What is the torque produced by 2 rotating bodies with a clutch I am trying to simulate a car engine etc, but I have failed to find any equations governing the torque created by $2$ different constant velocity shafts of different angular momenta joining together with some given slip or friction factor. I know $I_1w1 + I_2w_2 = I_3w_3$ which gives me the end angular velocities, but its not giving me the torque acting on each shaft at any moment in time whilst they are joining.
 A: A clutch has two sides and a degree of freedom between them. Let's call one side the engine side with known torque $\tau_E$ and unknown speed $\omega_E$. On the other side, the transmission side, the speed is known $\omega_T$, but the transmitted torque $\tau_T$ isn't.
The clutch itself has a critical torque $\tau_C$ which limits its ability to transfer torque. This limits, depends on the engagement of the clutch, and it varies between zero and some maximum value that is more than the maximum the engine can supply $\tau_{C_{\rm max}} >\tau_{E_{\rm max}}$.
So value of the maximum critical torque (torque rating) does not matter for the operation of the car. All it matters is that with 100% clutch in $\tau_C = 0$ and with 0% clutch in $\tau_C > \tau_{E_{\rm max}}$.

Here is a system diagram that will drive the equations

The components are shown in boxes, and the system variables connect the components.
First, the kinematics are valid regardless of how the clutch behaves
$$ \begin{aligned}
   \omega_W & = \frac{v}{R} & \dot \omega_W & = \frac{\dot v}{R} \\
   \omega_T & = \gamma\; \omega_W &  \dot \omega_T & = \gamma\; \dot \omega_W
\end{aligned} \tag{1} $$
Where $gamma$ is the total gear ratio, and $R$ the wheel radius.
And thus the engine torque is always known if the engine speed $\omega_E$ is known $$ \tau_E = f(\omega_E) \tag{2}$$
Considering that each component has each own mass moment of inertia, as well as the combined mass of the car $m$ and any drag force $F_D$ applied, the equations of motion for the car, wheel and transmission are:
$$ \begin{aligned}
 F  & = m \dot v + F_D \\
 \tau_W &= I_W \dot \omega_W + F\,R \\
 \tau_T &= I_T \dot \omega_T + \tfrac{1}{\gamma} \tau_W
\end{aligned} \tag{3}$$
Notice that all the accelerations can be written in terms of $\dot v$, such as $\dot \omega_W = \tfrac{1}{R} \dot v$ and $\dot \omega_T = \tfrac{\gamma}{R} \dot v$.
Then the two scenarios for the clutch:

*

*Clutch engaged - engine speed matches transmission speed, $\omega_E = \omega_T$ as well as $\dot \omega_E = \dot \omega_T$
$$ \begin{aligned} \tau_E &= \tau_C + I_E \dot \omega_T \\
  \tau_C & = I_C \dot \omega_T + \tau_T \end{aligned} \\ \tag{4a} $$
The above 5 equations in (3) & (4a) are solved for $\tau_C$, $\tau_T$, $\tau_W$, $F$ and finally $\dot{v}$
Notice that the engine speed is directly linked to the car speed $\omega_E = \omega_T = \tfrac{\gamma}{R} v $ and so engine torque is known $\tau_E = f(\omega_E)$.

*

*Clutch slipping - transmission torque limited to $\tau_C$
$$ \begin{aligned} \tau_E &= \tau_C + I_E \dot \omega_E \\
  \tau_C & = I_C \dot \omega_T + \tau_T \end{aligned} \tag{4b} $$
The above 5 equations in (3) & (4b) are solved for $\omega_E$, $\tau_T$ $\tau_W$, $F$ and finally $\dot{v}$
Here the engine speed is found from the numeric integral $\Delta \omega_E = \int \dot \omega_E \,{\rm d}t $ and the car acceleration $\dot v$ does not depend on engine torque.
The conditions of slipping the clutch are $\tau_E > \tau_C$, and for engaging the clutch $| \omega_E - \omega_T | \approx 0$.
A: In a system with two drive shafts connected by a clutch and a system of gears, the power being transmitted by each shaft will be the torque times the angular velocity. (The angular momentum of each shaft is not of concern.)  The ratio of the two angular velocities is determined by the system of gears. The clutch allows the second shaft to start at rest and be brought up to speed.  In bringing the second shaft up to speed, some energy will be lost to friction in the clutch, and some will be needed to increase the kinetic energy of the second shaft and the gears.  After that, (if you have friction-less bearings and gears) all of the power supplied to the first shaft should emerge from the second shaft.
A: 
I can see the Torque transferred to the engine from the wheels could be limited by torque clutch, but I dont know the Torque and I dont know the clutch join time!

Correct.  You need to make assumptions.  If you have an inexperienced driver that releases the clutch instantly, the maximum torque that can be applied may well exceed stress limits on parts.  The theoretical maximum depends on the specifics of the clutch.  The faster it engages, the greater the maximum torque can be.
It's like asking the force on an object that bounces.  It depends on the materials.  A soft ball on a piece of foam will have low force.  Two steel balls bouncing might have forces of several thousand newtons.  But the total momentum change might be identical.  The only difference is the time involved.
An automatic transmission's torque converter, or an experienced manual driver will manipulate things in a way to limit the total torque.  But this is a variable, not something that falls out of your equations.
