Chapter 18.1, Problem 18.54P

Determine the kinetic energy of the space probe of Prob. 18.34 in its motion about its mass center after its collision with the meteorite.

18.34    The coordinate axes shown represent the principal centroidal axes of inertia of a 3000-lb space probe whose radii of gyration are k_x = 1.375 ft, k_y = 1.425 ft, and k_z = 1.250 ft. The probe has no angular velocity when a 5-oz meteorite strikes one of its solar panels at point A and emerges on the other side of the panel with no change in the direction of its velocity, but with a speed reduced by 25 percent. Knowing that the final angular velocity of the probe is ω = (0.05 rad/s)i − (0.12 rad/s)j + ω_zk and that the x component of the resulting change in the velocity of the mass center of the probe is −0.675 in./s, determine (a) the component ω_z of the final angular velocity of the probe, (b) the relative velocity v₀ with which the meteorite strikes the panel.

Fig. P18.33 and P18.34

To determine

The kinetic energy $\left(T^{\prime }\right)$ of the space probe in its motion about its mass center after its collision with the meteorite.

Answer to Problem 18.54P

The kinetic energy $\left(T^{\prime }\right)$ of the space probe is $\underset{\_}{39.9\text{lb}\cdot \text{ft}}$.

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 $w^{\prime }$.

$\begin{array}{c}{m}^{\prime }=\frac{3000}{32.2}\\ =93.17\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}\end{array}$

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).

$\begin{array}{l}-5.2408=0.002426v_x\\ v_x=-\frac{5.2408}{0.002426}\\ v_x=-2160\text{ft/s}\end{array}$

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 –4840 ft/s for $v_y$.

$\begin{array}{l}0.02184\left(-4840\right)=145.58{\omega }_z\\ {\omega }_z=-\frac{0.02184\left(4840\right)}{145.58}\\ {\omega }_z=-0.726\text{rad/s}\end{array}$

Calculate the kinetic energy of motion of the probe relative to its mass center $\left(T^{\prime }\right)$ using the formula:

$\begin{array}{c}{T}^{\prime }=\frac{1}{2}\left(I_x{\omega }_x^2+I_y{\omega }_y^2+I_z{\omega }_z^2\right)\\ =\frac{1}{2}{m}^{\prime }\left(k_x^2{\omega }_x^2+k_y^2{\omega }_y^2+k_z^2{\omega }_z^2\right)\end{array}$

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 ${\omega }_y$, and $-0.726\text{rad/s}$ for ${\omega }_z$.

$\begin{array}{c}{T}^{\prime }=\frac{1}{2}\times 93.17\left[\left({1.375}^2\times {0.05}^2\right)+\left({1.425}^2\times -{0.12}^2\right)+\left({1.25}^2\times \left(-{0.726}^2\right)\right)\right]\\ =\frac{1}{2}\times 93.17\left(0.00473+0.02924+0.824\right)\\ =39.9\text{lb}\cdot \text{ft}\end{array}$

Thus, the kinetic energy $\left(T^{\prime }\right)$ of the space probe is $\underset{\_}{39.9\text{lb}\cdot \text{ft}}$.