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Using the moments of inertia calculated in the previous post for the hemisphere (and taking for granted those of the cylinder), we can now determine those of a spherocylinder.

**UPDATE 22/08/2014**: Corrected a misprint in the formula for \(I_x\). See comment below.]Using the moments of inertia calculated in the previous post for the hemisphere (and taking for granted those of the cylinder), we can now determine those of a spherocylinder.

The height of the cylinder is \(h\), while its radius (and that of the spherical caps) is \(R\).

\(I_z\) is easy to determine by summing the corresponding moments of the cylinder and of the two hemispheres: \(I_z= \rho \, m_c \frac{R^2}{2} + 2 \rho \, m_h \frac{2}{5} R^2\). Developing \(m_c\) and \(m_h\) and introducing the aspect ratio \(\gamma = 1+ \frac{h}{2R}\) yields:

\[I_z = \pi \rho R^5 \left [ (\gamma -1) + \frac{8}{15} \right ] \]

\[I_z = \pi \rho R^5 \left [ (\gamma -1) + \frac{8}{15} \right ] \]

\(I_x\) is the sum of the \(x\) moment for the cylinder and of twice the moment of a hemisphere around an axis distant from its center of mass by \(h/2 + z_{\text{CM}}\), calculated using the parallel axis theorem (see previous post).

\[ I_x = \pi \rho R^5 \left \lbrace \frac{\gamma -1}{6} \left [ 3+4 (\gamma - 1)^2) \right ] + \frac{4}{3} \left [ \frac{83}{320} +\left ( (\gamma - 1) +\frac{3}{8} \right )^2 \right ] \right \rbrace \]\(Oxyz\) is the principal reference frame on symmetry grounds.
I feel that there is a mistake in the final formula. The first (\gamma - 1)^2 term needs multiplication by a factor 4.

ReplyDeleteGood catch ! I made an error transcribing the formula from my notes... Thank you very much !

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Moderncalculator

It's a good post about mass moment of inertia. But if you illustrate it, then it'll a better post.

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