Optimization of a composite quadrupole mass at high-speed rotation

Abstract

An experiment to measure the speed of gravity is being planned. For this purpose, a numerical method was developed for the optimization of a composite quadrupole mass at high-speed rotation. The optimization calculations aim to obtain a quadrupole mass which must generate a periodic gravitational signal of 3200 Hz with maximum amplitude, taking into account its geometric features and the mechanical properties of the component materials. Considering the gravitational wave detector Mario Schenberg as the signal receiving device, an estimate was obtained in which the largest emitter-detector distance for detecting the gravitational signal is between the orders of magnitude $10^1$ and $10^2$ m. A simplified modeling of the emitter-detector system indicates that the gravitational signal amplitude h decreases approximately proportional to $r^{-5}$, where r is the emitter-detector distance. The results obtained in this work serve as reference for more detailed numerical simulations in the future.

Appendix: Justification for Eq. (25)

This appendix shows the arguments that result in Eq. (25). This equation is used to estimate the deformation suffered by the carbon fiber components in the vicinity of the xz plane. Initially, two approximate conditions are assumed:

• The densities $\rho _{\mathrm {c \, th}}$ and $\rho _{\mathrm {c \, lam}}$ are treated as a single density $\rho$ for the carbon fiber components. This argument can be seen as reasonable, since $\rho _{\mathrm {c \, lam}} / \rho _{\mathrm {c \, th}} \approx 0.95$. This approximation is used exclusively in the deduction of Eq. (25);

• The strain $\varepsilon _y$, which occurs perpendicularly and in the vicinity of the xz plane, is a continuous function of the length v (indicated in Fig. 5), even at the interfaces of the system components.

The following arguments are based on Fig. 5, which represents the cross section of a homogeneous cylindrical object of radius R. Such a object, idealized only for the deduction of Eq. (25), is used for the simplified treatment of the region composed of carbon fiber (thread or laminate) + epoxy, taking into account that the object has length s along the z-axis as well as the quadrupole mass (see Fig. 3).

We start the deduction by calculating the centrifugal force $df_c$ (in modulus) that acts on the mass element dm:

$$\begin{aligned} df_{c}=dm\,u\,\omega ^{2} \end{aligned}$$

(35)

The component of $df_c$ in the y-axis is given by

$$\begin{aligned} df_{c \, y}=df_{c}\frac{\textit{w}}{u}=dm\,\omega ^{2}\,\textit{w} . \end{aligned}$$

(36)

With Eq. (36), we determine the force $dF_y$, which acts in the y-direction on the xz plane, due to the sum of the forces $df_{c \, y}$ applied along the vertical mass (v.m.) with infinitesimal thickness dv (in light blue in Fig. 5). It is given by

$$\begin{aligned} dF_{y}=\int _{\mathrm {v.m.}} df_{c \, y}=\int _{\mathrm {v.m.}} \omega ^{2}\,\textit{w}\,dm . \end{aligned}$$

(37)

Defining the volumetric density

$$\begin{aligned} \rho =\frac{dm}{dV}=\frac{dm}{s\,d\textit{w}\,dv} \end{aligned}$$

(38)

the mass element dm is given by

$$\begin{aligned} dm=\rho \, s\,d\textit{w}\,dv . \end{aligned}$$

(39)

Applying Eq. (39) in Eq. (37), $dF_y$ is given by

$$\begin{aligned} dF_{y}=\int _{\textit{w}_{1}}^{\textit{w}_{2}}\omega ^{2}\textit{w}\rho s\,dv\,d\textit{w} = \omega ^{2}\rho s\,dv\int _{\textit{w}_{1}}^{\textit{w}_{2}}\textit{w}\,d\textit{w} . \end{aligned}$$

(40)

Remembering that the cylinder has radius R, the integration limits $\textit{w}_1$ and $\textit{w}_2$ are

$$\begin{aligned} \textit{w}_{1}=0\quad \mathrm {and} \quad \textit{w}_{2}=\sqrt{R^{2}-v^{2}} . \end{aligned}$$

(41)

Applying these limits in Eq. (40), we have

$$\begin{aligned} dF_{y}=\frac{1}{2}\omega ^{2}\rho s\,dv(R^{2}-v^{2}) . \end{aligned}$$

(42)

The area element dA, below the highlighted vertical mass and on the xz plane, is given by $dA = s \, dv$. Then Eq. (42) can be rewritten as

$$\begin{aligned} dF_{y}=\frac{1}{2}\omega ^{2}\rho (R^{2}-v^{2})dA . \end{aligned}$$

(43)

Therefore, the tensile stress $\sigma _y$, which acts in the y-direction on the area dA, is given by

$$\begin{aligned} \sigma _{y}=\frac{dF_{y}}{dA}=\frac{1}{2}\omega ^{2}\rho (R^{2}-v^{2}) . \end{aligned}$$

(44)

The strain $\varepsilon _y$, which occurs perpendicularly and in the vicinity of the area dA, is given by

$$\begin{aligned} \varepsilon _{y}=\frac{\sigma _{y}}{E} =\frac{1}{2E}\omega ^{2}\rho (R^{2}-v^{2}), \end{aligned}$$

(45)

where E is the Young’s modulus of the material under study. Eq. (45) shows that $\varepsilon _{y}$ is proportional to $(R^{2}-v^{2})$, that is,

$$\begin{aligned} \varepsilon _{y}(v)=c(R^{2}-v^{2}) \end{aligned}$$

(46)

where c is a coefficient to be determined so that the function $\varepsilon _y(v)$ is continuous along the xz plane.