# Force exerted on ground from a slipping wheel

I am trying to do some car calculations for a game.

I have a tyre (radius $r$, inertia $I$) rotating around it's axle with an angular velocity $\omega$ faster than the center of mass is moving across the surface. So the tyre is not rolling along the ground, but slipping. The surface has a friction coefficient $\mu$. The tyre is pressed down by the car with a weight $N$

I need to calculate two things.

1. The force the tyre exerts on the ground.
2. The resulting (de)acceleration of the tyre.

Imagine a car doing a wheelspin. I need to find the force the excess speed of the tyre transfers to the ground, giving an acceleration to the car, and thereby also de deacceleration of the tyre.

I have calculations for the external forces (engine, brakes, drag, lateral slipping) and this is the final piece to complete the puzzle. So help is much appreciated, as i have been trying to solve this in a myriad of ways, with no success.

//Edited - It is the wheels connected to the enigne i am trying to calculate. However i am in this case not interested in the torque coming from the engine, only actions from the extra energy stored in wheel from building up the spin.

The engine provides a torque $\tau$ on the tyre

• Usually free wheels on a car don't spin faster than the car is moving, only the driven wheels do. If that's true for your case, the moment of inertia of the wheel is much less important than the driving torque from the axle. You would need that information. – BowlOfRed Jun 26 '18 at 22:31
• It is the wheels that are connected to the engine i am referring to. They are the ones doing he wheel spin, and the ones i need to calculate. – Danfjo Jun 26 '18 at 23:00

• I have excluded weight transfer and suspension forces for this question. So the normal force remains tha same whether rolling or not, and the acceleration on the tyre comes from torque from engine - torque from ground? In that case lets say the engine provides a tourqe $\tau$ – Danfjo Jun 27 '18 at 6:02
• It makes sense. I just got all the forces wrong in my head. The $\delta\omega$ is a result of the frictions, not a contribution to them. Now i just have to model it right, and until it works, i am afraid to asy i am 100 % sure that i understand it :-P – Danfjo Jun 27 '18 at 7:56