## 28 May 2014

### Mass moment of inertia 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:
$I_z = \pi \rho R^5 \left [ (\gamma -1) + \frac{8}{15} \right ]$

$$I_x$$ 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.

1. I feel that there is a mistake in the final formula. The first (\gamma - 1)^2 term needs multiplication by a factor 4.

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

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