Question 12.4: The approximate mass properties of the Explorer 1 satellite ...
The approximate mass properties of the Explorer 1 satellite are
Mass: m=14 \mathrm{~kg}
Minor-axis moment of inertia: I_{3}=0.17 \mathrm{~kg}-\mathrm{m}^{2}
Major-axis moment of inertia: I_{1}=I_{2}=5 \mathrm{~kg}-\mathrm{m}^{2} Initially Explorer 1 was spinning at 750 \mathrm{rpm}\left(\omega_{0}=78.54 \mathrm{rad} / \mathrm{s}\right) about its minor axis.
Determine the initial and final rotational kinetic energies and the final spin rate \omega_{f}.
We use Eq. (12.89) to determine the initial kinetic energy:
T_{\text {rot }, 0}=\frac{1}{2} I_{3} \omega_{0}^{2}=524.3252 \mathrm{~kg}-\mathrm{m}^{2} / \mathrm{s}^{2} \quad(\text { or units of joules, J) }
T_{\mathrm{rot}}={\frac{1}{2}}I\omega^{2} (12.89)
Alternatively, we can compute rotational kinetic energy from the constant angular momentum:
H=I_{3} \omega_{0}=13.3518 \mathrm{~kg}-\mathrm{m}^{2} / \mathrm{s}=\text { constant }
Using Eq. (12.90)
T_{\text {rot }, 0}=\frac{H^{2}}{2 I_{3}}=524.3252 \mathrm{~kg}-\mathrm{m}^{2} / \mathrm{s}^{2} \text { (same result) }
T_{\mathrm{rot}}={\frac{H^{2}}{2I}} (12.90)
The final rotational kinetic energy can be determined using Eq. (12.90) and the maximum moment of inertia I_{1} :
T_{\text {rot }, f}=\frac{H^{2}}{2 I_{1}}=17.8271 \mathrm{~kg}-\mathrm{m}^{2} / \mathrm{s}^{2}
Rotational kinetic energy has decreased by nearly a factor of 30.
The final spin rate about the major axis can be computed from the constant angular momentum H or final kinetic energy T_{\text {rot, } f \text {. Using angular momentum, the final spin }} rate is
\omega_{f}=\frac{H}{I_{1}}=2.6704 \mathrm{rad} / \mathrm{s}(\text { or } 25.5 \mathrm{rpm})
Thus, the angular velocity of the “flat spin” shown in Figure 12.18 \mathrm{~b} is nearly 1 / 30 the initial spin rate along the minor axis. This 1 / 30 factor is the ratio of the minimum and maximum moments of inertia, or I_{3} / I_{1}.
