Question

A rigid body consists of three masses fastened as follows: \(m\) at \((a, 0,0), 2 m\) at \((0, a, a),\) and \(3 m\) at \((0, a,-a) .\) (a) Find the inertia tensor \(\mathbf{I} .\) (b) Find the principal moments and a set of orthogonal principal axes.

Short answer

The inertia tensor is \( \begin{bmatrix} 10ma^2 & 0 & 0 \\ 0 & 6ma^2 & -5ma^2 \\ 0 & -5ma^2 & 6ma^2 \end{bmatrix} \). Principal moments are \(5ma^2, 11ma^2, 10ma^2\). Principal axes: \([1, 0, 0]\), \([0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]\), \([0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}]\).

Step-by-step solution

Write the coordinates and mass of each point

We have three masses: \( m \) located at \((a, 0, 0)\), \( 2m \) located at \((0, a, a)\), and \( 3m \) located at \((0, a, -a)\). These coordinates will be used to calculate the inertia tensor.

Calculate moments of inertia

The moments of inertia for each axis \( I_{xx}, I_{yy}, I_{zz} \) need to be calculated. For \(I_{xx}\), sum \( y^2 + z^2 \) for each mass:\[I_{xx} = m(0^2 + 0^2) + 2m(a^2 + a^2) + 3m(a^2 + (-a)^2)\]\[I_{xx} = 0 + 4ma^2 + 6ma^2 = 10ma^2\]For \(I_{yy}\), sum \( x^2 + z^2 \) for each mass:\[I_{yy} = m(a^2 + 0^2) + 2m(0^2 + a^2) + 3m(0^2 + (-a)^2)\]\[I_{yy} = ma^2 + 2ma^2 + 3ma^2 = 6ma^2\]For \(I_{zz}\), sum \( x^2 + y^2 \) for each mass:\[I_{zz} = m(a^2 + 0^2) + 2m(0^2 + a^2) + 3m(0^2 + a^2)\]\[I_{zz} = ma^2 + 2ma^2 + 3ma^2 = 6ma^2\]

Calculate products of inertia

The products of inertia \(I_{xy}, I_{xz}, I_{yz}\) are calculated by summing the product of two different coordinates for each mass.For \( I_{xy} \):\[I_{xy} = -m(a \cdot 0) + (-2m)(0 \cdot a ) + (-3m)(0 \cdot a) = 0\]For \( I_{xz} \):\[I_{xz} = -m(a \cdot 0) + 2m(0 \cdot a) + 3m(0 \cdot (-a)) = 0\]For \( I_{yz} \):\[I_{yz} = m(0 \cdot 0) + (-2m)(a \cdot a) + 3m(a \cdot (-a)) = -2ma^2 - 3ma^2 = -5ma^2\]

Construct the inertia tensor

Combine the moments and products of inertia into the inertia matrix \[\mathbf{I} = \begin{bmatrix} I_{xx} & I_{xy} & I_{xz} \ I_{yx} & I_{yy} & I_{yz} \ I_{zx} & I_{zy} & I_{zz} \end{bmatrix} = \begin{bmatrix} 10ma^2 & 0 & 0 \ 0 & 6ma^2 & -5ma^2 \ 0 & -5ma^2 & 6ma^2 \end{bmatrix}\]

Find the principal moments (eigenvalues)

To find the principal moments, solve the characteristic equation for the inertia tensor, which is given by:\[\det(\mathbf{I} - \lambda \mathbf{I}) = 0\]Solving this for \(\lambda\) gives the eigenvalues of the inertia tensor: \\[\lambda_1 = 5ma^2, \quad \lambda_2 = 11ma^2, \quad \lambda_3 = 10ma^2\]

Determine the principal axes (eigenvectors)

Calculate the eigenvectors corresponding to each eigenvalue determined in the previous step. The eigenvectors provide the directions of the principal axes. Using an eigenvalue solver on the inertia tensor yields: Principal axis 1: \([1, 0, 0]\) Principal axis 2: \([0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]\) Principal axis 3: \([0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}]\)

Key concepts

Principal Moments

Principal moments are specific scalar values associated with the inertia tensor of an object. They are derived as the eigenvalues from that tensor. These values are crucial because they represent the moments of inertia about the principal axes of the object, which are directions where the inertia tensor is diagonalized.
The three principal moments obtained as eigenvalues give insight into how a rigid body behaves under rotation. In this exercise, the calculated principal moments were:

• \( \lambda_1 = 5m a^2 \)
• \( \lambda_2 = 11m a^2 \)
• \( \lambda_3 = 10m a^2 \)

These correspond to the distinct rotational characteristics of the body relative to each principal axis. They are integral in determining stability and dynamics of the system as each principal moment indicates how resistant the body is to changes in rotational velocity around one of its axes.

Products of Inertia

Products of inertia are off-diagonal terms in the inertia tensor, and they represent interactions between different rotational axes. These terms are generally present in systems where the axes are not principal axes and include contributions from cross products of coordinate values.
In the given exercise, the products of inertia were calculated to be:

• \(I_{xy} = 0\)
• \(I_{xz} = 0\)
• \(I_{yz} = -5ma^2\)

These values provide information on how the component interacts across different planes. A zero indicates no interaction, while a non-zero value, like in \(I_{yz}\), shows that rotations about the y- and z-axes influence each other. In practice, minimizing or eliminating products of inertia through proper alignment of the principal axes can simplify calculations and understanding of rotational dynamics.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors hold a key role in characterizing the inertia tensor. When dealing with rigid body rotations, finding these components allows for the simplification of rotational dynamics.

• Eigenvalues: They correspond to the principal moments of inertia. The eigenvalues, \(\lambda_1\), \(\lambda_2\), and \(\lambda_3\), were calculated as \(5ma^2\), \(11ma^2\), and \(10ma^2\) respectively. These values help to understand how mass distribution affects rotational resistance.
• Eigenvectors: These vectors give the directions of the principal axes. For the inertia tensor, eigenvectors are derived through solving linear equations associated with their respective eigenvalues. In this exercise, the principal axes were calculated as:
  • First axis: \([1, 0, 0]\)
  • Second axis: \([0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]\)
  • Third axis: \([0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}]\)
  These eigenvectors indicate the coordinate directions in which the moments of inertia are measured, ultimately simplifying down the dynamics of the system by aligning them with principal axes.