**Mechanical Practice Problem**
A Motor
==============
!!! WARNING
This is a practice electromechanical problem that isn't really optimized for use in an exam setting. By this I mean we didn't really do any problems with a rotational framework and that'd be a bit too much to put on you in an exam setting. The math on part 2 is also probably too long for a question, but you should be able to do it! It is at the same level of a first order circuit.
You have a standard DC motor that looks like the following:
![Figure [motor_example.png]: Simple DC motor with voltage $v_{SIG}$ applied to it and an angular velocity $\omega$ as an output. ](media/motor_example.png width="400px" )
Below is the electro-mechanical model of that standard DC motor. The left half of the circuit depicts the electrical equivalent of the motor (its coil), the right half is a circuit-equivalent model of the mechanical behavior of the motor (its rotation, momentum, etc), and the dependent sources represent how the two are inter-related.
![Figure [motor_equivalent.png]: Circuit Equivalent of the motor. ](media/motor_equivalent.png width="800px" )
The electrical portion of the circuit should be familiar. The mechanical portion is slightly different than what we've seen before. Because the motor operates in a rotational frame of reference there are the following notable differences from previous examples:
* Instead of a linear velocity, we have an angular velocity of the rotor $\omega$
* Instead of a linear acceleration, we have an angular acceleration of the rotor, $\alpha$
* Instead of a linear position, we have and angle of the rotor, $\theta$
* Instead of forces we have torques on the rotor, $\tau$
* Instead of a mass, we worry about moment of inertia of the rotor $J$.
* Generally speaking $\tau = J\times \alpha$, however we can assume the moment of inertia of our motor is always normal to our angular movement that means $\tau = J\alpha$ (which is analogous to $f =ma$ in our linear framework)
* $b$ is a rotational drag coefficient analogous to friction.
Rotational mechanics aside, you should see that this system is very similar to the one we had in lecture and in lab (the diving board magnet lab).
Answer the following questions about the motor:
Steady State Velocity
----------------------
If $v_{SIG}$ starts at 0V, the motor will have no angular velocity. When $v_{SIG}$ is then switched to a different voltage, the angular velocity of the motor will increase gradually before settling at a steady state velocity. Before we consider the time-dynamics of the system consider the case where the input voltage $v_{SIG}$ is set to a constant voltage $V_C$ for a very long time (far longer than the system's dynamics). Generate an expression for the steady state angular velocity $\omega$ of the motor as a function of system parameters and $V_C$ in this particular case.
In steady state, the following elements of the motor simplify away:
* On the electrical side, the inductor $L_e \to 0$
* For the capacitor equivalent on the motor side $\frac{1}{J} \to \infty$
This means our circuit simplifies to the following:
![Figure [simplified_1.png]: Simplified circuit ](media/simplified_1.png width="600px")
As a result we can say:
$$\omega = \tau\frac{1}{b}$$
and since
$$
Ki_m = \tau$$
and
$$
i_m = \frac{V_C-K\omega}{R_e}
$$
we can do some substitution:
$$
\omega = K\left(\frac{V_C-K\omega}{R_e}\right)\frac{1}{b}
$$
which simplifies to:
$$
\omega = \frac{KV_C}{bR_e} - \frac{K^2}{bR_e}\omega
$$
and therefore:
$$
\omega = \frac{\frac{KV_C}{bR_e}}{1+\frac{K^2}{bR_e}}
$$
or
$$
\omega = \frac{KV_C}{bR_e+K^2}
$$
Velocity Over Time
----------------------
If the motor starts at rest ($v_{SIG} = 0V$) and is the input voltage is then stepped to $V_F$, generate a time domain expression for motor speed as a function of time $\omega(t)$. For this problem assume that the motor's inductor is negligible and can be ignored. You can also use $\Omega_{SS}$ to be a stand-in for the steady state angular velocity you found as a function of voltage in the previous question.
We now must consider the entire circuit. To do this let's generate an expression for the left half of the circuit. KVL (and we're using KVL since it provides non-trivial information in a series circuit like this) tells us:
(to keep the equations a bit simpler I'm going to not explicitley write variables as functions of $t$ until the end, but before we go into that variables which will be functions of time are everything but the constants which means: $i_m(t)$, $v_{SIG}(t)$, $\omega(t)$!)
$$
v_{SIG} + i_mR_e + L_e\frac{di_m}{dt} + K\omega = 0
$$
If we can assume that $L_e$ is negligible (as stated), this simplifies to:
![Figure [no_inductor.png]: Simplified circuit ](media/no_inductor.png width="600px")
which has the following equation:
$$
v_{SIG} + i_mR_e + K\omega =0
$$
Then on the right side we have the following coming from KCL (this is a parallel circuit so that's usually where the non-trivial information can be extracted):
$$
Ki_m = J\frac{d\omega}{dt} + b\omega
$$
We can substitute this equation above with the one we got earlier via the current $i_m$. So:
$$
i_m = \frac{J}{K}\frac{d\omega}{dt} + \frac{b}{K}\omega
$$
so then:
$$
v_{SIG} + \left(\frac{J}{K}\frac{d\omega}{dt} + \frac{b}{K}\omega\right)R_e + K\omega=0
$$
and therefore:
$$
v_{SIG} + \frac{JR_e}{K}\frac{d\omega}{dt} + \left(\frac{bR_e}{K} + K\right)\omega=0
$$
This is a first order differential equation. The homogenous form of this will then take on the following:
$$
\frac{JR_e}{K}\frac{d\omega}{dt} + \left(\frac{bR_e}{K} + K\right)\omega = 0
$$
Reframing this we can see it says:
$$
\frac{d\omega}{dt} = -\frac{\left(\frac{bR_e}{K} + K\right)}{\frac{JR_e}{K}}\omega
$$
or when we simplify and write actually as a function of time:
$$
\frac{d\omega(t)}{dt} = -\frac{\left(bR_e + K^2\right)}{JR_e}\omega(t)
$$
The solution to this differential equation is of course a first order of the form:
$$
\omega(t) = A + Be^{-t\frac{\left(bR_e + K^2\right)}{JR_e}}
$$
If we're moving from a state where the motor has no velocity since the input voltage was originally 0V, that means
$$
\omega(0) = 0 = A + Be^{-0\cdot\frac{\left(bR_e + K^2\right)}{JR_e}}
$$
Or
$$
A = -B
$$
then if our final input voltage is $V_F$ that means:
$$
\omega(t\to\infty) = \Omega_{SS} = A + B\cdot(0)
$$
So
$$
A = \Omega_{SS}
$$
So:
$$
\omega(t) = \Omega_{SS}\left(1 - e^{-t\frac{\left(bR_e + K^2\right)}{JR_e}}\right)
$$
Follow-Up Questions
----------------------
Answer the following questions about how changing system parameters will change the behavior of the motor (it is a good idea to review the previous two sections before going on):
### Moment
What happens to the time dynamics of motor when its moment of Inertia is increased by a factor of 10? What happens to the steady-state velocity it runs at for a given input voltage?
If $J$ is increased it will **decrease the time constant by a factor of ten** since:
$$
=\frac{\left(bR_e + K^2\right)}{JR_e}
$$
This means the system will respond ten times slower to voltage steps.
If $J$ is increaesed by a var factor of ten it will **not** change the steady state angular velocity for a given voltage because (from part 1):
$$
\omega_{ss} = \frac{KV_C}{bR_e+K^2}
$$
This shoud make sense...more inertia in a motor's rotor will cause it to change a lot slower, but it will not change the final angular velocity a particular torque will impart. That will instead be based on the damping.
### Drag Coefficient
What happens to the time dynamics of motor when its moment of drag coefficient $b$ is increased by a factor of 10? What happens to the steady-state velocity it runs at for a given input voltage?
If $b$ is increased it will **increase the time constant by some amount** since:
$$
=\frac{\left(bR_e + K^2\right)}{JR_e}
$$
This means the system will respond faster...this might sound counter-intuitive but it means that the counteracting drag torque will come in and fight the driving torque sooner.
If $b$ is increaesed by a factor of ten it **will change** the steady state angular velocity for a given voltage because (from part 1):
$$
\omega_{ss} = \frac{KV_C}{bR_e+K^2}
$$
This should make sense...it means that for a given angular velocity, you have more drag being imparted on the motor which means more torque, which means it can't spin as fast in steady state.