Problem 39 - Thomas Calculus

Question

Center of mass, moments of inertia, and radii of gyration Find the center of mass, the moment of inertia and radii of gyration about the coordinate axes, and the polar moment of inertia and radius of gyration of a thin triangular plate bounded by the lines \(y=x, y=-x,\) and \(y=1\) if \(\delta(x, y)=y+1\)

Step-by-Step Solution

Answer

Center of mass is \((0, 3/4)\), moments of inertia can be calculated using integrals, and radii of gyration follow from those results.

Step 1: Define the Triangular Plate Area

The triangular plate is bounded by the lines \(y = x\), \(y = -x\), and \(y = 1\). These intersections define a triangle with vertices at \((1, 1)\), \((-1, 1)\), and \((0, 0)\).

Step 2: Find the Center of Mass

The center of mass \((\bar{x}, \bar{y})\) is calculated by integrating over the area. Use the density function \(\delta(x, y) = y + 1\). The mathematical setup involves double integrals:\[\bar{x} = \frac{1}{M} \int_{-1}^{1} \int_{-x}^{x} x(y + 1) \, dy \, dx\]\[\bar{y} = \frac{1}{M} \int_{-1}^{1} \int_{-x}^{x} y(y + 1) \, dy \, dx\]First, calculate the mass \(M\):\[M = \int_{-1}^{1} \int_{-x}^{x} (y + 1) \, dy \, dx\]Evaluating these integrals will give the coordinates of the center of mass.

Step 3: Calculate the Mass

Perform the integration to find the total mass \(M\):\[M = \int_{-1}^{1} \int_{-x}^{x} (y + 1) \, dy \, dx = 4/3\]

Step 4: Evaluate Center of Mass Coordinates

Substitute \(M = 4/3\) into the integrals for \(\bar{x}\) and \(\bar{y}\):\[\bar{x} = \frac{3}{4} \int_{-1}^{1} \int_{-x}^{x} x(y + 1) \, dy \, dx = 0\]\[\bar{y} = \frac{3}{4} \int_{-1}^{1} \int_{-x}^{x} y(y + 1) \, dy \, dx = \frac{3}{4}\]

Step 5: Moment of Inertia about Coordinate Axes

Calculate the moment of inertia about the x-axis \(I_x\) and y-axis \(I_y\):\[I_x = \int_{-1}^{1} \int_{-x}^{x} y^2 (y + 1) \, dy \, dx\]\[I_y = \int_{-1}^{1} \int_{-x}^{x} x^2 (y + 1) \, dy \, dx\]

Step 6: Radii of Gyration about Coordinate Axes

The radii of gyration about the x-axis \(k_x\) and y-axis \(k_y\) are calculated using:\[k_x = \sqrt{\frac{I_x}{M}}\]\[k_y = \sqrt{\frac{I_y}{M}}\]

Step 7: Polar Moment of Inertia and Radius of Gyration

Determine the polar moment of inertia \(J\) and radius of gyration \(k_p\):\[J = I_x + I_y\]\[k_p = \sqrt{\frac{J}{M}}\]

Key Concepts

Moment of Inertia

When we discuss the **moment of inertia**, we're referring to a property that quantifies how much an object will resist being rotated about an axis. It is similar to mass in linear motion but applies to rotational motion. For our triangular plate bounded by specific lines, the moment of inertia involves calculating the integral of squared distances, weighted by the density function.

For the x-axis, it is given by:

• \[I_x = \int_{-1}^{1} \int_{-x}^{x} y^2 (y + 1) \, dy \, dx\]

And for the y-axis:

• \[I_y = \int_{-1}^{1} \int_{-x}^{x} x^2 (y + 1) \, dy \, dx\]

These integrals are crucial as they take into account the distribution of mass throughout the triangular area and help us understand how the plate spins depending on where it's pivoted.

Radius of Gyration

The **radius of gyration** simplifies the moment of inertia into a discernible single radius value that indicates how mass is distributed about an axis. It essentially describes a hypothetical distance from the axis at which you can assume the entire mass is concentrated to produce the same moment of inertia.

Radii of gyration for axes are calculated using:

• For the x-axis: \[k_x = \sqrt{\frac{I_x}{M}}\]
• For the y-axis: \[k_y = \sqrt{\frac{I_y}{M}}\]

For the polar moment (which involves both axes), it is:

• \[k_p = \sqrt{\frac{J}{M}}\]

Here, \(J\) is the sum of \(I_x\) and \(I_y\), providing insight into the object's overall resistance to rotational motion about a central point.

Triangular Plate

In this problem, the **triangular plate** is our main area of focus. It is defined in the coordinate plane by the lines: \(y = x\), \(y = -x\), and \(y = 1\). These boundaries form an isosceles triangle with vertices at \((1,1)\), \((-1,1)\), and \((0,0)\).

Understanding the geometry of this triangular plate is crucial because it affects how calculations for physical properties are done. The symmetry around the y-axis often simplifies calculations, as seen with many of the integrals remaining manageable despite the triangular area being complexly bounded.

This kind of geometric setup also introduces practical applications, such as determining stability and rotational characteristics in engineering and physics fields.

Double Integrals

**Double integrals** are used to calculate the center of mass, mass, and moments of inertia in this problem. They help assess quantities over a two-dimensional area, such as our triangular plate.

To find the center of mass, these integrals take into account the shape of the area and the density function \(\delta(x, y) = y + 1\). For instance, the integration for mass \(M\) is:

• \[M = \int_{-1}^{1} \int_{-x}^{x} (y + 1) \, dy \, dx\]

This is the total weight of the object, calculated over its entire area.

• For \(\bar{x}\), the integral setup appears as: \[\bar{x} = \frac{1}{M} \int_{-1}^{1} \int_{-x}^{x} x(y + 1) \, dy \, dx\]
• For \(\bar{y}\): \[\bar{y} = \frac{1}{M} \int_{-1}^{1} \int_{-x}^{x} y(y + 1) \, dy \, dx\]

Double integrals provide a powerful tool in calculus for evaluating properties over complex shapes and are fundamental in physics and engineering.