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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?

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  • $\begingroup$ Surely to accelerate from a given speed then torque cannot be zero... $\endgroup$ – Solar Mike Oct 16 '18 at 12:29
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    $\begingroup$ 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 $\endgroup$ – Carl Witthoft Oct 16 '18 at 18:38
  • $\begingroup$ ^this. The key error is in the statement of the equation which doesn't include this. $\endgroup$ – Jonathan R Swift Oct 17 '18 at 6:27
  • $\begingroup$ 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. $\endgroup$ – Peppe Oct 17 '18 at 18:16
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I'm going to present a simplification, As I think this is a case of maths blinding. In Engineering, the intuitive approach is less prone to error in many cases.

The problem is the air... The air is not in equilibrium - air molecules are being smacked out of the way by the car body and doing this takes effort. If the car is traveling at a constant speed, such that it's not accelerating, and the forwards thrust from the engine equals the drag force from the air, then the car is said to be in equilibrium. The engine must produce torque with sufficient power to accelerate all of those air molecules out of the way. When you drive faster, you hit more air per second, and have to accelerate it to a higher velocity. This is why travelling at a constant 70mph burns more fuel than a constant 40mph.

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  • $\begingroup$ Maths wise - the only resistance term you have is rolling resistance, which you've said is only related to the radius of the wheel - it too will depend on speed (although with a more complex relationship than air resistance) so should not disappear when you take the derivatives $\endgroup$ – Jonathan R Swift Oct 16 '18 at 13:02
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    $\begingroup$ Rolling resistance is the larger cf air resistance up to about 40mph then air resistance is dominant.... $\endgroup$ – Solar Mike Oct 16 '18 at 13:03

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