Question

A lamina corresponding to a planar region \(R\) is given with a mass of 16 units. For each, compute \(I_{x}\) \(I_{y}\) and \(I_{0}\). \(R\) is the disk with radius 2 centered at the origin with density \(\delta(x, y)=4 / \pi\)

Short answer

\( I_x = 8 \), \( I_y = 8 \), \( I_0 = 16 \).

Step-by-step solution

Calculate the Area of the Disk

The region \( R \) is a disk with radius 2. The area \( A \) of a disk is given by the formula \( A = \pi r^2 \). Substituting \( r = 2 \), we have \( A = \pi \times 2^2 = 4\pi \).

Determine Density and Confirm Mass

The density function is given as \( \delta(x, y) = \frac{4}{\pi} \). To compute the total mass, we integrate density over the area or use the provided total mass. Given the total mass is 16 units, we confirm it satisfies \( \int_{R} \delta(x, y) \, dA = 16 \).

Formula for Moments of Inertia

The moments of inertia \( I_x \) and \( I_y \) for lamina are calculated by \( I_x = \int_{R} y^2 \delta \, dA \) and \( I_y = \int_{R} x^2 \delta \, dA \). The polar moment of inertia \( I_0 \) is \( I_0 = I_x + I_y \).

Convert to Polar Coordinates

Convert the equations from Cartesian to polar coordinates: \( x = r \cos \theta \), \( y = r \sin \theta \). The area element \( dA \) in polar coordinates is \( r \, dr \, d\theta \).

Calculate \( I_x \) using Polar Coordinates

Calculate \( I_x \):\[ I_x = \int_{0}^{2\pi} \int_{0}^{2} (r \sin\theta)^2 \cdot \frac{4}{\pi} \, r \, dr \, d\theta \].Simplify and integrate:\[ I_x = \frac{4}{\pi} \int_{0}^{2\pi} \int_{0}^{2} r^3 \sin^2\theta \, dr \, d\theta \].The integral of \( r^3 \) from 0 to 2 is \( \frac{16}{4} = 4 \), and \( \int_{0}^{2\pi} \sin^2 \theta \, d\theta = \pi \), leading to \( I_x = \frac{4}{\pi} \times 4 \times \pi = 8 \).

Calculate \( I_y \) using Polar Coordinates

The calculation for \( I_y \) is analogous to that of \( I_x \):\[ I_y = \int_{0}^{2\pi} \int_{0}^{2} (r \cos\theta)^2 \cdot \frac{4}{\pi} \, r \, dr \, d\theta \].This simplifies to \( I_y = 8 \), as \( \cos^2 \theta \) behaves symmetrically to \( \sin^2 \theta \).

Calculate \( I_0 \)

\( I_0 = I_x + I_y = 8 + 8 = 16 \).

Verify the Calculations

Re-check all integrations and confirm \( I_x = 8 \), \( I_y = 8 \), and \( I_0 = 16 \) based on symmetry and integral properties.

Key concepts

Lamina

A lamina is essentially a thin, flat surface or plate that has a certain mass and occupies a specific planar region. In physics and engineering, this concept is fundamental when analyzing the distribution of mass in a planar object and its rotational characteristics around specific axes.
In this context, the lamina corresponds to a disk centered at the origin. This type of problem often involves determining how mass is distributed across the lamina in order to compute quantities such as the moments of inertia. Here, the lamina has a specified mass of 16 units, which represents its total mass spread across the planar region in accordance with the given density function.
One of the key properties of a lamina is that despite its flatness, it can still have a non-negligible mass distribution. This is what makes it useful for calculating rotational inertia, which reflects how difficult it is to change the rotational motion of the lamina around an axis.

Polar Coordinates

Polar coordinates offer a convenient way to describe the position of a point in a plane, specifically when working with circular or radially symmetric regions, like the disk in this exercise. Instead of using the familiar Cartesian coordinates \( x, y \), polar coordinates use a radius \( r \) and an angle \( \theta \) to specify a point's location.
In mathematical terms, the transformation from Cartesian to polar coordinates is given by the equations: \( x = r \, \cos\theta \) and \( y = r \, \sin\theta \).
This method is particularly advantageous for integrating over circular regions due to the natural fit of polar coordinates in these cases. For the lamina in question, the region has rotational symmetry, which simplifies calculations. The area element in polar coordinates is determined by the expression \( r \, dr \, d\theta \), which helps neatly set up the integrals required to calculate moments of inertia.

Density Function

The density function in this exercise, \( \delta(x, y) = \frac{4}{\pi} \), describes how mass is distributed over the lamina. It assigns a numerical density to each point in the planar region covered by the lamina. This density remains constant across the disk.
Understanding density is critical as it is directly used to calculate the mass distribution necessary for determining the moments of inertia. With a given density function, the total mass of a lamina can be computed by integrating the density over the entire region. This relationship allows one to adjust density values to achieve a correct total mass when computing physical properties.
In this problem, the given density is uniform across the disk, which simplifies the integration; the mass property check verifies that the total mass equals the provided 16 units, confirming the correctness of the density function employed.

Planar Region

A planar region in geometry and physics is simply a flat, two-dimensional area occupied by a shape or object. In this exercise, the planar region is defined as a disk with a radius of 2 centered at the origin. The concept of a planar region is foundational because it defines the boundary within which any physical properties, such as mass or area, will be evaluated.
For calculations involving a disk-shaped planar region, it is often advantageous to use polar coordinates, as they are naturally suited to the circular symmetry of the shape. The disk's area can be easily calculated using the formula \( A = \pi r^2 \), reflecting the geometric properties of the shape.
This defined region is the basis for the computation of inertia-related properties, serving as the "footprint" over which density is distributed and integrated to evaluate the rotational characteristics of the lamina.