Problem 34 - A Course in Mathematical Methods for Physicists

Question

Consider the octant of a uniform sphere of density 5 grams per cubic centimeter and radius \(a\) lying in the first octant. a. Find the inertia tensor about the origin. b. What are the principal moments of inertia? c Find the principal axes of inertia, that is, the eigenvectors of the inertia tensor.

Step-by-Step Solution

Answer

The inertia tensor is \( \left( \frac{\rho \cdot a^5}{10} ,0,0 \right), \left(0, \frac{\rho \cdot a^5}{10} ,0 \right), \left(0,0, \frac{\rho \cdot a^5}{5} \right) \). The principal moments of inertia are \( \frac{\rho \cdot a^5}{10} , \frac{\rho \cdot a^5}{10} , \frac{\rho \cdot a^5}{5} \). The principal axes of inertia are \( \left(1,0,0 \right), \left(0,1,0 \right), \left(0,0,1 \right) \).

Step 1: Calculating the Inertia Tensor

The inertia tensor \(I\) for a volume relative to the origin is given by the integral: \(I = \int dm \left( r^2 \delta - r * r \right)\) where \( \delta \) is the identity tensor, \( r \) is the position vector from the origin, and dm is a small mass element. Since we are dealing with an octant of a sphere, we will be integrating over a volume from 0 to \( a \), 0 to \( \pi/2 \), and 0 to \( \pi/2 \) in spherical coordinates respectively. The mass element dm can be expressed as \( dm = \rho \cdot dV = \rho \cdot r^2 \cdot sin(\theta) \cdot dr \cdot d\theta \cdot d\phi \). Putting this into the previous formula, and after a lot of calculus, we obtain the inertia tensor as:\[ I=\rho \cdot \left( \frac{a^5}{10},0,0 \right),\left(0, \frac{a^5}{10},0 \right),\left(0,0, \frac{a^5}{5} \right) \]

Step 2: The Principal Moments of Inertia

The principal moments of inertia are the eigenvalues of the inertia tensor. They are given by the diagonal elements of the tensor. For this case, they are \( \frac{\rho \cdot a^5}{10} , \frac{\rho \cdot a^5}{10} , \frac{\rho \cdot a^5}{5} \)

Step 3: The Principal Axes of Inertia

The principal axes of inertia are given by the eigenvectors of the inertia tensor. In this case, since the inertia tensor is diagonal, the principal axes are the standard basis vectors \( \left(1,0,0 \right), \left(0,1,0 \right), \left(0,0,1 \right) \).

Key Concepts

Principal Moments of Inertia

The principal moments of inertia are incredibly important in understanding how an object rotates about different axes. They give us the eigenvalues of the inertia tensor. In simple terms, these are the numerical values that tell us how resistant an object is to rotational motion around its principal axes.
Consider a diagonal inertia tensor where the diagonal elements represent these moments. In this case, when we look at our example of the octant of a uniform sphere, the principal moments are

• \( \frac{\rho \cdot a^5}{10} \) (for two axes)
• \( \frac{\rho \cdot a^5}{5} \) (for the third axis)

These values come directly from the tensor elements, which show how mass is distributed with respect to the origin. They play a key role in predicting how the octant will rotate when forces are applied. The difference in value between the two smaller and the larger moment reflects how the mass is spread more widely in one direction compared to the other two directions.

Principal Axes of Inertia

The principal axes of inertia provide a straightforward visual understanding of the rotation properties of the object. Essentially, these axes give the directions along which the object can rotate without any tendency for wobbling.
In our example, identifying these axes is quite simple because the inertia tensor is diagonal. Thus, the principal axes correspond directly to the x, y, and z axes we use in theoretical physics:

• The x-axis, represented by the vector \( (1,0,0) \)
• The y-axis, represented by the vector \( (0,1,0) \)
• The z-axis, represented by the vector \( (0,0,1) \)

These standard basis vectors are crucial because they enable the object to rotate smoothly around them. There is no cross-coupling of movements, meaning there is no movement in one axis that inadvertently causes a rotation around another. This clarity of motion is essential for engineers when designing systems that involve spinning parts.

Eigenvectors

Eigenvectors are fundamental in linear algebra and physics because they offer directionality information about matrices, like the inertia tensor. In essence, an eigenvector of a matrix is a non-zero vector that only changes by a scalar factor when that matrix is applied to it.
For understanding rotations, eigenvectors represent the directions (axes) about which the principal moments of inertia are calculated. In practical terms, these vectors keep the axis of rotation stable, serving as the building blocks of rotational motion analysis.
For our octant sphere, because the inertia tensor is diagonal, determining the eigenvectors becomes straightforward. They become the simple unit vectors along each axis:

• \( (1,0,0) \)
• \( (0,1,0) \)
• \( (0,0,1) \)

This means that the rotational behavior of the sphere octant is neatly aligned with these axes, making them very intuitive to work with. The simplicity of diagonal matrices allows this crisp determination of the eigenvectors, which is highly beneficial in both theoretical explorations and real-world applications like simulating or designing rotating bodies.