Question 4.7.8: A satellite is considered as a system consisting of two equa......

Consider the kinematics of the satellite as shown in Figure 4E8b. Note that when the angular velocity is zero the system still has vibration, and therefore one still has motion, y and ÿ. Thus, if one assumes the mode shape or motion pattern as that given in Figure 4E8b, the acceleration in the y-direction is \ddot{y}-\Omega^{2}y\mathrm{~for~}\Omega\neq0. The following is concerned with the derivation of this quantity.

The velocity at point B of the cable is

\nu_{B}=\nu_{A}+\nu_{B/A}.

But the differentiation of the relative velocity with respect to time t is

{\frac{d\nu_{B/A}}{d t}}={\frac{d\Omega\times{\bf r}}{d t}}={\frac{d\Omega}{d t}}\times{\bf r}+\Omega\times{\frac{d{\bf r}}{d t}}.

Since Ω is constant and the first term on the rhs becomes zero, then

{\frac{d\nu_{B/A}}{d t}}=\Omega\times{\frac{d\mathbf{r}}{d t}}

where

{\frac{d\mathbf{r}}{d t}}=r_{x}{\frac{d \bf{ i}}{d t}}+r_{y}{\frac{d \bf{ j}}{d t}}={\bf{\Omega}}\times\left(r_{x}{\bf {i}+}r_{y}{\bf j}\right)={\bf{\Omega}}\times\mathbf{r}.

Note that Axyz is the rotating frame of reference in which the z-axis is perpendicular to the x-y plane and pointing outward from the x-y plane in accordance with the right-hand screw rule. Thus,

{\frac{d\nu_{B/A}}{d t}}={\bf{\Omega}}\times\left({\bf{\Omega}}\times{\bf r}\right).

Operating on this equation, one has

{\frac{d\nu_{B/A}}{d t}}={\bf{\Omega}}\times\left({\bf{\Omega}}\times{\bf{r}}\right)={\bf{\Omega}}\,k\times\left[{\bf{\Omega}}\,k\times\left(r_{x}\bf i+r_{y} \bf j\right)\right]

=\Omega k\times\left(\Omega r_{x} \bf j-\Omega r_{y} \bf i\right)=\Omega^{2}\left(-r_{x} \bf i-r_{y} \bf j\right).

Note that according to the symbol given in the example, r_{x}=x{\mathrm{~and~}}r_{v}=y. Since in the present example only lateral motion is required, only the component in the y direction is of interest.

This component is in addition to the acceleration term \frac{\partial^{2}y}{\partial t^{2}} or ÿ in Equation (4.23) in which Ω = 0. In other words, the total acceleration in the y direction is now

{\frac{\partial^{2}y}{\partial t^{2}}}=c^{2}{\frac{\partial^{2}y}{\partial x^{2}}}          (4.23)

{\frac{\partial^{2}y}{\partial t^{2}}}-\Omega^{2}y.

With reference to Equation (4.23), one has

T\frac{{\partial^{2}}y}{{\partial x}^{2}}=\rho\left(\frac{{\partial^{2}}y}{{\partial t}^{2}}-{\Omega^{2}}y\right).

Since the tension in the cable is T = mLΩ², this equation becomes

{\frac{\partial^{2}y}{\partial x^{2}}}={\frac{\rho}{m L\Omega^{2}}}\left({\frac{\partial^{2}y}{\partial t^{2}}}-\Omega^{2}y\right).

This is the required equation.