Problem 42 - Thomas Calculus

Question

Center of mass and moments of inertia for wire with variable density Find the center of mass and the moments of inertia about the coordinate axes of a thin wire lying along the curve $$ \mathbf{r}(t)=t \mathbf{i}+\frac{2 \sqrt{2}}{3} t^{3 / 2} \mathbf{j}+\frac{t^{2}}{2} \mathbf{k}, \quad 0 \leq t \leq 2 $$ if the density is \(\delta=1 /(t+1)\)

Step-by-Step Solution

Answer

Center of mass: \( (1, \frac{8\sqrt{2}}{15}, \frac{2}{3}) \); Moments of inertia: \( I_x = 5.85, I_y = 8, \) and \( I_z = 6.22. \)

Step 1: Arc Length Differential

The differential element of the arc length for a curve is given by \( ds = \sqrt{(x')^2 + (y')^2 + (z')^2} \, dt \). First, compute the derivatives of \( x(t) = t \), \( y(t) = \frac{2 \sqrt{2}}{3} t^{3/2} \), and \( z(t) = \frac{t^2}{2} \). We have \( x'(t) = 1 \), \( y'(t) = \sqrt{2} t^{1/2} \), and \( z'(t) = t \).

Step 2: Calculate ds

Substitute the derivatives into the arc length differential: \[ ds = \sqrt{1 + (\sqrt{2} t^{1/2})^2 + (t)^2} \, dt = \sqrt{1 + 2t + t^2} \, dt. \]Simplifying \( \sqrt{1 + 2t + t^2} = \sqrt{(t+1)^2} = t+1 \), we obtain \( ds = (t+1) \, dt \).

Step 3: Expressions for Center of Mass

For the center of mass, we calculate \( \bar{x}, \bar{y}, \bar{z} \) using \[ \bar{x} = \frac{1}{M} \int_0^2 x(t) \, \delta(t) \ ds, \]\[ \bar{y} = \frac{1}{M} \int_0^2 y(t) \, \delta(t) \ ds, \]\[ \bar{z} = \frac{1}{M} \int_0^2 z(t) \, \delta(t) \ ds, \]where \( M \) is the total mass given by \( \int_0^2 \delta(t) \ ds \).

Step 4: Compute Total Mass M

The total mass is \[ M = \int_0^2 \frac{1}{t+1} (t+1) \, dt = \int_0^2 1 \, dt = 2. \]

Step 5: Calculate Center of Mass (\( \bar{x}, \bar{y}, \bar{z} \))

Calculate each component:\[ \bar{x} = \frac{1}{2} \int_0^2 t \, dt = \frac{1}{2} \times \frac{t^2}{2} \Bigg|_0^2 = \frac{1}{2} \times 2 = 1, \]\[ \bar{y} = \frac{1}{2} \int_0^2 \frac{2\sqrt{2}}{3} t^{3/2} \, dt = \frac{1}{2} \times \frac{2\sqrt{2}}{3} \times \frac{2}{5} t^{5/2} \Bigg|_0^2 = \frac{8\sqrt{2}}{15}, \]\[ \bar{z} = \frac{1}{2} \int_0^2 \frac{t^2}{2} \, dt = \frac{1}{2} \times \frac{t^3}{6} \Bigg|_0^2 = \frac{4}{6} = \frac{2}{3}. \]

Step 6: Moments of Inertia Formulas

The moments of inertia \( I_x, I_y, I_z \) are calculated by:\[ I_x = \int_0^2 (y^2+z^2) \delta(t) \ ds, \]\[ I_y = \int_0^2 (x^2+z^2) \delta(t) \ ds, \]\[ I_z = \int_0^2 (x^2+y^2) \delta(t) \ ds. \]

Step 7: Calculate Moments of Inertia

Compute each moment:\[ I_x = \int_0^2 \left( \left(\frac{2\sqrt{2}}{3} t^{3/2}\right)^2 + \left(\frac{t^2}{2}\right)^2 \right) \delta(t) \ ds = \int_0^2 \left( \frac{8}{9} t^3 + \frac{t^4}{4} \right) \, dt \approx 5.85, \]\[ I_y = \int_0^2 (t^2 + \frac{t^4}{4}) \, dt = 8, \]\[ I_z = \int_0^2 \left( t^2 + \frac{8}{9} t^3 \right) \, dt = 6.22. \]

Step 8: Final Step: Solution Summary

The center of mass is \( (1, \frac{8\sqrt{2}}{15}, \frac{2}{3}) \) and the moments of inertia are approximately \( I_x = 5.85, I_y = 8, \) and \( I_z = 6.22. \)

Key Concepts

Moments of Inertia

Moments of inertia are a fascinating concept in physics and engineering that describe how the mass of an object is distributed relative to a given axis. Think of it as a measure of how difficult it is to rotate an object around that axis. The farther the mass is from the axis, the greater the moment of inertia. This principle is crucial in various applications, such as designing stable structures or rotating machinery.

To find the moments of inertia about different axes, we use integration. For a thin wire or curved object, we specifically calculate these moments by considering elements of mass along the curve. In this exercise, the wire's moments of inertia about the x, y, and z axes are determined using integrals. These involve expressions of the wire's coordinates and its variable density.

• For the x-axis: Integrate \( y^2 + z^2 \) over the wire.
• For the y-axis: Integrate \( x^2 + z^2 \).
• For the z-axis: Integrate \( x^2 + y^2 \).

Each integral incorporates the wire's density to account for varying mass along its length. The computed moments offer insight into how mass is distributed with respect to these axes, indirectly stating the wire's rotational stability.

Variable Density

In physics, variable density refers to a situation where the density of an object is not uniform throughout its volume or mass. This is common in real-world scenarios where materials might have varying compositions or thicknesses. For example, a wire might be thicker in some parts than others, leading to differences in how mass is distributed along its length.

In this exercise, the wire's density function is given as \( \delta = \frac{1}{t+1} \), where \( t \) represents a parameter along the wire's length. This implies that as \( t \) changes, so does the density. Instead of being constant, the density decreases as \( t \) increases.

• The density function plays a crucial role in calculations. It directly affects both the total mass and moments of inertia.
• To find the total mass, you integrate the density function over the given interval.
• The results influence how the center of mass and moments of inertia are computed, offering insights into mass distribution.

Understanding variable density is essential for accurate physical modeling, whether designing structures, predicting object behavior, or simply analyzing real-world phenomena.

Arc Length Differential

The arc length differential is an essential concept in calculus when dealing with curves. It represents an infinitesimally small segment of the curve, which we often denote as \( ds \). This differential helps calculate lengths of curves and becomes a tool to integrate over curves in 3D space.

The expression for \( ds \) on a curve parameterized by \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k} \) is given by \( ds = \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} \, dt \).

• First, compute the derivatives of the position functions: \( x'(t), y'(t), z'(t) \).
• Substitute these derivatives into the formula to get \( ds \).
• Simplifying this expression is crucial for integral calculations, especially if density is also a function of \( t \).

In this exercise, simplifying \( ds \) explains why the differential becomes \( (t+1) \, dt \), matching the structure of the variable density, simplifying further calculation steps.

Calculus Application

Calculus is a powerful tool that allows us to tackle complex physical problems by breaking them into manageable parts. When studying concepts like the center of mass or moments of inertia, calculus becomes indispensable.

For this problem, calculus helps to determine the center of mass of a wire with changing density along its curve by integrating the density function along its length. This application exemplifies how integral calculus is used in physical problem-solving.

• For the center of mass, integrate the position function weighted by density across the curve's length.
• Calculating total mass requires integrating the density itself, leading to a value that normalizes other calculations.
• Applying these principles aids in finding both the center of mass and understanding how inertia varies with different orientations.

Thus, calculus not only clarifies how mass and weight impact an object spatially but also enhances our understanding of how varied physical quantities interplay with dynamic systems. This layered application underscores the broad utility of calculus beyond straightforward numerical solutions.