2} \int_{0}^{\pi/2} \sin \theta \int_{2}^{2 + 2 \cos \theta} r \, dr \, d\theta\). Evaluate to find the moment about the x-axis.

Step 7: Final Calculation of Moments

Carry out the integrations to find specific values for \(\overline{x}\) and \(\overline{y}\). Substitute into \(\int_{2}^{2 + 2 \cos \theta} r^2 \, dr\) to find the exact values of the integrals for each component. Simplify according to trigonometric integrals.

Key Concepts

Polar Coordinates

Polar coordinates are a way to describe a location in a plane using the distance from a fixed point, called the pole, and an angle from a fixed direction. Instead of using the traditional x-y coordinate system, polar coordinates use two values: the radius \(r\) and the angle \(\theta\).

• The radius \(r\) is the distance from the pole to the point.
• The angle \(\theta\) is measured from a fixed direction, often the positive x-axis.

This system is incredibly useful for problems involving circular and spiral shapes. It simplifies the mathematical description of shapes like circles and cardioids, which are naturally expressed in terms of distance and angle.
In this particular problem, the lamina's boundaries are defined in polar coordinates—an outer circle with radius 2, and an inner boundary described by the cardioid equation \(r = 2 + 2 \cos \theta\). This setting is perfect for integrating based on angular and radial components.

Density Distribution

Density distribution describes how mass is spread across a given area. In many physical problems, knowing how density changes with position is crucial for calculating the center of mass. The density \(\delta(P)\) at a point \(P\) can depend on various factors, such as distance from a specific point. For our lamina, the density is inversely proportional to the distance from the pole, expressed as \(\delta(P) = \frac{k}{r}\). This relationship means:

• Closer points to the pole have higher density.
• Density decreases as distance from the pole increases.

This inverse relation is common in problems of gravitational fields or optical systems, where strength decreases over distance. Understanding this helps set up the proper integrals for mass and moments needed to find the center of mass accurately.

Integrals in Calculus

Integrals are fundamental in calculus, especially when dealing with areas, volumes, and curves described in non-linear ways. In our task, integrals help calculate key properties such as mass and the center of mass of the lamina. When setting up the integral in polar coordinates:

• The limits for \(\theta\) are from \(0\) to \(\frac{\pi}{2}\), encompassing the first quadrant.
• The radial limits for \(r\) come from the two curves: the constant radius 2 and the cardioid \(r = 2 + 2 \cos \theta\).

This structured approach allows us to compute the total mass \(m\) of the lamina and its center of mass by integrating over the area it covers.
Through expressions like \(\int_{0}^{\pi/2} \int_{2}^{2 + 2 \cos \theta} \frac{k}{r} \cdot r \, dr \, d\theta\), we're essentially 'summing up' the contribution of every infinitesimal area of the lamina, weighted by its density.

Moments of Inertia

Moments of inertia are crucial concepts in mechanics, relating to how mass is distributed relative to an axis of rotation or reference. For lamina's center of mass calculation, moments help balance the distribution of mass across the plane. To find the center of mass, moments \((\overline{x}, \overline{y})\) are computed using:

• The moment about the \(y\)-axis, which integrates \(x \delta(P)\).
• The moment about the \(x\)-axis, which integrates \(y \delta(P)\).

In polar coordinates again:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)

These components are part of the integral expressions needed to compute the lamina's balance point. Thus, moments of inertia in this context are not just about mass distribution relative to axes but essential for understanding how the overall mass center emerges from density variations and boundary conditions. Understanding these leads to precise computation of the object's balancing point and is foundational in any problem requiring center of mass determination.