Problem 79 - Thomas Calculus

Question

Variable density \(A\) solid is bounded below by the cone \(z=\sqrt{x^{2}+y^{2}}\) and above by the plane \(z=1 .\) Find the center of mass and the moment of inertia and radius of gyration about the \(z\) -axis if the density is a. \(\delta(r, \theta, z)=z\) b. \(\delta(r, \theta, z)=z^{2}\)

Step-by-Step Solution

Answer

For a), COM: \((0,0,\frac{2}{5})\), \(I_z = \frac{\pi}{12}\), \(k = \frac{1}{\sqrt{3}}\). For b), COM: \((0,0,\frac{5}{12})\), \(I_z = \frac{\pi}{14}\), \(k = \sqrt{\frac{5}{14}}\).

Step 1: Define the region and convert to cylindrical coordinates

The region is defined by the inequality \( \sqrt{x^2 + y^2} \leq z \leq 1 \). Converting to cylindrical coordinates, where \( x = r \cos\theta \), \( y = r \sin\theta \), and \( z = z \), the cone translates to \( z = r \). Thus, the solid is within: \( r \leq z \leq 1 \), for \( 0 \leq r \leq 1 \) and \( 0 \leq \theta < 2\pi \).

Step 2(a): Calculate mass for density \( \delta = z \)

Mass, \( M \), is given by integrating the density over the volume: \[ M = \int_0^{2\pi} \int_0^1 \int_r^1 z \, r \, dz \, dr \, d\theta \] First, integrate with respect to \( z \): \( \int_r^1 z \, dz = \frac{1}{2}(1^2 - r^2) \). The mass becomes: \[ M = \int_0^{2\pi} \int_0^1 \frac{1}{2}(1^2 - r^2)r \, dr \, d\theta = \int_0^{2\pi} \frac{1}{2} \int_0^1 (r - r^3) \, dr \, d\theta \] Evaluating: \( \int_0^1 (r - r^3) \, dr = \left[ \frac{r^2}{2} - \frac{r^4}{4} \right]_0^1 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \).Resulting in: \( M = \int_0^{2\pi} \frac{1}{2} \cdot \frac{1}{4} \, d\theta = \frac{1}{8} \int_0^{2\pi} \, d\theta = \frac{1}{8} \cdot 2\pi = \frac{\pi}{4} \).

Step 3(a): Calculate center of mass for \( \delta = z \)

The center of mass (COM) coordinates in cylindrical are given by:\[ \bar{x} = \frac{1}{M} \int_V x \delta \, dV, \; \bar{y} = \frac{1}{M} \int_V y \delta \, dV, \; \bar{z} = \frac{1}{M} \int_V z \delta \, dV \] Due to symmetry, \( \bar{x} = \bar{y} = 0 \). Calculate \( \bar{z} \):\[ \bar{z} = \frac{1}{M} \int_0^{2\pi} \int_0^1 \int_r^1 z \cdot z \, r \, dz \, dr \, d\theta \] Integrate \( z^2 \): \( \int_r^1 z^2 \, dz = \left[ \frac{z^3}{3} \right]_r^1 = \frac{1}{3} - \frac{r^3}{3} \). So, \[ \bar{z} = \frac{1}{\frac{\pi}{4}} \times 2\pi \times \int_0^1 \frac{r}{3}(1 - r^3) \, dr \] Evaluate: \( \int_0^1 (\frac{r}{3} - \frac{r^4}{3}) \, dr = \left[ \frac{r^2}{6} - \frac{r^5}{15} \right]_0^1 = \frac{1}{6} - \frac{1}{15} = \frac{5}{30} - \frac{2}{30} = \frac{1}{10} \).Thus, \( \bar{z} = \frac{4}{\pi} \times \frac{1}{10} = \frac{2}{5} \).The COM is \((0, 0, \frac{2}{5}) \).

Step 4(a): Calculate moment of inertia and radius of gyration for \( \delta = z \)

The moment of inertia about the \( z \)-axis is \( I_z = \int_V (x^2 + y^2) \delta \, dV = \int_V r^2 z \, dV \).\[ I_z = \int_0^{2\pi} \int_0^1 \int_r^1 r^2 z \, r \, dz \, dr \, d\theta \] Integrate \( z \, dz \) gives \( \int_r^1 z \, dz = \frac{1}{2} (1^2 - r^2) \).So, \[ I_z = \int_0^{2\pi} \frac{1}{2} \int_0^1 (r^3 - r^5) \, dr \, d\theta \] Evaluate: \( \int_0^1 (r^3 - r^5) \, dr = \left[ \frac{r^4}{4} - \frac{r^6}{6} \right]_0^1 = \frac{1}{4} - \frac{1}{6} = \frac{3}{12} - \frac{2}{12} = \frac{1}{12} \).Thus, \( I_z = \int_0^{2\pi} \frac{1}{2} \cdot \frac{1}{12} \, d\theta = \frac{1}{24} \cdot 2 \pi = \frac{\pi}{12} \).The radius of gyration, \( k \), is defined: \[ k = \sqrt{\frac{I_z}{M}} = \sqrt{\frac{\frac{\pi}{12}}{\frac{\pi}{4}}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \].

Step 2(b): Calculate mass for density \( \delta = z^2 \)

Using the density \( \delta = z^2 \), calculate mass:\[ M = \int_0^{2\pi} \int_0^1 \int_r^1 z^2 \, r \, dz \, dr \, d\theta \] Integrate with \( z^2 \, dz \):\( \int_r^1 z^2 \, dz = \left[ \frac{z^3}{3} \right]_r^1 = \frac{1}{3} - \frac{r^3}{3} \).Mass becomes:\[ M = \int_0^{2\pi} \int_0^1 \frac{1}{3}(1 - r^3)r \, dr \, d\theta = \int_0^{2\pi} \frac{1}{3} \int_0^1 (r - r^4) \, dr \, d\theta \] Evaluate:\( \int_0^1 (r - r^4) \, dr = \left[ \frac{r^2}{2} - \frac{r^5}{5} \right]_0^1 = \frac{1}{2} - \frac{1}{5} = \frac{5}{10} - \frac{2}{10} = \frac{3}{10} \).\( M = \int_0^{2\pi} \frac{1}{3} \cdot \frac{3}{10} \, d\theta = \frac{1}{10} \cdot 2\pi = \frac{\pi}{5} \).

Step 3(b): Calculate center of mass for \( \delta = z^2 \)

The center of mass, \( \bar{z} \), for \( \delta = z^2 \) is:\[ \bar{z} = \frac{1}{M} \int_0^{2\pi} \int_0^1 \int_r^1 z^3 \, r \, dz \, dr \, d\theta \] First integrate \( z^3 \, dz \):\( \int_r^1 z^3 \, dz = \left[ \frac{z^4}{4} \right]_r^{1} = \frac{1}{4} - \frac{r^4}{4} \).This gives the center of mass height:\[ \bar{z} = \frac{5}{\pi} \cdot 2\pi \times \int_0^1 \frac{3r}{4}(1 - r^4) \, dr \] Evaluate:\( \int_0^1 \left(\frac{3r}{4} - \frac{3r^5}{4}\right) \, dr = \left[ \frac{3r^2}{8} - \frac{3r^6}{24}\right]_0^{1} = \frac{3}{8} - \frac{3}{24} = \frac{9}{24} - \frac{3}{24} = \frac{1}{3} \).\( \bar{z} = \frac{5}{4} \cdot \frac{1}{3} = \frac{5}{12} \).Thus, COM is \((0, 0, \frac{5}{12}) \).

Step 4(b): Calculate moment of inertia and radius of gyration for \( \delta = z^2 \)

Calculate \( I_z \) for \( \delta = z^2 \):\[ I_z = \int_0^{2\pi} \int_0^1 \int_r^1 r^2 z^2 \, r \, dz \, dr \, d\theta \] Integrate \( z^2 \, dz \): \( \int_r^1 z^2 \, dz = \left[ \frac{z^3}{3} \right]_r^1 = \frac{1}{3} - \frac{r^3}{3} \). Now compute:\[ I_z = \int_0^{2\pi} \frac{1}{3} \int_0^1 (r^3 - r^6) \, dr \, d\theta \] Evaluate: \( \int_0^1 (r^3 - r^6) \, dr = \left[ \frac{r^4}{4} - \frac{r^7}{7} \right]_0^1 = \frac{1}{4} - \frac{1}{7} = \frac{7}{28} - \frac{4}{28} = \frac{3}{28} \).\( I_z = \int_0^{2\pi} \frac{1}{3} \cdot \frac{3}{28} \, d\theta = \frac{1}{28} \, \cdot 2\pi = \frac{\pi}{14} \).Radius of gyration:\[ k = \sqrt{\frac{I_z}{M}} = \sqrt{\frac{\frac{\pi}{14}}{\frac{\pi}{5}}} = \sqrt{\frac{5}{14}} \].

Key Concepts

Center of Mass

The center of mass is a critical concept in physics and engineering. It acts as the balance point of an object where all its mass can be considered to be concentrated. In this exercise, we are determining the center of mass for a solid bounded by a cone and a plane, using two different density functions:

• For density \( \delta = z \), the mass distribution is higher along the axis of the cone, leading to a center of mass located at \( (0, 0, \frac{2}{5}) \).
• For density \( \delta = z^2 \), the higher concentration of mass is even more towards the tip of the cone, resulting in a center of mass at \( (0, 0, \frac{5}{12}) \).

Both calculations utilize symmetry, allowing the simplification that the x and y components of the center of mass are zero. This aligns with the solid's rotational symmetry around the z-axis. The heights of the centers of mass from the base of the cone are determined through integration, taking into account the distribution of mass with variable density.

Moment of Inertia

Moment of inertia measures how difficult it is to rotate an object about an axis. It depends on both the geometry and mass distribution of the object. For our solid, we calculate the moment of inertia around the z-axis:

• With \( \delta = z \), the calculation involves integrating \( r^2z \), where \( r \) is the radial distance. The result is \( I_z = \frac{\pi}{12} \), indicating how mass farther from the axis contributes more to the moment of inertia.
• With \( \delta = z^2 \), we have \( I_z = \frac{\pi}{14} \). Here, the mass is distributed more towards the upper parts, which results in a different inertia value.

In both cases, these values illustrate how varying densities affect rotational dynamics, with denser regions creating larger moments of inertia.

Radius of Gyration

The radius of gyration provides a measure of the distribution of an object's mass around its axis of rotation and is used to simplify complex geometric configurations. Mathematically, it is represented as: \( k = \sqrt{\frac{I_z}{M}} \).

• For \( \delta = z \), \( k = \frac{1}{\sqrt{3}} \). This simplified measure helps understand how mass is spread out along the solid's volume with respect to the axis.
• For \( \delta = z^2 \), \( k = \sqrt{\frac{5}{14}} \), showing a slightly different spread due to the squared density function.

These values provide insight into structural stability and rotational properties, essential for engineering analysis and design.

Variable Density

A variable density indicates that mass distribution changes throughout the object. This can affect both physical properties and the calculation methods required.

• In the context of the solid bounded by a cone and plane, having densities \( \delta = z \) and \( \delta = z^2 \) shows how density increases with height, impacting center of mass, moment of inertia, and radius of gyration.
• Mathematically, handling variable density involves more complex integrations. It requires taking into account how density varies with location within the solid, leading to more nuanced results compared to uniform density situations.

Understanding and calculating properties with variable density is crucial in real-world applications where material density never remains constant across a volume, such as in atmospheric, material sciences, and fluid dynamics.