Problem 34 - Thomas Calculus

Question

Center of mass of a curved wire \(A\) wire of density \(\delta ( x , y , z ) = 15 \sqrt { y + 2 }\) lies along the curve \(\mathbf { r } ( t ) = \left( t ^ { 2 } - 1 \right) \mathbf { j } +\) \(2 t \mathbf { k } , - 1 \leq t \leq 1 .\) Find its center of mass. Then sketch the curve and center of mass together.

Step-by-Step Solution

Answer

The center of mass is at \((-0.6, 0)\).

Step 1: Parametrize the Curve

The given curve is already parametrized as \( \mathbf{r}(t) = (t^2 - 1) \mathbf{j} + 2t \mathbf{k} \) for \( -1 \leq t \leq 1 \). This expression gives the positioning of the wire in space along the \(y\)- and \(z\)-axes.

Step 2: Express Density in Terms of t

The density function is \( \delta(x, y, z) = 15 \sqrt{y + 2} \). Using the parameterization \( y = t^2 - 1 \), substitute it into the density function: \( \delta(t) = 15 \sqrt{(t^2 - 1) + 2} = 15 \sqrt{t^2 + 1} \).

Step 3: Determine the Differential Arc Length

The differential arc length \( ds \) of a parametrized curve is given by \( ds = \sqrt{(\frac{dy}{dt})^2 + (\frac{dz}{dt})^2} \, dt \). Here, \( \frac{dy}{dt} = 2t \) and \( \frac{dz}{dt} = 2 \), so \( ds = \sqrt{(2t)^2 + (2)^2} \, dt = \sqrt{4t^2 + 4} \, dt = 2 \sqrt{t^2 + 1} \, dt \).

Step 4: Setup for Center of Mass Calculation

The total mass \( m \) of the wire is given by \( m = \int_{-1}^{1} \delta(t) \, ds = \int_{-1}^{1} (15 \sqrt{t^2 + 1})(2 \sqrt{t^2 + 1}) \, dt = 30 \int_{-1}^{1} (t^2 + 1) \, dt \).

Step 5: Calculate Total Mass

Calculate the total mass: \[m = 30 \int_{-1}^{1} (t^2 + 1) \, dt = 30 \left[ \int_{-1}^{1} t^2 \, dt + \int_{-1}^{1} 1 \, dt \right] = 30 \left[ \frac{2}{3} + 2 \right] = 30 \times \frac{8}{3} = 80 \].

Step 6: Calculate Moment about y-axis (My)

The moment about y-axis \( M_y \) is given by \( M_y = \int_{-1}^{1} z \delta(t) \, ds = \int_{-1}^{1} (2t)(15 \sqrt{t^2 + 1})(2 \sqrt{t^2 + 1}) \, dt = 60 \int_{-1}^{1} t(t^2 + 1) \, dt \).

Step 7: Evaluate My and Find z-coordinate of Center of Mass

Calculate \( M_y \): \[M_y = 60 \int_{-1}^{1} (t^3 + t) \, dt = 60 \left[ \int_{-1}^{1} t^3 \, dt + \int_{-1}^{1} t \, dt \right] = 60 \left[ 0 + 0 \right] = 0\] Thus, the \( \bar{z} \)-coordinate of the center of mass is \( \bar{z} = \frac{M_y}{m} = \frac{0}{80} = 0 \).

Step 8: Calculate Moment about z-axis (Mz)

The moment about z-axis \( M_z \) is given by \( M_z = \int_{-1}^{1} y \delta(t) \, ds = \int_{-1}^{1} (t^2 - 1)(15 \sqrt{t^2 + 1})(2 \sqrt{t^2 + 1}) \, dt = 30 \int_{-1}^{1} (t^2 - 1)(t^2 + 1) \, dt \).

Step 9: Evaluate Mz and Find y-coordinate of Center of Mass

Calculate \( M_z \): \[M_z = 30 \int_{-1}^{1} (t^4 - 1) \, dt = 30 \left[ \int_{-1}^{1} t^4 \, dt - \int_{-1}^{1} 1 \, dt \right] = 30 \left[ \frac{2}{5} -2 \right] = 30 \left( \frac{-8}{5} \right) = -48\] Thus, the \( \bar{y} \)-coordinate of the center of mass is \( \bar{y} = \frac{M_z}{m} = \frac{-48}{80} = -0.6 \).

Step 10: Conclusion and Sketch

The center of mass of the wire is at \( ( \bar{y} , \bar{z} ) = (-0.6, 0) \). To sketch, plot the curve defined by \( \mathbf{r}(t) = (t^2 - 1) \mathbf{j} + 2t \mathbf{k} \) for \( -1 \leq t \leq 1 \) with point \( (-0.6, 0) \) marked on the graph as the center of mass.

Key Concepts

Parametrized Curves

Parametrized curves are a way of describing a path or shape using a parameter, typically represented by a variable "t." This method involves expressing the coordinates of points on the curve as functions of this parameter. In the exercise, the curve is defined by the parametrization \( \mathbf{r}(t) = (t^2 - 1) \mathbf{j} + 2t \mathbf{k} \).
Here, \( t \) is the parameter that varies over the interval from -1 to 1. The function \( t^2 - 1 \) describes the position along the y-axis, while \( 2t \) describes the position along the z-axis.

• Parametrized curves provide a clear, concise way to represent and analyze paths in space.
• They help in understanding the geometry and motion along the curve.

Given the curve's parametrization, it allows us to find relevant properties such as the direction, arc length, and density when used in physics and engineering problems.

Density Functions

Density functions typically describe how mass is distributed along a body. In this problem, the density \( \delta(x, y, z) \) depends on the position of the curve in 3D space.
For the given wire, the density function is \( \delta(x, y, z) = 15 \sqrt{y + 2} \). When dealing with parameterized curves, it is often necessary to express the density function in terms of the parameter. Given that \( y = t^2 - 1 \), you substitute this into the density function to obtain \( \delta(t) = 15 \sqrt{t^2 + 1} \).

• The transformed density function represents how density varies along the curve.
• It is essential for calculating the mass of the wire accurately.

By expressing the density in terms of \( t \), we connect the physical properties of the wire (its mass distribution) to its mathematical description through the parametrized curve.

Arc Length

The arc length of a curve is the distance along the curve between two points. It's a measure of the curve's length, and for a parametrized curve, it involves calculating small segments of the path known as differential arc lengths, denoted by \( ds \).
To compute \( ds \) for the given curve, you need the derivatives of the y- and z-components with respect to \( t \). These are \( \frac{dy}{dt} = 2t \) and \( \frac{dz}{dt} = 2 \), leading to \( ds = \sqrt{(2t)^2 + (2)^2} \, dt = 2 \sqrt{t^2 + 1} \, dt \).

• Arc length is crucial for integrating over a curve, such as when calculating mass.
• It reflects the true "path distance" along the curve, not just the straight-line distance between endpoints.

The differential arc length can be thought of as a tiny segment of the curve, which when summed up over an interval gives the total length of the curve.

Mass Calculation

Mass calculation involves integrating the density function over the length of the curve. This gives the total mass of the object, accounting for changes in density and geometry along the path.
For the wire, the mass \( m \) is calculated by integrating the product of the density function \( \delta(t) \) and the differential arc length \( ds \) over the interval \(-1 \leq t \leq 1\). The expression becomes:
\[m = \int_{-1}^{1} \delta(t) \, ds = \int_{-1}^{1} (15 \sqrt{t^2 + 1})(2 \sqrt{t^2 + 1}) \, dt = 30 \int_{-1}^{1} (t^2 + 1) \, dt\]

• This integral accounts for how the curved path and variable density affect the total mass.
• The computed mass is essential for determining the center of mass and other dynamics.

Thus, understanding how to set up and evaluate such integrals is key to solving complex physics problems that involve curved objects.