Question 13.3: The DC motor associated with the reaction wheel and spacecra...
The DC motor associated with the reaction wheel and spacecraft from Example 13.2 has the following parameters: motor torque constant K_{m}=0.043 \mathrm{N}-m/A, back-emf constant K_{b}=0.043 \mathrm{~V}-s / \mathrm{rad}, resistance R=2.8 \Omega, and friction coefficient b=6\left(10^{-5}\right) \mathrm{N}-m-s/rad. Determine the input voltage to the DC motor required to maintain a constant wheel spin rate of 3,000 rpm.
We can manipulate Eq. (13.40)
T_{m}={\frac{K_{m}}{R}}(e_{\mathrm{in}}-K_{b}\omega_{w}) (13.40)
to obtain the DC motor’s input voltage in terms of motor torque and the wheel’s angular velocity:
e_{\text {in }}=\frac{R}{K_{m}} T_{m}+K_{b} \omega_{w}
Equations (13.33) or (13.36)
\dot{\omega}_{w}=\frac{1}{I_{w}}(T_{m}-b\omega_{w})-\ddot{\phi} (13.33)
\dot{\omega}_{w}=\frac{I_{3}}{I_{\mathrm{sat}}I_{w}}(T_{m}-b\omega_{w}) (13.36)
show that the motor torque T_{m} is balanced by the wheel’s friction torque b \omega_{w} when the wheel reaches a constant angular velocity. Hence, the motor torque is
T_{m}=b \omega_{w}=\left[6\left(10^{-5}\right) \mathrm{N}-\mathrm{m}-\mathrm{s} / \mathrm{rad}\right](314.1593 \mathrm{rad} / \mathrm{s})=0.018850 \mathrm{~N}-\mathrm{m}
Note that the wheel’s angular velocity \omega_{w} must be expressed in rad/s (i.e., 3,000 rpm = 314.1593 \mathrm{rad} / \mathrm{s}). Therefore, the input voltage is
\begin{aligned} e_{\text {in }} & =\frac{R}{K_{m}} T_{m}+K_{b} \omega_{w} \\ & =(2.8 \Omega)(0.018850 \mathrm{~N}-\mathrm{m}) / 0.043 \mathrm{~N}-\mathrm{m} / \mathrm{A}+(0.043 \mathrm{~V}-\mathrm{s} / \mathrm{rad})(314.1593 \mathrm{rad} / \mathrm{s}) \\ & =14.736 \mathrm{~V} \end{aligned}
The back-emf voltage from the motor rotation is e_{b}=K_{b} \omega_{w}=13.509 \mathrm{~V}, and the motor current is i_{m}=\left(e_{\text {in }}-e_{b}\right) / R=0.438 \mathrm{~A}.