# In equilibrium is engine torque zero? Vehicle dynamics

Consider following equations for a driving wheel in a car.

$$J\dot{\omega} = T - F_x R - F_r R \\ m\dot{v} = F_x + F_r$$ where $$J$$, $$\omega$$, $$T$$, $$F_x$$, $$R$$, $$F_r$$, $$m$$, and $$v$$ are wheel inertia, angular velocity, torque input, traction force, tire radius, rolling resistance, mass and translation velocity respectively. Now assume that the vehicle (or the wheel) is in equilibrium state then both derivative should be zero. As it turns out you will get: $$0 = T$$ because: $$0 = F_x + F_r$$

Does that mean that if the vehicle is in equilibrium, even though you may press the pedal gas you torque is still zero? Imagine that you are riding your car on a slippery surface and you still press the pedal, it will of course transfer torque to the wheel. But the translation velocity will remain constant. Then the second diff equation is zero but not necessarily the first one. And you can have same reasoning on the first diff equation. So despite the fact that "equilibrium is when no changes are happening" what does equilibrium mean in this context? Does equilibrium in first equation mean that $$T = 0$$? That if you don't press gas pedal it leads to equilibrium state?

• Surely to accelerate from a given speed then torque cannot be zero... Oct 16 '18 at 12:29
• This is a simple physics question - so long as there's a resistive force of any kind, including intra-engine friction, you need to apply power to maintain speed Oct 16 '18 at 18:38
• ^this. The key error is in the statement of the equation which doesn't include this. Oct 17 '18 at 6:27
• First of all i dont know why I should get down voted for my topic. Secondly I was focused on the eq states and that is when the both equations are zero. It is not about the math itself. It is about what eq states mean in real life? Given the aerodynamic force is very very low(in very low velocities. The rolling resistance is strongly connected to the friction and so is the traction force and they are in turn very dependent on slip ratio of the tire. And since it is impossible for motion to take place with out slip the summation of these two forces will never be zero unless the tire is hanging. Oct 17 '18 at 18:16