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 and moments of inertia require solving integrals for mass and length quantities.

Step 1: Parametric Representation of Wire

The wire is represented by the vector function \(\mathbf{r}(t) = t \mathbf{i} + \frac{2 \sqrt{2}}{3} t^{3/2} \mathbf{j} + \frac{t^2}{2} \mathbf{k}\). The parameter \(t\) varies from \(0\) to \(2\).

Step 2: Expression for the Density

The linear density of the wire is given as \(\delta(t) = \frac{1}{t+1}\). This density is variable and dependent on the parameter \(t\).

Step 3: Differential Arc Length Element

The differential arc length element \(ds\) is calculated by finding the magnitude of the derivative \(\mathbf{r}'(t)\). Compute \(\mathbf{r}'(t) = \mathbf{i} + \frac{\sqrt{2}}{\sqrt{t}} \mathbf{j} + t \mathbf{k}\) and then \(\| \mathbf{r}'(t) \| = \sqrt{1 + \frac{2}{t} + t^2}\).

Step 4: Compute Total Length (L)

The total length of the wire is found by integrating the differential arc length from \(t = 0\) to \(t = 2\). \[ L = \int_0^2 \sqrt{1 + \frac{2}{t} + t^2} \ dt \]

Step 5: Compute Total Mass (M)

The total mass \(M\) is given by:\[ M = \int_{0}^{2} \delta(t) \cdot \|\mathbf{r}'(t)\| \ dt = \int_0^2 \frac{\sqrt{1 + \frac{2}{t} + t^2}}{t+1} \ dt \]

Step 6: Center of Mass Coordinates

The center of mass \((\bar{x}, \bar{y}, \bar{z})\) is found by using:\[ \bar{x} = \frac{1}{M} \int_0^2 t \frac{\sqrt{1 + \frac{2}{t} + t^2}}{t+1} \ dt \]\[ \bar{y} = \frac{1}{M} \int_0^2 \frac{2 \sqrt{2}}{3} t^{3/2} \frac{\sqrt{1 + \frac{2}{t} + t^2}}{t+1} \ dt \]\[ \bar{z} = \frac{1}{M} \int_0^2 \frac{t^2}{2} \frac{\sqrt{1 + \frac{2}{t} + t^2}}{t+1} \ dt \]

Step 7: Moments of Inertia Components

The moments of inertia about the coordinate axes are calculated as follows:\[ I_x = \int_{0}^{2} ( y(t)^2 + z(t)^2 ) \delta(t) \cdot \|\mathbf{r}'(t)\| \ dt \]\[ I_y = \int_{0}^{2} ( x(t)^2 + z(t)^2 ) \delta(t) \cdot \|\mathbf{r}'(t)\| \ dt \]\[ I_z = \int_{0}^{2} ( x(t)^2 + y(t)^2 ) \delta(t) \cdot \|\mathbf{r}'(t)\| \ dt \]

Key Concepts

Moments of Inertia

The moment of inertia of an object is a measure of its resistance to rotational motion about a particular axis. It plays a crucial role in dynamics and mechanical engineering. To find the moments of inertia for a wire with variable density, we calculate the integrals for each axis separately. These are:

• For the x-axis: \[I_x = \int_{0}^{2} ( y(t)^2 + z(t)^2 ) \delta(t) \cdot \|\mathbf{r}'(t)\| \, dt\]
• For the y-axis:\[I_y = \int_{0}^{2} ( x(t)^2 + z(t)^2 ) \delta(t) \cdot \|\mathbf{r}'(t)\| \, dt\]
• For the z-axis:\[I_z = \int_{0}^{2} ( x(t)^2 + y(t)^2 ) \delta(t) \cdot \|\mathbf{r}'(t)\| \, dt\]

Here, \(x(t), y(t)\) and \(z(t)\) correspond to the parametric equations of volume element positions.The product \(\delta(t) \cdot \|\mathbf{r}'(t)\|\)makes sure that both the variable density and the arc length are part of the moment calculations. By integrating, we find the total moment of inertia with respect to each coordinate axis.

Variable Density

Variable density means that the density of an object is not uniform but changes depending on another variable. For the exercise, the linear density is represented by \(\delta(t) = \frac{1}{t+1}\).It implies that the density decreases as the parameter \(t\)increases. Understanding how density varies is vital for correctly calculating physical properties like mass and center of mass.
Without considering variable density, calculations would incorrectly predict uniform distribution, leading to errors.
It introduces complexity into finding expressions for elements that change across the parameterized curve. By plugging \(\delta(t)\)into different integrals, we account for the deviation in mass distribution along the curve, yielding more accurate physical predictions.

Parametric Representation

A parametric representation uses a parameter to express a curve in space with a set of equations. In this problem, the wire is expressed as:

• \(\mathbf{r}(t) = t \mathbf{i} + \frac{2 \sqrt{2}}{3} t^{3/2} \mathbf{j} + \frac{t^2}{2} \mathbf{k}\)

These equations give the x, y, and z coordinates as functions of the parameter \(t\),providing a continuous path from \(t=0\)to \(t=2\).Each component (\(t\)for x, \(\frac{2 \sqrt{2}}{3} t^{3/2}\)for y, \(\frac{t^2}{2}\)for z) describes how the wire stretches through space. Differences in the path are easily analyzed through derivatives.
Using parametric equations simplifies calculations for geometrically complex objects like curves, enabling easier determination of the center of mass, arc length, and rotational inertia.

Differential Arc Length

To measure the total length of a curve, we divide it into small elements, each represented by the differential arc length \(ds\).First, take the derivative \(\mathbf{r}'(t)\)by differentiating each component function:

• \(\mathbf{r}'(t) = \mathbf{i} + \frac{\sqrt{2}}{\sqrt{t}} \mathbf{j} + t \mathbf{k}\)

Then calculate its magnitude:\[\| \mathbf{r}'(t) \| = \sqrt{1 + \frac{2}{t} + t^2}.\]This represents the rate of change of the curve with respect to \(t\),giving us the arc length increment.
Once we understand \(ds\),we integrate over the given range for \(t\),ensuring that small approximations cumulatively reveal the wire's entire length.
Iterative use in mass and inertia integrations, alongside variable density, gives a precise overview of the physical structure of the wire.