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14) The motor torque speed characteristic is given by the equation $\omega_m= 100 - T$ where $\omega_m$ is in rad/sec and T is in $N\cdot m$. The initial motor torque is $10~N\cdot m$. A step load torque of $50~N\cdot m$ is applied at $t = 0$. If the total inertia of the motor load combination is $5~kg\cdot m^2$, the value of the motor torque after 5 sec is:
a. 50.0 $N\cdot m$
b. 45.2 $N\cdot m$
c. 35.2 $N\cdot m$
d. 25.2 $N\cdot m$
Solution:
$$\begin{align}\omega_m &= 100 - T\label{eq:1}\\
T &= T_L + J\cdot \frac{d\omega_m}{dt}\label{eq:2}\\
\text{Fom \eqref{eq:2}}~\rightarrow~T &= T_L + J\cdot \frac{\omega_2-\omega_1}{t_2-t_1}\label{eq:3}\\
\text{where } J = 5 \text{, } \omega_1 = 90 \text{, } t_1 = 0 \text{, } t_2 = 5 \text{ and solving for } \omega_2 &= \left [ T - T_L + \omega_1\right ]\\
100 - T &= T - T_L + \omega_1 \\
T &= \frac{100 + T_L - \omega_1}{2}\\
T &= \frac{100 + 60 - 90}{2}\\
T &= 35~N\cdot m\\
\text{Therefore, the speed can be calculated from \eqref{eq:1} to be } \omega_2 &= 100 - 35 = 65 \frac{rad}{s}
\end{align}$$