Question

The perpendicular axis theorem applies to a lamina lying in the \(x y\) plane. It states that the moment of inertia of the lamina about the \(z\) axis is equal to the sum of the moments of inertia about the \(x\) and \(y\) axes. Suppose a thin circular disc of mass \(M\) and radius \(a\) lies in the \(x y\) plane and the \(z\) axis passes through its centre. The moment of inertia of the disc about this axis is \(\frac{1}{2} M a^{2}\). (a) Use this theorem to find the moment of inertia of the disc about the \(x\) and \(y\) axes. (b) Use the parallel axis theorem to find the moment of inertia of the disc about a tangential axis parallel to the plane of the disc.

Short answer

(a) The moment of inertia about the x and y axes is \(I_x = I_y = \frac{1}{4}Ma^2\). (b) The moment of inertia about a tangential axis parallel to the plane of the disc is \(I_t = \frac{3}{2}Ma^2\).

Step-by-step solution

(a) Moment of Inertia about the x and y axes

To find the moment of inertia about the x and y axes, we can use the perpendicular axis theorem. Since the disc is symmetric about the x and y axes, we can consider the moments of inertia about these axes to be equal (i.e., \(I_x = I_y\)). Using the given \(I_z\) and the perpendicular axis theorem, we get: \(I_z = I_x + I_y\) Substituting the given value of \(I_z\), we have: \(\frac{1}{2}Ma^2 = I_x + I_y\) And, because \(I_x = I_y\), we can rewrite it as: \(I_x + I_x = \frac{1}{2}Ma^2\) Solving for \(I_x\), we get: \(I_x = \frac{1}{4}Ma^2\) Since \(I_x = I_y\), we have: \(I_y = \frac{1}{4}Ma^2\)

(b) Moment of Inertia about a tangential axis

To find the moment of inertia about a tangential axis parallel to the plane of the disc, we will use the parallel axis theorem. Let \(I_t\) be the moment of inertia about the tangential axis. We will also need the distance from the center of mass to the tangential axis, which is equal to the radius of the disc, \(a\). Using the parallel axis theorem, we get: \(I_t = I_{cm} + Md^2\) \(I_t\) is the moment of inertia we want to find, and \(I_{cm}\) is the moment of inertia about the center of mass, which is perpendicular to the x and y axes (i.e., about the z axis). Using the given value of \(I_z\) as \(I_{cm}\), we get: \(I_t = \frac{1}{2}Ma^2 + Ma^2\) Solving for \(I_t\), we find: \(I_t = \frac{3}{2}Ma^2\) So, the moment of inertia of the disc about a tangential axis parallel to the plane of the disc is \(\frac{3}{2}Ma^2\).

Key concepts

Perpendicular Axis Theorem

The Perpendicular Axis Theorem is a significant principle used in physics to determine the moment of inertia of flat objects or laminas. Imagine you have a disc lying flat on the xy-plane. This theorem tells us how to find the moment of inertia about an axis perpendicular to the plane, which we'll call the z-axis.
It states:

• The moment of inertia about the z-axis, denoted as \(I_z\), is equal to the sum of the moments of inertia about the x and y axes, so \(I_z = I_x + I_y\).

In simple terms, if you know the inertia about the z-axis, you can find the inertia about the x and y axes easily using symmetry. Let's say you have a circular disc lying on the plane. This disc's moment of inertia about the z-axis is known to be \(\frac{1}{2} M a^2\), where \(M\) is the mass and \(a\) is the radius.
Because the disc is symmetric, \(I_x = I_y\). By using the perpendicular axis theorem, you can determine that the moment of inertia about each of these axes is \(\frac{1}{4} M a^2\). This is because when you add them up, the equation should give you \(I_z\). Calculating this way simplifies the problem using the symmetry that exists in the structure of the disc.

Parallel Axis Theorem

The Parallel Axis Theorem is a powerful tool for finding the moment of inertia when the axis of rotation is shifted away from the center of mass. Often, you'll need to know the inertia about an axis parallel to one through the center of mass. This theorem provides a straightforward way to calculate it.The formula is:

• \(I_t = I_{cm} + M d^2\)

Here, \(I_t\) is the moment of inertia about the tangential axis parallel to the plane of the disc, \(I_{cm}\) is the inertia about the center of mass, and \(d\) is the distance from the center of mass to the new axis.
For the disc, \(I_{cm}\) is the same as \(I_z\), which is \(\frac{1}{2} M a^2\). The distance \(d\) is simply the radius \(a\). So you simply plug in these known values into the formula.Computing these gives you:

• \(I_t = \frac{1}{2} M a^2 + M a^2 = \frac{3}{2} M a^2\)

This result shows how the disc's inertia changes when shifting to an axis not passing through its center.

Symmetry in Inertia Calculations

Symmetry plays a crucial role in simplifying moment of inertia calculations, especially with objects like discs. When an object is perfectly symmetric about multiple axes, the properties and calculations become predictable.
A symmetrical object, like the circular disc discussed, ensures that the moment of inertia is the same about both the x and y axes. This significantly reduces computational effort and complexities involved in the calculations, allowing us to use symmetry effectively.Key takeaways:

• Simplifies complex calculations: For symmetric objects, you can assume \(I_x = I_y\).
• Enhances understanding of rotational dynamics: Knowing that symmetry can simplify equations helps in predicting rotational behavior more accurately.
• Allows easy application of theorems: Symmetric properties align perfectly with the assumptions behind the Perpendicular and Parallel Axis Theorems.

Understanding these principles provides a foundation for tackling more complex shapes and rotational dynamics by recognizing patterns inherent in symmetry.