Chapter 18.1, Problem 18.34P

(a)

To determine

The component $\left({\omega }_z\right)$ of the final angular velocity of the probe.

Answer to Problem 18.34P

The component $\left({\omega }_z\right)$ of the final angular velocity of the probe is $\underset{\_}{-0.726\text{rad/s}}$.

Explanation of Solution

Given information:

The weight of the space probe $\left(w^{\prime }\right)$ is 3,000 lb.

The radius of gyration along x axis $\left(k_x\right)$ is 1.375 ft.

The radius of gyration along y axis $\left(k_y\right)$ is 1.425 ft.

The radius of gyration along z axis $\left(k_z\right)$ is 1.250 ft.

The weight of the meteorite (w) is 5 oz.

The angular velocity $\left(\omega \right)$ is $\left(\text{0}\text{.05rad/s}\right)i-\left(\text{0}\text{.12rad/s}\right)j+\left({\omega }_z\right)k$.

The change in velocity of the mass center of the probe $\left({v^{\prime }}_x\right)$ is -0.675 in./s.

The width of the side panel from center to point A (b) is 9 ft.

The length of the panel from center to point A (l) is 0.75 ft.

The speed is reduced by 25 percent.

Calculation:

Calculate the mass of the space probe $\left(m^{\prime }\right)$ using the formula:

$\begin{array}{l}{w}^{\prime }=m^{\prime }g\\ m^{\prime }=\frac{w^{\prime }}{g}\end{array}$

Here, g is the acceleration due to gravity.

Substitute $\text{32}{\text{.2ft/s}}^{\text{2}}$ for g and 3,000 lb for $w^{\prime }$.

$\begin{array}{c}{m}^{\prime }=\frac{3,000}{32.2}\\ =93.17\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}\end{array}$

Calculate the mass of the meteorite (m) using the formula:

$\begin{array}{l}w=mg\\ m=\frac{w}{g}\end{array}$

Substitute $\text{32}{\text{.2ft/s}}^{\text{2}}$ for g and 5 oz for w.

$\begin{array}{c}m=\frac{\text{5oz×}\frac{\text{1}}{\text{16}}\frac{\text{lb}}{\text{oz}}}{32.2}\\ =0.009705\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}\end{array}$

Write the relative position vector $\left(r_A\right)$ at the point (A) of impact as follows:

$r_A=bi+lk$

Substitute 9 ft for b and 0.75 ft.

$r_A=\left(9\text{ft}\right)i+\left(0.75\text{ft}\right)k$

Write the expression for the velocity $\left(v_0\right)$ along x, y and z axis as follows:

$v_0=v_{x}i+v_{y}j+v_{z}k$

Calculate the initial liner momentum of the meteorite using the relation:

$\text{Initial linear momentum}=m{v}_0$

Substitute $0.009705\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}$ for m and $v_{x}i+v_{y}j+v_{z}k$ for $v_0$.

$m{v}_0=\left(0.009705\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}\right)\times \left(v_{x}i+v_{y}j+v_{z}k\right)$

Calculate the moment about origin ${\left(H_A\right)}_O$ using the relation:

${\left(H_A\right)}_O=r_A\times m{v}_0$

Substitute $\left(0.009705\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}\right)\times \left(v_{x}i+v_{y}j+v_{z}k\right)$ for $m{v}_0$ and $\left(9\text{ft}\right)i+\left(0.75\text{ft}\right)k$ for $r_A$.

$\begin{array}{c}{\left(H_A\right)}_O=0.009705\left|\begin{array}{ccc}i& j& k\\ 9& 0& 0.75\\ v_x& v_y& v_z\end{array}\right|\\ =0.009705\left[\left(-0.75v_y\right)i-\left(9v_z-0.75v_x\right)j+\left(9v_y\right)k\right]\end{array}$

The speed is reduced to 25 percent.

Calculate the final liner momentum of the meteorite using the relation:

$\text{Final linear momentum}=0.75m{v}_0$

Substitute $0.009705\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}$ for m and $v_{x}i+v_{y}j+v_{z}k$ for $v_0$.

$\begin{array}{c}0.75m{v}_0=0.75\left(0.009705\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}\right)\times v_{x}i+v_{y}j+v_{z}k\\ =\left(0.007279\right)\times \left(v_{x}i+v_{y}j+v_{z}k\right)\end{array}$

Calculate the final linear momentum of meteorite and its moment about the origin using the relation:

$M=r_A\times \left(0.75m{v}_0\right)$

Substitute $\left(0.007279\right)\times \left(v_{x}i+v_{y}j+v_{z}k\right)$ for $0.75m{v}_0$ and $\left(9\text{ft}\right)i+\left(0.75\text{ft}\right)k$ for $r_A$.

$\begin{array}{c}M=0.007278\left|\begin{array}{ccc}i& j& k\\ 9& 0& 0.75\\ v_x& v_y& v_z\end{array}\right|\\ =0.007278\left[\left(-0.75v_y\right)i-\left(9v_z-0.75v_x\right)j+\left(9v_y\right)k\right]\end{array}$

The initial linear momentum of the space probe $\left(m^{\prime }{v^{\prime }}_0\right)$ is zero.

Calculate the final linear momentum of the space probe using the relation:

$\text{Final linear momentum}=m^{\prime }{v^{\prime }}_0$

Substitute ${v^{\prime }}_{x}i+{v^{\prime }}_{y}j+{v^{\prime }}_{z}k$ for ${v^{\prime }}_0$ and $93.17\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}$ for $m^{\prime }$.

$m^{\prime }{v^{\prime }}_0=93.17\left({v^{\prime }}_{x}i+{v^{\prime }}_{y}j+{v^{\prime }}_{z}k\right)$

Substitute -0.675 in./s for ${v^{\prime }}_x$.

$\begin{array}{c}{m}^{\prime }{v^{\prime }}_0=93.17\left(-\left(\frac{\text{0}\text{.675in}\text{.}}{\text{12}}\frac{\text{ft}}{\text{in}\text{.}}\right)i+{v^{\prime }}_{y}j+{v^{\prime }}_{z}k\right)\\ =93.17\left(-0.05625i+{v^{\prime }}_{y}j+{v^{\prime }}_{z}k\right)\end{array}$

Calculate the final angular momentum of the space probe $\left(H_A\right)$ using the formula:

$H_A=m^{\prime }\left(k_x^2{\omega }_{x}i+k_y^2{\omega }_{y}j+k_z^2{\omega }_{z}k\right)$

Substitute $93.17\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}$ for $m^{\prime }$, 1.375 ft for $k_x$, 1.425 ft for $k_y$, 1.25 ft for $k_z$, 0.05 rad/s for ${\omega }_x$, and –0.12 rad/s for ${\omega }_y$.

$\begin{array}{c}{H}_A=93.17\left[\left({1.375}^2\times 0.05\right)i+\left({1.425}^2\times -0.12\right)j+\left({1.25}^2\times {\omega }_z\right)k\right]\\ =93.17\left[0.09453i-0.243675j+1.5625{\omega }_{z}k\right]\\ =8.8074i-22.703j+145.58{\omega }_{z}k\end{array}$

Write the expression for the conservation of linear momentum of the probe plus the meteorite as follows:

$m{v}_0=0.75m{v}_0+m^{\prime }{v^{\prime }}_0$

Substitute $\left(0.009705\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}\right)\times \left(v_{x}i+v_{y}j+v_{z}k\right)$ for $m{v}_0$, $\left(0.007279\right)\times \left(v_{x}i+v_{y}j+v_{z}k\right)$ for $0.75m{v}_0$ and $93.17\left(-0.05625i+{v^{\prime }}_{y}j+{v^{\prime }}_{z}k\right)$ for $m^{\prime }{v^{\prime }}_0$.

$\begin{array}{l}\left\{\begin{array}{l}\left(0.009705\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}\right)\times \left(v_{x}i+v_{y}j+v_{z}k\right)-\\ \left[\left(0.007279\right)\times \left(v_{x}i+v_{y}j+v_{z}k\right)\right]\end{array}\right\}=\left[\begin{array}{l}93.17\\ \left(-0.05625i+{v^{\prime }}_{y}j+{v^{\prime }}_{z}k\right)\end{array}\right]\\ \left\{\begin{array}{l}0.009705v_{x}i+0.009705v_{y}j+\\ 0.009705v_{z}k-\\ \left(\begin{array}{l}0.007279v_{x}i+0.007279v_{y}j+\\ 0.007279v_{z}k\end{array}\right)\end{array}\right\}=\left(\begin{array}{l}-5.2408i+93.17{v^{\prime }}_{y}j+\\ 93.17{v^{\prime }}_{z}k\end{array}\right)\\ -5.2408i+93.17{v^{\prime }}_{y}j+93.17{v^{\prime }}_{z}k=\left(\begin{array}{l}0.009705v_{x}i+0.009705v_{y}j+\\ 0.009705v_{z}k-\\ 0.007279v_{x}i-0.007279v_{y}j-\\ 0.007279v_{z}k\end{array}\right)\\ -5.2408i+93.17{v^{\prime }}_{y}j+93.17{v^{\prime }}_{z}k=\left(\begin{array}{l}0.002426v_{x}i+\\ 0.002426v_{y}j+\\ 0.0024265v_{z}k\end{array}\right)\end{array}$ (1)

Equate the i component from the Equation (1).

$\begin{array}{l}-5.2408=0.002426v_x\\ v_x=-\frac{5.2408}{0.002426}\\ v_x=-2,160\text{ft/s}\end{array}$

Equate j component from the Equation (1).

$93.17{v^{\prime }}_y=0.002426v_y$

Equate k component from the Equation (1).

$93.17{v^{\prime }}_z=0.002426v_z$

Write the expression for the conservation of angular momentum about the origin as follows:

${\left(H_A\right)}_O=M+H_A$

Substitute $0.009705\left[\left(-0.75v_y\right)i-\left(9v_z-0.75v_x\right)j+\left(9v_y\right)k\right]$ for ${\left(H_A\right)}_O$, $0.007278\left[\left(-0.75v_y\right)i-\left(9v_z-0.75v_x\right)j+\left(9v_y\right)k\right]$ for M, and $8.8074i-22.703j+145.58{\omega }_{z}k$ for $H_A$.

$\begin{array}{l}\left\{\begin{array}{l}0.009705\\ \left[\left(-0.75v_y\right)i-\left(9v_z-0.75v_x\right)j+\left(9v_y\right)k\right]\end{array}\right\}=\left\{\begin{array}{l}0.007278\\ \left[\begin{array}{l}\left(-0.75v_y\right)i-\left(9v_z-0.75v_x\right)j\\ +\left(9v_y\right)k\end{array}\right]\end{array}\right\}+\\ \left(\begin{array}{l}8.8074i-22.703j+\\ 145.58{\omega }_{z}k\end{array}\right)\\ \left(\begin{array}{l}-0.00727875v_{y}i-0.087345v_{z}j+\\ 0.00727875v_{x}j+0.087345v_{y}k\end{array}\right)=\left(\begin{array}{l}-0.0054585v_{y}i-0.065502v_{z}j+\\ 0.0054585v_{x}j+0.065502v_{y}k\end{array}\right)+\\ \left(\begin{array}{l}8.8074i-22.703j+\\ 145.58{\omega }_{z}k\end{array}\right)\\ \left(\begin{array}{l}-0.00727875v_{y}i-0.087345v_{z}j+\\ 0.00727875v_{x}j+0.087345v_{y}k+\\ 0.0054585v_{y}i+0.065502v_{z}j-\\ 0.0054585v_{x}j-0.065502v_{y}k\end{array}\right)=\left(\begin{array}{l}8.8074i-22.703j+\\ 145.58{\omega }_{z}k\end{array}\right)\\ \left(\begin{array}{l}-0.00182v_{y}i-0.02184v_{z}j+\\ 0.00182v_{x}j+0.02184v_{y}k\end{array}\right)=\left(\begin{array}{l}8.8074i-22.703j+\\ 145.58{\omega }_{z}k\end{array}\right)\end{array}$ (2)

Equate i component from the Equation (2).

$\begin{array}{l}-0.00182v_y=8.8075\\ v_y=-\frac{8.8075}{0.00182}\\ v_y=-4840\text{ft/s}\end{array}$

Equate k component from the Equation (2).

$0.02184v_y=145.58{\omega }_z$

Substitute –4,840 ft/s for $v_y$.

$\begin{array}{l}0.02184\left(-4,840\right)=145.58{\omega }_z\\ {\omega }_z=-\frac{0.02184\left(4,840\right)}{145.58}\\ {\omega }_z=-0.726\text{rad/s}\end{array}$

Thus, the component $\left({\omega }_z\right)$ of the final angular velocity of the probe is $\underset{\_}{-0.726\text{rad/s}}$.

(b)

To determine

The relative velocity $\left(v_0\right)$ with which the meteorite strikes the panel.

Answer to Problem 18.34P

The relative velocity $\left(v_0\right)$ with which the meteorite strikes the panel is $\underset{\_}{-\left(2160\text{ft/s}\right)i-\left(\text{4840ft/s}\right)j+\left(860\text{ft/s}\right)k}$.

Explanation of Solution

Given information:

The weight of the space probe $\left(w^{\prime }\right)$ is 3,000 lb.

The radius of gyration along x axis $\left(k_x\right)$ is 1.375 ft.

The radius of gyration along y axis $\left(k_y\right)$ is 1.425 ft.

The radius of gyration along z axis $\left(k_z\right)$ is 1.250 ft.

The weight of the meteorite (w) is 5 oz.

The angular velocity $\left(\omega \right)$ is $\left(\text{0}\text{.05rad/s}\right)i-\left(\text{0}\text{.12rad/s}\right)j+\left({\omega }_z\right)k$.

The change in velocity of the mass center of the probe $\left({v^{\prime }}_x\right)$ is –0.675 in./s.

The width of the side panel from center to point A (b) is 9 ft.

The length of the panel from center to point A (l) is 0.75 ft.

The speed is reduced by 25 percent.

Calculation:

Find the velocity along z direction:

Equate j component from the equation (2).

$-0.02184v_z+0.00182v_x=-22.702$

Substitute -2160 ft/s for $v_x$.

$\begin{array}{l}-0.02184v_z+0.00182\left(-2160\right)=-22.702\\ -0.02184v_z=-22.702+3.9312\\ v_z=\frac{-22.702+3.9312}{-0.02184}\\ v_z=860\text{ft/s}\end{array}$

Calculate the relative velocity $\left(v_0\right)$ using the relation:

$v_0=v_{x}i+v_{y}j+v_{z}k$

Substitute -2160 ft/s for $v_x$, -4840 ft/s for $v_y$ and 860 ft/s for $v_z$.

$v_0=-\left(2160\text{ft/s}\right)i-\left(\text{4840ft/s}\right)j+\left(860\text{ft/s}\right)k$

Thus, the relative velocity $\left(v_0\right)$ with which the meteorite strikes the panel is $\underset{\_}{-\left(2160\text{ft/s}\right)i-\left(\text{4840ft/s}\right)j+\left(860\text{ft/s}\right)k}$.