# Does a bungee cord have a moment of inertia?

Does a bungee cord have a moment of inertia?

I'm trying to understand to what extent a body must be rigid in order for the moment of inertia to be defined for that body. Since the distance between points on the cord is variable, depending on how much the cord is stretched, I think of it as not being rigid.

Does this means that the moment of inertia (tensor) is simply not defined for the cord? Or does it take on some special value?

• Are you asking about mass moment of inertia as it pertains to rigid body rotations, or area moment as it pertains to bending problems. – ja72 Dec 12 '14 at 2:34
• You can compute the moment of inertia but I don't think it is a useful quantity since application of a torque will not lead to the expected angular acceleration as moment of inertia will change as the torque is applied and the bungee distorts. – Floris Dec 12 '14 at 3:51
• I'm curious about the mass moment of inertia – user44816 Dec 12 '14 at 13:55

If you have 1D rigid body that follows the curve defined by a 3D vector $\vec{r}(\alpha)$ where $\alpha$ is some parameter along the curve with cross sectional area $A$ and mass density $\rho$ then the mass of an infinitesimal part is $${\rm d} m = \rho A \| \frac{{\rm d} \vec{r}}{{\rm d} \alpha} \| \, {\rm d} \alpha$$

The mass moment of inertia of the same infinitesimal part is

$${\rm d}I = - [\vec{r}\times] [\vec{r} \times] {\rm d} m$$

where $[\vec{r}\times]$ is a skew symmetric matrix form of the cross product operator

$$[\begin{pmatrix} x\\y\\z \end{pmatrix} \times] = \begin{pmatrix} 0 & -z & y\\ z & 0 & -x \\ -y & x & 0 \end{pmatrix}$$

If you do the math you end up with

$${\rm d} I = \begin{pmatrix} y^2+z^2 & -x y & - x z \\ -x y & x^2+z^2 & -y z \\ -x z & -y z & x^2+y^2 \end{pmatrix} {\rm d}m$$

So the overall mass moment of inertia tension about the origin is $$m = \int \rho A \| \frac{{\rm d} \vec{r}}{{\rm d} \alpha} \| \, {\rm d} \alpha$$ $$I = \int \rho A \begin{pmatrix} y^2+z^2 & -x y & - x z \\ -x y & x^2+z^2 & -y z \\ -x z & -y z & x^2+y^2 \end{pmatrix} \| \frac{{\rm d} \vec{r}}{{\rm d} \alpha} \| \, {\rm d} \alpha$$

Example

A straight line of length $\ell$ defined by $\vec{r}(\alpha) = (\alpha \ell,0,0)$ with the length parameter $\alpha = 0 \ldots 1$. Each length segment has $$\| \frac{{\rm d} \vec{r}}{{\rm d} \alpha} \|= \| \frac{{\rm d} (\alpha \ell,0,0)}{{\rm d} \alpha} \| = \ell$$

$$m = \int_0^1 \rho A \ell \,{\rm d}\alpha$$ $$I = \int_0^1 \rho A \begin{pmatrix}0&0&0\\0 &\alpha^2 \ell^2 & 0\\ 0&0&\alpha^2\ell^2\end{pmatrix} \ell \,{\rm d}\alpha$$

which results in

$$m = \rho A \ell \\ I = \begin{pmatrix}0&0&0\\0 &\frac{m}{2} \ell^2 & 0\\ 0&0&\frac{m}{2}\ell^2\end{pmatrix}$$

Now given that the bungee cord is flexible, what you have is multiple masses linked together with articulations and the way to handle this situation is by computing the so-called Articulated Inertia Matrix.

• So the moment of inertia of a mass particle is a characteristic of that particle. This idea is extended to bodies by integrating over the body. But the quantity is useful/applicable only if the body is rigid, because of the assumption in the calculation that the distances between mass particles are constant. Is this right? – user44816 Dec 12 '14 at 14:06
• Absolutely correct @user44816. You said it much better that I could. – ja72 Dec 12 '14 at 14:33

You can certainly calculate the MOI of any given configuration of bungee cord. The MOI tensor is just some integrals over the mass distribution, which are nicely defined. Whether that is any use is another question. The usual reason to calculate the MOI is that the body is rigid and we can define its configuration with just a few (six) variables. If you want to calculate the motion of a non-rigid body, you usually have to break it up into pieces and consider the interactions of all the pieces.