25 May 2014

Mass moment of inertia of a hemisphere

I recently had to calculate the moments of inertia of various solid bodies (e. g. the spherocylinder). They can be obtained starting from the definition; this is the kind of calculation described in papers as "tedious but straightforward". I tried to simplify the process as much as possible, using the symmetry properties of the objects and the parallel axis theorem. The latter is of course useful for "composite" objects (combinations of simple units) but, less intuitively, also for sections of such units, as we'll see below for the (solid) hemisphere.

Center of mass

 First, let us find the center of mass \( \mathbf{r}_{\text{CM}}\) of the hemisphere shown on the left. Since \( x_{\text{CM}} = y_{\text{CM}} =  0\) by symmetry, we only need the height \(z_{\text{CM}}\):\[z_{\text{CM}} = \frac{1}{V} \int_{\text{obj}} z \, \text{d}^3\mathbf{r} \]where the volume integral is taken over the entire object.
Let us cut the hemisphere in horizontal slices and write the integral as: \( \displaystyle z_{\text{CM}} = \frac{3}{2 \pi R^3} \int_{0}^R \pi (R^2 - z^2) z \, \text{d}z \), with \(R\) the radius. The variable change \(z \rightarrow t=z/R\) yields: \( \displaystyle z_{\text{CM}} = \frac{3 R}{2} \int_{0}^1 t (1-t^2) \, \text{d}t \).
A new variable change \(t \rightarrow u=t^2\) further simplifies the integral to: \( \displaystyle z_{\text{CM}} = \frac{3 R}{4} \int_{0}^1 (1-u) \, \text{d}u =  \frac{3 R}{8} \).
The last integral is elementary, and can be solved term by term. We can also note that, as \(u\) goes from 0 to 1, \(1-u\) decreases linearly from 1 to 0, so that the integral is simply the area of a right isosceles triangle with leg size 1.

Moments of inertia

We would now like to determine the moment of inertia \(I_{\Omega}\) about an axis \(\Omega\), contained in the horizontal plane at height \( z_{\text{CM}}\). We can relate it to the moment of inertia \(I_{x}\) about axis \(x\) going through the center of the sphere via the parallel axis theorem: \( \displaystyle  I_{x} = I_{\Omega} + m z_{\text{CM}}^2 \), where \(m\) is the mass of the hemisphere.
This is interesting, since we can now stick together two hemispheres, and their total moment of inertia \(2 I_{x}\) will simply be that of the full sphere: \(2 I_{x} = \frac{2}{5} m_s R^2\), where the mass of the sphere \(m_s = 2m\).
Now putting it all together: \(2 I_{\Omega} \! + 2 m z_{\text{CM}}^2 \! = \frac{4}{5} m R^2\), or
\[I_{\Omega} = \frac{83}{320} m R^2\]It is even easier to determine the moment of inertia \(I_{z}\) of the hemisphere about axis \(z\), since it is simply half that of the full sphere (which is of course the same about any axis):\[I_{z} = \frac{2}{5} m R^2\]

Inertia tensor

We can show that \(I_{\Omega}\) and \(I_{z}\) are the eigenvalues of the inertia tensor \(\mathrm{I}\), i. e. that the latter is diagonal in the reference frame \(\Omega \Omega ' z\) (or in any other frame arbitrarily rotated around \(z\).) More specifically,
\[\mathrm{I} = \left (
I_{\Omega} & 0 & 0 \\
0 & I_{\Omega} & 0 \\
0 & 0 & I_{z}
\right ) \]
We know this because products \( x(z-z_{\text{CM}})\) and \(y(z-z_{\text{CM}})\) integrated over the hemisphere vanish by symmetry, so off-diagonal components in the third line and column are zero. In the subspace \(\Omega \Omega '\), \(\mathrm{I}\) is isotropic (rotation invariant), and thus diagonal with eigenvalues \(I_{\Omega} = I_{\Omega '}\).

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