Problem 39 - Thomas Calculus

Question

Two springs of constant density A spring of constant density \(\delta\) lies along the helix $$ \mathbf { r } ( t ) = ( \cos t ) \mathbf { i } + ( \sin t ) \mathbf { j } + t \mathbf { k } , \quad 0 \leq t \leq 2 \pi $$ a. Find \(I _ { z }\) b. Suppose that you have another spring of constant density \(\delta\) that is twice as long as the spring in part (a) and lies along the helix for \(0 \leq t \leq 4 \pi .\) Do you expect \(I _ { z }\) for the longer spring to be the same as that for the shorter one, or should it be different? Check your prediction by calculating \(I _ { z }\) for the longer spring.

Step-by-Step Solution

Answer

The longer spring has a moment of inertia \( I_z = 4\delta\pi \sqrt{2} \), twice that of the shorter spring.

Step 1: Define the Moment of Inertia Expression

The moment of inertia, \( I_z \), about the z-axis for a given mass distribution, is given by the integral:\[ I_z = \int r^2 \delta \, dm \]where \( r \) is the distance from the z-axis, which in this case is equivalent to \( \sqrt{x^2 + y^2} \). Here, \( \delta \) is the constant density.

Step 2: Express the Helix Parameterization

The helix is parameterized as \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} + t \mathbf{k} \). Here, the radial distance from the z-axis is given by:\[ r(t) = \sqrt{\cos^2 t + \sin^2 t} = 1 \]so the distance \( r \) simplifies to 1 for each \( t \).

Step 3: Compute the Mass Element

Since \(t\) varies as the parameter, we calculate the mass element using the differential of arc length:\[ dm = \delta \cdot ds = \delta \cdot \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} \, dt \]Where, \( \frac{dx}{dt} = -\sin t \), \( \frac{dy}{dt} = \cos t \), \( \frac{dz}{dt} = 1 \). Thus,\[ ds = \sqrt{\sin^2 t + \cos^2 t + 1} \, dt = \sqrt{2} \, dt \]

Step 4: Integrate Over the Interval [0, 2π]

For the first spring (\(0 \leq t \leq 2\pi\)):\[ I_z = \delta \int_{0}^{2\pi} (1)^2 \, \sqrt{2} \, dt = \delta \cdot \sqrt{2} \int_{0}^{2\pi} \, dt \]Evaluating the integral gives:\[ I_z = \delta \cdot \sqrt{2} \cdot [t]_{0}^{2\pi} = \delta \cdot \sqrt{2} \cdot 2\pi = 2\delta\pi \sqrt{2} \]

Step 5: Integrate Over the Interval [0, 4π]

For the longer spring (\(0 \leq t \leq 4\pi\)):\[ I_z = \delta \int_{0}^{4\pi} (1)^2 \, \sqrt{2} \, dt = \delta \cdot \sqrt{2} \int_{0}^{4\pi} \, dt \]Evaluating this integral gives:\[ I_z = \delta \cdot \sqrt{2} \cdot [t]_{0}^{4\pi} = \delta \cdot \sqrt{2} \cdot 4\pi = 4\delta\pi \sqrt{2} \]

Step 6: Compare the Results

Since the moment of inertia directly scales with the length of the spring, the moment of inertia for the longer spring is twice that of the shorter one calculated earlier. This matches our expectations.

Key Concepts

Spring Constant

The spring constant, often denoted by the symbol \( k \), is an important property of a spring. It tells us how stiff or flexible the spring is. The spring constant is defined as the force needed to compress or extend the spring by one unit of length. A high spring constant means the spring is very stiff, requiring more force to stretch or compress it.

In the context of integral calculus and moment of inertia, the spring constant can influence how a spring behaves under various forces.
It's important to understand how the density, shape, and span of a spring affect its stiffness. When dealing with helix shapes, the spring constant helps in understanding how these geometric properties will impact overall performance.

Helix Parameterization

To understand the helix parameterization, let's consider its mathematical description.
A helix is a three-dimensional curve that spirals around a central axis at a constant distance. It can be described using a parameter \( t \) with the functions of trigonometric equations:

\( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} + t \mathbf{k} \).

This description provides a way to "trace" the path of the helix as the parameter \( t \) changes from start to end. In simpler terms, as \( t \) varies, it "draws" the helix in 3D space. The parameterization is useful when evaluating properties like arc length or calculating integrals over the path of the helix. It allows us to determine various physical characteristics related to springs or other materials formed in a helical shape.

Integral Calculus

Integral calculus is a fundamental tool for solving problems involving continuous change. In the context of physics and engineering, it is extremely useful for finding quantities like area, volume, and in this case, moment of inertia.

When working with helix-shaped objects such as springs, integral calculus helps sum up infinitesimally small parts to calculate the total effect. Here, we use integral calculus to find the moment of inertia, \( I_z \), of springs parameterized by a helix.
The integral involves expressions like

• \( I_z = \int r^2 \delta \, dm \), which is crucial in understanding the distribution of mass.

Integral calculus handles these sums even when the variables change continuously, making it key in analyzing complex shapes and motions.

Arc Length Differential

The arc length differential is essential for working with curves, including helices. It represents an infinitesimally small part of the curve, allowing us to calculate lengths, masses, and moments of inertia along the curve accurately.

For a parameterized curve, the differential of arc length, \( ds \), is computed using:

• \( ds = \sqrt{(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2} \, dt \)
This formula finds the tiny piece of the arc's length for a small change in \( t \).
In the problem of a spring along a helix, the expression simplifies to \( ds = \sqrt{2} \, dt \), representing the consistency of the radial measure from the z-axis. Understanding this differential is crucial for calculating properties like mass or the moment of inertia, as seen in this exercise.