Problem 7 - Thomas Calculus

Question

Finding moments of inertia Find the moment of inertia about the \(x\) -axis of a thin plate of density \(\delta=1 \mathrm{gm} / \mathrm{cm}^{2}\) bounded by the circle \(x^{2}+y^{2}=4 .\) Then use your result to find \(I_{y}\) and \(I_{0}\) for the plate.

Step-by-Step Solution

Answer

The moment of inertia \( I_x = 4\pi \), \( I_y = 4\pi \), and \( I_0 = 8\pi \).

Step 1: Calculating Moment of Inertia about x-axis

The moment of inertia about the x-axis, denoted as \( I_x \), is given by the integral \( I_x = \int_{R} y^2 \, \delta \, dA \). Here, \( \delta = 1 \, \text{gm/cm}^2 \), and \( dA \) is the differential area element. The domain of integration, \( R \), is defined by the circle \( x^2 + y^2 = 4 \), implying the disk with radius 2 centered at the origin.

Step 2: Using Polar Coordinates

To evaluate this integral over the circle, it's convenient to change to polar coordinates. In polar coordinates, the relationship between \( x \) and \( y \) is \( x = r \cos\theta \) and \( y = r \sin\theta \), and the differential area element is given by \( dA = r \, dr \, d\theta \). With these, substitute \( y^2 = (r \sin\theta)^2 \).

Step 3: Setting up the Integral

Transforming the moment of inertia formula for polar coordinates, we have: \[ I_x = \int_0^{2\pi} \int_0^2 (r^2 \sin^2\theta) \, r \, dr \, d\theta = \int_0^{2\pi} \int_0^2 r^3 \sin^2\theta \, dr \, d\theta. \]

Step 4: Evaluating the Inner Integral

Integrate with respect to \( r \) first: \[ \int_0^2 r^3 \sin^2\theta \, dr = \sin^2\theta \int_0^2 r^3 \, dr. \] To solve this, integrate \( r^3 \): \[ \frac{r^4}{4} \bigg|_0^2 = \frac{16}{4} = 4. \] Hence the inner integral becomes \( 4 \sin^2\theta \).

Step 5: Evaluating the Outer Integral

Now, integrate with respect to \( \theta \): \[ I_x = \int_0^{2\pi} 4 \sin^2\theta \, d\theta. \] Using the identity \( \sin^2\theta = \frac{1 - \cos(2\theta)}{2} \), the integral becomes \( 2 \int_0^{2\pi} (1 - \cos(2\theta)) \, d\theta \). Separate it into two integrals: \( 2 \int_0^{2\pi} 1 \, d\theta - 2 \int_0^{2\pi} \cos(2\theta) \, d\theta \).

Step 6: Solving the Outer Integral

The first integral is \( 2 \int_0^{2\pi} 1 \, d\theta = 2 \cdot 2\pi = 4\pi \). The second integral, \( 2 \int_0^{2\pi} \cos(2\theta) \, d\theta \), evaluates to zero because the integral of cosine over one complete period is zero. Thus, \( I_x = 4\pi \).

Step 7: Finding Moment of Inertia about y-axis

By symmetry of the circle about the x and y axes, \( I_y = I_x = 4\pi \), as the form of the integrals would be identical except with \( x \) and \( y \) terms swapped, which do not affect the result in this case.

Step 8: Finding Polar Moment of Inertia

The polar moment of inertia \( I_0 \) is given by \( I_0 = I_x + I_y \). Thus, \( I_0 = 4\pi + 4\pi = 8\pi \).

Key Concepts

Polar Coordinates

Polar coordinates offer a distinct way to define points in the plane compared to Cartesian coordinates. Instead of using

• x and y
• we utilize a radius \( r \) and an angle \( \theta \).

Here:

• \( r \) is the distance from the origin to the point
• \( \theta \) represents the angle measured from the positive x-axis.

In the context of finding moments of inertia, using polar coordinates can give us a clearer picture, especially when dealing with circular shapes or problems involving symmetry.
For example, in the given solution, the circle is best described using:

• x = r \( \cos\theta \)
• y = r \( \sin\theta \)
• The differential area element \( dA \) becomes \( r \, dr \, d\theta \).

This transformation simplifies the integration processes as it aligns perfectly with the geometric nature of the problem.

Density

Density, in this context, refers to the mass property of a two-dimensional object like a thin plate or disk. It is specified by \( \delta \), which is measured in mass units per area, such as grams per square centimeter (\( \text{gm/cm}^2 \)).
For our problem:

• \( \delta = 1 \text{gm/cm}^2 \)

This means that each centimeter square of our circular disk holds one gram of mass.
In rotational dynamics, density is crucial because it influences how the mass is distributed over the object, which directly affects the moment of inertia. The moment of inertia \( I \), relates both the mass and distribution concerning the axis of rotation.

Integral Calculus

Integral calculus comes into play when calculating moments of inertia, especially when the object has a continuous mass distribution. The moment of inertia is essentially an integral that sums up all the infinitesimal mass elements of the object considering their distance from the axis.
The formula for our case is given by:

• \( I_x = \int_R y^2 \, \delta \, dA \)

In translating to polar coordinates, it becomes more manageable since it allows for easier integration over circular areas.
Our approach involved:

• Defining the limits as the entire circle from \( 0 \) to \( 2\pi \) in terms of \( \theta \) and 0 to the radius in terms of \( r \).
• Evaluating step-by-step, first with respect to \( r \), then \( \theta \), gives a fantastically neat result.

This showcases integral calculus's power in solving real-world mechanical problems involving continuous entities.

Symmetry in Mathematics

Symmetry is a fundamental concept that simplifies many mathematical problems. In mechanics, symmetry can help reduce complexity, particularly in calculating moments of inertia.
The circular shape considered in this exercise is perfectly symmetrical about both the x and y axes. This symmetry means that the shape and mass are uniformly distributed, which leads to:

• The moment of inertia about the x-axis \( I_x \) being equal to the moment of inertia about the y-axis \( I_y \).

Since nothing changes when swapped across either axis, we know immediately \( I_y = I_x \). This property is incredibly useful, reducing computation time and simplifying processes. Finally, this symmetry idea is extended to calculate the polar moment of inertia \( I_0 \),

• where both \( I_x \) and \( I_y \) are directly summed up,
• giving \( I_0 = 8\pi \).

Symmetry thus allows for elegant solutions to complex mechanics problems!