Chapter 18.3, Problem 18.129P

To determine

The precession axis, rates of precession, and the spin rate of the satellite after the impact.

Answer to Problem 18.129P

The precession along x, y, and z axis $\left({\theta }_x,{\theta }_y,\text{and}{\theta }_z\right)$ is $\underset{\_}{125°}$, $\underset{\_}{36.4°}$, and $\underset{\_}{80.7°}$ respectively.

The precession rate $\left(\dot{\varphi }\right)$ is $\underset{\_}{0.949\text{rad/s}}$.

The spin rate $\left(\dot{\psi }\right)$ is $\underset{\_}{0.16\text{rad/s}}$.

Explanation of Solution

Given information:

The weight of geostationary satellite (W) is 800 lb.

The angular velocity of the satellite $\left({\omega }_0\right)$ is $\left(1.5\text{rad/s}\right)\text{j}$.

The weight of a meteorite $W^{\prime }$ is 6 oz.

The travelling velocity of the meteorite ${\text{v}}_0$ is $-\left(1,600\text{ft/s}\right)\text{i}+\left(1,300\text{ft/s}\right)\text{j}+\left(4,000\text{ft/s}\right)\text{k}$.

The distance value b is 20 in..

The radii of gyration of the satellite along x, y, and z direction (${\overline{k}}_x$, ${\overline{k}}_y$, and ${\overline{k}}_z$) is 28.8 in., 32.4 in., and 28.8 in. respectively.

Calculation:

Determine the mass of the satellite (m).

$m=\frac{W}{g}$

Here, g is the acceleration due to gravity.

Substitute 800 lb for W and $32.2{\text{ft/s}}^{\text{2}}$ for g.

$\begin{array}{c}m=\frac{800}{32.2}\\ =24.845\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}\end{array}$

Determine the principal moment of inertia along x axis.

${\overline{I}}_x=m{\overline{k}}_x^2$

Substitute $24.845\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}$ for m and 28.8 in. for ${\overline{k}}_x$.

$\begin{array}{c}{\overline{I}}_x=24.845\times {\left(28.8\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\\ =143.11\text{lb}\cdot {\text{s}}^{\text{2}}\cdot \text{ft}\end{array}$

Determine the principal moment of inertia along y axis.

${\overline{I}}_y=m{\overline{k}}_y^2$

Substitute $24.845\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}$ for m and 32.4 in. for ${\overline{k}}_y$.

$\begin{array}{c}{\overline{I}}_y=24.845\times {\left(32.4\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\\ =181.12\text{lb}\cdot {\text{s}}^{\text{2}}\cdot \text{ft}\end{array}$

Determine the principal moment of inertia along z axis.

${\overline{I}}_z=m{\overline{k}}_z^2$

Substitute $24.845\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}$ for m and 28.8 in. for ${\overline{k}}_z$.

$\begin{array}{c}{\overline{I}}_z=24.845\times {\left(28.8\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\\ =143.11\text{lb}\cdot {\text{s}}^{\text{2}}\cdot \text{ft}\end{array}$

Determine the mass of the meteorite $m^{\prime }$.

$m^{\prime }=\frac{W^{\prime }}{g}$

Substitute 6 oz for $W^{\prime }$ and $32.2{\text{ft/s}}^{\text{2}}$ for g.

$\begin{array}{c}m=\frac{6\text{oz}\times \frac{0.0625\text{lb}}{1\text{oz}}}{32.2}\\ =0.0116\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}\end{array}$

Determine the initial moment of meteorite $\left(p_0\right)$.

$p_0=m^{\prime }{\text{v}}_0$

Substitute $0.0116\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}$ for $m^{\prime }$ and $-\left(1,600\text{ft/s}\right)\text{i}+\left(1,300\text{ft/s}\right)\text{j}+\left(4,000\text{ft/s}\right)\text{k}$ for ${\text{v}}_0$.

$\begin{array}{c}{p}_0=0.01164\times \left(-\left(1,600\text{ft/s}\right)\text{i}+\left(1,300\text{ft/s}\right)\text{j}+\left(4,000\text{ft/s}\right)\text{k}\right)\\ =\left(-18.62\text{lb}\cdot \text{s}\right)\text{i}+\left(15.13\text{lb}\cdot \text{s}\right)\text{j}+\left(46.56\text{lb}\cdot \text{s}\right)\text{k}\end{array}$

Consider that the position of the satellite mass center plus the meteorite is essentially that of the satellite alone.

Determine the position of the point B relative to the mass center.

${\text{r}}_{B/G}=x\text{i}-y\text{j}$

Here, x is the horizontal distance and y is the vertical distance.

Substitute 42 in. for x and 20 in. for y.

$\begin{array}{c}{\text{r}}_{B/G}=\left(42\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\text{i}-\left(20\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\text{j}\\ =\left(\text{3}\text{.5}\text{ft}\right)\text{i}-\left(1.667\text{ft}\right)\text{j}\end{array}$

The angular velocity of satellite before impact $\left({\text{ω}}_0\right)$ is $\left(1.5\text{rad/s}\right)\text{j}$.

The angular velocity of satellite before impact along x, y, and z axis is ${\left({\omega }_0\right)}_x=0$, ${\left({\omega }_0\right)}_y=1.5\text{rad/s}$, and ${\left({\omega }_0\right)}_z=0$.

Determine the angular momentum of satellite–meteorite system before impact ${\left({\text{H}}_G\right)}_0$.

${\left({\text{H}}_G\right)}_0={\overline{I}}_y{\omega }_0\text{j}+{\text{r}}_{B/G}\times p_0$

Substitute $181.12\text{lb}\cdot {\text{s}}^{\text{2}}\cdot \text{ft}$ for ${\overline{I}}_y$, $1.5\text{rad/s}$ for ${\omega }_0$, $\left(\text{3}\text{.5}\text{ft}\right)\text{i}+\left(1.667\text{ft}\right)\text{j}$ for ${\text{r}}_{B/G}$, and $\left(-18.62\text{lb}\cdot \text{s}\right)\text{i}+\left(15.13\text{lb}\cdot \text{s}\right)\text{j}+\left(46.56\text{lb}\cdot \text{s}\right)\text{k}$ for $p_0$.

$\begin{array}{c}{\left({\text{H}}_G\right)}_0=\left(181.12\times 1.5\right)\text{j}+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 3.5& -1.667& 0\\ -18.62& 15.13& 46.56\end{array}\right|\\ =\left\{\begin{array}{l}271.68\text{j}+\text{i}\left(-1.667\times 46.56-0\right)-\text{j}\left(3.5\times 46.56-0\right)+\\ \text{k}\left(3.5\times 15.13-\left(-1.667\times -18.62\right)\right)\end{array}\right\}\\ =271.68\text{j}-77.61\text{i}-162.96\text{j}+21.92\text{k}\\ =\left(-77.61\text{i}+108.72\text{j}+21.92\text{k}\right)\text{lb}\cdot \text{s}\cdot \text{ft}\end{array}$

Principle of impulse and momentum for satellite–meteorite system:

The value of moments about G is ${\left({\text{H}}_G\right)}_1={\left({\text{H}}_G\right)}_0={\text{H}}_G$.

The expression for ${\text{H}}_G$ is ${\text{H}}_G={\overline{I}}_x{\omega }_x\text{i}+{\overline{I}}_y{\omega }_y\text{j}+{\overline{I}}_z{\omega }_z\text{k}$.

Determine the angular velocity about x axis.

${\omega }_x=\frac{{\left({\text{H}}_G\right)}_x}{I_x}$

Substitute $-\text{77}\text{.61}\text{lb}\cdot \text{s}\cdot \text{ft}$ for ${\left({\text{H}}_G\right)}_x$ and $143.11\text{lb}\cdot {\text{s}}^{\text{2}}\cdot \text{ft}$ for $I_x$.

$\begin{array}{c}{\omega }_x=\frac{-77.61}{143.11}\\ =-0.5423\text{rad/s}\end{array}$

Determine the angular velocity about y axis.

${\omega }_y=\frac{{\left({\text{H}}_G\right)}_y}{I_y}$

Substitute $108.72\text{lb}\cdot \text{s}\cdot \text{ft}$ for ${\left({\text{H}}_G\right)}_y$ and $181.12\text{lb}\cdot {\text{s}}^{\text{2}}\cdot \text{ft}$ for ${\overline{I}}_y$.

$\begin{array}{c}{\omega }_y=\frac{108.72}{181.12}\\ =0.6\text{rad/s}\end{array}$

Determine the angular velocity about z axis.

${\omega }_z=\frac{{\left({\text{H}}_G\right)}_z}{I_z}$

Substitute $21.92\text{lb}\cdot \text{s}\cdot \text{ft}$ for ${\left({\text{H}}_G\right)}_z$ and $143.11\text{lb}\cdot {\text{s}}^{\text{2}}\cdot \text{ft}$ for ${\overline{I}}_z$.

$\begin{array}{c}{\omega }_z=\frac{21.92}{143.11}\\ =0.153\text{rad/s}\end{array}$

Determine the vector format of angular velocity using the relation;

$\text{ω}={\omega }_x\text{i}+{\omega }_y\text{j}+{\omega }_z\text{k}$

Substitute $-0.5423\text{rad/s}$ for ${\omega }_x$, $0.6\text{rad/s}$ for ${\omega }_y$, and $0.153\text{rad/s}$ for ${\omega }_z$.

$\text{ω}=\left(-0.5423\text{rad/s}\right)\text{i}+\left(0.6\text{rad/s}\right)\text{j}+\left(0.153\text{rad/s}\right)\text{k}$

Determine the magnitude of the angular velocity $\omega$.

$\omega =\sqrt{{\omega }_x^2+{\omega }_y^2+{\omega }_z^2}$

Substitute $-0.5423\text{rad/s}$ for ${\omega }_x$, $0.6\text{rad/s}$ for ${\omega }_y$, and $0.153\text{rad/s}$ for ${\omega }_z$.

$\begin{array}{c}\omega =\sqrt{{\left(-0.5423\right)}^2+{\left(0.6\right)}^2+{\left(0.153\right)}^2}\\ =0.823\text{rad/s}\end{array}$

Determine the magnitude of angular momentum $H_G$ using the relation.

$H_G=\sqrt{{\left({\text{H}}_G\right)}_x^2+{\left({\text{H}}_G\right)}_y^2+{\left({\text{H}}_G\right)}_z^2}$

Substitute $-\text{77}\text{.61}\text{lb}\cdot \text{s}\cdot \text{ft}$ for ${\left({\text{H}}_G\right)}_x$, $108.72\text{lb}\cdot \text{s}\cdot \text{ft}$ for ${\left({\text{H}}_G\right)}_y$, and $21.92\text{lb}\cdot \text{s}\cdot \text{ft}$ for ${\left({\text{H}}_G\right)}_z$.

$\begin{array}{c}{H}_G=\sqrt{{\left(-77.61\right)}^2+{\left(108.72\right)}^2+{\left(21.92\right)}^2}\\ =135\text{lb}\cdot \text{s}\cdot \text{ft}\end{array}$

Motion after impact:

The moment of inertia about x and y axis is equal, the body moves as an axisymmetrical body with the y axis as the symmetry axis.

The moment of inertia about the symmetry axis is $I={\overline{I}}_y=181.12\text{lb}\cdot {\text{s}}^{\text{2}}\cdot \text{ft}$.

The moment of inertia about a transverse axis through G is $I^{\prime }={\overline{I}}_x={\overline{I}}_z=143.11\text{lb}\cdot {\text{s}}^{\text{2}}\cdot \text{ft}$.

The precession is retrograde when the value of $I>I^{\prime }$.

Determine the angle $\theta$ about x axis.

$\cos {\theta }_x=\frac{{\left({\text{H}}_G\right)}_x}{H_G}$

Substitute $-\text{77}\text{.61}\text{lb}\cdot \text{s}\cdot \text{ft}$ for ${\left({\text{H}}_G\right)}_x$ and $135\text{lb}\cdot \text{s}\cdot \text{ft}$ for $H_G$.

$\begin{array}{l}\cos {\theta }_x=\frac{-77.61}{135}\\ {\theta }_x={\cos }^{-1}\left(-0.575\right)\\ {\theta }_x=125°\end{array}$

Determine the angle $\theta$ about y axis.

$\cos {\theta }_y=\frac{{\left({\text{H}}_G\right)}_y}{H_G}$

Substitute $108.72\text{lb}\cdot \text{s}\cdot \text{ft}$ for ${\left({\text{H}}_G\right)}_y$ and $135\text{lb}\cdot \text{s}\cdot \text{ft}$ for $H_G$.

$\begin{array}{l}\cos {\theta }_y=\frac{108.72}{135}\\ {\theta }_y={\cos }^{-1}\left(0.805\right)\\ {\theta }_y=36.4°\end{array}$

Determine the angle $\theta$ about z axis.

$\cos {\theta }_z=\frac{{\left({\text{H}}_G\right)}_z}{H_G}$

Substitute $21.92\text{lb}\cdot \text{s}\cdot \text{ft}$ for ${\left({\text{H}}_G\right)}_z$ and $135\text{lb}\cdot \text{s}\cdot \text{ft}$ for $H_G$.

$\begin{array}{l}\cos {\theta }_z=\frac{21.92}{135}\\ {\theta }_z={\cos }^{-1}\left(0.162\right)\\ {\theta }_z=80.7°\end{array}$

Thus, the precession along x, y, and z axis $\left({\theta }_x,{\theta }_y,\text{and}{\theta }_z\right)$ is $\underset{\_}{125°}$, $\underset{\_}{36.4°}$, and $\underset{\_}{80.7°}$ respectively.

The angle $\theta$ between the spin axis (y–axis) and the precession axis remains constant $\theta ={\theta }_y=36.4°$.

Determine the angle $\gamma$ between the angular velocity and the spin axis.

$\cos \gamma =\frac{{\omega }_y}{\omega }$

Substitute $0.6\text{rad/s}$ for ${\omega }_y$ and $0.823\text{rad/s}$ for $\omega$.

$\begin{array}{l}\cos \gamma =\frac{0.6}{0.823}\\ \gamma ={\cos }^{-1}\left(0.729\right)\\ \gamma =43.2°\end{array}$

Determine the value of $\left(\gamma -\theta \right)$.

Substitute $43.2°$ for $\gamma$ and $36.4°$ for $\theta$.

$\begin{array}{c}\left(\gamma -\theta \right)=43.2°-36.4°\\ =6.8°\end{array}$

Draw the free body diagram of precession and spin axis as in Figure (1).

Draw the free body diagram of triangle of vector addition as in Figure (2).

Write the relation between the angles using the sine law.

$\frac{\dot{\varphi }}{\sin \gamma }=\frac{\dot{\psi }}{\sin \left(\gamma -\theta \right)}=\frac{\omega }{\sin \theta }$

Determine the precession rate $\left(\dot{\varphi }\right)$ using law of sines.

$\dot{\varphi }=\frac{\omega \sin \gamma }{\sin \theta }$

Substitute $0.823\text{rad/s}$ for $\omega$, $43.2°$ for $\gamma$, and $36.4°$ for $\theta$.

$\begin{array}{c}\dot{\varphi }=\frac{0.823\times \sin 43.2°}{\sin 36.4°}\\ =0.949\text{rad/s}\end{array}$

Therefore, the precession rate $\left(\dot{\varphi }\right)$ is $\underset{\_}{0.949\text{rad/s}}$.

Determine the rate of spin $\dot{\psi }$ using the relation.

$\dot{\psi }=\frac{\omega \sin \left(\gamma -\theta \right)}{\sin \theta }$

Substitute $0.823\text{rad/s}$ for $\omega$, $6.8°$ for $\left(\gamma -\theta \right)$, and $36.4°$ for $\theta$.

$\begin{array}{c}\dot{\psi }=\frac{0.823\sin 6.8°}{\sin 36.4°}\\ =0.16\text{rad/s}\end{array}$

The precession is retrograde due to value of $\gamma$ is greater than the angle $\theta$.

Therefore, the spin rate $\left(\dot{\psi }\right)$ is $\underset{\_}{0.16\text{rad/s}}$.