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Can the Newton - Euler equations be used to model a rigid body attached on several springs?

Physics Asked by Tschibi on April 9, 2021

imagine a three dimensional rigid body with known Moment of Inertia (at the center of mass) $I_{text{cm}}$ which is suspended by several springs at different points on the surface of the body. The center of mass lies within the body itself. The springs are all connected to the same frame.

If I want to study the dynamics of such a system can I use the Newton-Euler equations (https://en.wikipedia.org/wiki/Newton%E2%80%93Euler_equations)?

Would it be easier to use the center of mass as frame or a fixed point outside of the body?

One Answer

Since you are doing vibrations, the velocity-related terms aren't important so the last term in the NE equations drops out. The choice of point of reference is up to you since it does not change the form of the equations.

$$ pmatrix{m vec{g} vec{c} times m vec{g} } + sum boldsymbol{f}_i = mathbf{I}, boldsymbol{ddot{x}} $$

where $boldsymbol{f}_i$ is the wrench from each spring, $mathbf{I}$ is the spatial mass matrix, and $boldsymbol{ddot{x}}$ is the (spatial) acceleration of the body displacement $boldsymbol{x}$. Also weight is included as applied at the center of mass $vec{c}$ along the gravity vector $vec{g}$.

Each spring should have spatial stiffness such that $$ boldsymbol{f}_i = - mathbf{K}_i, boldsymbol{x} $$

And the general mass matrix is

$$ mathbf{I} = begin{bmatrix} m [mathtt{1}] & -m [vec{c}times] m [vec{c}times] & mathcal{J}_{rm cm} - m [vec{c}times] [vec{c} times] end{bmatrix} $$

The tricky thing here is that all the points on the frame share the same spatial deflection $boldsymbol{x}$, but are interpreted as different vector deflections at different locations

$$ boldsymbol{x} = pmatrix{vec{x} vec{theta} } $$

For example, the deflection at the center of mass $vec{delta}_{rm cm}$ is given by $$ vec{x} = vec{delta}_{rm cm} + vec{r}_{rm cm} times vec{theta} $$

and the deflection at some other point i is $$ vec{x} = vec{delta}_i + vec{r}_i times vec{theta} $$

You see how the LHS is always the same, and the translational deflection $vec{delta}_i$ changes depending on the location $vec{r}_i$.

Answered by John Alexiou on April 9, 2021

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