Question

Show that the inertia matrix at the centre of a uniform solid cube with mass \(m\) and edges of length \(2 a\) is \(\frac{2}{3} m a^{2} I\). Find the principal axes and principal moments of inertia at a vertex.

Short answer

Answer: The principal moments of inertia of a uniform solid cube at a vertex are \(I'_x=\frac{5}{3}ma^2\), \(I'_y=\frac{5}{3}ma^2\), and \(I'_z=\frac{5}{3}ma^2\). The principal axes are at 45-degree angles along the coordinates x, y, and z in the first (positive-valued) octant.

Step-by-step solution

Calculate the Inertia Matrix at the center of mass

To find the inertia matrix at the center of mass, use the equation for the inertia tensor matrix components: \(I_{ij}=\int_{\text{cube}}\rho(\textbf{r})r^2\delta_{ij}-x_ix_j dV\) where \(i,j\) are the indices representing the Cartesian coordinates x, y, and z, \(\rho(\textbf{r})=m/V\) is the mass density (\(V\) being the volume of the cube), \(\delta_{ij}\) is the Kronecker delta, and the integral is over the volume of the cube.

Determine the Inertia Tensor Matrix components

We have that \(\rho(\textbf{r})=\frac{m}{(2a)^3}\). The integral for diagonal components (\(I_{xx}, I_{yy}, I_{zz}\)) is: \(I_{xx}=\int_{\text{cube}}\rho(\textbf{r})(y^2+z^2)dV\) \(I_{yy}=\int_{\text{cube}}\rho(\textbf{r})(x^2+z^2)dV\) \(I_{zz}=\int_{\text{cube}}\rho(\textbf{r})(x^2+y^2)dV\) Off-diagonal components have 0 integral, because of symmetry.

Evaluate the integrals

Without loss of generality, solving for \(I_{xx}\), we get: \(I_{xx}=\int_{-a}^{a}\int_{-a}^{a}\int_{-a}^{a}\rho(\textbf{r})(y^2+z^2)dxdydz\) \(I_{xx}=\rho\int_{-a}^{a}\int_{-a}^{a}\int_{-a}^{a}(y^2+z^2)dxdydz\) Now, the x, y, z integrals can be separated as they are independent: \(I_{xx}=\rho\int_{-a}^{a}dx\int_{-a}^{a}y^2dy\int_{-a}^{a}dz+\rho\int_{-a}^{a}dx\int_{-a}^{a}dy\int_{-a}^{a}z^2dz\) \(I_{xx}=2a\rho\int_{-a}^{a}y^2dy+2a\rho\int_{-a}^{a}z^2dz\) \(I_{xx}=2a\rho\left[\frac{(2a)^3}{3}+\frac{(2a)^3}{3}\right]=\frac{2}{3}ma^2\) Using symmetry, we get: \(I_{yy}=I_{zz}=\frac{2}{3}ma^2\)

Write the Inertia Matrix at the center of mass

The inertia matrix at the center of mass is: $I_{CM}=\begin{pmatrix} I_{xx} & 0 & 0 \\ 0 & I_{yy} & 0 \\ 0 & 0 & I_{zz} \\ \end{pmatrix} =\frac{2}{3}ma^2I $

Find the Principal axes and principal moments of inertia at a vertex

To find the principal axes and principal moments of inertia at a vertex, we need to move the origin from the center to a vertex of the uniform solid cube via parallel axis theorem: \(I'_{ij}=I_{ij}+m[(\delta_{ij}-1)a]^2\) \(j'\neq i: I'_{ji}=I_{ji}=0\) \(i=j\): \(I'_{xx}=I_{xx}+ma^2=\frac{5}{3}ma^2\), \(I'_{yy}=I_{yy}+ma^2=\frac{5}{3}ma^2\), \(I'_{zz}=I_{zz}+ma^2=\frac{5}{3}ma^2\) The principal moments of inertia at a vertex are: \(I'_x=\frac{5}{3}ma^2\), \(I'_y=\frac{5}{3}ma^2\), \(I'_z=\frac{5}{3}ma^2\) The principal axes are at 45-degree angles along the coordinates x, y, and z in the first (positive-valued) octant.

Key concepts

Principal Moments of Inertia

Principal moments of inertia are key concepts used to determine how an object's mass is distributed relative to an axis of rotation. Imagine spinning a figure skater with arms extended versus when they are pulled close; the distribution of mass changes and thus its resistance to rotational change.

In more technical terms, the principal moments of inertia are the eigenvalues of the inertia tensor matrix, and they represent the unique values when the inertia tensor no longer has any off-diagonal components—that is, when the tensor is diagonalized. This happens when the reference frame is aligned with the object's principal axes. In the exercise provided, the principal moments of inertia for a solid cube are all equal when calculated from the center of mass, indicating uniform mass distribution.

Inertia Tensor

The inertia tensor, also known as the inertia matrix, is a mathematical representation encapsulating the second moment of mass for a rigid body. If you think of a 3D graph where each point represents a 'bit' of mass, the inertia tensor gives us a way to summarize how all those points are spread out in space.

The inertia tensor takes into account how far each bit of mass is from the chosen axes of rotation and allows us to predict how the object will rotate. In our cube example, after calculating individual integrals for the x, y, and z components, the resulting tensor simplifies due to the symmetry of the cube. The inertia tensor is foundational for problems involving rotational dynamics, as it correlates the angular velocity of an object with the torque applied to it.

Mass Density

Mass density is fundamentally the amount of mass per unit volume of an object. It's quite like how you'd describe a sponge as light or lead as heavy—noting how tightly packed the material is. In the context of our cube, mass density is uniform, indicating that the mass is evenly distributed throughout the entire volume.

Calculating the mass density is a pivotal step in determining the inertia tensor components, as it serves to weight the contribution of each infinitesimal volume element to the overall inertia. In the exercise, the mass density \(\rho(\textbf{r})\) is a constant because the cube's mass distribution does not change with position within the cube.

Parallel Axis Theorem

The parallel axis theorem is a fascinating physics tool that lets us recalibrate our calculations for the moments of inertia when we shift our axis of rotation. Just as a tightrope walker adjusts their balance pole when shifting their position, we can adjust our inertia calculations when the rotation point moves.

More precisely, when you know the moment of inertia of an object about an axis through its center of mass, the parallel axis theorem allows you to find the moment of inertia about any parallel axis. It adds the inertia of the center of mass with an additional term accounting for the mass's distance from the new axis. This theorem is beautifully illustrated in our exercise where moving the axis from the center to the vertex changes the principal moments significantly, demonstrating the profound effect of axis location on rotational dynamics.