Consider an electrical motor on its own:

[![enter image description here][1]][1]

A voltage source exciting the stator resistance & inductance. The resultant current produces torque which produces an acceleration against an inertia, with some damping. 


We know:

**MOTOR: Electromagnetic & mechanical **

\$T = K_t i \$

torque equals constant multiplied by current

\$V = K_e \omega = K_e \dot{\Theta_a}\$

We know the backEMF voltage is proportional to velocity

\$T = J \ddot{\Theta_a} + D\dot{\Theta_a} \$

The Motor torque is that needed to accelerate an inertia with the damping torque subtracted.

\$ L_a\frac{di}{dt} + R_ai_a = e_a - K_e \dot{\Theta_a}\$

The voltage across the stator resistance and inductance equals the supply voltage minus the backEMF voltage (velocity dependant). With the stator current being that which flows through this network. 

**S-Domain** (for clarity)

\$s(Js + D)\Theta_a(s) = K_tI(s) \$ 

\$(L_as + R_a)I(s) = E_a(s) - Ke\Theta_a(s) \$





Exam Question
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The question authors query has additional complication via a load inertia & load damping all coupled via gearing.

Equally it does not include stator inductance (???) yet expects the final equation to include inertia and acceleration (???). One could argue in steady-state the inductor term drops out of the final equation (speed has settled, load has settled, backEMF has settled -> no change on the electrical side). HOWEVER, if steady-state is assumed then inertia & acceleration drops out. 

Personally I think the question has been stripped back too much, but if that is what has been shown... lets drop the \$ L_a\frac{di}{dt}\$ term.


By reflecting the inertia and the damping to the primary side of the gearbox:  

\$ J = J_a + (\frac{N_1}{N_2})^2J_L  \$

\$ D = D_a + (\frac{N_1}{N_2}) D_L \$

The question is using the subscript m to denote the total inertia and load as seen by the **m**otor. The **l**oad specific has a subscript L and the "motor specific" has a subscript a (to denote armature?) 

Combining: 

\$e_a = \Theta_a s (\frac{R(Js + D)}{K_t} + K_e )\$

\$e_a = \Theta_a s (\frac{R((J_a + (\frac{N_1}{N_2})^2J_L)s + (D_a + (\frac{N_1}{N_2}) D_L))}{K_t} + K_e )\$

\$e_a = \dot{\Theta_a}(\frac{R((J_a + (\frac{N_1}{N_2})^2J_L)s + (D_a + (\frac{N_1}{N_2}) D_L))}{K_t} + K_e )\$

Finally as it is needed to be in terms of \$\Theta_L\$ 

\$e_a = \dot{\Theta_a}\frac{N_1}{N_2}(\frac{R((J_a + (\frac{N_1}{N_2})^2J_L)s + (D_a + (\frac{N_1}{N_2}) D_L))}{K_t} + K_e )\$

since \$\Theta_m = \frac{N_2}{N_1}\Theta_L \Rightarrow  \Theta_L = \frac{N_1}{N_2}\Theta_m\$ 

  [1]: https://i.sstatic.net/jj7J9.png